STAT.
STAT.
AN ELEMENTARY TREATISE
ON
QUATEENIONS
SontlOtt : C. J. CLAY AND SONS,
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,
AVE MARIA LANE.
CAMBRIDGE : DEIGHTON, BELL, AND CO.
LEIPZIG : F. A. BROCKHAUS.
MATH.
STAT.
UMtABY
AN ELEMENTARY TREATISE
ON
QUATERNIONS
BY
P. G. TAIT, M.A., SEC. R.S.E.,
HONORARY FELLOW OF ST PETER S COLLEGE, CAMBRIDGE
PROFESSOR OF NATURAL PHILOSOPHY IN THE UNIVERSITY OF EDINBURGH
TCTpaKTVV
rrayciv devdov
THIRD EDITION, MUCH ENLARGED
CAMBRIDGE
AT THE UNIVERSITY PRESS
1890
[All Rights reserved.]
3
STAT.
LIPAACV
Sambrilige :
FEINTED BY C. J. CLAY, M.A. AND SONS,
AT THE UNIVERSITY PRESS.
ERRATA.
P. xxiv, 1. 18, for Vpp = read Vpp = 0.
for Vpp y read Vpp = y.
P. 232, 1. 15, for To=l read Tp = l.
Ox A a en
T
fTAT.
PREFACE.
TN the present edition this work has been very greatly en
* larged ; to the extent, in fact, of more than onethird. Had I
not determined to keep the book in moderate compass it might
easily have been doubled in size. A good deal of rearrangement
also has been thought advisable, especially with reference to the
elementary uses of q( )q~\ and of V. Prominent among the
additions is an entire Chapter, On the Analytical Aspect of Quater
nions, which I owe to the unsolicited kindness of Prof. Cayley.
As will be seen by the reader of the former Preface (reprinted
below) the point of view which I have, from the first, adopted
presents Quaternions as a Calculus uniquely adapted to Euclidian
space, and therefore specially useful in several of the most im
portant branches of Physical Science. After giving the necessary
geometrical and other preliminaries, I have endeavoured to de
velope it entirely from this point of view ; and, though one can
scarcely avoid meeting with elegant and often valuable novelties
to whatever branch of science he applies such a method, my chief
contributions are still those contained in the fifth and the two
last Chapters. When, twenty years ago, I published my paper
on the application of V to Greens and other Allied Theorems, I
was under the impression that something similar must have been
contemplated, perhaps even mentally worked out, by Hamilton as
the subject matter of the (unwritten but promised) concluding
section of his Elements. It now appears from his Life (Vol. ill.
p. 194) that such was not the case, and thus that I was not in any
way anticipated in this application (from rny point of view by far
the most important yet made) of the Calculus. But a bias in
such a special direction of course led to an incomplete because one
sided presentation of the subject. Hence the peculiar importance
of the contribution from an Analyst like Prof. Cayley.
T.Q.I. 781534 b
VI PREFACE.
It is disappointing to find how little progress has recently been
made with the development of Quaternions. One cause, which
has been specially active in France, is that workers at the subject
have been more intent on modifying the notation, or the mode of
presentation of the fundamental principles, than on extending the
applications of the Calculus. The earliest offender of this class
was the late M. Hoiiel who, while availing himself of my permis
sion to reproduce, in his Theorie des Quantite s Complexes, large parts
of this volume, made his pages absolutely repulsive by introducing
fancied improvements in the notation. I should not now have
referred to this matter (about which I had remonstrated with M.
Hoiiel) but for a remark made by his friend, M. Laisant, which
peremptorily calls for an answer. He says: "M. Tait...trouve
que M. Hoiiel a altere 1 ceuvre du maitre. Perfectionner n est .pas
detruire." This appears to be a parody of the saying attributed
to Louis XIV.: " Pardonner n est pas oublier": but M. Laisant
should have recollected the more important maxim " Le mieux est
Pennemi du bien." A line of Shakespeare might help him:
"...with taperlight
To seek the beauteous eye of heaven to garnish,
Is wasteful and ridiculous excess."
Even Prof. Willard Gibbs must be ranked as one of the retarders
of Quaternion progress, in virtue of his pamphlet on Vector
Analysis; a sort of hermaphrodite monster, compounded of the
notations of Hamilton and of Grassmann.
Apropos of Grassmann, I may advert for a moment to some
comparatively recent German statements as to his anticipations
&c. of Quaternions. I have given in the last edition of the Encyc.
Brit. (Art. QI}ATEKNIONS ; to which I refer the reader) all that is
necessary to shew the absolute baselessness of these statements.
The essential points are as follows. Hamilton not only published
his theory complete, the year before the first (and extremely
imperfect) sketch of the Ausdehnungslehre appeared; but had
given ten years before, in his protracted study of Sets, the very
processes of external and internal multiplication (corresponding to
the Vector and Scalar parts of a product of two vectors) which
have been put forward as specially the property of Grassmann.
The scrupulous care with which Hamilton drew up his account of
PREFACE. Vll
the work of previous writers (Lectures, Preface) is minutely detailed
in his correspondence with De Morgan (Hamilton s Life, Vol. in.).
Another cause of the slow headway recently made by Qua
ternions is undoubtedly to be ascribed to failure in catching the
"spirit" of the method: especially as regards the utter absence
of artifice, and the perfect naturalness of every step. To try to
patch up a quaternion investigation by having recourse to quasi
Cartesian processes is fatal to progress. A quaternion student loses
his selfrespect, so to speak, when he thus violates the principles
of his Order. Tannhauser has his representatives in Science as
well as in Chivalry ! One most insidious and dangerous form of
temptation to this dabbling in the unclean thing is pointed out in
500 below. All who work at the subject should keep before
them Hamilton s warning words (Lectures, 513):
"I regard it as an inelegance and imperfection in this calculus,
or rather in the state to which it has hitherto [1853] been un
folded, whenever it becomes, or seems to become, necessary to have
recourse to the resources of ordinary algebra, for the SOLUTION
OF EQUATIONS IN QUATERNIONS."
As soon as my occupation with teaching and with experimental
work perforce ceases to engross the greater part of my time, I
hope to attempt, at least, the full quaternion development of
several of the theories briefly sketched in the last chapter of
this book ; provided, of course, that no one have done it in the
meantime. From occasional glimpses, hitherto undeveloped, I feel
myself warranted in asserting that, immense as are the simplifi
cations introduced by the use of quaternions in the elementary
parts of such subjects as Hydrokinetics and Electrodynamics, they
are absolutely as nothing compared with those which are to be
effected in the higher and (from the ordinary point of view) vastly
more complex branches of these fascinating subjects. Complexity
is no feature of quaternions themselves, and in presence of their
attack (when properly directed) it vanishes from the subject
also : provided, of course, that what we now call complexity
depends only upon those spacerelations (really simple if rightly
approached) which we are in the habit of making all but incom
prehensible, by surrounding them with our elaborate scaffolding of
nonnatural coordinates.
viii PREFACE.
Mr Wilkinson has again kindly assisted me in the revision of
the proofs; and they have also been read and annotated by Dr
Plarr, the able French Translator of the second edition. Thanks
to their valuable aid, I may confidently predict that the present
edition will be found comparatively accurate.
With regard to the future of Quaternions, I will merely quote
a few words of a letter I received long ago from Hamilton :
" Could anything be simpler or more satisfactory ? Don t you
feel, as well as think, that we are on a right track, and shall be
thanked hereafter ? Never mind when."
The special form of thanks which would have been most
grateful to a man like Hamilton is to be shewn by practical
developments of his magnificent Idea. The award of this form of
thanks will, I hope, not be long delayed.
P. G. TA1T.
ADDITIONS, CHANGES, ETC. IN THE PRESENT EDITION.
(Only the more important are noticed, and they are indicated by the
sectional numbers.)
Chap. I. 31 (k), (m), 40, 43.
II. 51, 89.
III. 105, 108, 116, 119122, 1334.
IV. 140 (8) (12), 145149.
V. 174, 187, 1934, 196, 199.
VI. The whole.
VIII. 247, 250, 250*.
IX. 285, 286, 287.
X. 326, 336.
XI. 3578, 382, 3846.
XII. 390, 393403, 407, 458, 473 (a) (Z), 480, 489, 493, 499, 500,
503, 508511, 51213.
There are large additions to the number of Examples, some in fact
to nearly every Chapter. Several of these are of considerable importance ;
as they contain, or suggest, processes and results not given in the text.
PREFACE TO THE SECOND EDITION.
To the first edition of this work, published in 1867, the
following was prefixed :
THE present work was commenced in 1859, while I was a
Professor of Mathematics, and far more ready at Quaternion
analysis than I can now pretend to be. Had it been then
completed I should have had means of testing its teaching
capabilities, and of improving it, before publication, where found
deficient in that respect.
The duties of another Chair, and Sir W. Hamilton s wish that
my volume should not appear till after the publication of his
Elements, interrupted my already extensive preparations. I had
worked out nearly all the examples of Analytical Geometry in
Todhunter s Collection, and I had made various physical applica
tions of the Calculus, especially to Crystallography, to Geometrical
Optics, and to the Induction of Currents, in addition to those on
Kinematics, Electrodynamics, Fresnel s Wave Surface, &c., which
are reprinted in the present work from the Quarterly Mathematical
Journal and the Proceedings of the Royal Society of Edinburgh.
Sir W. Hamilton, when I saw him but a few days before his
death, urged me to prepare my work as soon as possible, his being
almost ready for publication. He then expressed, more strongly
perhaps than he had ever done before, his profound conviction of
the importance of Quaternions to the progress of physical science ;
and his desire that a really elementary treatise on the subject
should soon be published.
I regret that I have so imperfectly fulfilled this last request
X PREFACE TO THE SECOND EDITION.
of my revered friend. When it was made I was already engaged,
along with Sir W. Thomson, in the laborious work of preparing
a large Treatise on Natural Philosophy. The present volume has
thus been written under very disadvantageous circumstances,
especially as I have not found time to work up the mass of
materials which I had originally collected for it, but which I
had not put into a fit state for publication. I hope, however,
that I have to some extent succeeded in producing a thoroughly
elementary work, intelligible to any ordinary student ; and that
the numerous examples I have given, though not specially
chosen so as to display the full merits of Quaternions, will
yet sufficiently shew their admirable simplicity and naturalness
to induce the reader to attack the Lectures and the Elements;
where he will find, in profusion, stores of valuable results, and
of elegant yet powerful analytical investigations, such as are
contained in the writings of but a very few of the greatest
mathematicians. For a succinct account of the steps by which
Hamilton was led to the invention of Quaternions, and for
other interesting information regarding that remarkable genius,
I may refer to a slight sketch of his life and works in the
North British Review for September 1866.
It will be found that I have not servilely followed even so
great a master, although dealing with a subject which is entirely
his own. I cannot, of course, tell in every case what I have
gathered from his published papers, or from his voluminous
correspondence, and what I may have made out for myself.
Some theorems and processes which I have given, though wholly
my own, in the sense of having been made out for myself before
the publication of the Elements, I have since found there. Others
also may be, for I have not yet read that tremendous volume
completely, since much of it bears on developments unconnected
with Physics. But I have endeavoured throughout to point out
to the reader all the more important parts of the work which I
know to be wholly due to Hamilton. A great part, indeed, may
be said to be obvious to any one who has mastered the pre
liminaries ; still I think that, in the two last Chapters especially,
a good deal of original matter will be found.
PREFACE TO THE SECOND EDITION. XI
The volume is essentially a working one, and, particularly in
the later Chapters, is rather a collection of examples than a
detailed treatise on a mathematical method. I have constantly
aimed at avoiding too great extension ; and in pursuance of
this object have omitted many valuable elementary portions
of the subject. One of these, the treatment of Quaternion
logarithms and exponentials, I greatly regret not having given.
But if I had printed all that seemed to me of use or interest to
the student, I might easily have rivalled the bulk of one of
Hamilton s volumes. The beginner is recommended merely to
read the first five Chapters, then to work at Chapters VI,
VII, VIII* (to which numerous easy Examples are appended).
After this he may work at the first five, with their (more
difficult) Examples ; and the remainder of the book should
then present no difficulty.
Keeping always in view, as the great end of every mathe
matical method, the physical applications, I have endeavoured
to treat the subject as much as possible from a geometrical
instead of an analytical point of view. Of course, if we premise
the properties of i, j, k merely, it is possible to construct from
them the whole system f; just as we deal with the imaginary
of Algebra, or, to take a closer analogy, just as Hamilton
himself dealt with Couples, Triads, and Sets. This may be
interesting to the pure analyst, but it is repulsive to the
physical student, who should be led to look upon i, j, k, from
the very first as geometric realities, not as algebraic imagi
naries.
The most striking peculiarity of the Calculus is that mul
tiplication is not generally commutative, i.e. that qr is in general
different from rg, r and q being quaternions. Still it is to
be remarked that something similar is true, in the ordinary
coordinate methods, of operators and functions : and therefore
* [In this edition these Chapters are numbered VII, VIII, IX, respectively
Aug. 1889.]
t This has been done by Hamilton himself, as one among many methods he has
employed; and it is also the foundation of a memoir by M. Allegret, entitled Essai
sur le Calcul des Quaternions (Paris, 1862).
xil PREFACE TO THE SECOND EDITION.
the student is not wholly unprepared to meet it. No one is
puzzled by the fact that log. cos. x is not equal to cos. log. x,
or that A ~ is not equal to ^\A/. Sometimes, indeed, this
V dx dx l
rule is most absurdly violated, for it is usual to take cos 2 # as
equal to (cos#) 2 , while cos" 1 ^ is not equal to (cos a?)" 1 . No such
incongruities appear in Quaternions ; but what is true of operators
and functions in other methods, that they are not generally com
mutative, is in Quaternions true in the multiplication of (vector)
coordinates.
It will be observed by those who are acquainted with the
Calculus that I have, in many cases, not given the shortest or
simplest proof of an important proposition. This has been done
with the view of including, in moderate compass, as great a
variety of methods as possible. With the same object I have
endeavoured to supply, by means of the Examples appended
to each Chapter, hints (which will not be lost to the intelli
gent student) of farther developments of the Calculus. Many
of these are due to Hamilton, who, in spite of his great origi
nality, was one of the most excellent examiners any University
can boast of.
It must always be remembered that Cartesian methods are
mere particular cases of Quaternions, where most of the distinctive
features have disappeared ; and that when, in the treatment of
any particular question, scalars have to be adopted, the Quaternion
solution becomes identical with the Cartesian one. Nothing there
fore is ever lost, though much is generally gained, by employing
Quaternions in preference to ordinary methods. In fact, even
when Quaternions degrade to scalars, they give the solution
of the most general statement of the problem they are applied
to, quite independent of any limitations as to choice of particular
coordinate axes.
There is one very desirable object which such a work as this
may possibly fulfil. The University of Cambridge, while seeking
to supply a real want (the deficiency of subjects of examination for
mathematical honours, and the consequent frequent introduction
of the wildest extravagance in the shape of data for " Problems "),
PREFACE TO THE SECOND EDITION. Xlii
is in danger of making too much of such elegant trifles as Trilinear
Coordinates, while gigantic systems like Invariants (which, by the
way, are as easily introduced into Quaternions as into Cartesian
methods) are quite beyond the amount of mathematics which
even the best students can master in three years reading.
One grand step to the supply of this want is, of course, the
introduction into the scheme of examination of such branches
of mathematical physics as the Theories of Heat and Electricity.
But it appears to me that the study of a mathematical method
like Quaternions, which, while of immense power and compre
hensiveness, is of extraordinary simplicity, and yet requires
constant thought in its applications, would also be of great
benefit. With it there can be no "shut your eyes, and write
down your equations," for mere mechanical dexterity of analysis
is certain to lead at once to error on account of the novelty of
the processes employed.
The Table of Contents has been drawn up so as to give the
student a short and simple summary of the chief fundamental
formulae of the Calculus itself, and is therefore confined to an
analysis of the first five [and the two last] chapters.
In conclusion, I have only to say that I shall be much obliged
to any one, student or teacher, who will point out portions of the
work where a difficulty has been found; along with any inaccuracies
which may be detected. As I have had no assistance in the revision
of the proofsheets, and have composed the work at irregular in
tervals, and while otherwise laboriously occupied, I fear it may
contain many slips and even errors. Should it reach another
edition there is no doubt that it will be improved in many
important particulars/
To this I have now to add that I have been equally surprised
and delighted by so speedy a demand for a second edition and the
more especially as I have had many pleasing proofs that the
work has had considerable circulation in America. There seems
now at last to be a reasonable hope that Hamilton s grand
invention will soon find its way into the working world of science,
to which it is certain to render enormous services, and not be laid
T. Q. I. c
XIV PREFACE TO THE SECOND EDITION.
aside to be unearthed some centuries hence by some grubbing
antiquary.
It can hardly be expected that one whose time is mainly en
grossed by physical science, should devote much attention to the
purely analytical and geometrical applications of a subject like this;
and I am conscious that in many parts of the earlier chapters I
have not fully exhibited the simplicity of Quaternions. I hope,
however, that the corrections and extensions now made, especially
in the later chapters, will render the work more useful for my chief
object, the Physical Applications of Quaternions, than it could
have been in its first crude form.
I have to thank various correspondents, some anonymous, for
suggestions as well as for the detection of misprints and slips of
the pen. The only absolute error which has been pointed out to
me is a comparatively slight one which had escaped my own notice :
a very grave blunder, which I have now corrected, seems not to
have been detected by any of my correspondents, so that I cannot
be quite confident that others may not exist.
I regret that I have not been able to spare time enough to
rewrite the work ; and that, in consequence of this, and of the
large additions which have been made (especially to the later
chapters), the whole will now present even a more miscellaneously
jumbled appearance than at first.
It is well to remember, however, that it is quite possible to
make a book too easy reading, in the sense that the student may
read it through several times without feeling those difficulties
which (except perhaps in the case of some rare genius) must
attend the acquisition of really useful knowledge. It is better to
have a rough climb (even cutting one s own steps here and there)
than to ascend the dreary monotony of a marble staircase or a
wellmade ladder. Royal roads to knowledge reach only the
particular locality aimed at and there are no views by the way.
It is not on them that pioneers are trained for the exploration of
unknown regions.
But I am happy to say that the possible repulsiveness of my
early chapters cannot long be advanced as a reason for not attack
ing this fascinating subject. A still more elementary work than
PREFACE TO THE SECOND EDITION. XV
the present will soon appear, mainly from the pen of my colleague
Professor KELLAND. In it I give an investigation of the properties
of the linear and vector function, based directly upon the Kine
matics of Homogeneous Strain, and therefore so different in method
from that employed in this work that it may prove of interest to
even the advanced student.
Since the appearance of the first edition I have managed (at
least partially) to effect the application of Quaternions to line,
surface, and volume integrals, such as occur in Hydrokinetics,
Electricity, and Potentials generally. I was first attracted to
the study of Quaternions by their promise of usefulness in
such applications, and, though I have not yet advanced far in
this new track, I have got far enough to see that it is certain
in time to be of incalculable value to physical science. I have
given towards the end of the work all that is necessary to put
the student on this track, which will, I hope, soon be followed to
some purpose.
One remark more is necessary. I have employed, as the
positive direction of rotation, that of the earth about its axis, or
about the sun, as seen in our northern latitudes, i.e. that opposite
to the direction of motion of the hands of a watch. In Sir W.
Hamilton s great works the opposite is employed. The student
will find no difficulty in passing from the one to the other ; but,
without previous warning, he is liable to be much perplexed.
With regard to notation, I have retained as nearly as possible
that of Hamilton, and where new notation was necessary I
have tried to make it as simple and, as little incongruous with
Hamilton s as possible. This is a part of the work in which great
care is absolutely necessary; for, as the subject gains development,
fresh notation is inevitably required ; and our object must be to
make each step such as to defer as long as possible the revolution
which must ultimately come.
Many abbreviations are possible, and sometimes very useful in
private work ; but, as a rule, they are unsuited for print. Every
analyst, like every shorthand writer, has his own special con
tractions ; but, when he comes to publish his results, he ought
invariably to put such devices aside. If all did not use a common
XVI PREFACE TO THE SECOND EDITION.
mode of public expression, but each were to print as he is in the
habit of writing for his own use, the confusion would be utterly
intolerable.
Finally, I must express my great obligations to my friend
M. M. U. WILKINSON of Trinity College, Cambridge, for the care
with which he has read my proofs, and for many valuable sug
gestions.
P. G. TAIT.
COLLEGE, EDINBURGH,
October 1873.
CONTENTS.
CHAPTER I. VECTORS AND THEIR COMPOSITION . pp. 1 28
Sketch of the attempts made to represent geometrically the imaginary of
algebra. 113.
De Moivre s Theorem interpreted in plane rotation. 7, 8.
Curious speculation of Servois. 11.
Elementary geometrical ideas connected with relative position. 14.
Definition of a VECTOR. It may be employed to denote translation. Definition
of currency. 16.
Expression of a vector by one symbol, containing implicitly three distinct
numbers. Extension of the signification of the symbol = . 18.
The sign + defined in accordance with the interpretation of a vector as
representing translation. 19.
Definition of  . It simply reverses the currency of a vector. 20.
Triangles and polygons of vectors, analogous to those of forces and of simul
taneous velocities. 21.
When two vectors are parallel we have
a = xp. 22.
Any vector whatever may be expressed in terms of three distinct vectors, which
are not coplanar, by the formula
which exhibits the three numbers on which the vector depends. 23.
Any vector in the same plane with a and may be written
p = xa + y(3. 24.
The equation Tff p,
between two vectors, is equivalent to three distinct equations among
numbers. 25.
The Commutative and Associative Laws hold in the combination of vectors by
the signs + and  . 27.
XVlll CONTENTS.
The equation p = xfi,
where p is a variable, and j8 a fixed, vector, represents a line drawn through
the origin parallel to 8.
is the equation of a line drawn through the extremity of a and parallel
to 8. 28.
represents the plane through the origin parallel to a and /3, while
denotes a parallel plane through the point 7. 29.
The condition that p, a, /3 may terminate in the same line is
pp + qa + rp = Q,
subject to the identical relation
Similarly pp + q a + r p + S y = Q,
with jj + 2 + r + s = 0,
is the condition that the extremities of four vectors lie in one plane. 30.
Examples with solutions. Conditions that a vector equation may represent a
line, or a surface.
The equation p = 0i
represents a curve in space ; while
is a cone, and
is a cylinder, both passing through the curve. 31.
Differentiation of a vector, when given as a function of one number. 32 37.
If the equation of a curve be
P=0()
where s is the length of the arc, dp is a vectortangent to the curve, and its
length is ds. 38, 39.
Examples with solutions. 4043.
EXAMPLES TO CHAPTEK I. 28 30
CHAPTER II. PRODUCTS AND QUOTIENTS OF VECTORS . 31 57
Here we begin to see what a quaternion is. When two vectors are parallel
their quotient is a number. 45, 46.
When they are perpendicular to one another, their quotient is a vector perpen
dicular to their plane. 47, 64, 72.
When they are neither parallel nor perpendicular the quotient in general
involves four distinct numbers and is thus a QUATEKNION. 47.
A quaternion q regarded as the operator which turns one vector into another.
It is thus decomposable into two factors, whose order is indifferent, the
stretching factor or TENSOR, and the turning factor or VERSOR. These are
denoted by Tq, and Uq. 48.
CONTENTS. XIX
The equation p = qa
o
gives =<? or 8a~ l = q, "but not in general
a
a 1 8 = q. 49.
q or /Sa" 1 depends only on the relative lengths, and directions, of /3 and a.
50.
Reciprocal of a quaternion defined,
8 . I a
q =  gives or q~ 1 =  ,
a q 8
Tq
Definition of the Conjugate of a quaternion,
and qKq=Kq. q = (Tq)v. 52.
Eepresentation of versors by arcs on the unitsphere. 53.
Versor multiplication illustrated by the composition of arcs. The process
proved to be not generally commutative. 54.
Proof that K (qr) = Kr . Kq. 55.
Proof of the Associative Law of Multiplication
p.qr = pq.r. 5760.
[Digression on Spherical Conies. 59*.]
Quaternion addition and subtraction are commutative. 61.
Quaternion multiplication and division are distributive. 62.
Integral powers of a versor are subject to the Index Law. 63.
Composition of quadrantal versors in planes at right angles to each other.
Calling them i, j, k, we have
ijk=l. 6471.
A unitvector, when employed as a factor, may be considered as a quadrantal
versor whose plane is perpendicular to the vector. Hence the equations
just written are true of any set of rectangular unitvectors i, j, k. 72.
The product, and also the quotient, of two vectors at right angles to each other
is a third perpendicular to both. Hence
Ka =  a,
and (Ta) 2 = aKa= a 2 . 73.
Every versor may be expressed as a power of some unitvector. 74.
Every quaternion may be expressed as a power of a vector. 75.
The Index Law is true of quaternion multiplication and division. 76.
Quaternion considered as the sum of a SCALAR and a VECTOR.
a
Proof that SKq = Sq, VKq =  Vq, 2Kq = K2q . 79.
Quadrinomial expression for a quaternion
An equation between quaternions is equivalent to four equations between
numbers (or scalars). 80.
XX CONTENTS.
Second proof of the distributive law of multiplication. 81 83.
Algebraic determination of the constituents of the product and quotient of two
vectors. 83, 84.
Second proof of the associative law of multiplication. 85.
Proof of the formulae Sap = Spa,
S .qr = S .rq,
S . qrs = S . rsq = S . sqr,
2S.ap...<t> X )_
the upper sign belonging to the scalar if the number of factors is even.
8689.
Proof of the formulae
F. aVpy = ySap  pSya,
V.apy = aSpy  pSya + ySap,
V.apy= V.ypa,
V. VapVyd = aS . 1878  pS . 07$,
= 8S . apy  yS . a/35,
dS . apy = aS . pyd + pS . yad + yS . apd,
= VapSyd + VpySad + VyaSpd. 9092.
Hamilton s proof that the product of two parallel vectors must be a scalar, and
that of perpendicular vectors, a vector; if quaternions are to deal with
space indifferently in all directions. 93.
EXAMPLES TO CHAPTEK II. 5758
CHAPTER III. INTERPRETATIONS AND TRANSFORMA
TIONS OF QUATERNION EXPRESSIONS . . . 59 88
If 6 be the angle between two vectors, a and p, we have
Sap=TaTpcos0,
Ta
9496.
a Ta
Applications to plane trigonometry. 97.
The condition Sap =
shews that a is perpendicular to /3, while
Fa/3 = 0,
shews that a and p are parallel.
The expression S . apy
is the volume of the parallelepiped three of whose conterminous edges are
o, /3, 7. Hence
S.apy =
shews that a, p, y are coplanar.
CONTENTS. XXI
Expression of S . apy as a determinant. 98102.
Proof that (Tq) 2 = (Sq)* + ( TVq)*,
and T(qr) = TqTr. 103.
Simple propositions in plane trigonometry. 104.
Proof that  a/3a~ l is the vector reflected ray, when /3 is the incident ray and a
normal to the reflecting surface. 105.
Interpretation of apy when it is a vector. 106.
Examples of variety in simple transformations. 107.
The relation among the distances, two and two, of five points in space. 108.
De Moivre s Theorem, and Plane Trigonometry. 109 111.
Introduction to spherical trigonometry. 112 116.
Representation, graphic, and by quaternions, of the spherical excess. 117, 118.
Interpretation of the Operator
( ) T 1
in connection with rotation. Astronomical examples. 119 122.
Loci represented by different equations points, lines, surfaces, and volumes.
123126.
Proof that r~ l (?V)* V~ l = U (rq + KrKq). 127.
Proof of the transformation
BlQUATERNIONS. 130132.
Convenient abbreviations of notation. 133, 134.
EXAMPLES TO CHAPTER III ...... 89 93
CHAPTER IV. DIFFERENTIATION OF QUATERNIONS . 94105
Definition of a differential,
where dq is any quaternion whatever.
We may write dFq =f (q, dq),
where / is linear and homogeneous in dq ; but we cannot generally write
dFq = f(q)dq. 135138.
Definition of the differential of a function of more quaternions than one.
d (qr) = qdr + dq.r, but not generally d (qr) = qdr + rdq. 139.
Proofs of fundamental differential expressions :
Successive differentiation; Taylor s theorem. 141. 142.
XX11 CONTENTS.
If the equation of a surface be
F(p) = C,
the differential may be written
Sfdp = 0,
where v is a vector normal to the surface. 144.
Definition of Hamilton s Operator
. d . d , d
V = i~ +?  +k .
dx J dy dz
Its effects on simple scalar and vector functions of position. Its square
the negative of Laplace s Operator. Expressions for the condensation and
rotation due to a displacement. Application to fluxes, and to normals to
surfaces. Precautions necessary in its use. 145 149.
EXAMPLES TO CHAPTER IV. ...... 106, 107
CHAPTER V. THE SOLUTION OF EQUATIONS or THE
FIRST DEGREE ....... 108 141
The most general equation of the first degree in an unknown quaternion q,
may be written
SF. aqb + S.cq = d,
where a, &, c, d are given quaternions. Elimination of Sq, and reduction
to the vector equation
0/> = S.ctS/3/> = 7. 150,151.
General proof that 3 /> is expressible as a linear function of /o, 0/>, and 2 p.
152.
Value of for an ellipsoid, employed to illustrate the general theory.
153155.
Hamilton s solution of 0/ 9 = 7
If we write Sa^>p = S/o0V,
the functions and are said to be conjugate, and
W10 1 VX/JL = V<f> X0 V*.
Proof that ra, whose value may be written as
S . <>
is the same for all values of X, /A, v. 156158.
Proof that if m g = m m^g + w 2 # 2  s ,
, _ S (\0 /U,0V + Xytt0 j> + X0 /X I/ )
_ S (XM0V + X/uv + X0 V)
and flic,   pj  ,
S . \fJLV
(which, like m, are Invariants,)
then m g (0  g)~ l V\^ = (W0 1 gx + 9
Also that X = m 2<>
whence the final form of solution
W0 1 = m 1  Wo0 + 2 . 159, 160.
CONTENTS. XX111
Examples. 161173.
The fundamental cubic
3  ni 2 2 + TO 1  m= (0  ft) (0  <7 2 ) (0  3 ) = 0.
When is its own conjugate, the roots of the cubic are real; and the
equation
or (0<7)p = 0,
is satisfied by a set of three real and mutually perpendicular vectors.
Geometrical interpretation of these results. 174 178.
Proof of the transformation of the selfconjugate linear and vector function
<f>p=fp + hV. (i + ek) p(i ek)
where (0^)i = 0,
Another transformation is
0p = aaVap + bpSpp. 179181.
Other properties of 0. Proof that
Sp(<t>g)~ l p = 0, and Sp (<j>h) l p=
represent the same surface if
Proof that when is not selfconjugate
<t>p = <t> p+Vep. 182184.
Proof that, if q = a0ct + /30/3 + 707,
where a, , 7 are any rectangular unitvectors whatever, we have
Sq =  m 2 , Vq = e,
where Vcp = J (0  ) p.
This quaternion can be expressed in the important form
grV0/3. 185.
A nonconjugate linear and vector function of a vector differs from a self
conjugate one solely by a term of the form
Fep. 186.
Graphic determination of the conditions that there may be three real vector
solutions of
Fp0p = 0. 187.
Degrees of indeterminateness of the solution of a quaternion equation
Examples. 188 191.
The linear function of a quaternion is given by a symbolical biquadratic.
192.
Particular forms of linear equations. Differential of the nth root of a quaternion.
193196.
A quaternion equation of the rath degree in general involves a scalar equation
of degree ra 4 . 197.
Solution of the equation q~ = qa + b. 198.
EXAMPLES TO CHAPTER V. . ...... 142 145
XXIV CONTENTS.
CHAPTER VI. SKETCH OF THE ANALYTICAL THEORY
OF QUATERNIONS 146 159
CHAPTER VII GEOMETRY OF THE STRAIGHT LINE
AND PLANE . 160174
EXAMPLES TO CHAPTER VII 175177
CHAPTER VIII. THE SPHERE AND CYCLIC CONE . 178 198
EXAMPLES TO CHAPTER VIII. 199201
CHAPTER IX. SURFACES OF THE SECOND DEGREE . 202 224
EXAMPLES TO CHAPTER IX ...... . . . 224229
CHAPTER X. GEOMETRY OF CURVES AND SURFACES 230269
EXAMPLES TO CHAPTER X ........ 270 278
CHAPTER XL KINEMATICS ..... 279304
A. Kinematics of a Point. 354366.
If p = $t be the vector of a moving point in terms of the time, p is the vector
velocity, and p the vector acceleration.
<r p = (p (t) is the equation of the Hodograph.
p = iip + v 2 p" gives the normal and tangential accelerations.
VppsxQ if acceleration directed to a point, whence Vpp = y.
Examples. Planetary acceleration. Here the equation is
.. nUp
p= w
giving Vpp = y ; whence the hodograph is
p = ey l /j.Up.y 1 ,
and the orbit is the section of
ftTp = Se(y*eip)
by the plane Syp = Q.
Cotes Spirals, Epitrochoids, &c. 354366.
B. Kinematics of a Rigid System. 367375.
Rotation of a rigid system. Composition of rotations. If the position of a
system at time t is derived from the initial position by q( )q~ l , the
instantaneous axis is
Rodrigues coordinates. 367 375.
CONTENTS. XXV
C. Kinematics of a Deformable System. 376 386.
Homogeneous strain. Criterion of pure strain. Separation of the rotational
from the pure part. Extraction of the square root of a strain. A strain
is equivalent to a pure strain /^/^ ^ followed by the rotation 0/^/0 0.
Simple Shear. 376383.
Displacements of systems of points. Consequent condensation and rotation.
Preliminary about the use of V in physical questions. For displacement
<r, the strain function is
(pr = T SrV . a. 384 386.
D. Axes of Inertia. 387.
Moment of inertia. Binet s Theorem. 387.
EXAMPLES TO CHAPTER XI. 304308
CHAPTEK XII. PHYSICAL APPLICATIONS . . 309 409
A. Statics of a Rigid System. 389 403.
Condition of equilibrium of a rigid system is SS . /35ct = 0, where /3 is a vector
force, a its point of application. Hence the usual six equations in the
form 2/3 = 0, SFa/3 = 0. Central axis. Minding s Theorem, Ac. 389
403.
B. Kinetics of a Rigid System. 404 425.
For the motion of a rigid system
SS(ma/S)a=0,
whence the usual forms. Theorems of Poinsot and Sylvester. The
equation
2q = q<t>~ l (q l yq),
where 7 is given in terms of t and q if forces act, but is otherwise constant,
contains the whole theory of the motion of a rigid body with one point
fixed. Eeduction to the ordinary form
dt _ dw _ dx _ dy _ dz
~%~W~~X = ~Y = ~Z
Here, if no forces act, W, X, Y, Z are homogeneous functions of the third
degree in w, x, y, z. 404425.
C. Special Kinetic Problems. 426430.
Precession and Nutation. General equation of motion of simple pendulum.
Foucault s pendulum. 426430.
D. Geometrical and Physical Optics. 431 452.
Problem on reflecting surfaces. 431.
Fresnel s Theory of Double Eefraction. Various forms of the equation of
Fresnel s Wavesurface;
S.p(0p 2 )ip = l, T(p 2 0 1 )*p = 0, l = pp ~=F(TS)V\pVw.
The conical cusps and circles of contact. Lines of vibration, &c.
432452.
XXVI CONTENTS,
E. Electrodynamics. 453472.
Electrodynamics. The vector action of a closed circuit on an element of
current c^ is proportional to Varfi where
_ [Vada_ [dUa
J T J
the integration extending round the circuit. It can also be expressed
as  Vft, where ft is the spherical angle subtended by the circuit. This
is a manyvalued function. Special case of a circular current. Mutual
action of two closed circuits, and of solenoids. Mutual action of magnets.
Potential of a closed circuit. Magnetic curves. 453 472.
F. General Expressions for the Action between Linear Elements. 473.
Assuming Ampere s data .1, II, III, what is the most general expression for
the mutual action between two elements? Particular cases, determined
by a fourth assumption. Solution of the problem when I and II, alone,
are assumed. Special cases, including v. Helmholtz form. 473 (a)... (I).
G. Application of V to certain Physical Analogies. 474 478.
The effect of a currentelement on a magnetic particle is analogous to dis
placement produced by external forces in an elastic solid, while that
of a small closed circuit (or magnet) is analogous to the corresponding
vector rotation.
//. Elementary Properties of V. 479481.
SdpV=d=SdffV <r
where <r = i +.717 + fcf and V<r = i~ + j  + A ;
so that, if d<r = (f)dp,
J. Applications of V to Line, Surface and Volume Integrals.
482501.
Proof of the fundamental theorem for comparing an integral over a closed
surface with one through its content
Hence Green s Theorem. Examples in potentials and in conduction of
heat. Limitations and ambiguities. Determination of the displacement
in a fluid when the consequent rotation is given. 482 494.
Similar theorem for double and single integrals
fSvdp =ffS . UvVffds. 495497.
The fundamental forms, of which all the others are simple consequences, are
fdpu=ffdsV(Ui>V)u,
ffSVuds=ffUvuds. 498, 499.
The first is a particular case of the second. 500.
Expression for a surface in terms of an integral through the enclosed volume.
499.
CONTENTS. XXV11
K. Application of the V Integrals to Magnetic &c. Problems.
502506.
Volume and surface distributions due to a given magnetic field. Solenoidal
and Lamellar distributions. 502. Magnetic Induction and Vector
potential. 503. Ampere s Directrice. 504. Gravitation potential of
homogeneous solid in terms of a surfaceintegral. 505.
L. Application of V to the Stress Function. 507 511.
When there are no molecular couples the stressfunction is selfconjugate.
507. Properties of this function which depend upon the equilibrium
of any definite portion of the solid, as a whole. 508. Expression for
the stressfunction, in terms of displacement, when the solid is isotropic :
W =  n (SwV . <r + VSaxr)  (c  $n) wSVtr.
Examples. 509. Work due to displacement in any elastic solid. Green s
21 elastic coefficients. 511.
M. The Hydrokinetic Equations. 512, 513.
Equation of continuity, for displacement a :
.
dt dt
Equation for rate of change of momentum in unit volume:
* v * 512 
.
Term introduced by Viscosity:
ocV 2 <r + VSV<r. (Miscellaneous Examples, 36.)
Helmholtz s Transformation for Vortex motion:
Thomson s Transformation for Circulation:
 jj= \Q  it; 2 T , where/  (
dt L Ja J
N. Use of V in connection with Taylor s Theorem. 514 517.
Proof that
Applications and consequences. Separation of symbols of operation and their
treatment as quantities.
0. Applications of V in connection with the Calculus of Variations.
518527.
If A=fQTd P
we have dA= [QSUdp8p] +f{ 8QTdp + S.dp(Q Udp) } ,
whence, if A is a maximum or minimum,
~(Q P )VQ = 0.
Applications to Varying Action, Brachistochrones, Catenaries, &c. 518 526.
Thomson s Theorem that there is one, and but one, solution of the equation
S.V(e 2 VH) = 4irr. 527.
MISCELLANEOUS EXAMPLES ..... 409 __ 421.
QUATERNIONS.
CHAPTER I.
VECTORS, AND THEIR COMPOSITION.
1. FOR at least two centuries the geometrical representation
of the negative and imaginary algebraic quantities, 1 and ,J 1
has been a favourite subject of speculation with mathematicians.
The essence of almost all of the proposed processes consists in
employing such expressions to indicate the DIRECTION, not the
length, of lines.
2. Thus it was long ago seen that if positive quantities were
measured off in one direction along a fixed line, a useful and lawful
convention enabled us to express negative quantities of the same
kind by simply laying them off on the same line in the opposite
direction. This convention is an essential part of the Cartesian
method, and is constantly employed in Analytical Geometry and
Applied Mathematics.
c
3. Wallis, towards the end of the seventeenth century, proposed
to represent the impossible roots of a quadratic equation by going
out of the line on which, if real, they would have been laid off.
This construction is equivalent to the consideration of J 1 as a
directed unitline perpendicular to that on which real quantities
are measured.
4. In the usual notation of Analytical Geometry of two
dimensions, when rectangular axes are employed, this amounts
to reckoning each unit of length along Oy as + J 1, and on
Oy as J 1 ; while on Ox each unit is + 1, and on Ox it is 1.
T. Q. I. 1
2 QUATERNIONS. [5.
If we look at these four lines in circular order, i.e. in the order of
positive rotation (that of the northern hemisphere of the earth
about its axis, or opposite to that of the hands of a watch), they
give
1, N/1, 1, Vl.
In this series each expression is derived from that which precedes
it by multiplication by the factor J 1. Hence we may consider
J 1 as an operator, analogous to a handle perpendicular to the
plane of xy, whose effect on any line in that plane is to make it
rotate (positively) about the origin through an angle of 90.
5. Iu such a system, (which seems to have been first developed,
in 1805, by Buee) a point in the plane of reference is defined by a
single im&giiiai7 expression. Thus a + b J 1 may be considered
as a single quantity, denoting the point, P, whose coordinates are
a and b. Or, it may be used as an expression for the line OP
joining that point with the origin. In the latter sense, the ex
pression a + b V 1 implicitly contains the direction, as well as the
length, of this line ; since, as we see at once, the direction is
inclined at an angle tan" 1 b/a to the axis of oc, and the length is
Jo? + b z . Thus, say we have
the line OP considered as that by which we pass from one
J A
extremity, 0, to the other, P. In this sense it is called a VECTOR.
Considering, in the plane, any other vector,
OQ = a f + Vj^l
the addition of these two lines obviously gives
OR = a + a + (b + b ) J^l;
and we see that the sum is the diagonal of the parallelogram on
OP, OQ. This is the law of the composition of simultaneous
velocities ; and it contains, of course, the law of subtraction of one
directed line from another.
6. Operating on the first of these symbols by the factor J 1,
it becomes  b f a J  1 ; and now, of course, denotes the point
whose x and y coordinates are  b and a ; or the line joining this
point with the origin. The length is still J a* + V, but the angle
the line makes with the axis of x is tan 1 ( a/b) ; which is
evidently greater by ?r/2 than before the operation.
10.] VECTORS, AND THEIR COMPOSITION. 3
7. De Moivre s Theorem tends to lead us still further in the
same direction. In fact, it is easy to see that if we use, instead
of J 1, the more general factor cos a + J 1 sin a, its effect on
any line is to turn it through the (positive) angle a. in the plane
of x, y. [Of course the former factor, J 1, is merely the par
ticular case of this, when a = 7r/2.~
Thus (cos a + J 1 sin a) (a + b J 1 )
= a cos a b sin a + J 1 (a sin a f 6 cos a),
by direct multiplication. The reader will at once see that the new
form indicates that a rotation through an angle a has taken place,
if he compares it with the common formulae for turning the co
ordinate axes through a given angle. Or, in a less simple manner,
thus
Length = J (a cos a b sin a) 2 + (a sin a + b cos a) 2
= J a 2 + b* as before.
Inclination to axis of x
b
, tan a + 
_, a sin a 4 b cos a. a
= tan = : = tan .
a cos a b sin a n 6 ,
1 tan a
a
= a + tan" 1 b/a.
8. We see now, as it were, why it happens that
(cos a + v 1 sin a) m = cos ma. + J 1 sin ma.
In fact, the first operator produces m successive rotations in the
same direction, each through the angle a ; the second, a single
rotation through the angle ma.
9. It may be interesting, at this stage, to anticipate so far as to
remark that in the theory of Quaternions the analogue of
cos 6 + J ^1 sin 6
is cos f tff sin 0,
where w 2 =  1.
Here, however, & is not the algebraic J 1, but is any directed
unitline whatever in space.
10. In the present century Argand, Warren, Mourey, and
others, extended the results of Wallis and Buee. They attempted
12
4 QUATERNIONS. [ll.
to express as a line the product of two lines each represented by a
symbol such asa+6/s/ 1. To a certain extent they succeeded,
but all their results remained confined to two dimensions.
The product, II, of two such lines was defined as the fourth
proportional to unity and the two lines, thus
1 :a + bJ^I::a + V l J~^l : II,
or n = (aa  bV) + (a b + If a} J^l.
The length of II is obviously the product of the lengths of the
factor lines ; and its direction makes an angle with the axis of x
which is the sum of those made by the factor lines. From this
result the quotient of two such lines follows immediately.
11. A very curious speculation, due to Servois and published
in 1813 in Gergonne s Annales, is one of the very few, so far as has
been discovered, in which a wellfounded guess at a possible mode
of extension to three dimensions is contained. Endeavouring to
extend to space the form a + 6 J 1 for the plane, he is guided by
analogy to write for a directed unitline in space the form
p cos a + q cos ft + r cos 7,
where a, ft, 7 are its inclinations to the three axes. He perceives
easily that p, q, r must be nonreals : but, he asks, " seraientelles
imaginaires re ductibles a la forme generale A+Bjl?" The
i,j t k of the Quaternion Calculus furnish an answer to this question.
(See Chap. II.) But it may be remarked that, in applying the
idea to lines in a plane, a vector OP will no longer be represented
(as in 5) by
but by OP =pa + qb.
And if, similarly, OQ = pa + qb ,
the addition of these two lines gives for OR (which retains its
previous signification)
12. Beyond this, few attempts were made, or at least recorded,
in earlier times, to extend the principle to space of three dimen
sions; and, though many such had been made before 1843, none,
with the single exception of Hamilton s, have resulted in simple,
practical methods ; all, however ingenious, seeming to lead almost
at once to processes and results of fearful complexity.
1 4.] VECTORS, AND THEIR COMPOSITION. 5
For a lucid, complete, and most impartial statement of the
claims of his predecessors in this field we refer to the Preface to
Hamilton s Lectures on Quaternions. He there shews how his long
protracted investigations of Sets culminated in this unique system
of tridimensionalspace geometry.
13. It was reserved for Hamilton to discover the use and
properties of a class of symbols which, though all in a certain sense
square roots of 1, may be considered as real unit lines, tied down
to no particular direction in space ; the expression for a vector is,
or may be taken to be,
but such vector is considered in connection with an extraspatial
magnitude w, and we have thus the notion of a QUATERNION
w + p.
This is the fundamental notion in the singularly elegant, and
enormously powerful, Calculus of Quaternions.
While the schemes for using the algebraic J 1 to indicate
direction make one direction in space expressible by real numbers,
the remainder being imaginaries of some kind, and thus lead to
expressions which are heterogeneous ; Hamilton s system makes all
directions in space equally imaginary, or rather equally real, there
by ensuring to his Calculus the power of dealing with space
indifferently in all directions.
In fact, as we shall see, the Quaternion method is independent
of axes or any supposed directions in space, and takes its reference
lines solely from the problem it is applied to.
14. But, for the purpose of elementary exposition, it is best
to begin by assimilating it as closely as we can to the ordinary
Cartesian methods of Geometry of Three Dimensions, with which
the student is supposed to be, to some extent at least, acquainted.
Such assistance, it will be found, can (as a rule) soon be dispensed
with; and Hamilton regarded any apparent necessity for an oc
casional recurrence to it, in higher applications, as an indication
of imperfect development in the proper methods of the new
Calculus.
We commence, therefore, with some very elementary geometrical
ideas, relating to the theory of vectors in space. It will subsequently
appear how we are thus led to the notion of a Quaternion.
6 QUATERNIONS. [15.
15. Suppose we have two points A and B in space, and sup
pose A given, on how many numbers does B s relative position
depend ?
If we refer to Cartesian coordinates (rectangular or not) we find
that the data required are the excesses of B s three coordinates
over those of A. Hence three numbers are required.
Or we may take polar coordinates. To define the moon s
position with respect to the earth we must have its Geocentric
Latitude and Longitude, or its Right Ascension and Declination,
and, in addition, its distance or radiusvector. Three again.
16. Here it is to be carefully noticed that nothing has been
said of the actual coordinates of either A or B, or of the earth
and moon, in space ; it is only the relative coordinates that are
contemplated.
Hence any expression, as AB, denoting a line considered with
reference to direction and currency as well as length, (whatever
may be its actual position in space) contains implicitly three
numbers, and all lines parallel and equal to AB, and concurrent
with it, depend in the same way upon the same three. Hence, all
lines which are equal, parallel, and concurrent, may be represented
by a common symbol, and that symbol contains three distinct numbers.
In this sense a line is called a VECTOR, since by it we pass from
the one extremity, A, to the other, B , and it may thus be
considered as an instrument which carries A to B: so that a
vector may be employed to indicate a definite translation in space.
[The term " currency " has been suggested by Cayley for use
instead of the somewhat vague suggestion sometimes taken to
be involved in the word "direction." Thus parallel lines have
the same direction, though they may have similar or opposite
currencies. The definition of a vector essentially includes its
currency.]
17. We may here remark, once for all, that in establishing a
new Calculus, we are at liberty to give any definitions whatever
of our symbols, provided that no two of these interfere with, or
contradict, each other, and in doing so in Quaternions simplicity
and (so to speak) naturalness were the inventor s aim.
18. Let AB be represented by a, we know that a involves
three separate numbers, and that these depend solely upon the
position of B relatively to A. Now if CD be equal in length to AB
20.] VECTORS, AND THEIR COMPOSITION. 7
and if these lines be parallel, and have the same currency, we may
evidently write
where it will be seen that the sign of equality between vectors
contains implicitly equality in length, parallelism in direction,
and concurrency. So far we have extended the meaning of an
algebraical symbol. And it is to be noticed that an equation
between vectors, as
a = /3,
contains three distinct equations between mere numbers.
19. We must now define + (and the meaning of will follow)
in the new Calculus. Let A, B, G be any three points, and (with
the above meaning of = ) let
If we define + (in accordance with the idea ( 16) that a vector
represents a translation) by the equation
or
we contradict nothing that precedes, but we at once introduce the
idea that vectors are to be compounded, in direction and magnitude,
like simultaneous velocities. A reason for this may be seen in
another way if we remember that by adding the (algebraic) differ
ences of the Cartesian coordinates of B and A, to those of the
coordinates of G and B, we get those of the coordinates of C and
A. Hence these coordinates enter linearly into the expression for
a vector. (See, again, 5.)
20. But we also see that if C and A coincide (and G may be
any point)
AC=0,
for no vector is then required to carry A to G. Hence the above
relation may be written, in this case,
AB + BA = 0,
or, introducing, and by the same act defining, the symbol ,
BA=AB t
Hence, the symbol , applied to a vector, simply shews that its
currency is to be reversed.
8 QUATERNIONS. \_21.
And this is consistent with all that precedes ; for instance,
and AB = ACBC,
or =AC + CB,
are evidently but different expressions of the same truth.
21. In any triangle, ABC, we have, of course,
and, in any closed polygon, whether plane or gauche,
AB+BC+ ...... + YZ + ZA = Q.
In the case of the polygon we have also
AB + BC+ ...... + YZ = AZ.
These are the wellknown propositions regarding composition
of velocities, which, by Newton s second law of motion, give us
the geometrical laws of composition of forces acting at one point.
22. If we compound any number of parallel vectors, the result /
is obviously a numerical multiple of any one of them.
Thus, if A, B, G are in one straight line,
where x is a number, positive when B lies between A and C, other
wise negative : but such that its numerical value, independent
of sign, is the ratio of the length of BC to that of AB. This is
at once evident if AB and BC be commensurable ; and is easily
extended to incommensurables by the usual reductio ad absurdum.
23. An important, but almost obvious, proposition is that any
vector may be resolved, and in one way only, into three components
parallel respectively to any three given vectors, no two of which are
parallel, and which are not parallel to one plane.
Let OA, OB, OC be the three fixed c
vectors, OP any other vector. From P draw
PQ parallel to CO, meeting the plane BO A
in Q. [There must be a definite point Q,
else PQ, and therefore CO, would be parallel
to BOA, a case specially excepted.] From Q
draw QR parallel to BO, meeting OA in R.
Then we have OP=OR + RQ + QP ( 21),
and these components are respectively parallel to the three given
26.] VECTORS, AND THEIR COMPOSITION. 9
vectors. By 22 we may express OR as a numerical multiple
of OA, RQ of OB, and QP of OC. Hence we have, generally, for
any vector in terms of three fixed noncoplanar vectors, a, 0, 7,
OP = p = XQL + yft + 7,
which exhibits, in one form, the three numbers on which a vector
depends ( 16). Here x, y, z are perfectly definite, and can have
but single values.
24. Similarly any vector, as OQ, in the same plane with OA
and OB, can be resolved (in one way only) into components OR,
RQ, parallel respectively to A and OB ; so long, at least, as these
two vectors are not parallel to each other.
25. There is particular advantage, in certain cases, in em
ploying a series of three mutually perpendicular unitvectors as
lines of reference. This system Hamilton denotes by i,j, k.
Any other vector is then expressible as
p xi + yj + zk.
Since i, j, k are unitvectors, x, y, z are here the lengths of
conterminous edges of a rectangular parallelepiped of which p
is the vectordiagonal ; so that the length of p is, in this case,
Let w = f i + ijj + J&
be any other vector, then (by the proposition of 23) the vector
equation p = &
obyiously involves the following three equations among numbers,
a?=f, y = n, z =
Suppose i to be drawn eastwards, j northwards, and k upwards,
this is equivalent merely to saying that if two points coincide, they
are equally to the east (or west) of any third point, equally to the
north (or south) of it, and equally elevated above (or depressed below)
its level.
26. It is to be carefully noticed that it is only when a, /3, 7
are not coplanar that a vector equation such as
p = *&,
or XQL + y/3 f zy = f a + rj/3 + 7,
necessitates the three numerical equations
= fc V = V. * =
10 QUATERNIONS. [27.
For, if a, /3, 7 be coplanar ( 24), a condition of the following form
must hold
7 = aa. + b/3.
Hence p = (# + za) a
and the equation p = in
now requires only the two numerical conditions
x + za =
27. TAe Commutative and Associative Laws hold in the com
bination of vectors by the signs + and . It is obvious that, if we
prove this for the sign + , it will be equally proved for , because
before a vector ( 20) merely indicates that it is to be reversed
before being considered positive.
Let A, B, C, D be, in order, the corners of a parallelogram ; we
have, obviously,
And = ^ = .
Hence the commutative law is true for the addition of any two
vectors, and is therefore generally true.
Again, whatever four points are represented by A, B, G, D, we
have
or substituting their values for AD, BD, AC respectively, in these
three expressions,
And thus the truth of the associative law is evident.
28. The equation
p=3&
where p is the vector connecting a variable point with the origin,
/3 a definite vector, and x an indefinite number, represents the
straight line drawn from the origin parallel to @ ( 22).
The straight line drawn from A, where OA = a, arid parallel
to /3, has the equation
P = + */8 ................................. (1).
In words, we may pass directly from to P by the vector OP or
p ; or we may pass first to A, by means of OA or a, and then to P
along a vector parallel to /3 ( 16).
30.] VECTORS, AND THEIR COMPOSITION. 11
Equation (1) is one of the many useful forms into which
Quaternions enable us to throw the general equation of a straight
line in space. As we have seen ( 25) it is equivalent to three
numerical equations ; but, as these involve the indefinite quantity
x, they are virtually equivalent to but two, as in ordinary Geometry
of Three Dimensions.
29. A good illustration of this remark is furnished by the fact
that the equation
aft,
which contains two indefinite quantities, is virtually equivalent to:
only one numerical equation. And it is easy to see that it re
presents the plane in which the lines a. and j3 lie ; or the surface
which is formed by drawing, through every point of OA, a line
parallel to OB. In fact, the equation, as written, is simply 24
in symbols.
And it is evident that the equation
/i
P 7 + 2/ a + #
is the equation of the plane passing through the extremity of 7,
and parallel to a and ft.
It will now be obvious to the reader that the equation
where a t , a 2 , &c. are given vectors, and p lt p 9 , &c. numerical quan
tities, represents a straight line if p t , p 9 , &c. be linear functions of
one indeterminate number ; and a plane, if they be linear expres
sions containing two indeterminate numbers. Later ( 31 ()),
this theorem will be much extended.
Again, the equation
refers to any point whatever in space, provided a, ft, 7 are not
coplanar. (Ante, 23.)
30. The equation of the line joining any two points A and B,
where A = a and OB = ft, is obviously
or
These equations are of course identical, as may be seen by putting
1 y for x.
12 QUATE&NIONS. [31.
The first may be written
" tP
or
subject to the condition p + q + r = Q identically. That is A
homogeneous linear function of three vectors, equated to zero,
expresses that the extremities of these vectors are in one straight
line, if the sum of the coefficients be identically zero.
Similarly, the equation of the plane containing the extremities
A y B, C of the three noncoplanar vectors a, /3, 7 is
where x and y are each indeterminate.
This may be written
pp + qa + r/3 + sy = 0,
with the identical relation
which is one form of the condition that four points may lie in one
plane.
31. We have already the means of proving, in a very simple
manner, numerous classes of propositions in plane and solid
geometry. A very few examples, however, must suffice at this
stage ; since we have hardly, as yet, crossed the threshold of the
subject, and are dealing with mere linear equations connecting two
or more vectors, and even with them we are restricted as yet to
operations of mere addition. We will give these examples with a
painful minuteness of detail, which the reader will soon find to be
necessary only for a short time, if at all.
(a) The diagonals of a parallelogram bisect each other.
Let A BCD be the parallelogram, the point of intersection of
its diagonals. Then
AO + OB=AB=DC= DO + 00,
which gives AOOC = DOOB.
The two vectors here equated are parallel to the diagonals respect
ively. Such an equation is, of course, absurd unless
(1) The diagonals are parallel, in which case the figure
is not a parallelogram ;
(2) AO = OC, and DO = OB, the proposition.
31 (& )] VECTORS, AND THEIR COMPOSITION. 13
(6) To shew that a triangle can be constructed, whose sides
are parallel, and equal, to the bisectors of the sides of any
triangle.
Let ABC be any triangle, Aa, Bb, Cc the bisectors of the
sides.
Then Aa = AB + Tla = AB + BC, c
55 
~Cc  
Hence ~A~a + ~Bb + ~Cc = f ( + J9C + ) = ;
which (21) proves the proposition.
Also Aa=AB+BC
= AB$(CA+AB)
results which are sometimes useful. They may be easily verified
by producing A a to twice its length and joining the extremity
with B.
(b 1 ) The bisectors of the sides of a triangle meet in a point,
which trisects each of them.
Taking A as origin, and putting a, /3, 7 for vectors parallel, and
equal, to the sides taken in order BG, GA, AB; the equation of
Bb is ( 28 (1))
That of Cc is, in the same way,
p = (l H
At the point 0, where Bb and Cc intersect,
*
Since 7 and /S are not parallel, this equation gives
From these x = y = f .
Hence ZO = J (7  0) =  Aa. (See Ex. (6).)
This equation shews, being a vector one, that A a passes
through 0, and that AO : Oa :: 2:1.
14
(c) If
QUATERNIONS.
04 = 0,
OC = la + m/3,
be three given coplanar vectors, c the intersection of AB, OC, and
if the lines indicated in the figure be drawn, the points a v b v c l lie
in a straight line.
We see at once, by the process indicated in 30, that
c=
l + m
Hence we easily find
m/3
Im
m/3
Oa = 
ir
j
m l
ll2m l21m
These give
 (1  I  2m) ~0a l + (l2l m) ~Ob l  (m  I) Oc^ = 0.
But  (1  I  2m) + (1  21  m)  (m  I) = identically.
This, by 30, proves the proposition.
(d) Let OA = a, OB = /S, be any two vectors. If MP be a
given line parallel to OB ; and OQ, BQ, be drawn parallel to AP,
OP respectively ; the locus of Q is a straight line parallel to OA.
Let
Then
OM=eoi.
31 (e).] VECTORS, AND THEIR COMPOSITION. 15
Hence the equation of OQ is
and that of BQ is p = ft + z (ea. + acj3).
At Q we have, therefore,
These give xy = e, and the equation of the locus of Q is
p = e& + 2/X
i.e. a straight line parallel to 0.4, drawn through N in OjB pro
duced, so that
COR. If BQ meet MP in q, Pq = ; and if AP meet NQ in
p, Qp = a.
Also, for the point R we have pR = AP, QR = Bq.
Further, the locus of R is a hyperbola, of which MP and NQ
are the asymptotes. See, in this connection, 31 (k) below.
Hence, if from any two points, A and B, lines be drawn inter
cepting a given length Pq on a given line Mq ; and if, from R their
point of intersection, Rp be laid off= PA, and RQ = qB ; Q and p
lie on a fixed straight line, and the length of Qp is constant.
(e) To find the centre of inertia of any system of masses.
If OA = a, OB = a, l , be the vector sides of any triangle, the
vector from the vertex dividing the base AB in C so that
BC : CA :: m : m l
ma + m,GL
is
For AB is a x a, and therefore AC is
Hence 00 = OA + AC
_ ma. + m 1 1
m + ra t
This expression shews how to find the centre of inertia of two
masses ; m at the extremity of a, m l at that of a r Introduce m g
16
QUATERNIONS.
at the extremity of 2 , then the vector of the centre of inertia of the
three is, by a second application of the formula,
(m f
m
From this it is clear that, for any number of masses, expressed
generally by m at the extremity of the vector a, the vector of the
centre of inertia is
2 (ma)
This may be written 2m (a  ft) = 0.
Now a l  ft is the vector of m x with respect to the centre of inertia.
Hence the theorem, If the vector of each element of a mass, drawn
from the centre of inertia, be increased in length in proportion to the
mass of the element, the sum of all these vectors is zero.
(/) We see at once that
the equation
where t is an indeterminate
number, and a, ft given vec
tors, represents a parabola.
The origin, 0, is a point on
the curve, ft is parallel to
the axis, i.e. is the diameter
OB drawn from the origin,
and a is OA the tangent at the origin. In the figure
QP = at, OQ = ^.
2
The secant joining the points where t has the values t and t is
represented by the equation
**? (30)
V }
Write x for x (f  1) [which may have any value], then put tf = t,
and the equation of the tangent at the point (t) is
31 (g).~\ VECTORS, AND THEIR COMPOSITION. 17
In this put x = t, and we have
fit*
p ~ "T
or the intercept of the tangent on the diameter is equal in length
to the abscissa of the point of contact, but has the opposite
currency.
Otherwise : the tangent is parallel to the vector a + fit or
at + /3f or f + at + ^f or OQ + OP. But TP=TO + OP,
L 
hence TO = OQ.
(g) Since the equation of any tangent to the parabola is
*
+ fit),
let us find the tangents which can be drawn from a given point.
Let the vector of the point be
p=p + q/3 (24).
Since the tangent is to pass through this point, we have, as con
ditions to determine t and #,
f
2
by equating respectively the coefficients of a and fi.
Hence t=pJp* 2q.
Thus, in general, m> tangents can be drawn from a given point.
These coincide if p 2 = 2q ;
that is, if the vector of the point from which they are to be drawn
is p=pa + qP = pa + ^0,
i.e. if the point lies on the parabola. They are imaginary if
Zq >p z , that is, if the point be
r being positive. Such a point is evidently within the curve, as at
R, where OQ = ^/3, QP=pa, PR = rfi.
T. Q. I. 2
18 QUATERNIONS. [3 T (^)
(h) Calling the values of t for the two tangents found in (g)
^ and t 2 respectively, it is obvious that the vector joining the
points of contact is
which is parallel to 4/3  s >
or, by the values of ^ and 2 in (g),
a + p/3.
Its direction, therefore, does not depend on q. In words, If pairs of
tangents be drawn to a parabola from points of a diameter produced,
the chords of contact are parallel to the tangent at the vertex of the
diameter. This is also proved by a former result, for we must have
OT for each tangent equal to QO.
(i) The equation of the chord of contact, for the point whose
vector is
Rt 2
is thus p = 0^ +  + y (a + pft).
Suppose this to pass always through the point whose vector is
p = act + 6/3.
Then we must have ^ + y = a, \
or t^p + Jp* 2pa + 26.
Comparing this with the expression in (g), we have
q = pa  b ;
that is, the point from which the tangents are drawn has the vector
p=pa + (pab) fi
=  b/3 +p (a + aft), a straight line ( 28 (1)).
The mere form of this expression contains the proof of the usual
properties of the pole and polar in the parabola ; but, for the sake
of the beginner, we adopt a simpler, though equally general,
process.
Suppose a = 0. This merely restricts the pole to the particular
31 (/).] VECTORS, AND THEIR COMPOSITION. 19
diameter to which we have referred the parabola. Then the pole
is Q, where p = b/3;
and the polar is the line TU, for which
p=b/3+pot.
Hence the polar of any point is parallel to the tangent at the
extremity of the diameter on which the point lies, and its inter
section with that diameter is as far beyond the vertex as the pole
is within, and vice versa.
(j) As another example let us prove the following theorem.
If a triangle be inscribed in a parabola, the three points in which
the sides are met by tangents at the angles lie in a straight line.
Since is any point of the curve, we may take it as one corner
of the triangle. Let t and t t determine the others. Then, if
w,, sr 2 , OT 3 represent the vectors of the points of intersection of the
tangents with the sides, we easily find
t,"
tt l
f \ 1
These values give
j. 2 / 2
. J ir.=
v 2 t 2
Also ^   "*  " 1  = identically.
t tt.
Hence, by 30, the proposition is proved.
(k) Other interesting examples of this method of treating
curves will, of course, suggest themselves to the student. Thus
p = a. cos t + @ sin t
represents an ellipse, of which the given vectors a and ft are semi
conjugate diameters. If t represent time, the radiusvector of this
ellipse traces out equal areas in equal times. [We may anticipate
so far as to write the following :
2 Area = TJVpdp = TVap . fdt ;
which will be easily understood later.]
22
20 QUATERNIONS. [31 (/).
/?
Again, p = at + or p = a tan x + (3 cot x
t
evidently represents a hyperbola referred to its asymptotes. [If
t represent time, the sectorial area traced out is proportional to
log t t taken between proper limits.]
Thus, also, the equation
p = a. (t + sin t) + /3 cos t
in which a and ft are of equal lengths, and at right angles to one
another, represents a cycloid. The origin is at the middle point of
the axis (2/9) of the curve. [It may be added that, if t represent
time, this equation shews the motion of the tracing point, provided
the generating circle rolls uniformly, revolving at the rate of a
radian per second.]
When the lengths of a, ft are not equal, this equation gives the
cycloid distorted by elongation of its ordinates or abscissae : not a
trochoid. The equation of a trochoid may be written
p = a (et + sin f) + ft cos t,
e being greater or less than 1 as the curve is prolate or curtate.
The lengths of a and ft are still taken as equal.
But, so far as we have yet gone with the explanation of the
calculus, as we are not prepared to determine the lengths or in
clinations of vectors, we can investigate only a very limited class of
the properties of curves, represented by such equations as those
above written.
(1) We may now, in extension of the statement in 29, make
the obvious remark that
(where, as in 23, the number of vectors, a, can always be reduced
to three, at most) is the equation of a curve in space, if the
numbers p v p z , &c. are functions of one indeterminate. In such
a case the equation is sometimes written
P =<t>(t).
But, if p v p 2 , &c. be functions of two indeterminates, the locus of
the extremity of p is a surface; whose equation is sometimes written
[It may not be superfluous to call the reader s attention to the
fact that, in these equations, </> (t) or c/> (t, u) is necessarily a vector
expression, since it is equated to a vector, p,]
32.] VECTORS, AND THEIR COMPOSITION. 21
(m) Thus the equation
p = a cos t + /3 sin t 4 yt ..................... (1)
belongs to a helix, while
p = a cos t + @ sin t + 7^ .................. (2)
represents a cylinder whose generating lines are parallel to 7, and
whose base is the ellipse
p = a cos + /3 sin t.
The helix above lies wholly on this cylinder.
Contrast with (2) the equation
p = u(acost + /3sin t + 7) ..................... (3)
which represents a cone of the second degree : made up, in fact,
of all lines drawn from the origin to the ellipse
p = a cos t + sin t + 7.
If, however, we write
p = u (a cos t + @ sin 4 yt),
we form the equation of the transcendental cone whose vertex is
at the origin, and on which lies the helix (1).
In general
p = u<f> (t)
is the cone whose vertex is the origin, and on which lies the curve
while p = (f>(t) + UOL
is a cylinder, with generating lines parallel to a, standing on the
same curve as base.
Again, p = pa + q/3 + ry
with a condition of the form
op 2 + bf + cr 2 = I
belongs to a central surface of the second order, of which a, (3, 7
are the directions of conjugate diameters. If a, b, c be all positive,
the surface is an ellipsoid.
32. In Example (/") above we performed an operation equi
valent to the differentiation of a vector with reference to a single
numerical variable of which it was given as an explicit function.
As this process is of very great use, especially in quaternion investi
gations connected with the motion of a particle or point ; and as it
will afford us an opportunity of making a preliminary step towards
22 QUATERNIONS. [33.
overcoming the novel difficulties which arise in quaternion differen
tiation ; we will devote a few sections to a more careful, though
very elementary, exposition of it.
33. It is a striking circumstance, when we consider the way
in which Newton s original methods in the Differential Calculus
have been decried, to find that Hamilton was obliged to employ
them, and not the more modern forms, in order to overcome the
characteristic difficulties of quaternion differentiation. Such a thing
as a differential coefficient has absolutely no meaning in quaternions,
except in those special cases in which we are dealing with degraded
quaternions, such as numbers, Cartesian coordinates, &c. But a
quaternion expression has always a differential, which is, simply,
what Newton called a fluxion.
As with the Laws of Motion, the basis of Dynamics, so with the
foundations of the Differential Calculus ; we are gradually coming
to the conclusion that Newton s system is the best after all.
34. Suppose p to be the vector of a curve in space. Then,
generally, p may be expressed as the sum of a number of terms,
each of which is a multiple of a constant vector by a function of some
one indeterminate ; or, as in 31 (1), if P be a point on the curve,
OP= P = <HO.
And, similarly, if Q be any other point on the curve,
06 = ^ = ^ + 8^=0(0 = *^+^),
where St is any number whatever.
The vectorchord PQ is therefore, rigorously,
35. It is obvious that, in the present case, because the vectors
involved in c/> are constant, and their numerical multipliers alone vary,
the expression * (t + $t) is, by Taylor s Theorem, equivalent to
*, .
*. .
Hence, Sp = ^ ( +
And we are thus entitled to write, when Bt has been made inde
finitely small,
T . . /&p\ dp d<f) (t) , , A
Limit =  = ^ = </> (*)
36.]
VECTORS, AND THEIR COMPOSITION.
23
In such a case as this, then, we are permitted to differentiate,
or to form the differential coefficient of, a vector, according to the
ordinary rules of the Differential Calculus. But great additional
insight into the process is gained by applying Newton s method.
36. Let OP be
and OQ 1
Pl = (f> (t + dt),
where dt is any number whatever.
The number t may here be taken
as representing time, i.e. we may
suppose a point to move along the
curve in such a way that the value
of t for the vector of the point P of
the curve denotes the interval which
has elapsed (since a fixed epoch) when the moving point has
reached the extremity of that vector. If, then, dt represent any
interval, finite or not, we see that
will be the vector of the point after the additional interval dt.
But this, in general, gives us little or no information as to the
velocity of the point at P. We shall get a better approximation
by halving the interval dt, and finding Q 2 , where OQ 2 = $ (t + cft),
as the position of the moving point at that time. Here the vector
virtually described in \dt is PQ 2 . To find,_on this supposition,
the vector described in dt, we must double PQ 2 , and we find, as a
second approximation to the vector which the moving point would
have described in time dt, if it had moved for that period in the
direction and with the velocity it had at P,
= 2
The next approximation gives
 8
And so on, each step evidently leading us nearer the sought truth.
Hence, to find the vector which would have been described in time
dt had the circumstances of the motion at P remained undisturbed,
we must find the value of
dp = Tq =
(t 4 i dt}  4> (t)\
\ X /
24 QUATERNIONS. [37.
We have seen that in this particular case we may use Taylor s
Theorem. We have, therefore,
dp = ^ = 00 X {< (0 \ dt + f (t) A ^ + &C.J
= (#> (t) dt.
And, if we choose, we may now write
s *<
37. But it is to be most particularly remarked that in the
whole of this investigation no regard whatever has been paid to
the magnitude of dt. The question which we have now answered
may be put in the form A point describes a given curve in a given
manner. At any point of its path its motion suddenly ceases to be
accelerated. What space will it describe in a definite interval ? As
Hamilton well observes, this is, for a planet or comet, the case
of a celestial Atwood s machine.
38. If we suppose the variable, in terms of which p is expressed,
to be the arc, s, of the curve measured from some fixed point, we
find as before
dp = fi(s) ds.
From the very nature of the question it is obvious that the length
of dp must in this case be ds, so that </> () is necessarily a unit
vector. This remark is of importance, as we shall see later ; and
it may therefore be useful to obtain afresh the above result without
any reference to time or velocity.
39. Following strictly the process of Newton s Vllth Lemma,
let us describe on Pg 2 an arc similar to PQ 2 , and so on. Then
obviously, as the subdivision of ds is carried farther, the new arc
(whose length is always ds) more and more nearly (and without
limit) coincides with the line which expresses the corresponding
approximation to dp.
40. As additional examples let us take some wellknown
plane curves; and first the hyperbola (31
Here db1 )<&
41.] VECTORS, AND THEIR COMPOSITION. 25
This shews that the tangent is parallel to the vector
f
In words, if the vector (from the centre) of a point in a hyperbola
be one diagonal of a parallelogram, two of whose sides coincide with
the asymptotes, the other diagonal is parallel to the tangent at the
point, and cuts off a constant area from the space between the
asymptotes. (For the sides of this triangular area are t times the
length of a, and I/t times the length of ft, respectively ; the angle
between them being constant.)
Next, take the cycloid, as in 31 (&),
p = a (t + sin t)+ft cos t.
We have
dp = {(!+ cos t) ft sin t} dt.
At the vertex
t = 0, cos t = l, sin t = 0, and dp = 2adt.
At a cusp
t = TT, cos = 1, sin = 0, and dp = 0.
This indicates that, at the cusp, the tracing point is (instan
taneously) at rest. To find the direction of the tangent, and the
form of the curve in the vicinity of the cusp, put t = TT + T, where
powers of T above the second are omitted. We have
dp = ftrdt + J" dt,
2*
so that, at the cusp, the tangent is parallel to ft. By making the
same substitution in the expression for p, we find that the part of
the curve near the cusp is a semicubical parabola,
or, if the origin be shifted to the cusp (p = TTOL  ft\
41. Let us reverse the first of these questions, and seek the
envelop of a line which cuts off from two fixed axes a triangle of
constant area.
If the axes be in the directions of a and ft, the intercepts may
evidently be written at and ^. Hence the equation of the line is
(30)
//? \
p = a.t 4 x ( j at) .
26 QUATERNIONS. [42.
The condition of envelopment is, obviously, (see Chap. IX.)
This gives = la 
(
x f + a dt +  ctf
*
Hence (1 as) dt tdx = 0,
x , dx
and   dt + = 0.
6 C
From these, at once, x = J, since cfcc and cfa are indeterminate.
Thus the equation of the envelop is
the hyperbola as before ; a, /3 being portions of its asymptotes.
42. It may assist the student to a thorough comprehension
of the above process, if we put it in a slightly different form.
Thus the equation of the enveloping line may be written
which gives dp = = CLd {t(l  as)} + @d .
\*/
Hence, as a is not parallel to /3, we must have
d
and these are, when expanded, the equations we obtained in the
preceding section.
43. For farther illustration we give a solution not directly
employing the differential calculus. The equations of any two of
the enveloping lines are
* Here we have opportunity for a remark (very simple indeed, but) of the
utmost importance. We are not to equate separately to zero the coefficients of dt
and dx; for we must remember that this equation is of the form
where p and q are numbers ; and that, so long as a and /3 are actual and nonparallel
vectors, the existence of such an equation requires ( 24)
43] VECTORS, AND THEIR COMPOSITION. 27
(R \
p = at + x (  at ) ,
\i /
t and t t being given, while x and x l are indeterminate.
At the point of intersection of these lines we have ( 20),
These give, by eliminating x v
t n _ .\
i \L x)
or x .
Hence the vector of the point of intersection is
_ aflt + ff
P= tt + t
and thus, for the ultimate intersections, where <  = 1,
t
+ ^ ) as before.
COR. If. instead of the ultimate intersections, we consider
the intersections of pairs of these lines related by some law, we
obtain useful results. Thus let
OL + /3
;
or the intersection lies in the diagonal of the parallelogram on a, 0.
If ^ = mt, where m is constant,
8
mtot +
P =
i
But we have also x =
m
1
mf 1
Hence the locus of a point which divides in a given ratio a line
cutting off a given area from two fixed axes, is a hyperbola of which
these axes are the asymptotes.
28 QUATERNIONS. [44.
If we take either
W
tt^t + tfj) = constant, or * = constant,
r + c,
the locus is a parabola ; and so on.
It will be excellent practice for the student, at this stage, to
work out in detail a number of similar questions relating to the
envelop of, or the locus of the intersection of selected pairs from, a
series of lines drawn according to a given law. And the process
may easily be extended to planes. Thus, for instance, we may
form the general equation of planes which cut off constant tetra
hedra from the axes of coordinates. Their envelop is a surface of
the third degree whose equation may be written
p = CCOL + y/3 + 7 ;
where ocyz = a 3 .
Again, find the locus of the point of intersection of three of
this group of planes, such that the first intercepts on /3 and 7, the
second on 7 and a, the third on a and /3, lengths all equal to one
another, &c. But we must not loiter with such simple matters as
these.
44. The reader who is fond of Anharmonic Ratios and Trans
versals will find in the early chapters of Hamilton s Elements of
Quaternions an admirable application of the composition of vectors
to these subjects. The Theory of Geometrical Nets, in a plane,
and in space, is there very fully developed; and the method is
shewn to include, as particular cases, the corresponding processes of
Grassmann s Ausdehnungslehre and Mobius Barycentrische Calcul.
Some very curious investigations connected with curves and surfaces
of the second and third degrees are also there founded upon the
composition of vectors.
EXAMPLES TO CHAPTER I.
1. The lines which join, towards the same parts, the extremities
of two equal and parallel lines are themselves equal and parallel.
(Euclid, I. xxxiii.)
2. Find the vector of the middle point of the line which joins
the middle points of the diagonals of any quadrilateral, plane or
VECTORS, AND THEIR COMPOSITION. 29
gauche, the vectors of the corners being given ; and so prove that
this point is the mean point of the quadrilateral.
If two opposite sides be divided proportionally, and two new
quadrilaterals be formed by joining the points of division, the mean
points of the three quadrilaterals lie in a straight line.
Shew that the mean point may also be found by bisecting the
line joining the middle points of a pair of opposite sides.
3. Verify that the property of the coefficients of three vectors
whose extremities are in a line ( 30) is not interfered with by
altering the origin.
4. If two triangles ABC, abc, be so situated in space that Aa,
Bb, Cc meet in a point, the intersections of AB, ab, of BO, be, and
of CA, ca, lie in a straight line.
5. Prove the converse of 4, i. e. if lines be drawn, one in each
of two planes, from any three points in the straight line in which
these planes meet, the two triangles thus formed are sections of a
common pyramid.
6. If five quadrilaterals be formed by omitting in succession
each of the sides of any pentagon, the lines bisecting the diagonals
of these quadrilaterals meet in a point. (H. Fox Talbot.)
7. Assuming, as in 7, that the operator
cos 6 + J 1 sin 6
turns any radius of a given circle through an angle 6 in the
positive direction of rotation, without altering its length, deduce
the ordinary formulae for cos (A f B), cos (A B), sin (^L + B), and
sin (A B), in terms of sines and cosines of A and B.
8. If two tangents be drawn to a hyperbola, the line joining
the centre with their point of intersection bisects the lines join
ing the points where the tangents meet the asymptotes : and the
secant through the points of contact bisects the intercepts on
the asymptotes.
9. Any two tangents, limited by the asymptotes, divide each
other proportionally.
10. If a chord of a hyperbola be one diagonal of a parallelogram
whose sides are parallel to the asymptotes, the other diagonal passes
through the centre.
30 QUATERNIONS.
11. Given two points A and B, and a plane, C. Find the
locus of P, such that if AP cut C in Q, and BP cut G in R, QR
may be a given vector.
12. Shew that p = # 2 a + ?/ 2 /3 + (a? + y)V
is the equation of a cone of the second degree, and that its section
by the plane
_ pa. + q/3 + ry
^ p + q + r
is an ellipse which touches, at their middle points, the sides of the
triangle of whose corners a, /3, 7 are the vectors. (Hamilton,
Elements, p. 96.)
13. The lines which divide, proportionally, the pairs of opposite
sides of a gauche quadrilateral, are the generating lines of a hyper
bolic paraboloid. (Ibid. p. 97.)
14. Shew that p = x s a + y s /3 + z s y,
where # + 2/ + 2 = 0,
represents a cone of the third order, and that its section by the
plane
__ pa 4 q/3 + ry
p + q + r
is a cubic curve, of which the lines
pCL + q/3 p
p =   , &c.
p + q
are the asymptotes and the three (real) tangents of inflection. Also
that the mean point of the triangle formed by these lines is a
conjugate point of the curve. Hence that the vector a + /3 + y is
a conjugate ray of the cone. (Ibid. p. 96.)
CHAPTER II.
PRODUCTS AND QUOTIENTS OF VECTORS.
45. WE now come to the consideration of questions in which
the Calculus of Quaternions differs entirely from any previous
mathematical method ; and here we shall get an idea of what a
Quaternion is, and whence it derives its name. These questions
are fundamentally involved in the novel use of the symbols of
multiplication and division. And the simplest introduction to
the subject seems to be the consideration of the quotient, or ratio,
of two vectors.
46. If the given vectors be parallel to each other, we have
already seen ( 22) that either may be expressed as a numerical
multiple of the other; the multiplier being simply the ratio of
their lengths, taken positively if they have similar currency,
negatively if they run opposite ways.
47. If they be not parallel, let OA and OB be drawn parallel
and equal to them from any point ; and the question is reduced
to finding the value of the ratio of two vectors drawn from the
same point. Let us first find upon how many distinct numbers this
ratio depends.
We may suppose OA to be changed into OB by the following
successive processes.
1st. Increase or diminish the length of OA till it becomes
equal to that of OB. For this only one number is required, viz.
the ratio of the lengths of the two vectors. As Hamilton remarks,
this is a positive, or rather a signless, number.
2nd. Turn OA about 0, in the common plane of the two
vectors, until its direction coincides with that of OB, and (remem
32 QUATERNIONS. [48.
bering the effect of the first operation) we see that the two vectors
now coincide or become identical. To specify this operation three
numbers are required, viz. two angles (such as node and inclination
in the case of a planet s orbit) to fix the plane in which the rotation
takes place, and one angle for the amount of this rotation.
Thus it appears that the ratio of two vectors, or the multiplier
required to change one vector into another, in general depends upon
four distinct numbers, whence the name QUATERNION.
A quaternion q is thus defined as expressing a relation
between two vectors a, /3. By what precedes, the vectors a, /3,
which serve for the definition of a given quaternion, must be in a
given plane, at a given inclination to each other, and with their
lengths in a given ratio ; but it is to be noticed that they may be
any two such vectors. [Inclination is understood to include sense,
or currency, of rotation from a to /3.]
The particular case of perpendicularity of the two vectors, where
their quotient is a vector perpendicular to their plane, is fully
considered below ; 64, 65, 72, &c.
48. It is obvious that the operations just described may be
performed, with the same result, in the opposite order, being per
fectly independent of each other. Thus it appears that a quaternion,
considered as the factor or agent which changes one definite vector
into another, may itself be decomposed into two factors of which
the order is immaterial.
The stretching factor, or that which performs the first operation
in 47, is called the TENSOR, and is denoted by prefixing T to the
quaternion considered.
The turning factor, or that corresponding to the second operation
in 47, is called the VERSOR, and is denoted by the letter U pre
fixed to the quaternion.
49. Thus, if OA = a, OB = {3, and if q be the quaternion
which changes a. to /3, we have
which we may write in the form
a ^ *
if we agree to define that
50.]
RODUCTS AND QUOTIENTS OF VECTORS.
33
Here it is to be particularly noticed that we write q before a to
signify that a is multiplied by (or operated on by) q, not q
multiplied by ot.
This remark is of extreme importance in quaternions, for, as we
shall soon see, the Commutative Law does not generally apply to
the factors of a product.
We have also, by 47, 48,
q = Tq.Uq=Uq. Tq,
where, as before, Tq depends merely on the relative lengths of
a and /3, and Uq depends solely on their directions.
Thus, if t and /9 X be vectors of unit length parallel to a. and (3
respectively,
T =
= U/3J Ua^
As will soon be shewn, when a is perpendicular to /3, i.e. when the
versor of the quotient is quadrantal, it is a unitvector.
50. We must now carefully notice that the quaternion which
is the quotient when (3 is divided by a in no way depends upon
the absolute lengths, or directions, of these vectors. Its value
will remain unchanged if we substitute for them any other pair
of vectors which
(1) have their lengths in the same ratio,
(2) have their common plane the same or parallel,
and (3) make the same angle with each other.
Thus in the annexed figure
if, and only if,
^w !J;=S>
(2) plane AOB parallel to plane Afl^B v
/o\ / A (~\~R / A C\ 2?
\*J/ Z. xl L/O == . 4TL.\J.D .
[Equality of angles is understood to include
concurrency of rotation. Thus in the annexed
figure the rotation about an axis drawn upwards
from the plane is negative (or clock wise) from /
OA to OB, and also from O l A l to 0^.]
T. Q. I.
34 QUATERNIONS. [51.
It thus appears that if
f} = qaL, & = qy,
the vectors a, /3, 7, S are parallel to one plane, and may be repre
sented (in a highly extended sense) as proportional to one another,
thus :
j3 : a = 8 : 7.
And it is clear from the previous part of this section that this
may be written not only in the form
a :=7 : 8
but also in either of the following forms :
7 : a = B : .
a : 7 = : 3.
While these proportions are true as equalities of ratios, they
do not usually imply equalities of products.
Thus, as the first of these was equivalent to the equation
^= S =g, or /3or l = bf l = q ,
OL 7
the following three imply separately, (see next section)
a _ ?_! 7_S a _/3 i
jg* 0 7~S~
or, if we please,
afiT 1 = y& 1 = q~\ 7~ l = S/3" 1 = r, a<f l = /3S 1 = r~ l ;
where r is a new quaternion, which has not necessarily anything
(except its plane), in common with q.
But here great caution is requisite, for we are not entitled to
conclude from these that
ctS = /3y, &c.
This point will be fully discussed at a later stage. Meanwhile
we may merely state that from
we are entitled to deduce a number of equivalents such as
a/rT^ = 7, or a = 7S~ 1 /9, or yS 1 ^ = a ^y, &c.
51. The Reciprocal of a quaternion <? is defined by the
equation
51.] PRODUCTS AND QUOTIENTS OF VECTORS. 35
R
Hence if = 7 or
we must have 5 ==s ~ =s *?~ 1
For this gives ~. ft = q~ l . qy,
and each member of the equation is evidently equal to a.
Or thus :
= q
Operate by q~ l ,
?  /3 = .
Operate on {3 l ,
Or, we may reason thus : since q changes OA to 07?, t?" 1 must
change OB to O.A, and is therefore expressed by ^ ( 49).
The tensor of the reciprocal of a quaternion is therefore the
reciprocal of the tensor ; and the versor differs merely by the
reversal of its representative angle. The versor, it must be
remembered, gives the plane and angle of the turning it has
nothing to do with the extension.
[Remark. In 49 51, above, we had such expressions as
 = /3oT l . We have also met with a" 1 /#. Cayley suggests that this
also may be written in the ordinary fractional form by employing
the following distinctive notation :
(It might, perhaps, be even simpler to use the solidus as
recommended by Stokes, along with an obviously correlative
type : thus,
I have found such notations occasionally convenient for private
work, but I hesitate to introduce changes unless they are abso
lutely required. See remarks on this point towards the end of the
Preface to the Second Edition reprinted above.]
32
J
36 QUATERNIONS. [5
2.
52. The Conjugate of a quaternion q, written Kq, has the
same tensor, plane, and angle, only the angle is taken the reverse
way; or the versor of the conjugate is the reciprocal of the versor
of the quaternion, or (what comes to the same thing) the versor of
the reciprocal.
Thus, if OA, OB, OA , lie in one plane, and if
OA = OA, and Z A OB = Z BOA, we have
OB ,OB
 = q, and =. = coniugate oi q = Kq.
OA OA
By last section we see that
Hence qKq = Kq . q = (Tq)\
This proposition is obvious, if we recollect that
the tensors of q and Kq are equal, and that the
J versors are such that either annuls the effect of the other ; while
the order of their application is indifferent. The joint effect of
these factors is therefore merely to multiply twice over by the
common tensor.
53. It is evident from the results of 50 that, if a. and ft be
of equal length, they may be treated as of unitlength so far as
their quaternion quotient is concerned. This quotient is therefore
a versor (the tensor being unity) and may be represented indif
ferently by any one of an infinite number of concurrent arcs of
given length lying on the circumference of a circle, of which the
two vectors are radii. This is of considerable importance in the
proofs which follow.
r\ 7>
Thus the versor = may be represented
in magnitude, plane, and currency of rota
tion ( 50) by the arc AB, which may in
this extended sense be written AB.
And, similarly, the versor =4 is repre
OA t
sented by AJBj which is equal to (and concurrent with) AB if
i.e. if the versors are equal, in the quaternion meaning of the
word.
55] PRODUCTS AND QUOTIENTS OF VECTORS. 37
54. By the aid of this process, when a versor is represented as
an arc of a great circle on the unitsphere, we can easily prove
that quaternion multiplication is not generally commutative.
^^ (VR
Thus let q be the versor AB or _ _ ,
where is the centre of the sphere.
Take BC = AB, (which, it must be re
membered, makes the points A, B, C, lie
in one great circle), then q may also be
OC
represented by = .
In the same way any other versor r may be represented by
^_^ ^^ /\ 7)
DB or BE and by = or = .
3 OD OB
[The line OB in the figure is definite, and is given by the
intersection of the planes of the two versors.]
Now rOD = OB, and qOB=OC.
Hence qrOD = OC,
t\ri
or qr = T=T , and may therefore be represented by the arc DC of
a great circle.
But rq is easily seen to be represented by the arc AE.
For qOA=OB, and rOB=OE 9
OE
whence rqOA OE. and rq =
OA
Thus the versors rq and qr, though represented by arcs of equal
length, are not generally in the same plane and are therefore
unequal: unless the planes of q and r coincide.
Remark. We see that we have assumed, or defined, in the
above proof, that q . m = qr . a. and r.qz = rq.oLm the special case
when qcn, ra, q . rot, and r . qx are all vectors.
55. Obviously CB is Kq, BD is Kr, and CD is K (qr). But
CD = BI) . CB, as we see by applying both to OC. This gives us
the very important theorem
K (qr) = Kr . Kq, \/
i.e. the conjugate of the product of two versors is the product of their
38 QUATERNIONS. [56.
conjugates in inverted order. This will, of course, be extended to
any number of factors as soon as we have proved the associative
property of multiplication. ( 58 below.)
56. The propositions just proved are, of course, true of quater
nions as well as of versors; for the former involve only an additional
numerical factor which has reference to the length merely, and not
the direction, of a vector ( 48), and is therefore commutative with
all other factors.
57. Seeing thus that the commutative law does not in general
hold in the multiplication of quaternions, let us enquire whether
the Associative Law holds generally. That is if p, q, r be three
quaternions, have we
p . qr=pq. r ?
This is, of course, obviously true if^>, q, r be numerical quantities,
or even any of the imaginaries of algebra. But it cannot be con
sidered as a truism for symbols which do not in general give
pq = qp.
We have assumed it, in definition, for the special case when r,
qr, and pqr are all vectors. ( 54.) But we are not entitled to
assume any more than is absolutely required to make our
definitions complete.
58. In the first place we remark that p, q, and r may be
considered as versors only, and therefore represented by arcs of /
great circles on the unit sphere, for their tensors may obviously
( 48) be divided out from both sides, being commutative with the
versors.
LetAB=p,EI) = CA = q, and FE = r.
Join BC and produce the great circle till it meets EF in H, and
make KH=FE = r, and HG = CB =pq ( 54).
Join GK. Then JtG = HG . KH = pq . r.
59 ".] PRODUCTS AND QUOTIENTS OF VECTORS. 39
Join FD and produce it to meet AB in M. Make
LM = FD, and MN = AB,
and join NL. Then LN= MN .Lbl=p.qr.
Hence to shew that p . qr = pq . r
all that is requisite is to prove that LN, and KG, described as
above, are equal arcs of the same great circle, since, by the figure,
they have evidently similar currency. This is perhaps most easily
effected by the help of the fundamental properties of the curves
known as Spherical Conies. As they are not usually familiar to
students, we make a slight digression for the purpose of proving
these fundamental properties ; after Chasles, by whom and Magnus
they were discovered. An independent proof of the associative
principle will presently be indicated, and in Chapter VIII. we shall
employ quaternions to give an independent proof of the theorems
now to be established.
59.* DEF. A spherical conic is the curve of intersection of a
cone of the second degree with a sphere, the vertex of the cone being
the centre of the sphere.
LEMMA. If a cone have one series of circular sections, it has
another series, and any two circles belonging to different series lie
on a sphere. This is easily proved as follows.
Describe a sphere, A, cutting the cone in one circular section,
C, and in any other point whatever, and let the side OpP of the
cone meet A in p, P ; P being a point in G. Then PO.Op is
constant, and, therefore, since P lies in a plane, p lies on a sphere,
a, passing through 0. Hence the locus, c, of p is a circle, being
the intersection of the two spheres A and a.
Let OqQ be any other side of the cone, q and Q being points in
c, C respectively. Then the quadrilateral qQPp is inscribed in a
circle (that in which its plane cuts the sphere A) and the exterior
40 QUATERNIONS. [60.
angle at p is equal to the interior angle at Q. If OL, OM be the
lines in which the plane POQ cuts the cyclic planes (planes through
parallel to the two series of circular sections) they are obviously
parallel to pq, QP, respectively ; and therefore
z LOp = z Opq = z OQP = z M OQ.
Let any third side, OrR, of the cone be drawn, and let the
plane OPR cut the cyclic planes in 01, Om respectively. Then,
evidently,
^IOL= Z qpr,
Zl/0m=
and these angles are independent of the position of the points p
and P, if Q and R be fixed points.
In the annexed section of the above spacediagram by a sphere
whose centre is 0, IL, Mm are the great circles which represent
the cyclic planes, PQR is the spherical conic which represents the
cone. The point P represents the line OpP, and so with the
others. The propositions above may now be stated thus,
Arc PL = arc MQ ;
and, if Q and R be fixed, M m and IL are constant arcs whatever be
the position of P.
60. The application to 58 is now obvious. In the figure of
that article we have
= GB, LM
Hence L, (7, G, D are points of a spherical conic whose cyclic
planes are those of AB, FE. Hence also KG passes through L,
and with LM intercepts on AB an arc equal to AB. That is, it
passes through j\ r , or KG and LN are arcs of the same great circle :
and they are equal, for G and L are points in the spherical
conic.
62.]
PRODUCTS AND QUOTIENTS OF VECTORS.
41
Also, the associative principle holds for any number of
quaternion factors. For, obviously,
qr . st = qrs . t = &c., &c.,
since we may consider qr as a single quaternion, and the above
proof applies directly.
61. That quaternion addition, and therefore also subtraction,
is commutative, it is easy to shew.
For if the planes of two quaternions,
q and r, intersect in the line OA, we
may take any vector OA in that line,
and at once find two others, OB and
00, such that
OB = qOA,
and CO = rOA.
And (q + r)OA**OB+OC=OC+OB=(r + q) OA,
since vector addition is commutative ( 27).
Here it is obvious that (q + r) OA, being the diagonal of the
parallelogram on OB, OC, divides the angle between OB and OG
in a ratio depending solely on the ratio of the lengths of these
lines, i.e. on the ratio of the tensors of q and r. This will be useful
to us in the proof of the distributive law, to which we proceed.
62. Quaternion multiplication, and therefore division, is
distributive. One simple proof of this depends on the possibility,
shortly to be proved, of representing any quaternion as a linear
function of three given rectangular unit vectors. And when the
proposition is thus established, the associative principle may readily
be deduced from it.
[But Hamilton seems not to have noticed that we may employ
for its proof the properties of Spherical Conies already employed
7
42 QUATERNIONS. [62.
in demonstrating the truth of the associative principle. "For
continuity we give an outline of the proof by this process.
Let BA, CA represent the versors of q arid r, and be the great
circle whose plane is that of p.
Then, if we take as operand the vector OA, it is obvious that
U (q + r) will be represented by some such arc as D A where
B, D, C are in one great circle; for (q + r) OA is in the same plane
as qO A and rOA, and the relative magnitude of the arcs BD and
DC depends solely on the tensors of q and r. Produce BA, DA,
CA to meet be in b, d, c respectively, and make
Eb = BA t Fd = DA, Gc = CA.
Also make bj3 = d& = cy=p. Then E, F, G, A lie on a spherical
conic of which BC and be are the cyclic arcs. And, because
b/3 = d8 = cry, (3E, SF, 7$, when produced, meet in a point H
which is also on the spherical conic ( 59*). Let these arcs meet
BC in J, L, K respectively. Then we have
LH=F8=pU(q
Also LJ
and KL = CD.
And, on comparing the portions of the figure bounded respectively
by HKJ and by ACB we see that (when considered with reference
to their effects as factors multiplying OH and OA respectively)
pU(q 4 r) bears the same relation to pUq and pUr
that U(q + r) bears to Uq and Ur.
But T(q + r)U(q + r) = q + r=TqUq+TrUr.
Hence T (q + r~) .pU(q + r) = Tq .pUq + Tr .pUr;
or, since the tensors are mere numbers and commutative with all
other factors,
In a similar manner it may be proved that
(q + r)p = qp + rp.
And then it follows at once that
(p + q) (r + s) = pr + ps + q r + qs,
where, by 61, the order of the partial products is immaterial.]
Y
66.] PRODUCTS AND QUOTIENTS OF VECTORS. 43
63. By similar processes to those of 53 we see that versors,
and therefore also quaternions, are subject to the indexlaw
at least so long as m and n are positive integers.
The extension of this property to negative and fractional
exponents must be deferred until we have defined a negative or
fractional power of a quaternion.
64. We now proceed to the special case of quadrantal versors,
from whose properties it is easy to deduce all the foregoing
results of this chapter. It was, in fact, these properties whose
invention by Hamilton in 1843 led almost intuitively to the
establishment of the Quaternion Calculus. We shall content
ourselves at present with an assumption, which will be shewn
to lead to consistent results ; but at the end of the chapter we
shall shew that no other assumption is possible, following for this
purpose a very curious quasimetaphysical speculation of Hamilton.
65. Suppose we have a system of three mutually perpendicular
unitvectors, drawn from one point, which we may call for shortness
i, j, k. Suppose also that these are so situated that a positive
(i.e. lefthanded) rotation through a right angle about i as an axis
brings j to coincide with k. Then it is obvious that positive
quadrantal rotation about j will make k coincide with i; and, about
k, will make i coincide with j.
For defmiteness we may suppose i to be drawn eastwards, j
northwards, and k upwards. Then it is obvious that a positive
(lefthanded) rotation about the eastward line (i) brings the north
ward line (j) into a vertically upward position (k) ; and so of the
others.
\J 66. Now the operator which turns j into k is a quadrantal
versor ( 53) ; and, as its axis is the vector i, we may call it i.
Thus = i, or k = ?j (1).
Similarly we may put  = j, or i = jk (2),
k
and j=&, or J = H (3).
[It may be here noticed, merely to shew the symmetry of the
system we arc explaining, that if the three mutually perpendicular
44
QUATERNIONS.
[67
vectors i, j, k be made to revolve about a line equally inclined to
all, so that i is brought to coincide with j, j will then coincide
with k, and k with i : and the above equations will still hold good,
only (1) will become (2), (2) will become (3), and (3) will become
(i).]
67. By the results of 50 we see that
i.e. a southward unit vector bears the same ratio to an upward
unitvector that the latter does to a northward one ; and therefore
we have
or j =ik
Similarly
and
k
=j, or k=ji.
= K, or i =
.(4).
(5);
(6).
G8. By (4) and (1) we have
j = ^k = i(ij) (by the assumption in 54) = i"j.
Hence i 2 =  1 (7).
Arid in the same way, (5) and (2) give
J 2 = 1 (8),
and (6) and (3) & 2 = l (9).
Thus, as the directions of i, j, k are perfectly arbitrary, we see
that the square of every quadrantal versor is negative unity.
[Though the following proof is in principle exactly the same as
the foregoing, it may perhaps be of use to the student, in shewing
him precisely the nature as well as the simplicity of the step we
have taken.
Let ABA be a semicircle, whose centre
is 0, and let OB be perpendicular to AOA .
CYR
Then = , = q suppose, is a quadrantal
UA
versor, and is evidently equal to  ;
50, 53.
Hence
OA OH OA
^=. = = =^
OB OA OA
69]
PRODUCTS AND QUOTIENTS OF VECTORS.
45
69. Having thus found that the squares of i, j, k are each
equal to negative unity ; it only remains that we find the values of
their products two and two. For, as we shall see, the result is such
as to shew that the value of any other combination whatever of
i,j, k (as factors of a product) may be deduced from the values of
these squares and products.
Now it is obvious that
k i
("Us VH>f
utM
^^
r W
T, fv
(i.e. the versor which turns a westward unitvector into an upward
one will turn the upward into an eastward unit) ;
or k=j(i) = ji* ........................ (10).
Now let us operate on the two equal vectors in (10) by the
same versor, i t and we have
ik = i (ji) = iji.
But by (4) and (3)
ik = J = H.
Comparing these equations, we have
or, by 54 (end),
and symmetry gives
ij = k,\
jk i t \
hi =. }
(11).
The meaning of these important equations is very simple ; and
is, in fact, obvious from our construction in 54 for the multiplica
tion of versors ; as we see by the annexed figure, where we must
remember that i, j, k are quadrantal versors whose planes are at
right angles, so that the figure represents
a hemisphere divided into quadrantal
triangles. [The arrowheads indicate the
direction of each vector arc.]
Thus, to shew that ij = k, we have,
being the centre of the sphere, N, E,
8, W the north, east, south, and west,
and ^the zenith (as in 65) ;
60 JOW=OZ,
whence ij OW = iOZ=OS = kOW.
* The negative sign, being a mere numerical factor, is evidently commutative
with j ; indeed we may, if necessary, easily assure ourselves of the fact that to turn
the negative (or reverse) of a vector through a right (or indeed any) angle, is the
same thing as to turn the vector through that angle and then reverse it.
J
1
4fi QUATERNIONS. [70.
70. But, by the same figure,
iON=OZ,
whence jiON =jOZ = OE =  OW =  kON.
71. From this it appears that
and similarly kj = i, J (12)
a jj
and thus, by comparing (11),
jf c = kj = i, ((11), (12)).
ki = ik=j,
These equations, along with
?=f=V = l ((7), (8), (9)),
contain essentially the whole of Quaternions. But it is easy to see
that, for the first group, we may substitute the single equation
ijk=l, (13)
since from it, by the help of the values of the squares of i,j, k, all
the other expressions may be deduced. We may consider it proved
in this way, or deduce it afresh from the figure above, thus
kON= OW,
jkON= JOW=OZ,
ijkON =
72. One most important step remains to be made, to wit the
assumption referred to in 64. We have treated i, j, k simply as
quadrantal versors ; and i, j, k as unitvectors at right angles to
each other, and coinciding with the axes of rotation of these versors.
But if we collate and compare the equations just proved we have
f =i. (7) .a
li = l, (9)
(*> = 1; (11)
Ki= k (i)
{> = * 02)
b i = k, (5)
with the other similar groups symmetrically derived from them.
75] PRODUCTS AND QUOTIENTS OF VECTORS. 47
Now the meanings we have assigned to i, j, k are quite inde
pendent of, and not inconsistent with, those assigned to i, j, k.
And it is superfluous to use two sets of characters when one will
suffice. Hence it appears that i, j, k may be substituted for i, j, k ;
in other words, a unitvector when employed as a factor may be con
sidered as a quadrantal versor whose plane is perpendicular to the
vector. (Of course it follows that every vector can be treated as the
product of a number and a quadrantal versor.) This is one of the
main elements of the singular simplicity of the quaternion calculus.
73. Thus the product, and therefore the quotient, of two perpen
dicular vectors is a third vector perpendicular to both.
Hence the reciprocal ( 51) of a vector is a vector which has
the opposite direction to that of the vector, arid its length is the
reciprocal of the length of the vector.
The conjugate ( 52) of a vector is simply the vector reversed.
Hence, by 52, if a be a vector
(Ta) 2  aKcL = a (  a) =  a 2 .
74. We may now see that every versor may be represented by
a power of a unitvector.
For, if a be any vector perpendicular to i (which is any definite
unitvector),
id, = /3, is a vector equal in length to a, but perpendicular to
both i and a. ; ^ /
i Z CL =  *,
*a fa A _/7r
i 4 a = ift = i*a = a.
Thus, by successive applications of i, a. is turned round i as an axis
through successive right angles. Hence it is natural to define i m as
a versor which turns any vector perpendicular to i through m right
angles in the positive direction of rotation about i as an axis. Here
m may have any real value whatever, whole or fractional, for it is
easily seen that analogy leads us to interpret a negative value of m
as corresponding to rotation in the negative direction.
75. From this again it follows that any quaternion may be
expressed as a power of a vector. For the tensor and versor
elements of the vector may be so chosen that, when raised to the
same power, the one may be the tensor and the other the versor
of the given quaternion. The vector must be, of course, perpen
dicular to the plane of the quaternion.
48
QUATERNIONS.
[ 7 6.
76. And we now see, as an immediate result of the last two
sections, that the indexlaw holds with regard to powers of a
quaternion ( 63).
77. So far as we have yet considered it, a quaternion has been
regarded as the product of a tensor and a versor : we are now to
consider it as a sum. The easiest method of so analysing it seems
to be the following.
T OB
Let   represent any quaternion.
OA
Draw
BG perpendicular to OA, produced if neces
sary.
Then, 19, OB = OC + CB.
But, 22, OC=xOA,
where a? is a number, whose sign is the same
as that of the cosine of Z A OB.
Also, 73, since CB is perpendicular to OA,
A w^*
w H*
14^"
> .
where 7 is a vector perpendicular to OA and CB, i.e. to the plane
of the quaternion; and, as the figure is drawn, directed towards the
reader.
Hence
OB
OA
OA
Thus a quaternion, in general, may be decomposed into the sum
of two parts, one numerical, the other a vector. Hamilton calls
them the SCALAR, and the VECTOR, and denotes them respectively
by the letters $ and V prefixed to the expression for the
quaternion.
78. Hence q = Sq+ Vq, and if in the above example
OB
Vq.OA*.
then OB
The equation above gives
GBVq.OA.
* The points are inserted to shew that S and V apply only to q, and not to qOA
So.] PRODUCTS AND QUOTIENTS OF VECTORS. 49
79. If, in the last figure, we produce BC to D, so as to double
its length, and join OD, we have, by 52,
so that OI)=OC+CD = SKq.OA + VKq.OA.
Hence OC = SKq.OA,
and CD = VKq.OA.
Comparing this value of OC with that in last section, we find
SKq = Sq, ................................. (1)
or the scalar of the conjugate of a quaternion is equal to the scalar of
the quaternion.
Again, CD = CB by the figure, and the substitution of their
values gives
VKq = Vq, ........................ (2)
or the vector of the conjugate of a quaternion is the vector of the
quaternion reversed.
We may remark that the results of this section are simple con
sequences of the fact that the symbols S, V, K are commutative*.
Thus SKq = KSq = Sq,
since the conjugate of a number is the number itself; and
VKq = KVq =  Vq 73).
Again, it is obvious that,
and thence
80. Since any vector whatever may be represented by
cci + yj + zk
where a?, y, z are numbers (or Scalars), and i, j, k may be any three
noncoplanar vectors, 23, 25 though they are usually under
stood as representing a rectangular system of unitvectors and
* It is curious to compare the properties of these quaternion symbols with those
of the Elective Symbols of Logic, as given in BOOLE S wonderful treatise on the
Laws of Thought; and to think that the same grand science of mathematical
analysis, by processes remarkably similar to each other, reveals to us truths in the
science of position far beyond the powers of the geometer, and truths of deductive
reasoning to which unaided thought could never have led the logician.
T. Q. T. 4
5 QUATERNIONS. [ 8 I .
since any scalar may be denoted by w ; we may write, for any
quaternion q, the expression
q = w + a?i + yj+zk( 78).
Here we have the essential dependence on four distinct numbers,
from which the quaternion derives its name, exhibited in the most
simple form.
Arid now we see at once that an equation such as
? =?
where q = w + otfi + y j + z k,
involves, of course, the four equations
w = w, x = x, y = y, z = z.
81. We proceed to indicate another mode of proof of the dis
tributive law of multiplication.
We have already defined, or assumed ( 61), that
7_ff + 7
a a a
or /3cr 1 + 7 a 1 = (/3 + 7 )a 1 ,
and have thus been able to understand what is meant by adding
two quaternions.
But, writing a for of 1 , we see that this involves the equality
(13 + 7) a = @OL + 7 ;
from which, by taking the conjugates of both sides, we derive
oL (ff + y) = OL p + a y (5:>).
And a combination of these results (putting {3 + y for a! in the
latter, for instance) gives
= 13/3 + 7/5 + J3y f 77 by the former.
Hence the distributive principle is true in the multiplication of
vectors.
It only remains to shew that it is true as to the scalar and
vector parts of a quaternion, and then we shall easily attain the
general proof.
Now, if a be any scalar, a any vector, and q any quaternion,
(a + a) q = aq + aq
For, if /3 be the vector in which the plane of q is intersected by
82.] PRODUCTS AND QUOTIENTS OF VECTORS. 51
a plane perpendicular to a, we can find other two vectors, 7 and S,
one in each of these planes such that
7 13
*=/ *?
And, of course, a may be written ~ ; so that
aq.
And the conjugate may be written
?X + ) = ? + ? * ( 55).
Hence, generally,
(a + a) (6 + ) = a& + a/3 f 6a + a/3 ;
or, breaking up a and 6 each into the sum of two scalars, and a, fi
each into the sum of two vectors,
(a, + a 2 + a, + * 2 ) (6, + & + , + s )
= (a, + a,) (6, + 6 8 ) + (a, f a,) (
(by what precedes, all the factors on the right are distributive, so
that we may easily put it in the form)
+ K + ,) (6 2 + /3 a ) + (a 2 + 2 ) (
Putting 0, + ^=^, a 2 + 2 = ^,
we have (^ + q) (r + s) =pr + ps + qr + ^5.
82. Cayley suggests that the laws of quaternion multiplication
may be derived more directly from those of vector multiplication,
supposed to be already established. Thus, let a. be the unit vector
perpendicular to the vector parts of q and of q . Then let
p = q.*, a = a.q,
as is evidently permissible, and we have
pa = q . OLOL = q ; a<r = act. . q q ,
so that q . q = px . acr = p . cr.
The student may easily extend this process.
For variety, we shall now for a time forsake the geometrical
mode of proof we have hitherto adopted, and deduce some of our
42
52 QUATERNIONS. [83.
next steps from the analytical expression for a quaternion given in
80, and the properties of a rectangular system of unitvectors as
in 71.
We will commence by proving the result of 77 anew.
83. Let a = xi + yj + zk,
ft = x i + yj + zk.
Then, because by 71 every product or quotient of i,j t k is reducible
to one of them or to a number, we are entitled to assume
q =  = v + & + vj + &,
where o>, f, 77, f are numbers. This is the proposition of 80.
[Of course, with this expression for a quaternion, there is no
necessity for a formal proof of such equations as
p + (q+r) = (p + q) + r,
where the various sums are to be interpreted as in 61.
All such things become obvious in view of the properties of
i,j,k.]
84. But it may be interesting to find co, , 77, f in terms of
x, y, z, x, y , z .
We have ft qa,
or x i + yj + z k = (a) + %i + rjj + %k) (xi + yj + zk)
as we easily see by the expressions for the powers and products of
i>j> k> gi ven i n ^1 But the student must pay particular attention
to the order of the factors, else he is certain to make mistakes.
This ( 80) resolves itself into the four equations
= f x + vjy + z,
z = coz + f y  TJX.
The three last equations give
j/ xx + yy + zz = w ( 2 + if + 2 )
which determines o>.
Also we have, from the same three, by the help of the first,
f .r + w + ty = :
87.] PRODUCTS AND QUOTIENTS OF VECTOES. 53
which, combined with the first, gives fy *M~~
yz zy zx xz xy yx
and the common value of these three fractions is then easily seen
to be
1_
x* + y 2 + z z
It is easy enough to interpret these expressions by means of
ordinary coordinate geometry : but a much simpler process will
be furnished by quaternions themselves in the next chapter, and, in
giving it, we shall refer back to this section.
85. The associative law of multiplication is now to be proved
by means of the distributive ( 81). We leave the proof to the
student. He has merely to multiply together the factors
w + xi + yj + zk, w + xi + yj + z k, and w" 4 x"i + y"j + z"k t
as follows :
First, multiply the third factor by the second, and then multiply
the product by the first ; next, multiply the second factor by the
first and employ the product to multiply the third: always re
membering thatjthe multiplier in any product is placed before the
multiplicand. He will find the scalar parts and the coefficients of
i,j, k, in these products, respectively equal, each to each.
86. With the same expressions for a, /3, as in section 83, we
have
a/3 = (xi + yj 4 zk) (xi 4 y j + zk)
=  (ocx +yy + zz) + (yz  zy ) i + (zx  xz )j + (xy  yx ) k.
But we have also
/3a = (xx 4 yy 4 zz) (yz zy) i (zx xz )j (xy yx) k.
The only difference is in the sign of the vector parts.
Hence Safi = S/3a, (1)
Fa = F/3a, (2)
also a/3 + /3a = 2$a/3, (3)
a/3a = 2Fa#, (4)
and, finally, by 79, a/3 = ^T./3a (5).
87. If a = /3 we have of course ( 25)
/y  /y 77 /j /y ^
* ^j y ~~y > z z ,
54 QUATERNIONS.
and the formulae of last section become
which was anticipated in 73, where we proved the formula
and also, to a certain extent, in 25.
88. Now let q and r be any quaternions, then
S.qr = S.(Sq+Vq) (Sr+Vr),
= S . (8q Sr + Sr. Vq + Sq . Vr + VqVr),
= SqSr + S. VqVr,
since the two middle terms are vectors.
Similarly, S.rq = SrSq + S . Vr Vq.
Hence, since by (1) of 86 we have
8. VqVr = S. VrVq,
we see that S.qr = S.rq, (1)
a formula of considerable importance.
It may easily be extended to any number of quaternions,
because, r being arbitrary, we may put for it rs. Thus we have
S . qrs = S . rsq,
= S . sqr
by a second application of the process. In words, we have the
theorem the scalar of the product of any number of given
quaternions depends only upon the cyclical order in which they are
arranged.
89. An important case is that of three factors, each a vector.
The formula then becomes
But S.OL@y = SOL (S/3y
= SOL Vfiy, since a8/3y is a vector,
= aF 7 /3, by (2) of 86,
=  SOL (Syj3 + Vy/3)
= S. ay $.
Hence the scalar of the product of three vectors changes sign when
the cyclical order is altered.
9O.] PRODUCTS AND QUOTIENTS OF VECTORS. 55
By the results of 55, 73, 79 we see that, for any number
of vectors, we have
K . ct/3y ... </>x = %</> V@<*
(the positive sign belonging to the product of an even number of
vectors) so that
S . j3 . . . (f>x = S x<t>  &<*
Similarly
F. ... x = *F.x ....
Thus we may generalize (3) and (4) of 86 into
2&a...^Fa0...tax*...a,
2V. a/3 ... <fo = a/3 ... <x + % c  ,
the upper sign still being used when the number of factors is
even.
Other curious propositions connected with this will be given
later (some, indeed, will be found in the Examples appended to
this chapter), as we wish to develop the really fundamental
formulae in as compact a form as possible.
90. By (4) of 86,
Hence 2V.otV/3y = V.a(8yyl3)
(by multiplying both by a, and taking the vector parts of each
side)
= F (a/3y + flay j3ay  ay {3)
(by introducing the null term @ay pay).
That is
2 F . aV/3y = F. (a/3 + 0a) 7  F (/3a 7 + /5Fa 7 + S*y./3+ Vay . )
(if we notice that F (F 7 . /3) =  F . /3Fa 7? by (2) of 86).
Hence F . a F/fy = ySz{3  jSSya .................. (1),
a formula of constant occurrence.
Adding aS{3y to both sides, we get another most valuable
formula
V.a0y = aSfal3Sya + y8oL0 ............ (2);
and the form of this shews that we may interchange 7 and a
without altering the righthand member. This gives
F . OLj3y = V . yj3a,
a formula which may be greatly extended. (See 89, above.)
50 QUATERNIONS. [91.
Another simple mode of establishing (2) is as follows :
K . a/37 =  7/3a,
.. 2 V . a/3 7  a/3 7  K . a/3 7 (by 79 (2))
= a/37 f 7/3a
= a (Py + 7/3)  (7 4 7) /3 + 7 (a/3 + /3a)
91. We have also
FFa/3FyS =  FFySFa/3 by (2) of 86 :
 &S( 7 Fa/3  ySS Fa/3 = SS . a/3 7  yS . a/
=  /3SoL FyS + aS/3 1 78 =  /3S . a 7 S + a
all of these being arrived at by the help of 90 (1) and of 89 ;
and by treating alternately Fa/3 and FyS as simple vectors.
Equating two of these values, we have
$S . a/3 7 = aS . /3jS + 08 . 7S + 7$ . a/3S ......... (3),
a very useful formula, expressing any vector whatever in terms
of three given vectors. [This, of course, presupposes that a, /3, 7
are not coplanar, 23. In fact, if they be coplanar, the factor
8. a/37 vanishes, and thus (3) does not give an expression for 8.
This will be shewn in 10 i below.]
92. That such an expression as (3) is possible we knew already
by 23. For variety we may seek another expression of a similar
character, by a process which differs entirely from that employed
in last section.
a, ft, 7 being any three noncoplanar vectors, we may derive
from them three others Fa/3, V(B^ y Vya. and, as these will not be
coplanar, any other vector 8 may be expressed as the sum of the
three, each multiplied by some scalar. It is required to find this
expression for 8.
Let 8 = # Fa/3 + 7/F/37 + zVy*
Then SyS = xS . yOL/3 = xS . a/3 7 ,
the terms in y and z going out, because
7 F/3 7 = S . 7/37 = 80v* = 7 2 S/3 = 0,
for 7 2 is ( 73) a number.
Similarly S/3S = zS . /3 7 a = zS . a/3 7 ,
and $a = yS . ifty.
Thus $8 . a/3 7 = Fa/3# 7 8 + F/3 7 /Sfa8 + Vy*S/3B ......... (4).
93] PRODUCTS AND QUOTIENTS OF VECTORS. 57
93. We conclude the chapter by shewing (as promised in 64)
that the assumption that the product of two parallel vectors is
a number, and the product of two perpendicular vectors a third
vector perpendicular to both, is not only useful and convenient,
but absolutely inevitable, if our system is to deal indifferently with
all directions in space. We abridge Hamilton s reasoning.
Suppose that there is no direction in space preeminent, and
that the product of two vectors is something which has quantity,
so as to vary in amount if the factors are changed, and to have its
sign changed if that of one of them is reversed ; if the vectors be
parallel, their product cannot be, in whole or in part, a vector
inclined to them, for there is nothing to determine the direction in
which it must lie. It cannot be a vector parallel to them ; for by
changing the signs of both factors the product is unchanged,
whereas, as the whole system has been reversed, the product
vector ought to have been reversed. Hence it must be a number.
Again, the product of two perpendicular vectors cannot be wholly
or partly a number, because on inverting one of them the sign of
that number ought to change ; but inverting one of them is simply
equivalent to a rotation through two right angles about the other,
and (from the symmetry of space) ought to leave the number
unchanged. Hence the product of two perpendicular vectors must
be a vector, and a simple extension of the same reasoning shews
that it must be perpendicular to each of the factors. It is easy to
carry this farther, but enough has been said to shew the character
of the reasoning.
EXAMPLES TO CHAPTER II.
1. It is obvious from the properties of polar triangles that any
mode of representing versors by the sides of a spherical triangle
must have an equivalent statement in which they are represented
by angles in the polar triangle.
Shew directly that the product of two versors represented
by two angles of a spherical triangle is a third versor represented
by the supplement of the remaining angle of the triangle ; and
determine the rule which connects the directions in which these
angles are to be measured.
58 QUATERNIONS.
2. Hence derive another proof that we have not generally
pq = qp.
3. Hence shew that the proof of the associative principle,
57, may be made to depend upon the fact that if from any point
of the sphere tangent arcs be drawn to a spherical conic, and also
arcs to the foci, the inclination of either tangent arc to one of the
focal arcs is equal to that of the other tangent arc to the other
focal arc.
4. Prove the formulae
2$ . a/3y = a/37  7 /3a,
2F.a/37 = a/37 + 7a.
5. Shew that, whatever odd number of vectors be represented
by a, j3, 7, &c., we have always
F. a/3y$efr = V. rteSypa, &c.
6. Shew that
8 . Fa/3F/3 7 F 7 a = (S. a/3 7 ) 2 ,
F. Vot{3V/3yVyoL=Voi{3(ry 2 SoL/3S/3vSvoi) + ...... ,
and F ( Fa/3 F . Vj3y Fya) = ((3 Say  a/3 7 ) 8 . ay.
7. If a, (3, 7 be any vectors at right angles to each other, shew
that
(a 3 f /3 3 h y 3 ) S . a/3 7 = a 4 F/3 7 + /3 4 F>a + 7 4 Fa/3.
(a" 1 " 1 + P n ~ l + 7 in  1 ) 8 . a/fy = 2 " V/3y + /3 2w Fya + 7 2n Fa/3.
8. If a, /3, 7 be noncoplanar vectors, find the relations among
the six scalars, x, y, z and f, 77, f, which are implied in the
equation XOL + yfi + 27 = f F/37 + 77 Fya + f FayS.
9. If a, /3, 7 be any three noncoplanar vectors, express any
fourth vector, S, as a linear function of each of the following sets of
three derived vectors.
F.yaft F.a^7, F. ya,
and F. Fa/3F/3 7 F 7 a, F. F/3 7 Fya Fa/3, F. FyaFa/3F/3 7 .
10. Eliminate p from the equations
where a, /3, 7, 8 are vectors, and a, 6, c, d scalars.
11. In any quadrilateral, plane or gauche, the sum of the
squares of the diagonals is double the sum of the squares of the
lines joining the middle points of opposite sides.
CHAPTER III.
INTERPRETATIONS AND TRANSFORMATIONS OF
QUATERNION EXPRESSIONS.
94. AMONG the most useful characteristics of the Calculus of
Quaternions, the ease of interpreting its formulae geometrically,
and the extraordinary variety of transformations of which the
simplest expressions are susceptible, deserve a prominent place.
We devote this Chapter to some of the more simple of these,
together with a few of somewhat more complex character but of
constant occurrence in geometrical and physical investigations.
Others will appear in every succeeding Chapter. It is here,
perhaps, that the student is likely to feel most strongly the
peculiar difficulties of the new Calculus. But on that very account
he should endeavour to master them, for the variety of forms
which any one formula may assume, though puzzling to the
beginner, is of the utmost advantage to the advanced student, not
alone as aiding him in the solution of complex questions, but
as affording an invaluable mental discipline.
95. If we refer again to the figure of 77 we see that
OC=OBcosAOB,
CB^OBainAOB.
Hence, if OA = a, OB = /3, and Z A OB = 0, we have
OB=T0, OA = Ta,
00 = Tfi cos 0, OB = T{3 sin 6.
a /3 00 T/3
Hence S  = TTT = m~ cos 6.
a OA TOL
cv 1 i mtrfi CB J (3 . A
Similarlv TV   777 = 7f P sin 0.
a OA To.
60 QUATERNIONS. [96.
Hence, if ?; be a unitvector perpendicular to a and /3, and such
that positive rotation about it, through the angle 6, turns a
towards /3, or
UOA OA
R
we have V  ~ sin 6 . 77. (See, again, 84.)
96. In the same way, or by putting
S + V
a a.
we may shew that
Sa/3 =  TOL T8 cos 0,
and Fa/3  2 7 a T sin 6 . 7;
where rj = UVaft =U( V/3a) =UV /3 .
Thus the scalar of the product of two vectors is the continued
product of their tensors and of the cosine of the supplement of the
contained angle.
The tensor of the vector of the product of two vectors is the con
tinued product of their tensors and the sine of the contained angle ;
and the versor of the same is a unitvector perpendicular to both,
and such that the rotation about it from the first vector (i. e. the
multiplier) to the second is lefthanded or positive.
Hence also T VOL ft is double the area of the triangle two of whose
sides are a, j3.
97. (a) In any plane triangle ABC we have
Hence AC 2 = 8. AC AC = S . A
With the usual notation for a plane triangle the interpretation
of this formula is
1} Z = be cos A ab cos G,
or 6 = a cos C + c cos A .
99] INTERPRETATIONS AND TRANSFORMATIONS. 61
(b) Again we have, obviously,
V.ABAC=V.AB(AB+BC!)
= V.ABBC )
or cb sin A = ca sin B,
sin A sin B sin G
whence = > = .
a b c
These are truths, but not truisms, as we might have been led
to fancy from the excessive simplicity of the process employed.
98. From 96 it follows that, if a and /3 be both actual (i. e.
real and nonevanescent) vectors, the equation
shews that cos 6 = 0, or that a. is perpendicular to (3. And, in fact,
we know already that the product of two perpendicular vectors is
a vector.
Again : if Fa/3 = 0,
we must have sin 6 = 0, or a is parallel to j3. We know already
that the product of two parallel vectors is a scalar.
Hence we see that
is equivalent to a =
where 7 is an undetermined vector; and that
is equivalent to a = xj3,
where x is an undetermined scalar.
99. If we write, as in SS 83, 84,
JO
a = isc +jy +kz,
j3 = ix +jy 4 kz ,
we have, at once, by 86,
Sa/3 = xx yy zz
(IT IT II II % %
r r r r r r
where r = Jx* 4 y* + z\ r = Jx * + y"* + z*.
Tr n , (yz zy . zx xz . xy yx
Also VOL 8 rr \ ~ i + /?+ 7  A
rr rr rr
62 QUATERNIONS. [lOO.
These express in Cartesian coordinates the propositions we have
just proved. In commencing the subject it may perhaps assist
the student to see these more familiar forms for the quaternion
expressions ; and he will doubtless be induced by their appearance
to prosecute the subject, since he cannot fail even at this stage to
see how much more simple the quaternion expressions are than
those to which he has been accustomed.
100. The expression S . a/3y
may be written SV (aft) 7,
because the quaternion a/3y may be broken up into
of which the first term is a vector.
But, by 96,
8 V (a/3) y = TaT/3 sin
Here Trj = 1, let < be the angle between 77 and 7, then finally
S .a/3y =  TOL Tft Ty sin cos <j>.
But as rj is perpendicular to a and {3, Ty cos <j> is the length of the
perpendicular from the extremity of 7 upon the plane of a, ft. And
as the product of the other three factors is ( 96) the area of the
parallelogram two of whose sides are a, ft, we see that the mag
nitude of S . a/37, independent of its sign, is the volume of the
parallelepiped of which three coordinate edges are a, ft, y : or six
times the volume of the pyramid which has a, ft, y for edges.
101. Hence the equation
8. a/37 = 0,
if we suppose a, ft, y to be actual vectors, shews either that
sin (9 = 0,
or cos (/> = 0,
i. e. two of the three vectors are parallel, or all three are parallel to
one plane.
This is consistent with previous results, for if 7 =pft we have
S.a/3y = pS.a/3* = 0;
and, if 7 be coplanar with a, ft, we have 7 =poi 4 qft, and
S.aj3y = S.aij3 (pa + q/3) = 0.
IO3] INTERPRETATIONS AND TRANSFORMATIONS. 63
102. This property of the expression 8 . afiy prepares us to
find that it is a determinant. And, in fact, if we take a, ft as in
83, and in addition 7 = ix" +jy" + kz",
we have at once
8 . a/3y =  x" (yz f  zy )  y" (zx f  xz )  z" (xy f  yx ),
x y z
x y z
x" y" z
The determinant changes sign if we make any two rows change
places. This is the proposition we met with before ( 89) in the
form 8. afty =  S . pay = 8 . j3ya, &c.
If we take three new vectors
a 1 = ix +jx + kx",
I 3 1 = iy +jy + %">
y^iz+jz + kz",
we thus see that they are coplanar if a, /3, y are so. That is, if
8 . afiy = 0,
then S.afitf^Q.
103. We have, by 52,
(Tqf = qKq = (Sq + Vq) (Sq  Vq) ( 79),
by algebra,
If q = a/3, we have Kq = /3a, and the formula becomes
*$.$* = 2 /3 2 = (Sap)*  ( Vapy.
In Cartesian coordinates this is
(^ + 2/ 2 + /)( ^2 + ^ 2+/2)
= (xaf + yy + zzj + (yz  zyj + (zx f  xzj + (xy f  yxj.
More generally we have
(T(qr)) z =
If we write q w + OL = W + IX +jy + kz,
r = w f P = w 4 ix +jy + kz \
this becomes
(w 2 f x* + y 1 + ^ 2 ) (w 2 + x 2 + y"* + / 2 )
= (low xx f yy zz)* + (wx + wx + yz zy J
+ (wy + w y + zx xz ) z + (wz + w z + xy yx ^,
a formula of algebra due to Euler.
64 QUATERNIONS. [104.
104. We have, of course, by multiplication,
(a + ft) 2 = a 2 + a/3 + /3a + /3 2 = a 2 + 2Sa/3 + ft 2 ( 86 (3)).
Translating into the usual notation of plane trigonometry, this
becomes c 2 = a 2 2ab cos G + 6 2 ,
the common formula.
Again, F . (a + ft) (a  /3) =  Fa/3 + F/3a =  2 Fa/3 ( 86 (2)).
Taking tensors of both sides we have the theorem, the paral
lelogram whose sides are parallel and equal to the diagonals of a
given parallelogram, has double its area ( 96).
Also S (a + ft) (a  ft) = a 2  ft 2 ,
and vanishes only when a 2 = ft 2 , or Ten = Tft ; that is, the diagonals
of a parallelogram are at right angles to one another, when, and
only when, it is a rhombus.
Later it will be shewn that this contains a proof that the angle
in a semicircle is a right angle.
105. The expression p = afta~ l
obviously denotes a vector whose tensor is equal to that of ft.
But we have S . ftap = 0,
so that p is in the plane of a, ft.
Also we have Sap = Saft,
so that ft and p make equal angles with a, evidently on opposite
sides of it. Thus if a be the perpendicular to a reflecting surface
and ft the path of an incident ray, p will be the path of the
reflected ray.
Another mode of obtaining these results is to expand the above
expression, thus, 90 (2),
p = 2 a" 1 Sa ft ft
= 2 1 a/3  of 1 (Saft + Fa/3 )
 a 1 (Saft  Fa/3),
so that in the figure of 77 we see that if A = a, and OB = ft, we
have OD = p = afta~ l .
Or, again, we may get the result at once by transforming the
equation to ^ = K (a" 1 p)  K ? .
IO6.] INTERPRETATIONS AND TRANSFORMATIONS. 65
106. For any three coplanar vectors the expression
p = ct/3y
is ( 101) a vector. It is interesting to determine what this vector
is. The reader will easily see that if a circle be described about
the triangle, two of whose sides are (in order) a and /3, and if from
the extremity of /3 a line parallel to 7 be drawn, again cutting the
circle, the vector joining the point of intersection with the origin
of a. is the direction of the vector a/3y. For we may write it in the
form
p = a /3*/3 7 =  (T/9) Vy =  ( W 1 7,
which shews that the versor [ 5] which turns yS into a direction
parallel to a, turns 7 into a direction parallel to p. And this ex
presses the longknown property of opposite angles of a quadri
lateral inscribed in a circle.
Hence if a, /3, y be the sides of a triangle taken in order, the
tangents to the circumscribing circle at the angles of the triangle
are parallel respectively to
a/3y, ftya, and ya(3.
Suppose two of these to be parallel, i. e. let
afiy = xfiya = xay/3 ( 90),
since the expression is a vector. Hence
which requires either
x=l, Vyj3 = or 711/3,
a case not contemplated in the problem ;
or a? = l, S/3y=0,
i. e. the triangle is rightangled. And geometry shews us at once
that this is correct.
Again, if the triangle be isosceles, the tangent at the vertex is
parallel to the base. Here we have
or x (a f 7) = a (a + 7) 7 ;
whence x 7* = a 2 , or Ty = Tea, as required.
As an elegant extension of this proposition the reader may
T. Q. I. 5
66 QUATERNIONS. [107.
prove that the vector of the continued product a/rtyS of the vector
sides of any quadrilateral inscribed in a sphere is parallel to the
radius drawn to the corner (a, 8). [For, if 6 be the vector from 8,
a to /3, 7, a/9e and 78 are (by what precedes) vectors touching the
sphere at a, 8. And their product (whose vector part must be
parallel to the radius at a, 8) is
OL/3e . 78 = e 2 . a
107. To exemplify the variety of possible transformations
even of simple expressions, we will take cases which are of
frequent occurrence in applications to geometry.
Thus T(p+a)=:T(pQi),
[which expresses that if
Ol=a, 01 = a, and OP = p,
we have AP = A P,
and thus that P is any point equidistant from two fixed points,]
may be written (p + a) 2 = (p a) 2 ,
or p 2 + 2Sap + a 2  p 9  2Sap + a 2 ( 104),
whence Sap = 0.
This may be changed to
a/3 + pa. = 0,
or dp + Kaip = 0,
OL
or finally, TVU?=l,
all of which express properties of a plane.
Again, Tp  TOL
may be written T = l,
or finally, T. (p + a) (p  a) = 2TVap.
All of these express properties of a sphere. They will be
interpreted when we come to geometrical applications.
I08.] INTERPRETATIONS AND TRANSFORMATIONS. 67
108. To find the space relation among five points.
A system of five points, so far as its internal relations are
concerned, is fully given by the vectors from one to the other four.
If three of these be called a, /?, y, the fourth, S, is necessarily
expressible as xa. + yfi + zy. Hence the relation required must be
independent of x, y, z.
But SaiS = ax 2 + ySa/3 +
Sy$ = xSya + ySy/3 f z
SB8 = tf = xS8
The elimination of a?, y, z gives a determinant of the fourth order,
which may be written
SoLOi Sa/3 Say S
S/3a S{3j3 S{3y 8/38
Syot Sy(3 Syy SyS
Now each term may be put in either of two forms, thus
S/3y = 4 {P* + y*(p y) 2 } =  TpTy cos py.
If the former be taken we have the expression connecting the
distances, two and two, of five points in the form given by Muir
(Proc. R. S. E. 1889) ; if we use the latter, the tensors divide out
(some in rows, some in columns), and we have the relation among
the cosines of the sides and diagonals of a spherical quadrilateral.
We may easily shew (as an exercise in quaternion manipulation
merely) that this is the only condition, by shewing that from it
we can get the condition when any other of the points is taken as
origin. Thus, let the origin be at a, the vectors are a, P a,
y a, 8 a. But, by changing the signs of the first row, and first
column, of the determinant above, and then adding their values
term by term to the other rows and columns, it becomes
S( )() S( a)(P~a) S( a)( 7 a) S( a) (8a)
S(ya)(a) S(y a)(/3  a) 8 (y  a) (7  a)
which, when equated to zero, gives the same relation as before.
[See Ex. 10 at the end of this Chapter.]
52
68
QUATERNIONS.
[108.
An additional point, with e = x a. f y ft + 2 % gives six additional
equations like (1) ; i. e.
Sae = a/a 2 + tfSa/3 + z Say,
= x Sya
+ yj + s
+ySe/3
from which corresponding conclusions may be drawn.
Another mode of solving the problem at the head of this
section is to write the identity
where the ms are undetermined scalars, and the as are given
vectors, while is any vector whatever.
Now, provided that the number of given vectors exceeds four, we
do not completely determine the ms by imposing the conditions
2m = 0, 2ma = 0.
Thus we may write the above identity, for each of five vectors
successively, as
2m (a a a ) 2 = 2ma 2 ,
2m (a a 2 ) 2 = 2ma 2 ,
2m (a  5 ) 2 = 2ma 2 .
Take, with these, 2m = 0,
and we have six linear equations from which to eliminate the ms.
The resulting determinant is
_a* a
a a a a 2 1 2ma 2 = 0.
W. . tA <*_ X
1 1 ..10
This is equivalent to the form in which Cayley gave the
relation among the mutual distances of five points. (Camb. Math.
Journ. 1841.)
110.] INTERPRETATIONS AND TRANSFORMATIONS. 69
109. We have seen iu 95 that a quaternion may be divided
into its scalar and vector parts as follows :
where is the angle between the directions of a and ft and e UV
is the unitvector perpendicular to the plane of a. and /3 so situated
that positive (i. e. lefthanded) rotation about it turns a towards ft
Similarly we have ( 96)
= TaTj3 ( cos + e sin 0),
arid e having the same signification as before.
110. Hence, considering the versor parts alone, we have
U = cos + e sin 0.
a
Similarly U j. = cos (f> + e sin ;
</> being the positive angle between the directions of 7 and ft and e
the same vector as before, if a, ft 7 be coplanar.
Also we have
U ^ = cos (6 + <) + e sin (0 + 0).
But we have always
and therefore U.U= U;
pa a
or cos (</> + 6) + e sin (< + 0) = (cos </> f e sin (/>) (cos + e sin 6)
= cos (/> cos 6 sin </> sin 6 + e (sin </> cos + cos $ sin 0),
from which we have at once the fundamental formulae for the
cosine and sine of the sum of two arcs, by equating separately the
scalar and vector parts of these quaternions.
And we see, as an immediate consequence of the expressions
above, that
cos mO + e sin mO = (cos 6 f e sin 0) m .
if m be a positive whole number. For the lefthand side is a versor
70 QUATERNIONS. [ill.
which turns through the angle m6 at once, while the righthand
side is a versor which effects the same object by m successive turn
ings each through an angle 6. See 8, 9.
111. To extend this proposition to fractional indices we have
r\
only to write  for 0, when we obtain the results as in ordinary
n
trigonometry.
From De Moivre s Theorem, thus proved, we may of course
deduce the rest of Analytical Trigonometry. And as we have
already deduced, as interpretations of selfevident quaternion trans
formations ( 97, 104), the fundamental formulae for the solution
of plane triangles, we will now pass to the consideration of spherical
trigonometry, a subject specially adapted for treatment by qua
ternions ; but to which we cannot afford more than a very few
sections. (More on this subject will be found in Chap. XI. in con
nexion with the Kinematics of rotation.) The reader is referred to
Hamilton s works for the treatment of this subject by quaternion
exponentials.
112. Let a, /3, 7 be unitvectors drawn from the centre to the
corners A, B, C of a triangle on the unitsphere. Then it is evident
that, with the usual notation, we have ( 96),
Sa/3 = cos c, S/3y = cos a, Sya. = cos b,
TVQLJ3 = sin c, TV/3y = sin a, T Fya = sin b.
Also [7 Fa/3, UV/3y, UVyct are evidently the vectors of the corners
of the polar triangle.
Hence S . UVajS UVj3y = cos B t &c.,
TV. U Fa/3 UVfa = sin B, &c.
Now ( 90 (1)) we have
Remembering that we have
SVafiVPy = TVaQTVPyS .
we see that the formula just written is equivalent to
sin a sin c cos B = cos a cos c f cos b,
or cos b cos a cos c + sin a sin c cos J5.
I 1 5.] INTERPRETATIONS AND TRANSFORMATIONS. 71
1 1 3. Again, V . Fa/3 F/3 7 =  /3a/3 7 ,
which gives
TV. FaF/3 7 = TS . a/3 7 = TS . aF/3 7 = TS . /3F 7 a = TS . 7 Fa/3,
or sin a sin c sin B = sin a sin p a = sin 6 sin p & = sin c sin ^) c ;
where p a is the arc drawn from A perpendicular to BC, &c.
Hence sin p a = sin c sin J9,
sin a sin c .
sn p = sn a sn .
114. Combining the results of the last two sections, we have
Vap . V/3y = sin a sin c cos B /3 sin a sin c sin B
= sin a sin c (cos jB /3 sin B).
Hence ^7 . Fa/3 F/3 7 = (cos 5  /3 sin ))
and U . Fy/3 F/3a = (cos B + ft sin B) }
These are therefore versors which turn all vectors perpendicular to
OB negatively or positively about OB through the angle B.
[It will be shewn later ( 119) that, in the combination
(cos B+/3smB)( ) (cos B  /3 sin B) t
the system operated on is made to rotate, as if rigid, round the
vector axis ft through an angle 2B.]
As another instance, we have
sin B
tan B =
cos B
S.Fa/3F/3 7
F.Fa/3F/3 7
P S.Fa/3F/3 7
(1)
The interpretation of each of these forms gives a different theorem
in spherical trigonometry.
115. Again, let us square the equal quantities
F a/3 7 and a#/3 7 
7 2 QUATERNIONS. [ I I 6.
supposing a, @, 7 to be any unitvectors whatever. We have
 ( F, a/3 7 ) 2  8*0y + V + fy
But the lefthand member may be written as
T 2 . afiy $ 2 . a/3<y,
wfience
or 1 cos 2 tt cos 2 6 cos 2 c + 2 cos a cos 6 cos c
= sin 2 a sm 2 j9 a = &c.
= sin 2 a sin 2 6 sin 2 = &c.,
all of which are wellknown formulae.
116. Again, for any quaternion,
so that, if n be a positive integer,
f = (Sqy + n (Sq) 1 Vq + W f= (Sg) M ( Vqf + . . .
From this at once
F. " = F ,S  T*V + &c
If q be a versor we have
q = cos u + sin u t
so that
S.q n = (cos u) n 2 (cos u) n ~ 2 (sin w) 2 + . . .
= cos nu ;
r n . ^TTi . 7^2 1
F . ^ n = 6 sin M w (cos uf 1 19*3 ^ C S U ^ * ^^ ^) 2 +
= 6 sin nu ;
as we might at once have concluded from 110.
Such results may be multiplied indefinitely by any one who has
mastered the elements of quaternions.
I I 7.] INTERPRETATIONS AND TRANSFORMATIONS. 73
117. A curious proposition, due to Hamilton, gives us a
quaternion expression for the spherical excess in any triangle.
The following proof, which is very nearly the same as one of his,
though by no means the simplest that can be given, is chosen here
because it incidentally gives a good deal of other information.
We leave the quaternion proof as an exercise.
Let the unitvectors drawn from the centre of the sphere to
A, B, C, respectively, be a, j3, 7. It is required to express, as an
arc and as an angle on the sphere, the quaternion
The figure represents an orthographic projection made on a
plane perpendicular to 7. Hence C is the centre of the circle DEe.
Let the great circle through A, B meet DEe in E, e, and let DE be
a quadrant. Thus DE represents 7 ( 72). Also make EF = AB
= /3a~\ Then, evidently,
which gives the arcual representation required.
Let DF cut Ee in G. Make Ca = EG, and join D, a, and a, F.
Obviously, as D is the pole of Ee, Da is a quadrant ; and since
EG = Ca, Ga EG, a quadrant also. Hence a is the pole of DG,
and therefore the quaternion may be represented by the angle
DaF.
Make Cb = Ca, and draw the arcs Pa/3, P6a from P, the pole of
74 QUATERNIONS. [ IJ 7
A B. Comparing the triangles Ebi and ea/3, we see that Ecu = e/3.
But, since P is the pole of AB, Ffia is a right angle: and therefore
as Fa is a quadrant, so is P/3. Thus AB is the complement of Eot
or fte, and therefore
Join 6J. and produce it to c so that Ac = bA; join c, P, cutting
AB in o. Also join c, 5, and 5, a.
Since P is the pole of AB, the angles at o are right angles ;
and therefore, by the equal triangles baA, coA, we have
aA = Ao.
But a/3 = 2AB,
whence oB = 5/3,
and therefore the triangles coB and Bafi are equal, and c, 5, a lie
on the same great circle.
Produce cA and cB to meet in H (on the opposite side of the
sphere). H and c are diametrically opposite, and therefore cP,
produced, passes through H.
Now Pa = Pb = P^T, for they differ from quadrants by the
equal arcs a/3, ba, oc. Hence these arcs divide the triangle Hab
into three isosceles triangles.
But Z PHb + Z PHa = Z aHb = Z 6ca.
Also Z Pa& = TT  Z ca&  Z
Z P&a = Z Pa& = TT 
Adding, 2 Z Pa& = 2?r Z ca& Z c6a Z 6ca
= TT (spherical excess of abc).
But, as Z Pa/3 and Z Dae are right angles, we have
angle of /Ba^y = Z FaD = Z /3ae = Z Pa&
=  (spherical excess of a&c).
[Numerous singular geometrical theorems, easily proved ab
initio by quaternions, follow from this : e.g. The arc AB, which
bisects two sides of a spherical triangle abc, intersects the base at
the distance of a quadrant from its middle point. All spherical
triangles, with a common side, and having their other sides
bisected by the same great circle (i.e. having their vertices in a
I 1 9] INTERPRETATIONS AND TRANSFORMATIONS. 75
small circle parallel to this great circle) have equal areas, &c.
&c.]
118. Let Oa = a. , Ob = /3 , Oc = 7 , and we have
^  Ga . cA . Be
a/
= Ca.BA
= EG.FE=FG.
But FG is the complement of DF. Hence the angle of the
quaternion
is Aa/" iAe spherical excess of the triangle whose angular points are
at the extremities of the unitvectors a , /3 , 7 .
[In seeking a purely quaternion proof of the preceding proposi
tions, the student may commence by shewing that for any three
unitvectors we have
The angle of the first of these quaternions can be easily assigned ;
and the equation shews how to find that of ffa. 1 ^.
Another easy method is to commence afresh by forming from
the vectors of the corners of a spherical triangle three new vectors
thus :
Then the angle between the planes of a, @ and 7 , a ; or of ft, 7
and a , j3 ; or of 7, a and /3 , 7 ; is obviously the spherical excess.
But a still simpler method of proof is easily derived from the
composition of rotations.]
119. It may be well to introduce here, though it belongs
rather to Kinematics than to Geometry, the interpretation of the
operator
9( )<?"
By a rotation, about the axis of q, through double the angle of q )
the quaternion r becomes the quaternion qrq~ l . Its tensor and
angle remain unchanged, its plane or axis alone varies.
7G QUATERNIONS. [l2O.
A glance at the figure is sufficient for
the proof, if we note that of course
T . qrq~ l = Tr, and therefore that we need
consider the versor parts only. Let Q
be the pole of q, n/
AB = q, AB = q~\ BC = r.
Join C A, and make AC=C A. Join
CB. c
Then CB is qrq 1 , its arc CB is evidently equal in length to that
of r, B C ; and its plane (making the same angle with B B that
that of B C does) has evidently been made to revolve about Q, the
pole of q, through double the angle of q.
It is obvious, from the nature of the above proof, that this
operation is distributive ; i. e. that
q (r 4 s) q 1 = qrq~ l + qsq \
If r be a vector, = p, then qpq~ l (which is also a vector) is the
result of a rotation through double the angle of q about the axis
of q. Hence, as Hamilton has expressed it, if B represent a rigid
system, or assemblage of vectors,
qBq*
is its new position after rotating through double the angle of q
about the axis of q.
120. To compound such rotations, we have
r . qBq~ l . r 1 = rq . B . (rq)~ l .
To cause rotation through an angle fold the double of the angle
of q we write cfBf*.
To reverse the direction of this rotation write q^Bq*.
To translate the body B without rotation, each point of it moving
through the vector a, we write a + B.
To produce rotation of the translated body about the same axis,
and through the same angle, as before,
Had we rotated first, and then translated, we should have had
a + qBq~\
121.] INTERPRETATIONS AND TRANSFORMATIONS. 77
From the point of view of those who do not believe in the
Moon s rotation, the former of these expressions ought to be
qaq 1 + B
instead of
qaq~ l + qSq~\
But to such men quaternions are unintelligible.
121. The operator above explained finds, of course, some
of its most direct applications in the ordinary questions of
Astronomy, connected with the apparent diurnal rotation of the
stars. If X be a unitvector parallel to the polar axis, and h the
hour angle from the meridian, the operator is
/ h h\f \fh h\
(^cosg  X sin J ^ J (^cos g + X sin ^J ,
or L~ l ( ) L,
the inverse going first, because the apparent rotation is negative
(clockwise).
If the upward line be i, and the southward j, we have
X = i sin I j cos I,
where I is the latitude of the observer. The meridian equatorial
unit vector is
fjb = i cosl+j sin I;
and X, /t, k of course form a rectangular unit system.
The meridian unitvector of a heavenly body is
8 = i cos (I d) + j sin (I d),
= X sin d + jj, cos d,
where d is its declination.
Hence when its hourangle is h t its vector is
v = L I SL.
The vertical plane containing it intersects the horizon in
so that
/ A i \
tan (azimuth) = 7 .................. (1).
[This may also be obtained directly from the last formula (1)
of 114.]
78 QU ATERNIONS. [ I 2 2 .
To find its Amplitude, i.e. its azimuth at rising or setting,
the hourangle must be obtained from the condition
& 8 = (2).
These relations, with others immediately deducible from them,
enable us (at once and for ever) to dispense with the hideous
formulae of Spherical Trigonometry.
122. To shew how readily they can be applied, let us
translate the expressions above into the ordinary notation. This
is effected at once by means of the expressions for X, //,, L, and
S above, which give by inspection
= x sin d + (fjb cos h k sin h) cos d,
and we have from (1) and (2) of last section respectively
. sin h cos d
tan (azimuth) = j. T . = , (1),
cos I sin d sin / cos d cos h
cos h + tan I tan d = (2).
In Capt. Weir s ingenious Azimuth Diagram, these equations
are represented graphically by the rectangular coordinates of a
system of confocal conies : viz.
x = sin h sec I
y = cos h tan I
The ellipses of this system depend upon / alone, the hyperbolas
upon h. Since (1) can, by means of (3), be written as
tan (azimuth) = =
tan d y
we see that the azimuth can be constructed at once by joining
with the point 0, tan d, the intersection of the proper ellipse and
hyperbola.
Equation (2) puts these expressions for the coordinates in the
form
x = sec lj\ tan 2 j tan 8 d \
y = tan 2 1 tan d j
The elimination of d gives the ellipse as before, but that of I
gives, instead of the hyperbolas, the circles
x* + 7/ 2 y (tan d cot d) = 1.
The radius is
J (tan d 4 cot d) ;
and the coordinates of the centre are
0, J (tanrf cot d).
1 2 4] INTERPRETATIONS AND TRANSFORMATIONS. 79
123. A scalar equation in p, the vector of an undetermined
point, is generally the equation of a surface; since we may use
in it the expression p = oca.,
where x is an unknown scalar, and a any assumed unitvector.
The result is an equation to determine x. Thus one or more
points are found on the vector XOL, whose coordinates satisfy the
equation ; and the locus is a surface whose degree is determined
by that of the equation which gives the values of x.
But a vector equation in p, as we have seen, generally leads to
three scalar equations, from which the three rectangular or other
components of the sought vector are to be derived. Such a vector
equation, then, usually belongs to a definite number of points in
space. But in certain cases these may form a line, and even a
surface, the vector equation losing as it were one or two of the
three scalar equations to which it is usually equivalent.
Thus while the equation cup ft
gives at once p = a" 1 /?,
which is the vector of a definite point (since by making p a vector
we have evidently assumed
Sa/3 = 0);
the closely allied equation Vap = j3
is easily seen to involve Sa{3 = 0,
and to be satisfied by p = of */8 + xa,
whatever be x. Hence the vector of any point whatever in the
line drawn parallel to a from the extremity of a~ l (3 satisfies the
given equation. [The difference between the results depends
upon the fact that Sap is indeterminate in the second form, but
definite (= 0) in the first.]
124. Again, Vap .Vp/3 = ( Fa/3) 2
is equivalent to but two scalar equations. For it shews that Vap
and V/3p are parallel, i.e. p lies in the same plane as a and /3, and
can therefore be written ( 24)
p = xa. + yfl,
where x and y are scalars as yet undetermined.
We have now Vap = yVa/3,
8 QUATERNIONS. [125.
which, by the given equation, lead to
ajy = l, or y = ,
sc
or finally p = xa.+ ft;
CO
which ( 40) is the equation of a hyperbola whose asymptotes are
in the directions of a and ft.
125. Again, the equation
though apparently equivalent to three scalar equations, is really
equivalent to one only. In fact we see by 91 that it may be
written
whence, if a be not zero, we have
and thus ( 101) the only condition is that p is coplanar with a, ft.
Hence the equation represents the plane in which a and ft lie.
126. Some very curious results are obtained when we extend
these processes of interpretation to functions of a quaternion
q = w + p
instead of functions of a mere vector p.
A scalar equation containing such a quaternion, along with
quaternion constants, gives, as in last section, the equation of a
surface, if we assign a definite value to w. Hence for successive
values of w, we have successive surfaces belonging to a system ;
and thus when w is indeterminate the equation represents not a
surface, as before, but a volume, in the sense that the vector of any
point within that volume satisfies the equation.
Thus the equation (Tqf = a 8 ,
ooo
or w p = a,
or
represents, for any assigned value of w, not greater than a, a sphere
whose radius is Jd 2 w\ Hence the equation is satisfied by the
vector of any point whatever in the volume of a sphere of radius a,
whose centre is origin.
128.] INTERPRETATIONS AND TRANSFORMATIONS. 81
Again, by the same kind of investigation,
where q = iu + p, is easily seen to represent the volume of a sphere
of radius a described about the extremity of ft as centre.
Also S(q*) = a 2 is the equation of infinite space less the space
contained in a sphere of radius a about the origin.
Similar consequences as to the interpretation of vector equa
tions in quaternions may be readily deduced by the reader.
127. The following transformation is enuntiated without proof
by Hamilton (Lectures, p. 587, and Elements, p. 299).
To prove it, let r~ l (r 2 q*) 2 q~ l = t,
then Tt = 1,
and therefore Kt = t~\
But ( ? Y) 4 = rfy,
or r*q* = rtqrtq,
or rq = tqrt.
Hence KqKr = t~ l KrKqr\
or KrKq = tKqKrt.
Thus we have U (rq KrKq) = tU(qr KqKr) t,
or, if we put s = U (qr KqKr),
Ks= tst.
Hence sKs = (Ts) 2 =l = stst,
which, if we take the positive sign, requires
**=1,
or t = s 1 = UKs,
which is the required transformation.
[It is to be noticed that there are other results which might
have been arrived at by using the negative sign above ; some
involving an arbitrary unitvector, others involving the imaginary
of ordinary algebra.]
128. As a final example, we take a transformation of Hamil
ton s, of great importance in the theory of surfaces of the second
order.
Transform the expression
T. Q. I.
82 QUATERNIONS. [128.
in which a, /3, 7 arc any three mutually rectangular vectors, into
the form
v ?
which involves only two vectorconstants, i, K.
[The student should remark here that t, K, two undetermined
vectors, involve six disposable constants : and that a, ft, 7, being
a rectangular system, involve also only six constants.]
[T(ip + pK)} 2 = (ip + />*) (pi + KP) ( 52, 55)
f
Hence (%>) + (S/3 P )* + (>S 7P ) 2 = /V^C
K L
But ^ 2 (Sapf + /T (/>) + 7 " 2 (^) 2 = p 2 (i 25, 73).
Multiply by /3 2 and subtract, we get
The left side breaks up into two real factors if /3 2 be intermediate
in value to a 2 and 7 2 : and that the right side may do so the term
in p 2 must vanish. This condition gives
/3 2 = ,42X8 ; an d the identity becomes
 l ) (P
Hence we must have

where jp is an undetermined scalar.
To determine p, substitute in the expression for /3 2 , and we find
= p* + ( a2  r)
I2Q.] INTERPRETATIONS AND TRANSFORMATIONS. 83
Thus the transformation succeeds if
.
P +
which gives p +  = + 2 A /
p y * 2  7 2
Hence
6 2 ) /I 2 \
Tf = t * J ^  7*) =
*) = * ,/*,
or (**  i*)" 1 = TaTy.
Toc+Ty 1 TaTy
Again, ft = ,  = 
/ 2 3. rt)
and therefore
Thus we have proved the possibility of the transformation, and
determined the transforming vectors i, K.
129. By differentiating the equation
(Sap? + (Sto? + (Hyp? =
we obtain, as will be seen in Chapter IV, the following,
+ S/3 P S/3 P
_ _
(K i )
where p also may be any vector whatever.
This is another very important formula of transformation ; and
it will be a good exercise for the student to prove its truth by
processes analogous to those in last section. We may merely
observe, what indeed is obvious, that by putting p = p it becomes
the formula of last section. And we see that we may write, with
the recent values of i and K in terms of a, /3 y, the identity
i J
62
84 QUATERNIONS. [ 1 30.
130. In various quaternion investigations, especially in such
as involve imaginary intersections of curves and surfaces, the old
imaginary of algebra of course appears. But it is to be particularly
noticed that this expression is analogous to a scalar and not to a
vector, and that like real scalars it is commutative in multipli
cation with all other factors. Thus it appears, by the same proof
as in algebra, that any quaternion expression which contains this
imaginary can always be broken up into the sum of two parts, one
real, the other multiplied by the first power of V 1. Such an
expression, viz.
q = q + >f^lq",
where q and q" are real quaternions, is called by Hamilton a
BIQUATERNION. [The student should be warned that the term
Biquaternion has since been employed by other writers in the
sense sometimes of a "set" of 8 elements, analogous to the
Quaternion 4 ; sometimes for an expression q + Oq" where 6 is not
the algebraic imaginary. By them Hamilton s Biquaternion is
called simply a quaternion with nonreal constituents.] Some
little care is requisite in the management of these expressions, but
there is no new difficulty. The points to be observed are : first,
that any biquaternion can be divided into a real and an imaginary
part, the latter being the product of V  1 by a real quaternion ;
second, that this V  1 is commutative with all other quantities in
multiplication ; third, that if two biquaternions be equal, as
g fVlg W + Vl/ ,
we have, as in algebra, q = /, q" = r" ;
so that an equation between biquaternions involves in general
eight equations between scalars. Compare 80.
131. We have obviously, since \l  1 is a scalar,
8 (q + V  1 q") = Sq + V  1 Sq",
V (q + V ~1" q") = Vq + V 7 T Vq".
Hence ( 103)
= (Sq + V  1 Sq"+ Vq + V^l Vq") (Sq + V  1 Sq"  Vq
 V  1 Vq")
= (Sq> + V^l Sq"T ~(Vq + V  1 Vq J,
= (TqJ  (Tq J + 2 V^LSf . q Kq".
132.] INTERPRETATIONS AND TRANSFORMATIONS. 85
The only remark which need be made on such formulae is this, that
the tensor of a biquaternion may vanish while both of the component
quaternions are finite.
Thus, if Tq = Tq",
and S.q Kq" = 0,
the above formula gives
The condition S . q Kq =
may be written
or
where a is any vector whatever.
Hence Tq = Tq" = TKq" = ~ ,
and therefore
Tq (Uq f  V^T U* . Uq ) = (1  V^l Ua) q
is the general form of a biquaternion whose tensor is zero.
132. More generally we have, q, r, q, r being any four real
and nonevanescent quaternions,
(q + V^lg ) (r + V~lr ) = qr  q r + V^ 1 (qr + q r).
That this product may vanish we must have
qr = q r ,
and qr =  qr.
Eliminating r we have qq ~ 1 qr = qr )
which gives (q ~ l q) 2 = ~ 1.
i.e. q = qa.
where a is some unitvector.
And the two equations now agree in giving
r = ar ,
so that we have the biquaternion factors in the form
q (OL + V^T) and  (a  V~^T) r ;
and their product is
q(QL + V^T) (a  V ^1) r,
which, of course, vanishes.
86 QUATERNIONS. [133.
[A somewhat simpler investigation of the same proposition
may be obtained by writing the biquaternions as
q (q" 1 q + V^ ) and (rr ~ l + V^l) r ,
or q (q" + V ^1) and (r" + V ^1) r,
and shewing that
q" r" = a, where To. 1.]
From this it appears that if the product of two livectors
p + crV^T and p +er Vl
is zero, we must have
o 1 p = p f a  1 =Ua,
where a may be any vector whatever. But this result is still more
easily obtained by means of a direct process.
133. It may be well to observe here (as we intend to avail our
selves of them in the succeeding Chapters) that certain abbreviated
forms of expression may be used when they are not liable to confuse,
or lead to error. Thus we may write
T>q for (Tqf,
just as we write cos 2 for (cos 0) 2 ,
although the true meanings of these expressions are
T(Tq) and cos (cos 0).
The former is justifiable, as T (Tq) = Tq, and therefore T q is not
required to signify the second tensor (or tensor of the tensor) of q.
But the trigonometrical usage is defensible only on the score of
convenience, and is habitually violated by the employment of
cos" 1 x in its natural and proper sense.
Similarly we may write
S 2 q for (Sq) z , &c,
but it may be advisable not to use
Sq*
as the equivalent of either of those just written; inasmuch as it
might be confounded with the (generally) different quantity
although this is rarely written without the point or the brackets.
The question of the use of points or brackets is one on which
no very definite rules can be laid down. A beginner ought to use
1 34.] INTERPRETATIONS AND TRANSFORMATIONS. 87
them freely, and he will soon learn by trial which of them are
absolutely necessary to prevent ambiguity.
In the present work this course has been adopted : the
earlier examples in each part of the subject being treated with
a free use of points and brackets, while in the later examples
superfluous marks of the kind are gradually got rid of.
It may be well to indicate some general principles which
regulate the omission of these marks. Thus in 8 .a@ or V. a/3
the point is obviously unnecessary : because So. = 0, and Va. = a,
so that the S would annihilate the term if it applied to a alone,
while in the same case the V would be superfluous. But in S.qr
and V . qr, the point (or an equivalent) is indispensable, for Sq . r,
and Vq.r are usually quite different from the first written
quantities. In the case of K, and of d (used for scalar differen
tiation), the omission of the point indicates that the operator acts
only on the nearest factor : thus
Kqr = (Kq) r = Kq . r, dqr = (dq) r=dq.r;
while, if its action extend farther, we write
K . qr = K (qr), d . qr = d (qr} } &c.
In more complex cases we must be ruled by the general
principle of dropping nothing which is essential. Thus, for
instance
V(pK(dq)V(Vq.r))
may be written without ambiguity as
V.pKdqVVqr,
but nothing more can be dropped without altering its value.
Another peculiarity of notation, which will occasionally be
required, shows which portions of a complex product are affected
by an operator. Thus we write
VSar
if V operates on a and also on r, but
if it operates on r alone. See, in this connection, the last Example
at the end of Chap. IV. below.
134. The beginner may expect to be at first a little puzzled
with this aspect of the notation ; but, as he learns more of the
subject, he will soon see clearly the distinction between such an
expression as
88 QUATERNIONS. [134.
where we may omit at pleasure either the point or the first V
without altering the value, and the very different one
Sa/3. F/3 7 ,
which admits of no such changes, without alteration of its value.
All these simplifications of notation are, in fact, merely examples
of the transformations of quaternion expressions to which part of
this Chapter has been devoted. Thus, to take a very simple ex
ample, we easily see that
S. Fa/3 V/3y=S Fa/3 V/3y = .a/3F/3 7  SaV./SVpy =  SOL V. ( F/3 7 )/3
= SOL V. ( Vyfl/3 = S.a F( 7 /3)/3 = S. F( 7 /3)/9a = S F 7 /3 F/3a
= S . Vy/3V/3ot, &c., &c.
The above group does not nearly exhaust the list of even the simpler
ways of expressing the given quantity. We recommend it to the
careful study of the reader. He will find it advisable, at first, to
use stops and brackets pretty freely; but will gradually learn to
dispense with those which are not absolutely necessary to prevent
ambiguity.
There is, however, one additional point of notation to which
the reader s attention should be most carefully directed. A very
simple instance will suffice. Take the expressions
.2 and &.
7 OL ya.
The first of these is
/3 7  1 . 7 a 1 = ,9a 1 ,
and presents no difficulty. But the second, though at first sight
it closely resembles the first, is in general totally different in
value, being in fact equal to
For the denominator must be treated as one quaternion. If,
then, we write
^  ft
7 a
we have
7
so that, as stated above,
q =
We see therefore that
ay
INTERPRETATIONS AND TRANSFORMATIONS. 89
EXAMPLES TO CHAPTER III.
1. Investigate, by quaternions, the requisite formulae for
changing from any one set of coordinate axes to another ; and
derive from your general result, and also from special investiga
tions, the usual expressions for the following cases :
(a) Rectangular axes turned about z through any angle.
(6) Rectangular axes turned into any new position by rota
tion about a line equally inclined to the three.
(c) Rectangular turned to oblique, one of the new axes
lying in each of the former coordinate planes.
2. Point out the distinction between
a a
and find the value of their difference.
= 1, then U
OL + ft Fa/3
If ZW = 1, then U .
a. \a
Shew also that
aift 1 + tfa/3
, a  B Fa
and
provided a and ft be unitvectors. If these conditions are not
fulfilled, what are the true values ?
3. Shew that, whatever quaternion r may be, the expression
ar + rft,
in which a and ft are any two unit vectors, is reducible to the
form
where I and m are scalars.
4. If Tp = To. = Tft = 1, and S . a/3p = 0, shew by direct trans
formations that
S.U(poi} U(pp)=
Interpret this theorem geometrically.
( JO
QUATERNIONS.
5. If 8*0 = 0, Ta = T/3 = I, shew that
W7T  m
(1 + a" 1 ) = 2 cos "  a* = 2Sa * . a* {3.
6. Put in its simplest form the equation
pS. Fa/3F/3 7 F 7 a = ttFFyaFa/3 + bV. Fa/3F/3 7 + cFF/3 7 F 7 a ;
and shew that a = S . (Byp, &c.
7. Shew that any quaternion may in general, in one way only,
be expressed as a homogeneous linear function of four given
quaternions. Point out the nature of the exceptional cases. Also
find the simplest form in which any quaternion may generally be
expressed in terms of two given quaternions.
8. Prove the following theorems, and exhibit them as proper
ties of determinants :
(a) S.(* + 0)(0 + y)(y + a ) = 2S.*0y,
(b) S . Fa/3 F/3 7 Vya = (8. a/3 7 ) 2 ,
(c) 8. V(a + /3) (0 + y) F(/3 + 7 ) (7 + a) V(y f a) (a + 0)
(d) S.V(Va(3V{3y)V( F/3 7 Fya) F ( F 7 a Fa/3) =  (S . a/3 7 ) 4 ,
(e) >Sf.ae?16(/Sf.a/3 7 ) 4 ,
where 3 = F(F(a + 0) (0 + 7) F(^ + 7 ) ( 7 + a)),
9. Prove the common formula for the product of two determi
nants of the third order in the form
Soft W, Sy/3,
Say, S/3 7l S Wl
10. Shew that, whatever be the eight vectors involved,
W t >W! S/3 7l S/33,
Vt tyft ^7% ^
SfSa, fifS^ S8 yi SSS,
If the single term tfaaj be changed to Sai a,, the value of the
determinant is
INTERPRETATIONS AND TRANSFORMATIONS. 91
State these as propositions in spherical trigonometry.
Form the corresponding null determinant for any two groups
of five quaternions : and give its geometrical interpretation.
11. If, in 102, a, /3, 7 be three mutually perpendicular
vectors, can anything be predicated as to c^, j3 v y l ? If a, /3, 7 be
rectangular unitvectors, what of 15 /3 t , y l ?
12. If a, /3, 7, a , /3 , 7 be two sets of rectangular unitvectors,
shew that
t &c., &c.
13. The lines bisecting pairs of opposite sides of a quadri
lateral (plane or gauche) are perpendicular to each other when the
diagonals of the quadrilateral are equal.
14. Shew that
(a) S . q* = 2S*q  T 2 q,
(b) S.q* = S?q3SqrVq,
(c) 2 /3V + S 2 . a/3y = V 2 . a@y,
(d)
(e)
and interpret each as a formula in plane or spherical trigonometry.
15. If q be an undetermined quaternion, what loci are repre
sented by
(a) ( ? O = a 2 ,
(6) ( 3 ay = o 4 ,
(c) S.(ga) = o ,
where a is any given scalar and a. any given vector ?
16. If q be any quaternion, shew that the equation
<2 2 = ? 2
is satisfied, not alone by Q = q, but also by
Q = Jl(Sq.UVqTVq).
(Hamilton, Lectures, p. 673.)
1 7. Wherein consists the difference between the two equations
92 QUATERNIONS.
What is the full interpretation of each, a being a given, and p an
undetermined, vector ?
18. Find the full consequences of each of the following
groups of equations, as regards both the unknown vector p and
the given vectors a, /3, 7 :
S*p =0, Sap =0,
(") * fl n. (&) S.afip = 0, (c) S.aft, =0,
8 to pm 8$p = 0; s!#yp = <X
19. From 74, 110, shew that, if 6 be any unitvector, and m
mir mir
any scalar, e = cos  + e sin .
 2i
Hence shew that if a, ft, 7 be radii drawn to the corners of a tri
angle on the unitsphere, whose spherical excess is m right angles,
Also that, if A, B, G be the angles of the triangle, we have
2C 2 ZA
7 7r /3 7r ct 7r = 1.
20. Shew that for any three vectors a, /3, 7 we have
= 2.
(Hamilton, Elements, p. 388.)
21. If a v 2 , a s , x be any four scalars, and p lf p 2 , p :i any three
vectors, shew that
(S . Pl p 2 p 3 f + (S, . a, Vp 2 p 3 Y + x* (2 V Pl p 2 }*  a? (2 . a, (p 2  p,)) 8
+ 2H (x 2 + S Pl p 2 + a^ 2 ) = 2H (x* 4 p z ] + 2Ha 2
+ S {(x* + oj 2 + /3 X 2 ) (( Vp 2 p s ) 2 + 2a 2 a 3 ( 2 + 8p z p 9 ) a? (p 3 p 3 } 2 )} ;
where Ha 2 = a*a*a*.
Verify this formula by a simple process in the particular case
22. Eliminate p from the equations
V.0pap = t ^7/0 = 0;
and state the problem and its solution in a geometrical form.
INTERPRET ATIONS AND TRANSFORMATIONS. 93
23. If p, q, r, s be four versors, such that
qp =  S r = a,
rq = ps = /3,
where a and /3 are unitvectors ; shew that
8(V. VsVqV. VrVp) = 0.
Interpret this as a property of a spherical quadrilateral.
24. Shew that, if pq, rs, pr, and qs be vectors, we have
S(V.VpVsV.VqVr)=0.
25. If a, /3, 7 be unitvectors,
 (3 (S
26. If i, j, k, i , j , k y be two sets of rectangular unitvectors,
shew that
8 . Vii Vjj Vkk = (SijJ  (Sji) 2
and find the values of the vector of the same product.
27. If a, /3, 7 be a rectangular unitvector system, shew that,
whatever be X, /A, v,
and
are coplanar vectors. What is the connection between this and
the result of the preceding example ?
CHAPTER IV.
DIFFERENTIATION OF QUATERNIONS.
135. IN Chapter I. we have already considered as a special
case the differentiation of a vector function of a scalar independent
variable : and it is easy to see at once that a similar process is
applicable to a quaternion function of a scalar independent variable.
The differential, or differential coefficient, thus found, is in general
another function of the same scalar variable ; and can therefore be
differentiated anew by a second, third, &c. application of the same
process. And precisely similar remarks apply to partial differentia
tion of a quaternion function of any number of scalar independent
variables. In fact, this process is identical with ordinary differ
entiation.
136. But when we come to differentiate a function of a vector,
or of a quaternion, some caution is requisite ; there is, in general
(except, of course, when the independent variable is a mere scalar),
nothing which can be called a differential coefficient ; and in fact
we require (as already hinted in 33) to employ a definition of a
differential, somewhat different from the ordinary one but, coinciding
with it when applied to functions of mere scalar variables.
137. If T = F (q) be a function of a quaternion q,
where n is a scalar which is ultimately to be made infinite, is
defined to be the differential of r or Fq.
Here dq may be any quaternion whatever, and the righthand
member may be written
1 39.] DIFFERENTIATION OF QUATERNIONS. 95
where f is a new function, depending on the form of F; homo
geneous and of the first degree in dq ; but not, in general, capable
of being put in the form
138. To make more clear these last remarks, we ma,y observe
that the function
f(<t,dq),
thus derived as the differential of F (q), is distributive with respect
to dq. That is
r and s being any quaternions.
For
*)  Fq\
w/ ]
And, as a particular case, it is obvious that if x be any scalar,
t r).
139. And if we define in the same way
dF (q, r, s )
as being the value of
dq dr ds
n n n
where q, r, s, ... dq, dr, ds, are any quaternions whatever; we
shall obviously arrive at a result which may be written
f(q, r, s,...dq, dr, ds, ),
where / is homogeneous and linear in the system of quaternions
dq, dr, ds, and distributive with respect to each of them. Thus,
in differentiating any power, product, &c. of one or more quater
nions, each factor is to be differentiated as if it alone were variable ;
and the terms corresponding to these are to be added for the com
plete differential. This differs from the ordinary process of scalar
differentiation solely in the fact that, on account of the noncom
mutative property of quaternion multiplication, each factor must in
general be differentiated in situ. Thus
d (qr) dq.r\ qdr, but not generally = rdq + qdr.
96 QUATERNIONS. [ T 4 O 
140. As Examples we take chiefly those which lead to results
which will be of constant use to us in succeeding Chapters. Some
of the work will be given at full length as an exercise in quaternion
transformations.
(1)
The differential of the lefthand side is simply, since Tp is a
scalar,
2TpdTp.
That of f is C4(WT? 2
r ^ 1 V r n f
Hence Tp dTp =  Spdp,
or dTp = 
dTp Q dp
or ^=3^.
Tp p
(2) Again, p = Tp Up
dp = dTp. Up + TpdUp,
dp dTp dUp
whence = rn + j r
p Tp Up
d Up Tr dp
Hence jj^ = V .
Up p
This may be transformed into F 2 or  /^ , &c.
p lp
(3)
qKdq + dqKq 1
= Cao w + 2 dqKdq
\ n n
= qKdq + dqKq,
= qKdq + IT (qKdq) ( 55),
= 28. qKdq = 2S. dqKq.
140.] DIFFERENTIATION OF QUATERNIONS. 97
Hence dTq = S . dqUKq = S. dqUq~ l = TqS^ ,
since Tq = TKq, and UKq = Uq~\
[If q = p, a vector, Kq = Kp = p, and the formula becomes
Again, dq = Tqd Uq + UqdTq,
which gives dq = dTq dUq ,
q Tq Uq
whence, as
dTq
we have ydq = dUq_
q Uq
= qdq + dq.q
= ZS.qdq + 2Sq . Vdq + 2Sdq . Vq.
If q be a vector, as p, Sq and Sdq vanish, and we have
d(p*) = 2Spdp,SiS in (1).
(5) Let q = r*.
This gives dr* = dq. But
dr = d (q 2 ) = qdq + dq . q.
This, multiplied by q and into Kq, gives the two equations
qdr = q*dq + qdq . g,
and drKq = dg ?V + qdq . Kq.
Adding, we have
qdr + dr.Kq = (q* + Tq 2 + 2Sq .q)dq = 4<Sq . qdq ;
whence dq, i.e. dr , is at once found in terms of dr. This process
is given by Hamilton, Lectures, p. 628. See also 193 below, and
No. 7 of the Miscellaneous Examples at the end of this work.
(6) ?<f = l>
. . dq~ l = q~ l dq . q 1 .
If q is a vector, = p suppose,
dq+ = dp l = p l dp.p 1
T. Q. I.
9 8 QUATERNIONS. [ 1 4
P P
dp ~ dp
P P 1 P
(7) q = Sq+Vq,
But dq = $d# 4
Comparing, we have
dSq = Sdq, dVq=Vdq.
Since Kq = Sq Vq, we find by a similar process
(8) In the expression qaq~\ where a is any constant quaternion,
q may be regarded as a mere versor, so that
Thus S.dqKq = ,
and hence dg^ 1 =  (J^ 1 ,
as well as
q~ l dq =  dq~*q,
are vectors. But, if a = a + a, where a is a scalar, q^q 1 = a, i.e.
constant, so that we are concerned only with d (q<*q~ l ).
Hence d (qvq 1 ) = dq aq~ l  qctq" l dqq l t
= dqq l . qaq~ l  qaq~ l . dqq~\
= 2V. d a 1 = 2V.
(9) With the restriction in (8) above we may write
q = cos u + 6 sin u,
where T0 = I; S0d0 = 0.
Hence q 1 = cosu0 sin u ;
 q~ l dq = dq~ l q = {  (sin u+6 cos u) du  dO sin u} (cos u + sin ?<
= 0 dud6 sin w (cos w + ^ sin w) ;
= ^ q~ l = 0du + d0 sin u (cos w  ^ sin u).
Both forms are represented as linear functions of the rectangula
system of vectors
0, d0 t OdB.
140.] DIFFERENTIATION OF QUATERNIONS. 99
If the plane of q be fixed, 6 is a constant unit vector, and
dq q 1 = dq 1 q = Odu.
(10) The equation (belonging to a family of spheres)
pct
gives 8 dp {(p + a)e 2 (pa)} = 0;
or, by elimination of e,
sdp{( P +ar(p*r}=o,
whose geometrical interpretation gives a wellknown theorem.
If we confine our attention to a plane section through the
vector a, viz.
= U; or
where /3F7aFap ;
we have
dp\\ V. $ {(p + a) 1  (p  a) 1 } or F. dp F/3 {(p + )  (p  a) 1 } = 0.
(11) Again, from
pct
(which is the equation of the family of tores produced by the
rotation of a group of circles about their common chord) we have
Also this gives VU. (p + a) (p  a) = j3 = Jl  e*U. VOL P .
We obtain from the first of these, by differentiation,
or 8 . &dp {(p + a) 1  (p  a)" 1 } = 0.
If we consider /3 to be constant, we limit ourselves to a meridian
section of the surface (i.e. a circle) and the form of the equation
shews that, as /3 is perpendicular to the plane of a, p (and, of
course, dp),
v.d p {(p+*r(p*r}=o
We leave to the reader the differentiation of the vector form of
the equation above.
These results are useful, not only as elementary proofs of
geometrical theorems but, as hints on the integration of various
simple forms.
72
100 QUATERNIONS. [141.
(12) As a final instance, take the equation
P~YP = >
where p stands for dp/ds, s being the arc of a curve.
By 38, a is a unit vector, and the expression shews by its
form that it belongs to a plane curve. Let (3 be a vector in its
plane, and perpendicular to a. Operate by S . /3 and we get
whose integral is
P 2
the tensor of /3 being the constant of integration.
141. Successive differentiation of course presents no new
difficulty.
Thus, we have seen that
d (f) = dq.q + qdq.
Differentiating again, we have
and so on for higher orders.
If q be a vector, as p, we have, 140 (1),
Hence d 2 (p 2 ) = 2 (dpY + 2Spd*p, and so on.
Similarly d*Up= d (^ Vpd^j .
l_ 2dT P _2Spdp
Tp*~ ~ Tp s : Tp*
and d . Vpdp = V . pd 2 p.
Up /jr . X2 UpVptfp ZUpVpdpSpdp
Hence d* Up =  ( Vpdpf   f
2 p  2 VpdpSpdp}*.
142. If the first differential of ^ be considered as a constant
quaternion, we have, of course,
tfq = 0, d 3 q = 0, &c.
and the preceding formulaa become considerably simplified.
* This may be farther simplified ; but it may be well to caution the student that
we cannot, for such a purpose, write the above expression as
r.
I 44.] DIFFERENTIATION OF QUATERNIONS, 1.6 f
Hamilton has shewn that in this case Taylors Theorem admits
of an easy extension to quaternions. That is, we may write
f(q + xdq) =/(<?) + xdf (q) + <Ff(g) + ......
if d 2 q = ; subject, of course, to particular exceptions and limita
tions as in the ordinary applications to functions of scalar variables.
Thus, let
f (q) = q 3 > an d we have
d f(q) = q*dq + qdq.q + dq. q\
= 2dq . qdq + 2q (dq)* + 2 (dq) z q,
and it is easy to verify by multiplication that we have rigorously
(q + ocdqY = q* + x (q*dq + qdq .q + dq.q*) +
x* (dq .qdq + q (dq)* + (dqfq) + ** (dq) 3 ;
which is the value given by the application of the above form of
Taylor s Theorem.
As we shall not have occasion to employ this theorem, and as
the demonstrations which have been found are all too laborious for
an elementary treatise, we refer the reader to Hamilton s works,
where he will find several of them.
143. To differentiate a function of a function of a quaternion
we proceed as with scalar variables, attending to the peculiarities
already pointed out.
144. A case of considerable importance in geometrical and
physical applications of quaternions is the differentiation of a scalar
function of p, the vector of any point in space.
Let F(p) = C,
where F is a scalar function and C an arbitrary constant, be the
equation of a series of surfaces. Its differential,
f(p, dp) = 0,
is, of course, a scalar function : and, being homogeneous and linear
in dp, 137, may be thus written,
where v is a vector, in general a function of p.
This vector, v, is shewn, by the last written equation, to have
the direction of the normal to the given surface at the extremity
of p. It is, in fact, perpendicular to every tangent line dp ;
1 36, 98.
1:02 QUATERNIONS. [H5
145. This leads us directly to one of the most remarkable
operators peculiar to the Quaternion Calculus ; viz.
__ . d . d j d
V = i^ r + <) r + k r ..................... (1),
dx J dy dz
to whose elementary properties we will devote the remainder of
the chapter. The above definition is that originally given by
Hamilton, before the calculus had even partially thrown off its
early Cartesian trammels. Since i, j, k stand for any system of
rectangular unit vectors, while a?, y, z are Cartesian coordinates
referred to these as axes, it is implied in (1) that V is an Invariant.
This will presently be justified. Meanwhile it is easy to see that
if p be the vector of any point in space, so that
p = ix + jy + kz t
we have V/> = 3 ................................. (2),
"p ............ (4),
of which the most important case is
vJL a P .UP
Tp (Tp)*~ Tp 9 "
A second application gives
v ^=$ v ^= ............... (G) 
Again
so that VUp = ~ ........................... (7).
By the definition (1) we see that
\ a (d\* fd\*\
} +(3) + b~ r ............... ( 8 )
dxj \dyj \dz) }
the negative of what has been called Laplace s Operator. Thus
(6) is merely a special case of Laplace s equation for the potential
in free space.
Again we see by (2), a being any constant vector,
SaV . p  VSap = F. F Vp = 0,
from which
VTap + FaV. p = (SaV.p  otSVp) + (aSVp  VSap) = 0.
146.] DIFFERENTIATION OF QUATERNIONS. 103
[The student should note here that, in expanding the terms of
the vector function on the left by the formula (1) of 90, the
partial terms are written so that V is always to the left of (though
not necessarily contiguous to) its subject, p.]
146. By the help of these elementary results, of which (3)
and (7) are specially noteworthy, we easily find the effect of V
upon more complex functions.
For instance, taking different modes of operating, we have with
a = ia +jb + kc
SaV . p = VSap =  V (ax + by + cz) =  (ia + jb + kc) =  ...(!),
or thus VSoip = iSai + jSaj + kScnk a ;
while  VaV .p = WoLp = WpoL = V (poiSpOL)
= 3aa = 2a ...... , .............. (2),
or V Vap = i Vai + j VOLJ + k Vcnk = 2 a.
From the latter of these we have
Vap _ 2a _
"
_
Tp* " Tp> Tp 5 Tp 5 Tp 5
[where note that the first of these values is obtained thus,
 1
, q ,
The order is of vital importance.]
This, in its turn, gives
 g
Tf" ~W~ T?
where 8 is. a symbol of variation. This is a result of great physical
importance, especially in electrodynamics. We may alter the
righthand member (by 145, (5)) so as to write the whole in the
form
......... (4 ).
And it is easy to see that 8 may be substituted for V in the left
hand member. [The reason for this may be traced in the result of
145(6).]
As an addition to these examples, note that (by (2) of 148,
below)
Voif)
104 QUATERNIONS. [147.
which may be contrasted with (4) above. The altered position of
the point produces a complete change in the meaning of the left
hand member.
Finally, we see that
V2Sp = 2a (5),
a result which will be found useful in next Chapter.
147. Still more important are the results obtained from the
operator V when it is applied to
and functions of this vector. (Here , 77, f are functions of x, y, z,
so that a is a vector whose value is definite for each point of
space.)
We have at once
"_*?
/ dz
Those who are acquainted with mathematical physics will
recognize at a glance the importance of this expression. For, if a
denote the displacement (or the velocity) of a point originally
situated at p, it is clear that
\dx dy dz)
represents the consequent condensation of the group of points (say
particles of a fluid) originally in the neighbourhood of p, while
 ...(4),
dz) J \dz dx) \dx dyj
represents double the (vector) axis of rotation of the same group.
Other, and more purely quaternion, methods will be employed
later, to deduce these results afresh, and to develop their applica
tions. They are introduced here in their semiCartesian form
merely to shew the importance of the operator V.
148. Let us recur to the equation of 144, viz.
F( P } = G .......................... (1).
Ordinary complete differentiation gives
dF . dF , dF 1
dF = j dx 4 T dy + j dx,
dx dy dz
1 49] DIFFERENTIATION OF QUATERNIONS. 105
k
or, what is obviously the same,
dF=SdpVF ........................ (2),
which we may write, if we please, as
SdpV.F.
Here the point is obviously unnecessary, but we shall soon
come to cases in which it, or some equivalent, is indispensable.
Thus it appears that the operator
is equivalent to total differentiation as involved in the passage
from p to p + dp. Hence, of course, as in 144
dF (p)=0 = Svdp =  SdpVF,
and thus (as dp may have any of an infinite number of values)
v = VF ........................... (3).
If we pass from one surface of the series (1) to a consecutive
one by the vector Sp, we have
Hence v~ l SC is a value of Sp ; so that the tensor of v is, at every
point, inversely as the normal distance between two consecutive
surfaces of the series.
Thus, if (1) be the equation of a series of equipotential surfaces,
v, as given by (3), represents the vector force at the point p ; if
(1) be a set of isothermals, v (multiplied by the conductivity, a
scalar) is the flux of heat, &c.
149. We may extend the result (2) of 148 to vector functions
by multiplying both sides into a constant vector, a, and adding
three such expressions together. Thus if
cr = 0^+^ + 7^,
we obtain at once
d<r = S(dpV)<r = SdpV.(r ............... (4).
But here the brackets, or the point, should (at first, at least) be
employed ; otherwise we might confound the expression with
which, as equating a vector to a scalar, is an absurdity (unless both
sides vanish). See again, 148.
Finally, from (2) and (4), we have for any quaternion
dq = SdpV.q ........................ (5).
106 QUATERNIONS. [l49
The student s attention is particularly called to the simple
processes we have adopted in obtaining (4) and (5) from (2) of
148 ; because, in later chapters, other and more complex results
obtained by the same processes will frequently be taken for
granted ; especially when other operators than 8 (dpV) are em
ployed. The precautions necessary in such matters are twofold,
(a) the operator must never be placed anywhere after the operand ;
(6) its commutative or non commutative character must be carefully
kept in view.
EXAMPLES TO CHAPTER IV.
1. Shew that
(a) d.SUq = S.UqV = S TVUq,
(b) d.VUq=V. Uq l
(c) d.
(d) d.a. x = a? + l dx,
i
(e) d z .Tq={S 2 .dqq 1 S. (dqq 1 )*} Tq =  Tq V* ^ .
2. If Fp = 2.SoLpS/3p + gp*
give dFp = Svdp,
shew that v = 2 F . ap/3 + (g + 2a/3) p.
3. Find the maximum and minimum values of Tp, when
(a) (p) 2  a 2 ;
(b) (pa? = 
(c) p 2  SapS/3p =  a 2 ;
(d) p*  SoLpS/Bp =  a 2 ,)
S 7P = V. }
Point out the differences, geometrical and analytical, between
(a), (c) on the one hand, and (6), (d) on the other.
State each of the problems in words.
DIFFERENTIATION OF QUATERNIONS. 107
4. With V as in 145, shew that
8V (qaq~ l ) = 28 . Vq uf 1 = 28 . aq 1 Vq.
where q is a function of p, and a any constant vector.
5. Shew that, if a, /3, 7 be a constant rectangular unitvector
system,
qaq~ l d . qa.q~ l + qftq~ l d . qj3q~ l + qyq~ l d . qfiq 1 = 4>dqq~\
6. Integrate the differential equations :
(a) /3 + xSftp = 7,
(6) q + aq = b,
(c) 6 = VoiO. Proc. R. S. E. 1870.
7. Shew that
(a) fVadpSfip = V.a ( P S/3p + V . PjVpdp).
(b) fdp Vap = (pVapV. afVpdp) + 8 . afVpdp.
(c) fV . V*dp V(3p = i (*S . PfVpdp + (38 . afVpdp  P 8 . #>),
(d) fS . Vadp V(3p = J (Szp8j3p p 2 Sz/3 8 . afifVpdp).
When these integrals are taken round a closed plane curve we
have
jVpdp = 2Ay,
where A is the area, and 7 a unit vector perpendicular to its plane.
In this case
JdpVap = A Vya + 2 A Syot,
fV.VadpVpp = A (z8j3Y +
8. State, in words, the geometrical theorems involved in the
equations of 140, (10), (11), (12).
9. Shew (by means of 91) that
where V, V t , operate respectively on a, <r l ; but, after the
operations are performed, we put
v = v * = *.
CHAPTER V.
THE SOLUTION OF EQUATIONS OF THE FIRST DEGREE.
150. WE have seen that the differentiation of any function
whatever of a quaternion, q, leads to an equation of the form
dr=f(q,dq),
where /is linear and homogeneous in dq. To complete the process
of differentiation, we must have the means of solving this equation,
so as to be able to exhibit directly the value of dq.
This general equation is not of so much practical importance
as the particular case in which dq is a vector; and, besides, as we
proceed to shew, the solution of the general question may easily be
made to depend upon that of the particular case ; so that we shall
commence with the latter.
The most general expression for the function / is easily seen
to be
dr =f(q, dq) = 2 V. adqb + S . cdq,
where a, b, and c may be any quaternion functions of q whatever,
including even scalar constants. Every possible term of a linear
and homogeneous function is reducible to this form, as the reader
may see at once by writing down all the forms he can devise.
Taking the scalars of both sides, we have
Sdr = S.cdq = SdqSc + S.Vdq Vc.
But we have also, by taking the vector parts,
Vdr = 2F. adqb = Sdq . 2Fa& + 2F. a(Vdq) b.
Eliminating Sdq between the equations for Sdr and Vdr it is
obvious that a linear and vector expression in Vdq will remain.
Such an expression, so far as it contains Vdq, may always be
reduced to the form of a sum of terms of the type aS . /3 Vdq, by
the help of formulae like those in 90, 91. Solving this, we have
Vdq, and Sdq is then found from the preceding equation.
152.] SOLUTION OF EQUATIONS. 109
151. The problem may now be stated thus.
Find the value of p from the equation
aS/3p + fl^ftp + ... = 2. aSfa = 7,
where a, ft, a lt /3 lf ...y are given vectors.
The most general form of the lefthand member requires but
three distinct or independent terms. These, however, in con
sequence of the form of the expression, involve scalar constants
only; since the whole can obviously be reduced to terms of the
forms AiSip, BiSjp, CjSjp, &c. and there are only nine such forms.
In fact we may write the most general form either as
iSaip +J8j3p + kSyp,
or as afiip + faSjp + yfikp,
according as we arrange it by the vector, or by the scalar, factors
of the several terms. But the form
is that which, as committing us to no special system of vectors of
reference, is most convenient for a discussion of its properties.
If we write, with Hamilton,
4,p = 2.a80p ........................... (1),
the given equation may be written
<t>p = 7,
or p = ^ry,
and the object of our investigation is to find the value of the
inverse function 0" 1 .
It is important to remark that the definition (1) shews < to be
distributive, so that
A particular case of this is
(f) (xp} = xcj)p,
where x is a scalar.
Also, by the statement above, it is clear that <, in its most
general form, essentially involves nine independent scalars.
152. We have seen that any vector whatever may be expressed
linearly in terms of any three noncoplanar vectors. Hence, we
should expect d priori that a vector such as </></>$/9, or <f> 3 p, for
instance, should be capable of expression in terms of p, (ftp, and
[This is, of course, on the supposition that p, <f)p, and ftp are
110 QUATERNIONS. [152.
not generally coplanar. But it may easily be seen to extend to
that case also. For if these vectors be generally coplanar, so are <pp,
2 p, and cj) s p, since they may be written cr, (fxr, and <V. And thus,
of course, <f?p can be expressed as above. If in a particular case,
we should have, for some definite vector p, $p = gp where g is a
scalar, we shall obviously have <f> z p = g*p and (f> 3 p = g 3 p, so that the
equation will still subsist. And a similar explanation holds for
the particular case when, for some definite value of p, the three
vectors p, <pp, (/> 2 p are coplanar. For then we have an equation of
the form
which gives <fip = A$p + Bfip
= AB P +(A +B 2 )<f>p,
so that </> 3 /j is in the same plane.]
If, then, we write
<pp = xp + y<j>p + z<Fp ........................ (1),
and bear in mind the distributive character of the operator </>, it is
evident (if only ex absurdo) that x t y, z are quantities independent
of the vector p.
[The words above, "it is evident," have been objected to by
more than one correspondent. But, on full consideration, I not
only leave them where they are, but put them in Italics. For
they are, of course, addressed to the reader only ; and it is to be
presumed that, before he reaches them, he has mastered the
contents of at least the more important previous sections which
bear on this question, such as 23, 151. If, with these sections
in his mind, and a homogeneous linear equation such as (1) before
him, he does not see the " evidence," he has begun the study of
Quaternions too soon. A formal demonstration, giving the values
of x, y, z, will however be found in 156 9 below.]
If any three vectors, as i, j, k, be substituted for p, they will in
general enable us to assign the values of the three coefficients on
the right side of the equation, and the solution of the problem of
1 5 1 is complete. For by putting $T l p for p and transposing, the
equation becomes
~
that is, the unknown inverse function is expressed in terms of
direct operations. Should x vanish, while y remains finite, we
must substitute $~*p for p, and have
1 54.] SOLUTION OF EQUATIONS. Ill
and if x and y both vanish
 z$ l p = p.
[We may remark here that it is in general possible to determine
x,y,z by putting one known vector for p in (1). The circumstances
in which some particular vector does not suffice will be clear from
the theory to be given below.]
153. To illustrate this process by a simple example we shall
take the very important case in which $ belongs to a central
surface of the second order ; suppose an ellipsoid ; in which case it
will be shewn (in Chap. IX.) that we may write
(frp =  cfiSip  tfjSjp  c*kSkp,
where i,j, k are parallel to the principal diameters, and the semi
lengths of these are I/a, 1/6, 1/c.
Here we have
<l>k = c 2 &, tfk = cVj, <pk = ck.
Hence, putting separately i, j, k for p in the equation (1) of
last section, we have
 a 6 = x + 7/a 2 + za*,
 6 6 = x + yV + zb\
c 6 = x + yc* + 2C 4 .
Hence a 2 , 6 2 , c 2 are the roots of the cubic
which involves the conditions
y = + c
x =  a 2 b*c 2 .
Thus, with the above value of (/>, we have
(a 2
154. Putting </>~V in place of p (which is any vector what
ever) and changing the order of the terms, we have the desired
inversion of the function (j> in the form
aWc/TV  (a 2 6 2 + 6V + cV) a  (a 2 + 6 2 + c 2 ) fa + (/>V,
where the inverse function is expressed in terms of the direct
112 QUATERNIONS. [ T 5 5
function. For this particular case the solution we have given
is complete, and satisfactory; and it has the advantage of pre
paring the reader to expect a similar form of solution in more
complex cases.
155. It may also be useful as a preparation for what follows,
if we put the equation of 153 in the form
= cf> (p) = <p p  (a 2 + 6 2 + c 2 ) <pp + (a 2 6 2 + 6V + cV) <j>p  a Wp
= {<#> 3  (a 2 + 6 2 + c 2 ) c/> 2 + (a 2 6 2 + 6V + cV)  a 2 6 2 c 2 ) p
= {((/, a 2 ) (</>6 2 )(0c 2 )} P ........................... (2).
This last transformation is permitted because (151) (f> is com
mutative with scalars like a 2 , i.e. (f> (a?p) = cffyp. The explanation
of its meaning must, however, be deferred to a later section.
( 177.)
Here we remark that the equation
V.p<f>p = 0, or <t>p = gp,
where </> is as in 153, and g is some undetermined scalar, is
satisfied, not merely by every vector of nulllength, but by the
definite system of three rectangular vectors Ai, Bj, Ck whatever
be their tensors, the corresponding particular values of g being
a 2 , V, c\
156. We now give Hamilton s admirable investigation.
The most general form of a linear and vector function of a
vector may of course be written as
where q and r are any constant quaternions, either or both of which
may degrade to a scalar or a vector.
Hence, operating by S . cr where a is any vector whatever,
S<r<l>p = S<T?,V.qpr = Sp2V.r(rq = Sp<l> (r ............ (1),
if we agree to write </>V = lV .rcrq,
and remember the proposition of 88. The functions <p and <
are thus conjugate to one another, and on this property the whole
investigation depends.
157. Let X, fi be any two vectors, such that
Operating by 8 . X and S. fi we have
= 0, Sfi^p = 0.
SOLUTION OF EQUATIONS.
113
But, introducing the conjugate function </> , these become
SpfiX = 0, Spffj, = 0,
and give p in the form mp = V$ \<f> /z,
where ra is a scalar which, as we shall presently see, is independent
of \, p, and p.
But our original assumption gives
hence we have m^" 1 FXyu, = F^ Xc//^ (2),
and the problem of inverting <fi is solved.
It remains to find the value of the constant m, and to express
the vector
as a function of FX/t.
158. To find the value of m, we may operate on (2) by S. <f>v,
where v is any vector not coplanar with X and /j,, and we get
mS . <f> v<l)~ l FX/4 = mS . zx/x/T 1 FXyu, (by (1) of 156)
= mS . X//Z/ = S . tfr XfiiJbfiv, or
wi = FT^: ~~ ~ (3).
S. \fjiV
[That this quantity is independent of the particular vectors
X, /*, v is evident from the fact that if
X = x\ + yfM + zv t fji = ^X + y^ + z^v, and v = % 2 \ + y^ + zp
be any other three vectors (which is possible since X, //,, v are not
coplanar), we have
from which we deduce
8 . <f> X <f> V<V = os y z
and
* y* z *
x y z
*i 2/i ^
^2 2/2 ^2
S.\flV t
so that the numerator and denominator of the fraction which ex
presses m are altered in the same ratio. Each of these quantities
is in fact an Invariant, and the numerical multiplier is the same
for both when we pass from any one set of three vectors to another.
T. Q. I. 8
114 QUATERNIONS. [l59
A still simpler proof is obtained at once by writing X + xp for X
in (3), and noticing that neither numerator nor denominator is
altered.]
159. We have next to express
as a function of FX//,. For this purpose let us change </> to </> g,
where g is any scalar. It is evident that < becomes </> g, and
our equation (2) becomes
= (rac/T 1 gx + f f] V*>p, suppose.
In this equation (see (3) above)
_ S.($g)\(fg)iL(<l>  g) v
S.\/J,V
= mm 1 g + m 2 g 2 g 3 ............................ (4)
is what m becomes when is changed into <f) g ; m l and m z being
two new scalar constants whose values are
S .
1=
S .
If, in these expressions, we put X + osfi for X, we find that the terms
in x vanish identically ; so that they also are invariants.
Substituting for m gt and equating the coefficients of the
various powers of g after operating on both sides by </> g, we
have two identities and the following two equations,
[The first determines ^, and shews that we were justified in
treating V (0 X//. + Xc/> ^) as a linear and vector function of V . X/A.
The result might have been also obtained thus,
S .
S . v% V\JJL = S .
= mjSKfjiv S .
= 8,v
l62.] SOLUTION OF EQUATIONS. 115
and all three (the utmost generality) are satisfied by
% = w 2 </>.]
160. Eliminating ^ from these equations we find
or mc/T 1 = m, ??? 2 (/> 4 < 2 ......................... (5),
which contains the complete solution of linear and vector equa
tions.
161. More to satisfy the student as to the validity of the
above investigation, about whose logic he may at first feel some
difficulties, than to obtain easy solutions, we take a few very
simple examples to begin with : we treat them with all desirable
prolixity, as useful practice in quaternion analysis ; and we append
for comparison easy solutions obtained by methods specially
adapted to each case. The advanced student need therefore pay
but little attention to the next ten sections.
162. Example I.
Let <f>p= V.ap/3 = y.
Then <>= F.pa = 0p.
Hence m = \ S ( V. a\j3 V. ap/3 V. ai/).
>O . i^lLV
Now A, p, v are any three noncoplanar vectors ; and we may
therefore put for them a, ft, 7 , if the latter be noncoplanar.
With this proviso
B ^ ( a / 32 F a 7/ 3 + a2 ^ ^ F af y + a2 8 7)
Hence we have by (5) above
. <F l v = a 2 /3 2 Sa/3 . p =  ^^ + F. 7 /3 + F. a (F. a
which is one form of solution.
82
116 QUATERNIONS. [ 1 6 3 .
By expanding the vectors of products we may easily reduce it
to the form
*2/D2CV /O rs,2/Q2 i , /O2 Ct n , Q^CfQ
a p ootp . p = a p y + ap bay + pa >py,
a" 1 Say + (3~ l S/3y  y
p= SaW~ 
163. To verify this solution, we have
V.apft = ^g (ftSay + aSfty  V. ay/3) = y,
which is the given equation.
164. An easier mode of arriving at the same solution, in this
simple case, is as follows :
Operating by 8. a and 8. ft on the given equation
we obtain afSffp = Say,
ff8*p = S
and therefore aS/3p = of 1
But the given equation may be written
aS/3p  pSa/3 + @Sap = y.
Substituting and transposing we get
pSa/3 = a 1 Say + P^SjSy  y,
which agrees with the result of 162.
[Note that, at first sight, one might think that the value of p
should have a term with an arbitrary scalar factor added. But the
notation p is limited to a vector. Had the equation been written
V.aq(3 = y
we should have had
aq/3 = x + y, or q = xa 1 ^ 1 + a~ l y/3~\
But because q is to be a vector
Sq = Q, or xSa/3 + S.ay/3 = )
and, with this value of x, the expression for q takes the form given
above for p.]
165. If a, ft, y be coplanar, the above mode of solution is
applicable, but the result may be deduced much more simply.
For ( 101) 8 . a/By = 0, and the equation then gives 8 . a/3p = 0,
so that p also is coplanar with a, ft, y.
l66.] SOLUTION OF EQUATIONS. 117
Hence the equation may be written
and at once
p =
and this, being a vector, may be written
This formula is equivalent to that just given, but not equal to
it term by term. [The student will find it a good exercise to
prove directly that, if a, ft, y are coplanar, we have
 7) = a^
The conclusion that
.a/3p =
in this case, is not necessarily true if
But then the original equation becomes
aSpp+pSap = y,
which is consistent with
This equation gives
2 /3 2 <? 2 fi\  ^ a ^ ^ a @ i Q ^^ ^ ay ^
Say <
by comparison of which with the given equation we find
Sap and S/3p.
The value of p remains therefore with an indeterminate vector
part, parallel to aft ; i.e. it involves one arbitrary scalar.
166. Example II.
Let <f)p = V . aftp = y.
Suppose a, ft, 7 not to be coplanar, and employ them as X, yu,, v
to calculate the coefficients in the equation for </>~ 1 . We have
S . o"</>/9 = S . <raftp = S . p V . o~a/3 = S . p<f) cr.
Hence ftp = V . pa ft = V . ftap.
We have now
118 QUATERNIONS.
s8(*./3.V.to + *.l3afl.
Hence by (5) of 1 GO
= (2 (S*/3) 2 + a 2 /? 2 ) 7  3Sa/3F. a/3 7 + F. a/3F. a/3 7 ,
which, by expanding the vectors of products, takes easily the
simpler form
. p = a 2 /3 2 7  a/
167. To verify this, operate by F.a/3 on both sides, and we
have
or F . a/3p = 7 .
168. To solve the same equation without employing the
general method, we may proceed as follows :
ry = F. {3p = pSoL/3 + F. F
Operating by 8. Fa/3 we have
Divide this by Sa/3, and add it to the given equation. We thus
obtain
Hence p = r
a form of solution somewhat simpler than that before obtained.
To shew that they agree, however, let us multiply by a 2 y# 2 $a/3,
and we get
. p =
169.] SOLUTION OF EQUATIONS. 119
In this form we see at once that the righthand side is a vector,
since its scalar is evidently zero ( 89). Hence we may write
a 2 /3 2 a/3 .p=V. /3a 7 a/3  Fa/3S . a/3 7 .
But by (3) of 91,
 jS . a/3 Fa/3 + a . /3 ( Fa/3) 7 + /3S . F(a/3) a 7 + Fa/3 . a/3 7 = 0.
Add this to the righthand side, and we have
a 2 /3 2 Sa/3 . p = 7 ((Sa(3) 2  S . a/3 Fa/3)  a (Sot{3S/3y S.0( Fa/3) 7)
+ /3 (SajSSoLy + S . F(a) a 7 ).
But (Sa/3) 2  . a/3 Fa/3 = (Sa/3) 2  ( Fa) 2  a 2 /3 2 ,
tfa/3>S 7  8 . j3 ( Fa/3) 7  SoL(3S/3y  S/3aS/3y + /3 2 a 7 = /3 2 /Sfa 7 ,
>Sfa/3>Sfa7 + S . F (a/3) a 7 = Sa/BSoLy + /S Y
and the substitution of these values renders our equation identical
with that of 166.
[If a, /3, 7 be coplanar, the simplified forms of the expression
for p lead to the equation
801/3 . p^y = 7  a 1 Sa 7 + 2/3>Sfa 1 /3 1 ASfa 7  /T S/ty,
which, as before, we leave as an exercise to the student.]
169. Example III. The solution of the equation
Vep = 7
leads to the vanishing of some of the quantities m. Before, how
ever, treating it by the general method, we shall deduce its solu
tion from that of
F. OL/3p = 7
already given. Our reason for so doing is that we thus have an
opportunity of shewing the nature of some of the cases in which
one or more of m, m^ m 2 vanish; and also of introducing an
example of the use of vanishing fractions in quaternions. Far
simpler solutions will be given in the following sections.
The solution of the last written equation is, 166,
a 2 /3 2 Sa/3 , p = a 2 /3 2 7  a/3 2 a 7  {B^Spy + 2/3a/3a 7 .
If we now put QL/3 = e + e
where e is a scalar, the solution of the firstwritten equation will
evidently be derived from that of the second by making e gradually
tend to zero.
120 QUATERNIONS. [ 1 7<D.
We have, for this purpose, the following necessary transforma
tions :
a/3 2 Say + /3a 2 S/3y = a/3 . ft Say + ft a . aSfty,
= (e + e) ftSay + (ee) aSfty,
= e (PSay + aSfty) + eV. 7 Fa/3,
e (ffSoiy + a fifty) + eVye.
Hence the solution becomes
(e 2  e 2 ) ep = (e 2 e*)ye ({BSay + otSjSy) eVye + ZeftSay,
= ( _ 6 2) ry + eV. y Va/3 eVye,
= (e 2  e 2 ) y + e Vye + ye 2  eSye,
= e*y + eVye  eSy.
Dividing by e, and then putting e = 0, we have
Q /
Now, by the form of the given equation, we see that
Sye = 0.
Hence the limit is indeterminate, and we may put for it x, where
x is any scalar. Our solution is, therefore,
or, as it may be written, since >Sfye = 0,
p = e 1 (y + x).
The verification is obvious for we have
ep = y + x.
170. This suggests a very simple mode of solution. For we
see that the given equation leaves Sep indeterminate. Assume,
therefore,
Sep = x
and add to the given equation. We obtain
ep = x + 7,
or p = e 1 (7 + x\
if, and only if, p satisfies the equation
Vep = 7.
171. To apply the general method, we may take e, 7 and 67
(which is a vector) for X, //,, v.
72.]
We find
Hence
SOLUTION OF EQUATIONS.
p = Vpe.
m = 0,
121
or
Hence
= 0,
That is, p = 
= e 1 7 + a?e, as before.
Our warrant for putting a?e, as the equivalent of c/)" 2 is this :
The equation 0V =
may be written V . eFecr = ae 2 eSecr.
Hence, unless a = 0, we have a \\ e = xe.
[Of course it is well to avoid, when possible, the use of expressions
such as 0~ 2 &c. but the student must be prepared to meet them ;
and it is well that he should gain confidence in using them, by
verifying that they lead to correct results in cases where other
modes of solution are available.]
172. Example IV. As a final example let us take the most
general form of (/>, which, as has been shewn in 151, may be
expressed as follows :
0p = aSl3p + a^p + ajSfij) = 7.
Here $p = fiSap + ftSfl^p + 2 2 p,
and, consequently, taking a, a v a 2 , which are in this case non
coplanar vectors, for X, ^, v, we have
8 . OLOL^
Sao.
~8
21
where
A =
122 QU ATEBNIONS. [172.
=  8 .
Hence the value of the determinant is
 (SaaS . Ffl^a, Va^ + Sa^S . Va 2 a Va^ + Sa^aS . VOL^ Va^)
= S.a( Va^S . a A) {by 92 (4)} =  (S . aatf.
The interpretation of this result in spherical trigonometry is
very interesting. (See Ex. (9) p. 90.)
By it we see that
Similarly,
m < = s~i^ s  [a (/3/Saa + /Si<Sa a + /8 srv )
s c< 2 ) + &c.]
+ . . .)
. a ( F/3/3.S . Faa, Fa.o
o . aa ct
+ S.a l (V/3(3 l S.Vaa l Va 2 a+...)
or, taking the terms by columns instead of by rows,
y 4
+ .... I
=  8 ( Fact, F/3& + Fa x a 2 F/9 A +
Again,
or, grouping as before,
= ^ ^ ^f [^ ( Vaafiw, + Va^Saa, + Fa^aa) + ...],
1 74] SOLUTION OF EQUATIONS. 123
*i a ) +]( 92 (4)),
And the solution is, therefore,
. Fa otj F/3&
[It will be excellent practice for the student to work out in
detail the blank portions of the above investigation, and also to
prove directly that the value of p we have just found satisfies the
given equation.]
173. But it is not necessary to go through such a long process
to get the solution though it will be advantageous to the student
to read it carefully for if we operate on the proposed equation by
8 . a^g, 8 . 2 a, and 8 . aa t we get
S . o^aa^/^/o = 8 . 2 7,
8 . aLOL^SPyp = S . OLOL t y.
From these, by 92 (4), we have at once
pS . aot^S . /3/3 A = Vft^S . l7 + F/S^S . a^y
The student will find it a useful exercise to prove that this is
equivalent to the solution in 172.
To verify the present solution we have
(*SP P + afifo + o^S&p) S . awfi . 0J3&
174. It is evident, from these examples, that for special cases
we can usually find modes of solution of the linear and vector
equation which are simpler in application than the general process
of 160. The real value of that process however consists partly in
its enabling us to express inverse functions of </>, such as(< g)~ l
for instance, in terms of direct operations, a property which will be
of great use to us later ; partly in its leading us to the fundamental
cubic
(f> 3 w 2 2 + m^ m = 0,
which is an immediate deduction from the equation of 160, and
whose interpretation is of the utmost importance with reference to
the axes of surfaces of the second order, principal axes of inertia,
124 QUATERNIONS. [175.
the analysis of strains in a distorted solid, and various similar
enquiries.
We see, of course, that the existence of the cubic renders it a
mere question of ordinary algebra to express any rational function
whatever of (f> in a rational threeterm form such as
In fact it will be seen to follow, from the results of 177 below,
that we have in general three independent scalar equations of the
form
which determine the values of A, B, C without ambiguity. The
result appears in a form closely resembling that known as
Lagrange s interpolation formula.
175. When the function <f> is its own conjugate, that is, when
Sp<f)(T = 8(7(j)p
for all values of p and cr, the vectors for which
form in general a real and definite rectangular system. This, of
course, may in particular cases degrade into one definite vector, and
any pair of others perpendicular to it ; and cases may occur in
which the equation is satisfied for every vector.
To prove this, suppose the roots of
( 159 (4)) to be real and different, then
<M = #iPij
0/> 2 = 9*P* \
4>P*=fffJ
where p v p 2 , p 3 are three definite vectors determined by the
constants involved in <j).
Hence, operating on the first by Sp 2 , and on the second by Sp v
we have
The first members of these equations are equal, because $ is its
own conjugate.
Thus (fc&)%p, = <>;
which, as g l and g t2 are by hypothesis different, requires
<>> =o.
1 77.] SOLUTION OF EQUATIONS. 125
Similarly Sp,p 3 = 0, Sp^ = 0.
If two roots be equal, as g v g 3 , we still have, by the above proof,
Sp^ = and Sp^ = 0. But there is nothing farther to determine
ps and p 9 , which are therefore any vectors perpendicular to p lf
If all three roots be equal, every real vector satisfies the equation
(</><7)p = 0.
176. Next as to the reality of the three directions in this
case.
Suppose g + h \l 1 to be a root, and let p + cr V 1 be the
corresponding value of p, where g and h are real numbers, p and cr
real vectors, and \/" 1 the old imaginary of algebra.
Then <f>(p + (T V~l) = (g + h\/  1) (p + a V^l),
and this divides itself, as in algebra, into the two equations
(f)p = gp ho;
c/>cr = hp + go:
Operating on these by 8 . cr, S.p respectively, and taking the
difference of the results, remembering our condition as to the
nature of c/>
we have h (<r 2 + p 2 ) = 0.
But, as cr and p are both real vectors, the sum of their squares
cannot vanish, unless their tensors separately vanish. Hence h
vanishes, and with it the impossible part of the root.
The function c/> need not be selfconjugate, in order that the
roots of
ra, =
may be all real. For we may take g lt # 2 , g 3 any real scalars, and
a, /3 } 7 any three real, noncoplanar, vectors. Then if be such that
S.a@y.<l>p= g^S . /3yp + g^S . yctp + g^S . afip,
we have obviously
((/> gj * = 0, (< &) /3 = 0, (</>  ffa ) 7 = 0.
Here c/> is selfconjugate only if a, /3, 7 form a rectangular
system.
177. Thus though we have shewn that the equation
g 3 m^g* 4 m^y m =
has three real roots, in general different from one another, when c/>
126 QUATERNIONS. [178.
is selfconjugate, the converse is by no means true. This must be
most carefully kept in mind.
In all cases the cubic in may be written
(*sO(*fc)(*0 i ) = o (i),
and in this form we can easily see its meaning, provided the values
of g are real. For there are in every such case three real (and in
general noncoplanar) vectors, p v p 2 , p s for which respectively
(*^)P! = O, &)ft = 0, W><7 8 )p 8 = 0.
Then, since any vector p may be expressed by the equation
pS PiPiPs = Pi S PiPaP + P* S PsPiP +P* S  PiPzP ( 91 )>
we see that when the complex operation, denoted by the lefthand
member of the symbolic equation, (1), is performed on p, the
first of the three factors makes the term in p l vanish, the second
and third those in p^ and p s respectively. In other words, by the
successive performance, upon a vector, of the operations <f>g v <j> g z ,
$ g a , it is deprived successively of its resolved parts in the direc
tions of p v p v p 3 respectively ; and is thus necessarily reduced to
zero, since p t , p 2 , p a are (because we have supposed g v g z , g 3 to be
distinct) distinct and noncoplanar vectors.
178. If we take p v p 2 , p 3 as rectangular unit vectors, we have
~P= Pl$PlP +
whence ^ =  g lPl S Pl p 
or, still more simply, putting i,j, k for p v p 2 , p 3 , we find that any
selfconjugate function may be thus expressed
fo^ffiiSipgJ&ipgJcSkp (2),
provided, of course, i, j, k be taken as the roots of the equation
Vpj>p = 0.
A rectangular unitvector system requires three scalar quan
tities, only, for its full specification. g v </ 2 , g 3 are other three.
Thus any selfconjugate function involves only six independent
scalars.
179. A very important transformation of the selfconjugate
linear and vector function is easily derived from this form.
We have seen that it involves, besides those of the system i,j, k,
three scalar constants only, viz. g v g^, g y Let us enquire, then,
whether it can be reduced to the following form
<l>p=fp + hV.(i + ek) P (iek) (3),
1 8 I.] SOLUTION OF EQUATIONS. 127
which also involves but three scalar constants/, h, e, in addition to
those of i, j y k, the roots of
= 0.
Substituting for p the equivalent
p = iSip  jSjp kSkp,
expanding, and equating coefficients of i,j, k in the two expressions
(2) and (3) for <f>p, we find
These give at once
Hence, as we suppose the transformation to be real, and therefore
e* to be positive, it is evident that g l <7 2 and # 2 g. A have the same
sign ; so that we must choose as auxiliary vectors in the last term
of <j)p those two of the rectangular directions i, j, k for which the
coefficients g have respectively the greatest and least values.
We have then
and /
180. We may, therefore, always determine definitely the
vectors X, //,, and the scalar/, in the equation
= + V.
when (j) is selfconjugate, and the corresponding cubic has not equal
roots ; subject to the single restriction that
T.\fL
is known, but not the separate tensors of X and //,. This result is
important in the theory of surfaces of the second order, and in that
of Fresnel s WaveSurface, and will be considered in Chapters IX.
and XII.
181. Another important transformation of <f> when self
conjugate is the following,
<f)p = aaVap + b/SSffp,
where a and b are scalars, and a and /3 unitvectors. This, of
128 QUATERNIONS. [182.
course, involves six scalar constants, and belongs to the most
general form
<I>P= 9lPl S PlP ~ 9*P$P*P  93P* S P3P>
where p l} p 2 , p 3 are the rectangular unitvectors for which p and (f>p
are parallel. We merely mention this form in passing, as it
belongs to the focal transformation of the equation of surfaces of
the second order, which will not be farther alluded to in this work.
It will be a good exercise for the student to determine a, /3, a and
6, in terms of g v g 2) g 3y and p v p 2 , p 3 .
182. We cannot afford space for a detailed account of the
singular properties of these vector functions, and will therefore
content ourselves with the enuntiation and proof of one or two of
the more important.
In the equation m<f^V\i^= F^ X^V ( 157),
substitute \ for <fi\ and /it for (jb /it, and we have
Change </> to (/> g, and therefore (/> to (/> g, and m to m g , we have
, V (*  g) ^ (f  0)~V = (*  9) V\n ;
a formula which will be found to be of considerable use.
183. Again, by 159,
Similarly S.pft h)~ l p = Sp^p  S PXP + hp 2 .
Hence
wi. rv / , x^, m,, ci /i T \i 7 \ f 9 wiSpd) p]
jt8.p(4>grpj*8.p(4>hrp=te h) Y  ^^
That is, the functions
^S.ptfgrp, and ^S.pQhrp
are identical, i.e. when equated to constants represent the same series
of surfaces, not merely when
9**k*
but also, whatever be g and h, if they be scalar functions of p which
satisfy the equation
mS . p<f>~ l p =
185.] SOLUTION OF EQUATIONS. 129
This is a generalization, due to Hamilton, of a singular result ob
tained by the author*.
184. It is easy to extend these results ; but, for the benefit of
beginners, we may somewhat simplify them. Let us confine our
attention to cones, with equations such as
 (?)> = 0,1
These are equivalent to mSp^p gSpXP+9*P* = >
mSp^~ l p  hSpxp + h p* = 0.
Hence
m (1  x) Sp^p (g hx) Spxp + (g*  Wx] p* = 0,
whatever scalar be represented by x.
That is, the two equations (1) represent the same surface if this
identity be satisfied. As particular cases let
(1) x = \, in which case
(2) g hx=0, in which case
or mSp~ l ^~ l p gh=0.
(3) ^ = ^, giving
or m(h + g) Sp^p + ghSpxp = 0.
185. In various investigations we meet with the quaternion
= < + 0/8 + 707 ..................... (1),
where a, 0, 7 are three unitvectors at right angles to each other.
It admits of being put in a very simple form, which is occasionally
of considerable importance.
We have, obviously, by the properties of a rectangular unit
system
q = /#70a + 70/3 + a/307.
As we have also
S./8 7 = l (71 (13)),
* Note on the Cartesian equation of the Wave Surface. Quarterly Math. Journal,
Oct. 1859.
T. Q. I. 9
130 QUATERNIONS. [l86.
a glance at the formulae of 159 shews that
Sq = m a ,
at least if <j> be selfconjugate. Even if it be not, still (as will be
shewn in 186) the term by which it differs from a selfconjugate
function is of such a form that it disappears in Sq.
We have also, by 90 (2),
Vq = OL
. yea + 7$ . ae/3 ( 186)
=  (aae + /3S/36 + jSye) = e.
[We may note in passing that the quaternion (1) admits of
being expressed in the remarkable forms
or
, __ d ~ d d
where ( 145) V = a, + /5 y + 7 ^ ,
cfo? cZy ^
and p = a^ + ^7 + 72.
We will recur to this towards the end of the work.]
Many similar singular properties of <j> in connection with a rect
angular system might easily be given ; for instance,
) = mV.
which the reader may easily verify by a process similar to that just
given, or (more directly) by the help of 157 (2). A few others
will be found among the Examples appended to this Chapter.
186. To conclude, we may remark that, as in many of the
immediately preceding investigations we have supposed <f> to be
selfconjugate, a very simple step enables us to pass from this to
the nonconjugate form.
For, if < be conjugate to </>, we have
Spficr =
and also Spfya =
Adding, we have
so that the function (< + < ) is selfconjugate.
Again, Spfp = Spflp,
which gives Sp ($ <}> ) p = 0.
Hence (< $ ) p= Yep,
1 87.] SOLUTION OF EQUATIONS. 131
where, if < be not selfconjugate, e is some real vector, and
therefore
Thus every nonconjugate linear and vector function differs from
a conjugate function solely ly a term of the form
Vep.
The geometric signification of this will be found in the Chapter on
Kinematics.
The vector e involves, of course, three scalar constants. Hence
( 151, 178) the linear and vector function involves, in general, nine.
187. Before leaving this part of the subject, it may be well
to say a word or two as to the conditions for three real vector
solutions of the equation
Vp$p = 0.
This question is very fully treated in Hamilton s Elements, and
also by Plarr in the Trans. K S. E. For variety we adopt a semi
graphic method*, based on the result of last section. By that
result we see that the equation to be solved may be written as
<f)p = &p + Vep = xp ........................ (1)
where w is a given selfconjugate function, e a given vector, and x
an unknown scalar.
Let a v 2 , a 3 and g v g 2 , g 3 (the latter taken in descending order
of magnitude), be the vector and scalar constants of OT, so that
( 177)
(&)*! = (), &C.
We have obviously, by operating on (1) with S . o^ &c., three
equations of the form
fy{( S P 1 flOa 1 Fe 1 )=0 ..................... (2).
Eliminating p (whose tensor is not involved) we have
S K0> *K V**J {(g 2  x) ,  FeJ {(?.  x) a  F6 3 } = 0,
or (xg l )(xg 2 )(xg 3 )xe* + Sevre = ............ (3).
From each value of x found from this equation the corre
sponding value of p is given by (2) in the form
P H V {(9,  ) ,  FeoJ {(g,  x) 2  Fea 2 },
II (&  ) (0,  *0 8 + (02  ) ^ V  (01 ~ ) a 2^i e ~ ^^S 6
II (0i ~ ) (02  ) 8 + ^ a 3 (w  0) e
* Proc. ^. S. E., 187980.
92
132 QUATERNIONS. [l88.
The simplest method of dealing with (3) seems to be to find the
limiting value of Te, Ue being given, that the roots may be all real.
They are obviously real when Te = 0. It is clear from the pro
perties of or that the extreme values of S . UetzUe (which will be
called f ) are g l and g y
Trace the curve
and draw the (unique) tangent to it from the point x = J; , y = 0,
f having any assigned value from g 3 to g^. Let this tangent make
an angle with the axis of x. Suppose a simple shear to be
applied to the figure so as to make this tangent turn round the
point f, 0, and become the x axis, while the y axis is unchanged.
The value of y will be increased by (x f ) tan 0. Comparing this
with (3) we see that tan 6 is the desired limiting value of (Tef.
188. We have shewn, at some length, how a linear and vector
equation containing an unknown vector is to be solved in the most
general case ; and this, by 150, shews how to find an unknown
quaternion from any sufficiently general linear equation containing
it. That such an equation may be sufficiently general it must
have both scalar and vector parts : the first gives one, and the
second three, scalar equations ; and these are required to determine
completely the four scalar elements of the unknown quaternion.
Thus Tq = a
being but one scalar equation, gives
q a Ur,
where r is any quaternion whatever.
Similarly Sq = a
gives q = a + 0,
where is any vector whatever. In each of these cases, only one
scalar condition being given, the solution contains three scalar in
determinates. A similar remark applies to the following :
TVq = a
gives q = x + ad ;
and SUq = cos A,
gives q = x0 2A/7r ,
in each of which x is any scalar, and 6 any unit vector.
I 9 2.] SOLUTION OF EQUATIONS. 133
189. Again, the reader may easily prove that
V.aVq = /3,
where a is a given vector, gives, by putting Sq = x,
Vaq = @ + xa.
Hence, assuming Saq = y,
we have aq = y + XOL + ft,
or q = x + yen 1 + a" 1 /?.
Here, the given equation being equivalent to two scalar con
ditions, the solution contains two scalar indeterminates.
190. Next take the equation
Vq = /3.
Operating by S. a" 1 , we get
Sq = Sa 1 ^,
so that the given equation becomes
or
From this, by 170, we see that
Vq = QL l (x+<xVar*p),
whence q = Sa^ft + a" 1 (x + aFa 1
and, the given equation being equivalent to three scalar conditions,
but one undetermined scalar remains in the value of q.
This solution might have been obtained at once, since our
equation gives merely the vector of the quaternion aq, and leaves
its scalar undetermined.
Hence, taking x for the scalar, we have
aq = Saq + Vaq
191. Finally, of course, from
= &
which is equivalent to four scalar equations, we obtain a definite
value of the unknown quaternion in the form
2 =  .
192. Before taking leave of linear equations, we may mention
134 QUATERNIONS. [ I 93
that Hamilton has shewn how to solve any linear equation con
taining an unknown quaternion, by a process analogous to that
which he employed to determine an unknown vector from a linear
and vector equation ; and to which a large part of this Chapter has
been devoted. Besides the increased complexity, the peculiar fea
ture disclosed by this beautiful discovery is that the symbolic
equation for a linear quaternion function, corresponding to the cubic
in (/> of 174, is a biquadratic, so that the inverse function is given
in terms of the first, second, and third powers of the direct function.
In an elementary work like the present the discussion of such a
question would be out of place : although it is not very difficult to
derive the more general result by an application of processes
already explained. But it forms a curious example of the well
known fact that a biquadratic equation depends for its solution
upon a cubic. The reader is therefore referred to the Elements of
Quaternions, p. 491.
193. As an example of the solution of the linear equation in
quaternions, let us take the problem of finding the differential of
the n th root of a quaternion. This comes to finding dq in terms of
dr when
q = r.
[Here n may obviously be treated as an integer ; for, if it were
fractional, both sides could be raised to the power expressed by the
denominator of the fraction.]
This gives
q n ~ l dq + q n  2 dq.q+...+ dq.q n  1 = </>(dq) = dr (1),
and from this equation dq is to be found ; < being now a linear
and quaternion function.
Multiply by q, and then into q, and subtract. We obtain
q n dq dq. q n = qdr dr.q,
or 2V.Vq n Vdq=2V. VqVdr (2).
But, from the equation
q = Sq+ Vq,
we have at once Vr = Vq n = Q n Vq,
where = *
[The value of Q n is obvious from 116, but we keep the
present form.]
1 94.] SOLUTION OF EQUATIONS. 135
With this (2) becomes
Q n V.VqVdq= V.VqVdr,
whence Q n Vdq = Vdr + x Vq,
x being an undetermined scalar.
Adding another such scalar, so as to introduce Sdq and Sdr, we
have
Q n dq = (y + xVq) + dr ..................... (3).
Substitute in (1) and we have
Q n dr = nf 1 (y + xVq} + $ (dr),
or, by (3) again,
Q n dr = nq n  1 (Q n dq  dr) + j> (dr) ;
so that, finally,
.............. (4).
Thus dq is completely determined.
It is interesting to form, in this case, an equation for <. This
is easily done by eliminating dr from (4) by the help of (1). We
thus obtain
This might have been foreseen from the nature of $, as defined in
(1) ; because it is clear that its effect on a scalar, or on a vector
parallel to the axis of q (which is commutative with q), is the same
as multiplication by n(f~ l ; while for any vector in the plane of q
it is equivalent to the scalar factor Q n .
It is left to the student to solve the equation (1) by putting it
in the form
dr = p + qpq 1 + (fpq~* + . . . + q n ~ l pq~ n+l
or dr  qdrq~ l =p q n pq~ n ,
where p = dqq n ~ l .
The nature of the operator q ( ) q~ l was considered in 119
above.
194. The question just treated involves the solution of a
particular case only of the following equation :
<!>(q)=2aqa = b ........................ (1),
where a, a , &c. are coplanar quaternions.
Let = r + ,
136 QUATERNIONS.
where r is a quaternion coplanar with the as, and p a vector in
their plane. Then, for any a,
ra = ar,
while pa = Ka . p.
Thus the given equation takes the form
(/> (q) = 2 (aa ) . r + 2 (a#a) . p,
so that the functional equation becomes in its turn
{(/>  2 (aa )} (02 (aJTa )J = 0.
If a be the unitvector perpendicular to the plane of the as, we
have
b =
and the required solution is obviously
q = (2 aa ) 1 (Sb  aSab)  (2 a^Ta )" 1 a Fa Vb.
In the case ( 193) of the differential of the n ih root of a quaternion,
s, we have
2(aa )=?wr 1 ,
2 (aKaf) = 2(8. s n ~ l + T*sS. s n ~ 3 +...)
The last expression (in which, it must be noticed, the last term is
not to be doubled when n is odd) is the Q n of the former solution,
though the form in which it is expressed is different. It will
be a good exercise for the student to prove directly that they are
equal.
195. The solution of the following frequentlyoccurring par
ticular form of linear quaternion equation
aq+qb = c,
where a, b, and c are any given quaternions, has been effected by
Hamilton by an ingenious process, which was applied in 140 (5)
above to a simple case.
Multiply the whole by Ka, and (separately) into b, and we
have
T 2 a . q f Ka . qb = Ka . c,
and a . qb + qb* = cb.
Adding, we have
q (Ta + 6 2 f 2Sa . b) = Ka . c + cb,
from which q is at once found.
196.] SOLUTION OF EQUATIONS. 137
To this form any equation such as
a qb + c qd = e
can of course be reduced, by multiplication by c " 1 and into b ~ l .
196. To shew some of the characteristic peculiarities in the
solution of quaternion equations even of the first degree when they
are not sufficiently general, let us take the very simple one
aq qb,
and give every step of the solution, as practice in transformations.
Apply Hamilton s process ( 195), and we get
qtf = aqb.
These give q (Ta + 6 2  2bSa) = 0,
so that the equation gives no real finite value for q unless
or = a
where j3 is some unitvector. This gives Sa = Sb.
By a similar process we may evidently shew that
a. being another unitvector.
But, by the given equation,
Ta = Tb,
or S*a + T 2 Va = S*b + T 2 Vb;
from which, and the above values of a and b, we see that we may
write
Sa Sb
Thus we may write
a = a + a, 6 =
where a and /3 are unitvectors.
If, then, we separate q into its scalar and vector parts, thus
q = u + p,
the given equation becomes
(a + a)( + p) = (w + /C ))(a + /9) ............... (1).
Multiplying out we have
u (a0)=pj3 a/o,
which gives 8 (a  /3) p = 0,
and therefore p = Vy (a  {3),
where 7 is an undetermined vector.
138 QUATERNIONS.
We have now
u (a. ft) = p/3 ap
Having thus determined u, we have
 7 ( + 0) + 7 (  /
=  27  27/3.
Here, of course, we may change the sign of 7, and write the solu
tion of
aq = qb
in the form q = ay + 7/3,
where 7 is any vector, and
a=UVa, /3=UVb.
To verify this solution, we see by (1) that we require only to
shew that
aq = q@.
But their common value is evidently
 7 + ay/3.
An apparent increase of generality of this solution may be
obtained by writing
q = ar + r/3
where r is any quaternion. But this is easily seen to be equiva
lent to adding to 7 (which is any vector] a term of the form xVa/3.
It will be excellent practice for the student to represent the
terms of this equation by versorarcs, as in 54, and to deduce the
above solution from the diagram annexed :
b
The vector of the intersection of the plane of q, with that of aq
198.] SOLUTION OF EQUATIONS. 139
and qb, is evidently symmetrically situated with regard to the
great circles of a and b. Hence it is parallel to
(af /3) Fa/3, i.e. to a/3.
Let 7 be any vector in the plane of a.
Then #x 7 ()>
oc cty+ry/3,
because 8a.y = 0, and thus ya. = ay.
Another simple form of solution consists in writing the equa
tion as
a = qbq~\
and applying the results of 119.
197. No general quaternion method of solving equations of
the second or higher degrees has yet been found ; in fact, as will be
shewn immediately, even those of the second degree involve (in
their most general form) algebraic equations of the sixteenth degree.
Hence, in the few remaining sections of this Chapter we shall con
fine ourselves to one or two of the simpler forms for the treatment
of which a definite process has been devised. But first, let us
consider how many roots an equation of the second degree in an
unknown quaternion must generally have.
If we substitute for the quaternion the expression
w + ix + jy + kz (80),
and treat the quaternion constants in the same way, we shall have
on development ( 80) four equations, generally of the second
degree, to determine w, x, y, z. The number of roots will therefore
be 2 4 or 16. And similar reasoning shews us that a quaternion
equation of the mth degree has m 4 roots. It is easy to see, how
ever, from some of the simple examples given above ( 188 190,
&c.) that, unless the given equation is equivalent to four inde
pendent scalar equations, the roots will contain one or more
indeterminate quantities.
198. Hamilton has effected in a simple way the solution of
the quadratic
q 2 = qa + b,
or the following, which is virtually the same (as we see by taking
the conjugate of each side),
b.
140 QUATERNIONS. [l99
He puts q = J (a + w + p),
where w is a scalar, and p a vector.
Substituting this value in the first written form of the equation,
we get
a 2 + (w + pf + 2iua + ap + pa = 2 (a 2 + wa + /oa) + 46,
or (w + /o) 2 + ap  pa = a 2 + 46.
If we put Fa = a, $ (a 2 + 46) = c, F(a 2 + 46) = 2y, this becomes
which, by equating separately the scalar and vector parts, may be
broken up into the two equations
W 1 + /> = (!,
V(iu+ a.) p =7.
The latter of these can be solved for p by the process of 168 ; or
more simply by operating at once by $ . a, which gives the value
of 8 (w + a) p. If we substitute the resulting value of p in the
former we obtain, as the reader may easily prove, the equation
(w*  a 2 ) (w 4  cw* + 7 2 )  ( Fa 7 ) 2 = 0.
The solution of this scalar cubic gives six values of w, for each of
which we find a value of p, and thence a value of q.
Hamilton shews (Lectures, p. 633) that only two of these values
are real quaternions, the remaining four being biquaternions, and
the other ten roots of the given equation being infinite.
Hamilton farther remarks that the above process leads, as the
reader may easily see, to the solution of the two simultaneous
equations
q + r = a,
qr = b;
and he connects it also with the evaluation of certain continued
fractions with quaternion constituents. (See the Miscellaneous
Examples at the end of this volume.)
199. The equation <f = aq + qb,
though apparently of the second degree, is easily reduced to the
first degree by multiplying by, and into, q~ l , when it becomes
1 = q~ l a + bq~ l ,
and may be treated by the process of 195.
The equation
200.] SOLUTION OF EQUATIONS. 141
where a and /3 are given vectors, is easily seen to require for a
real (i.e. a non biquaternion) solution that q shall be a vector.
Hence we may write it as
whence, at once,
a + Vj3p = xp.
Assume S{3p = y,
and we have
 (y  a ) =  ) p,
or ( a} + l3)(ya)=( a ?F)p.
The condition that p is a vector gives
xy  Sa/3 = 0,
so that the value of p, containing one scalar indeterminate, is
To determine p completely we require one additional scalar
condition.
If we have, for instance,
Syp = e,
x is given by the cubic equation
 8. *.
oc
But if the condition be that p is a vectorradius of the unit
sphere (a result which will be required below) we have the
biquadratic
_
This gives two real values of # 2 , but they have opposite signs ;
so that there are always two, and only two, real values of x.
200. The equation q m = aqb,
where a and b are given quaternions, gives
and, by 54, it is evident that the planes of q and aqb must coin
cide. A little consideration (after the manner of the latter part
of 196) will shew that the solution depends upon drawing two
arcs which shall intercept given arcs upon each of two great
circles; while one of them bisects the other, and is divided by
it in the proportion of m : 1. The equation treated in 196 is
the special case of this when m = l.
142 QUATERNIONS.
EXAMPLES TO CHAPTER V.
1. Solve the following equations :
(6) apPp = pap/3.
(d) S . afip + /3Sap  a Vftp = 7.
(e) p + ap(3 = a/3.
Do any of these impose any restriction on the generality of a and /3?
2. Suppose p = ix +jy + kz,
and ~ ( t ) P = aiBip + bj&jp + ckSkp ;
put into Cartesian coordinates the following equations :
(a) T<t>p=l.
(c) S . p ((/>
(d) Tp = I
3. If X, /A, v be aw/ three noncoplanar vectors, and
q = Fwz^ . <i>X f FfX . (fa/A f K Xyu< . cpz ,
shew that g is necessarily divisible by S.\pv.
Also shew that the quotient is
m 2 2e,
where Fe/o is the noncommutative part of </>.
Hamilton, Elements, p. 442.
4. Solve the simultaneous equations :
= 0,1
=0,
=0.
=0,
S.aipicp=0
SOLUTION OF EQUATIONS. 143
5. If fo = 1,13 Sap + Vrp,
where r is a given quaternion, shew that
m = 2 (5 . W .S . 0J3J3J + 2S (r Va^ . V/B J3J + SrZS . */3r
 2 (SarSjSr) + SrTr\
and m0V = 2 (Fa^flf.fl&cr) + 27. *V(V/3(r.r) + V<rrSr VrSar.
Lectures, p. 561.
6. If [pq] denote pq qp,
(pqr) S.p[qr],
[pqr] (pqr) + [rq] Sp + [pr] Sq + [qp] Sr,
and (pqrs) S.p [qrs] ;
shew that the following relations exist among any five quaternions
0=p (qrst) + q (rstp) + r (stpq) + s (tpqr) + t (pqrs),
and # (prst) = [rst] Spq  [stp] Srq + [tpr] Ssq [prs] Stq.
Elements, p. 492.
7. Shew that if <f>, ty be any linear and vector functions, and
a, /3, 7 rectangular unitvectors, the vector
6=V (</>^a + 0/3^/3 + 7 f 7)
is an invariant. [This will be immediately seen if we write it in
the form 6 = V. </>V^p,
which is independent of the directions of a, /3, 7. But it is good
practice to dispense with V, when possible.]
If *p =
and typ =
shew that this invariant may be expressed as
SFi^K or SF^.
Shew also that ^^V ^^ p = Vdp.
The scalar of the same quaternion is also an invariant, and may be
written as
8. Shew that if $p = aSap + /380p +
where a, /3, 7 are any three vectors, then
where x = V/3y, &c.
144 QUATERNIONS.
9. Shew that any selfconjugate linear and vector function may
in general be expressed in terms of two given ones, the expression
involving terms of the second order.
Shew also that we may write
+ z = a (w + xY + b (a + x) (to + y) + c (to + y)\
where a, b, c, x, y, z are scalars, and ta and o> the two given func
tions. What character of generality is necessary in tzr and o> ? How
is the solution affected by nonselfconjugation in one or both ?
10. Solve the equations :
(a) q* = 5qi+IOj.
(6) g 2 = 2g + t. "
(c) qaq = bq + c.
(d) aq = qr = rb.
11. Shew that
12. If (/> be selfconjugate, and a, /3, 7 a rectangular system,
13. 0^ and ^< give the same values of the invariants m t
Wj, m 2 .
14. If ft be conjugate to </>, (j><f) is selfconjugate.
15. Shew that ( Fa<9) 2 + ( F/3<9) 2 + ( Fy0) 2 = 26> 2
if a, /3, 7 be rectangular unitvectors.
1 6. Prove that V 2 ($g)p =  pV 2 g + 2V g.
17. Solve the equations :
(a) $* = &;
(^ * + *::
where one, or two, unknown linear and vector functions are given
in terms of known ones. (Tait, Proc. R. 8. E. 187071.)
18. If </> be a selfconjugate linear and vector function, f and TJ
two vectors, the two following equations are consequences one of
the other, viz. :
SOLUTION OF EQUATIONS. 145
From either of them we obtain the equation
This, taken along with one of the others, gives a singular theorem
when translated into ordinary algebra. What property does it give
of the surface
= I? [Ibid.]
19. Solve the equation
qaq = /3q@.
Shew that it has a vector solution, involving the trisection of an
angle : and find the condition that it shall admit of a real
quaternion solution.
20. Solve
bqaq = qbqa,
and state the corresponding geometrical problem ; shewing that
when a and b are equal vectors, q is equal to each.
21. Given (/>, a selfconjugate linear and vector function, and
a vector e ; find the cubic in ty, where
typ = <f>p + Vep.
22. Investigate the simplest expressions for any linear and
vector function in terms of given ones: and point out what
degree of generality is necessary in the latter.
Why cannot the conjugate of a linear and vector function be
generally expressed in powers of the function itself ?
T. Q. I. 10
CHAPTEE VI.
SKETCH OF THE ANALYTICAL THEORY OF QUATERNIONS.
(By PROF. CAYLEY.)
(a) Expression, Addition, and Multiplication.
BY what precedes we are led to an analytical theory of the
Quaternion q = w + ix + jy + kz, where the imaginary symbols
i, jy k are such that
i 2 = 1, f = 1, k 2 = 1, jk kj = i, ki =  ik =j, ij ji = k.
The Tensor Tq is = Jw* + x* + f + z 2 ] and
the Versor Uq is = . = (w \ix+jy + kz),
x/w 2 + a? +
which, or the quaternion itself when Tq = 1, may be expressed in
the form
cos 8 f sin 8 (ia +jb + kc) where a 2 + 6 2 + c 2 = 1 ;
such a quaternion is a Unit Quaternion. The squared tensor
w 2 + 3? f y* + 2 2 is called the Norm.
The scalar part Sq is = w, and the vector part Fg, or say a
Vector, \$ = ix\jy + kz. The Length is = ^sc* + 2/ 2 + z 2 , and the
quotient ._. g ___ ^=^ (^ + J2/ + ^z), or say a vector ix + jy + kz
v x \ y \ z
where a? + if + z 2 = 1, is a Unit Vector.
The quaternions w + ix +jy + &z and w ix jy kz are said
to be Conjugates, each of the other. Conjugate quaternions have
the same norm ; and the product of the conjugate quaternions is
the norm of either of them. The conjugate of a quaternion is
denoted by q, or Kq.
ANALYTICAL THEORY. 147
Quaternions q = w + ix+jy + kz, q w 4 ix 4 jy 4 kz are
added by the formula
q + q = w + w + i(x + x) 4 j (y + y ) + k(z + z\
the operation being commutative and associative.
They are multiplied by the formula
qq = ww xx yy zz
+ i (iox 4 xw 4 yz zy f )
+j (wy + yw 4 zx z x)
+ k (wz f 4 zw 4 xy x y\
where observe that the norm is
= O 2 4 x* 4 2/ 2 4 s 8 ) (uP + x* 4 7/ /2 4 **)
the product of the norms of q and q.
The multiplication is not commutative, q q =}= qq ; but it is
associative, qq q" = q q q" = ##V > ^ Ct ^ n combination with
addition it is distributive, q (q f 4 q } qq 4 qq , &c.
(6) Imaginary Quaternions. Nullitats.
The components w, x, y, z of a quaternion are usually real,
but they may be imaginary of the form a + b J 1, where J 1 is
the imaginary of ordinary algebra: we cannot (as in ordinary
algebra) represent this by the letter i, but when occasion requires
another letter, say 0, may be adopted (the meaning, = J 1, being
explained). An imaginary quaternion is thus a quaternion of the
form (w 4 0w^ +i(x + Ooc^) +j(y + Oy^) 4 k (z 4 6z^, or, what is the
same thing, if q, q^ be the real quaternions w+ix+jy+kz t
w l + ix l +jy l + kz l , it is a quaternion q\Oq l \ this algebraical
imaginary 6 = J 1 is commutative with each of the symbols
i, j, k : or, what comes to the same thing, it is not in general
necessary to explicitly introduce 6 at all, but we work with the
quaternion w 4 ix +jy 4 kz, in exactly the same way as if w, x, y, z
were real values. A quaternion of the above form, q + Oq^, was
termed by Hamilton a " biquaternion " but it seems preferable to
speak of it simply as a quaternion, using the term biquaternion
only for a like expression q 4 Oq l , wherein 6 is not the J 1 of
ordinary algebra.
It may be noticed that, for an imaginary quaternion, the squared
tensor or norm w* 4 x* 4 y 2 + z 1 may be = ; when this is so, the
quaternion .is said to be a " Nullitat " ; the case is one to be
separately considered.
102
148
QUATERNIONS.
(c) Quaternion as a Matrix.
Quaternions have an intimate connection with Matrices.
Suppose that 6, = J 1, is the J 1 of ordinary algebra, and in
place of i, j, k consider the new imaginaries x, y, z, w which are
such that
x = i ( 1 6i\ or conversely 1 = x + w,
2 = I ( j  0fy> j = (y z \
so that a, b, c, d being scalars, ax + by +cz + dw denotes the
imaginary quaternion
We obtain for x, y, z, w the laws of combination
x y z w
JA
that is x* = x, xy = y, xz= 0, xw
&c.,
and consequently for the product of two linear forms in (x, y, z, w)
we have
(ax + by + cz + dw) (a x + b y t c z + d w)
(aa + be) x + (ab + bd) y 4 (ca + dc ) z + (cb + dd) w ;
and this is precisely the form for the product of two matrices, viz.
we have
(a, c ) (& , d)
a b
c d
a, b
c , d
= (a, b)
(c, d)
aa + be , ab + bd
I ca + dc , cb + dd
and hence the linear form ax + by + cz + dw, and the matrix
a, b
c, d
may be regarded as equivalent symbols. This identifica
tion was established by the remark and footnote " Peirce s Linear
Associative Algebra," Amer. Math. Jour. t. 4 (1881), p. 132.
(d) The Quaternion Equation
In ordinary algebra, an equation of the first degree, or linear
equation with one unknown quantity x, is merely an equation of
the form ax = b, and it gives at once x = a~ l b.
ANALYTICAL THEORY. 149
But the case is very different with quaternions; the general
form of a linear equation with one unknown quaternion q is
AtfBt + A z qB 2 K . .= C, or say ^AqB = C,
where G and the several coefficients A and B are given qua
ternions.
Considering the expression on the lefthand side, and assuming
(f=w + i+jy + kz, it is obvious that the expression is in effect of
the form
S W + OLX +/3 y + 7 z
+ i (S^v + QLjX + /^y + 7^)
4
where the coefficients 8, a, /3, 7 &c. are given scalar magnitudes :
if then this is equal to a given quaternion (7, say this is
X + i\ +j\ + k\,
we have for the determination of w, x, y, z the four equations
7 3 s  X a ,
and we thence have w, x, y, z, each of them as a fraction with a
given numerator, and with the common denominator
A=
S, a,
8 a 3 ft 7,
viz. this is the determinant formed with the coefficients 8, a, ft 7,
&c. Of course if A = 0, then either the equations are inconsistent,
or they reduce themselves to fewer than four independent
equations.
The number of these coefficients is = 16, and it is thus clear
that, whatever be the number of the terms A^qB^ A 2 qB> 2) &c. we
only in effect introduce into the equation 16 coefficients. A single
term such as A 1 qB J may be regarded as containing seven coefficients,
for we may without loss of generality write it in the form
g (1 + ia + jb + kc) q(l+ id +je + kf\
and thus we do not obtain the general form of linear equation
150 QUATERNIONS.
by taking a single term A l qB l (for this contains seven coefficients
only) nor by taking two terms A t qB v A 2 qB z (for these contain 14
coefficients only) ; but we do, it would seem, obtain the general
form by taking three terms (viz. these contain 21 coefficients,
which must in effect reduce themselves to 16) : that is, a form
A 1 qS l + A^qB t2 + A 3 qB 3 is, or seems to be, capable of representing
the above written quaternion form with any values whatever of the
16 coefficients S, a, (3, 7 &c. But the further theory of this reduc
tion to 16 coefficients is not here considered.
The most simple case of course is that of a single term, say we
have AqB = C : here multiplying on the left by A~ l and on the
right by B~ l , we obtain at once q = A~ l CB~ l .
(e) The Nivellator, and its Matrix.
In the general case, a solution, equivalent to the foregoing, but
differing from it very much in form may be obtained by means
of the following considerations.
A symbol of the above form 24 ( ) B, operating upon a
quaternion q so as to change it into 24 (q) B, is termed by
Prof. Sylvester a "Nivellator:" it may be represented by a single
letter, say we have < = 2.4 ( ) B ; the effect of it, as has just
been seen, is to convert the components (w, x t y, 2), into four linear
functions (w v x v y v z^ which may be expressed by the equation
(w v x v y lt ^)   8, a,
(w, x, y t
2 > 2 > & 7 2
3 a s> ft> 7 3
or say by the multiplication of (w, x, y, z) by a matrix which may
be called the matrix of the nivellator; and the theory of the
solution of the linear equation in quaternions thus enters into
relation with that of the solution of the linear equation in
matrices.
The operation denoted by c/> admits of repetition : we have for
instance
and similarly for more than two terms, and for higher powers.
Considering < in connexion with its matrix M, we have
M 2 (w, x, y, z) for the components of <t>*(q), M* (w, x, y, z) for
ANALYTICAL THEORY.
151
those of c/> 3 (q), and so on. Hence also we have the negative
powers <$>~ l , &c. of the operation (. The mode in which <"* can
be calculated will presently appear: but assuming for the moment
that it can be calculated, the given equation is (f> (q) C, that
is we have q = (f>~ 1 ((7), the solution of the equation.
A matrix M of any order satisfies identically an equation of
the same order : viz. for the foregoing matrix M of the fourth
order we have
BM, v, 0, y
viz. this is
*.,
*3>
/3 3 ,
7,
ryM
where h is the before mentioned determinant
S, a, ft, 7 i , say this is, h = A.
Jf, in its operation on the components (w, x, y, z) of q, exactly
represents (f> in its operation on q : we thus have
viz. this means that operating successively with on the arbitrary
quaternion Q we have identically
f (Q)  ff (Q) +f<F (Q) g<l>(Q) + hQ = 0;
where observe that the coefficients e, /, g, h have their foregoing
values, calculated by means of the minors of the determinant : but
that their values may also be calculated quite independently of
this determinant: viz. the equation shews that there is an identical
linear relation connecting the values 4 (Q), </> 8 (Q), < 2 (Q), < (Q)
and Q : and from the values (assumed to be known) of these
quantities, we can calculate the identical equation which connects
them. But in whatever way they are found, the coefficients
e, f, g, h are to be regarded as known scalar functions.
Writing in the equation c/T 1 Q in place of Q, we have
<t> s (Q)  <><? (Q) +/* (Q) 9Q + A*" (Q) = o,
viz. this equation gives (/T 1 (Q) as a linear function of Q, <p (Q), 2 (Q)
and < 3 (Q) : and hence for the arbitrary quaternion Q writing the
value 0, we have q, = ^T 1 (6 ) given as a linear function of
152
(7, <(0), 2 (C) and
given linear equation.
QUATERNIONS.
(C) : we have thus the solution of the
(/) The Vector Equation
The theory is similar if, instead of quaternions, we have vectors.
As to this observe in the first place that, even if A, q, B are each
of them a vector, the product AqB will be in general, not a vector,
but a quaternion. Hence in the equation %AqB = 0, if C and the
several coefficients A and B be all of them vectors, the quantity q
as determined by this equation will be in general a quaternion :
and even if it should come out to be a vector, still in the process
of solution it will be necessary to take account, not only of the
vector components, but also of the scalar part ; so that there is
here no simplification of the foregoing general theory.
But the several coefficients A, B may be vectors so related to
each other that the sum ^ApB, where p is an arbitrary vector,
is always a vector 1 ; and in this case, if be also a vector, the
equation ^ApB = G will determine p as a vector : and there is
here a material simplification. Writing p = ix +jy + kz, then
is in effect of the form
viz. we have these three linear functions of (x t y, z) to be equalled
to given scalar values \ t , X 2 , \ 3 , and here #, y, z have to be
determined by the solution of the three linear equations thus
obtained. And for the second form of solution, writing as before
(f> = 2A ( )B, then </> is connected with the more simple matrix
^=1 !, &, 7,
2 8 > 7 2
s> & 7 3
and it thus (instead of a biquadratic equation) satisfies the cubic
equation
a,M,
7,
 M,
P,,
M
= 0,
say
1 Thus, if A, B are conjugate quaternions, ApB is a vector a: this is in fact
the form which presents itself in the theory of rotation.
ANALYTICAL THEORY. 153
We have therefore for < the cubic equation
.**+/< 0 = 0,
and thus cf)~ l (Q) is given as a linear function of Q, c#>(Q), <J> 2 (Q), or,
what is the same thing, <f>~ l (C) as a linear function of C, $(C), <t>*(C)
and (this being so) then for the solution of the given equation
(f) (p) = (7, we have p <f>~ 1 (C), a given linear function of C, <f> (C),
<?(C).
(g) Nullitats.
Simplifications and specialities present themselves in particular
cases, for instance in the cases Aq + qB=C, and Aq = qB, which
are afterwards considered.
The product of a quaternion into its conjugate is equal to
the squared tensor, or norm; aa = T*(a) , and thus the reciprocal
of a quaternion is equal to the conjugate divided by the norm ;
hence if the norm be = 0, or say if the quaternion be a nullitat,
there is no reciprocal. In particular, 0, qua quaternion, is a
nullitat.
The equation aqb = c, where a, b, c are given quaternions,
q the quaternion sought for, is at once solvable ; we have
q a~ l cb~ 1 , but the solution fails if a, or 6, or each of them, is a
nullitat. And when this is so, then whatever be the value of q,
we have aqb a nullitat, and thus the equation has no solution
unless also c be a nullitat.
If a and c are nullitats, but b is not a nullitat, then the equa
tion gives aq = cb~ l , which is of the form aq = c , and similarly if
b and c are nullitats but a is not a nullitat, then the equation
gives qb = a~ 1 c, which is of the form qb = c: thus the forms to
be considered are aq = c, qb = c, and aqb = c, where in the first
equation a and c, in the second equation b and c, and in the
third equation a, b, c, are nullitats.
The equation aq = c, a and c nullitats, does not in general
admit of solution, but when it does so, the solution is indeter
minate; viz. if Q be a solution, then Q + aR (where R is an
arbitrary quaternion) is also a solution. Similarly for the equation
qb = c, if Q be a solution, then Q + Sb (S an arbitrary quaternion)
is also a solution : and in like manner for the equation aqb = c, if
Q be a solution then also Q iaR + Sb (R, S arbitrary quaternions)
is a solution.
154 QUATERNIONS.
(h) Conditions of Consistency, ivheti some Coefficients
are Nullitats.
Consider first the equation aq = c; writing a
(a* 4 a* + a* + a 3 = 0) and C = c 4 +ic 1 +jc 2 + ^c 3 (c 4 2 + c 1 2 + c 2 2 f c 3 2 =0);
also q = w 4 la; 4 jy 4 &, the equation gives
c 4  4 w  a^x  a z y  a 3 z,
c l = cijW 4 a 4 x a 3 y 4 a/,
c 2 = a 2 w 4 a 8 # 4 a$ a^z,
c s = a s w  a 2 # 4 a^y 4 a^z,
equations which are only consistent with each other when two
of the c s are determinate linear functions of the other two c s ;
and when this is so, the equations reduce themselves to two
independent equations. Thus from the first, second and third
equations, multiplying by a l a 3 2 a 4 , a 4 a 3 a^, and a 2 2 a 3 2 ,
and adding, we obtain
(a^  8 a 4 ) c 4  (a 3 a 4 4 a^) c t  (a 2 2 4 a*) C 2 = ;
similarly from the first, second and fourth equations, multiplying
by a 3 a 4 aji^ a^a 3 + a 2 a 4 , a 2 2 a 3 2 , and adding, we have
 (a^ + a 3 a 4 ) c 4  (a,a 8  a 2 a 4 ) c,  (a 2 2 + a 3 2 ) c 3  0,
and when these two equations are satisfied, the original equations
are equivalent to two independent equations ; so that we ha,ve for
instance a solution Q = w + ix where c 4 = a^w a^jc y c t = a^w 4 a 4 #,
, . a.c. 4 tt.c, a,c. 4 a.c, , , , .
that is w = n o , x= V  2 4  1 ; and the general solution is
af + a* a 4 * + a*
then obtained as above.
The equations connecting the as and the c s may be presented
in a variety of different forms, all of them of course equivalent in
virtue of the relations <x 4 2 4 a* + a* f a 3 2 = 0, c 4 2 4 c* + c 2 2 + c s 2 = ;
viz. writing
A l = a* f a 4 2 =  a 2 2 a s 2 , F^ a 2 a 3 + a t a 4 , F^ = a z a 3 a^,
A 2 = a^ + a* =  a 3  a* , F 2 = a 3 a 1 4 a 2 a 4 , F^ = a g a 1  a 2 a 4 ,
A 3 = a* + a* =  a*  a*, F 3 = a^ + a s a 4 , F 3 = a x a 2  a 3 a 4 ,
then the relation between any three of the c s may be expressed
in three different forms, with coefficients out of the sets A l ,A 2 ,A 3 ;
F v F z , F 3 ; F{, F^ y F s . Obviously the relation between the c s is
satisfied if c = : the equation then is aq = 0, satisfied by q = aR,
R an arbitrary quaternion.
ANALYTICAL THEORY. 155
We have a precisely similar theory for the equation qb=c;
any two of the c s must be determinate linear functions of the
other two of them; and we have then only two independent
equations for the determination of the w, an, y, z.
In the case of the equation aqb = c (a, b, c all nullitats) the
analysis is somewhat more complicated, but the final result is a
simple and remarkable one ; from the condition that a, b are
nullitats, it follows that ab, aib, ajb, akb are scalar (in general
imaginary scalar) multiples of one and the same nullitat, say of
ab: the condition to be satisfied by c then is that c shall be a
scalar multiple of this same nullitat, say c = \ab; the equation
aqb = \ab has then a solution q = X, and the general solution is
q = X j a R + Sb, where R, S are arbitrary quaternions.
(i) The Linear Equations, aq qb = 0, and aqqb = c.
The foregoing considerations explain a point which presents
itself in regard to the equation aqqb = 0, (a, b given quaternions,
q a quaternion sought for): clearly the equation is not solvable
(otherwise than by the value q = 0) unless a condition be satisfied
by the given quaternions a, b; but this condition is not (what
at first sight it would appear to be) T 2 a = T 2 b. The condition
(say ft) may be satisfied although T*a=t= T 2 b } and being satisfied,
there exists a determinate quaternion q, which must evidently
be a nullitat (for from the given equation aq = qb we have
(To,  T*b) T*q = 0, that is T 2 q = 0). If in addition to the con
dition ft we have also T z a T*b = 0, then (as will appear) we
have an indeterminate solution q, which is not in general a
nullitat.
Take the more general equation aq qb = c: this may be
solved by a process (due to Hamilton) as follows: multiplying
on the left hand by a and on the right hand by b, we have
daq aqb dc, aqb qb* = cb, whence subtracting
aaq (a + a) qb + qb 2 = ac cb,
or since ad, a + d are scalars q {ad (a 4 d) b + 6 2 } = ac cb: viz.
this is an equation of the form qB = C (B, C given quaternions),
having a solution q = CB ~ 1 .
Suppose c = 0, then also (7 = 0; and unless B is a nullitat, the
equation qB = (representing the original equation aq = qb), has
only the solution g = 0; viz. the condition in order that the
156 QUATERNIONS.
equation aq = qb may have a solution other than q = 0, is B = nul
litat, that is aa (a + a) b + 6 2 = nullitat ; viz. we must have
+ b? + 26 4 (ib, +j\ + kb 3 )  6, a  &,  6 3 8 = nullitat,
that is
^ + < + a, 8 4 a 8 8  2</. 4 & 4 + 6 4 8  &/  6 8 8  6 3 2
+ 2 (6 4  a 4 ) (t 6, + j/> 2 4 &6 3 ) = nullitat.
The condition ft thus is
that is
j(a 4  6 4 ) 8 + < + a./ 4 a 3 2  6 1 3 6 8 8 6 8 7 + 4(a 4 6 4 ) 8 (6 1 s +6;+6 8 s )=0,
or, as this may also be written,
( 4  6 4 ) 4 4 2 ( 4  6 4 ) 2 ( ai 8 + < + 3 2 + b: 4 6, 8 + 6 3 2 )
+ + < + <  6 t 2  6.;  6 8 8 )* = 0.
Writing herein
< + a, 9 + a? + <  ^L 8 , 6 4 2 + 6 a 2 + 6 2 2 + 6 3 2 = ^ 2 ,
the condition is
(a 4  6 4 ) 4 + 2 ( 4  6 4 ) 8 (4* + 2  a 4 8  & 4 ") + (^L 2  B 2  a* + 6 4 2 ) 2 = 0,
which is easily reduced to
4 (a,  6 4 ) (a 4 ^ 2  6 4 ^1 2 ) + (A*  BJ = 0,
and, as already noticed, this is different from T 2 a T 2 b = 0, that is
A*& = 0.
If the equation A* B* = Q is satisfied, then the condition 1
reduces itself to a 4 6 4 = 0; we then have a = a 4 a, 6 = a 4 + /5,
where a, y5 are vectors, and the equation is therefore aq = q/3
where (since A 2 B\ = a x 2 + a 2 2 + ct 3 2  b*  b*  6 3 2 , = 0), the tensors
are equal, or we may without loss of generality take a, /3 to be
given unit vectors, viz. we have a 2 = ] , /3 2 = 1 : and this being
so, we obtain at once the solution q = X (a + /3) + //,(!  a/3) (X, IJL,
arbitrary scalars): in fact this value gives
aq = X ( 1 + a/3) + p (a + 0) = q@.
Reverting to the general equation oq qb = c, the conjugate of
ad(a\d)b + b 2 is ad (a + a)b + b \ and we thus obtain the
solution
2 {* K 6 4 ) K^ 2  M a ) + (^ 2 ^ 2 ) 2 } = (c  cb) [ad ( a + a)b + 6 2 },
ANALYTICAL THEORY. 157
but this solution fails if ad (a + a)b + 6 2 is a imllitat : supposing
it to be so, the equation is only solvable when C satisfies the
condition which expresses that the equation qB = C is solvable
when B, C are nullitats.
The equation aq qb = c, could it is clear be in like manner
reduced to the form Aq= C.
(j) The Quadric Equation <f 2aq+b = 0.
We consider the quadric equation (f 2aq + 6 = 0; a and b
given quaternions, q the quaternion sought for. The solution
which follows is that given by Prof. Sylvester for a quadric
equation in binary matrices.
In general if q be any quaternion, = w 4 ix + jy + kz, then
(qw) 2 4 x?+ y*+z* = 0, that is q*  2qw + n? 4 a? + y 2 + z~ = 0, or say
(f 2q (seal, q) + norm q = : viz. this is an identical relation
connecting a quaternion with its scalar and its norm.
Writing as above q = w + i&+jy + kz t and t = w* + a? f?/ 2 + 2
for the norm, we thus have
cf  2wq + t = 0,
and combining this with the given equation
q*2aq + b = 0,
we find 2 (a  w) q(bt) = 0, that is 2q = (a  w) 1 (b  t),
an expression for q in terms of the scalar and norm w, t, and of the
known quaternions a and b.
2q as thus determined satisfies the identical equation
(2#) 2  2 (2q) seal, {(a  lu) 1 (&*)} + norm {(a  w)~ l (b  1)} = 0,
and we have
\i/7 *\) seal, {(a w) (& $)!
seal, {(a w) l (bt)\ = ,
norm (a  w)
\i/i ,M norm (6^)
norm\(aw) l (bt)}= ^
norm (a w)
(a the conjugate of a).
The equation thus becomes
4g 2 norm (a w) ^q (seal, (d w)(b t)} f norm (b t) = :
this must agree with
(f 2qw + t =0,
158 QUATERNIONS.
or say the function is = 4X (q z 2qw + t) ; we thus have
norm (a w) X,
seal, (a w}(b t} 2Xw,
norm (b t)  4X,
three equations for the determination of X, w, t ; and then, w, t
being determined, the required value of q is *2q = (a w)~ l (b t)
as above.
To develope the solution let the values of a, b, c, f, g, h be
denned as follows : viz.
norm (ax + by 4 z) = (a, .b, c, f, g, h$#, y, z) z ,
viz. writing a = a 4 + ia l +ja 2 4 ka a ,
b = b 4 + i\ 4 j6 2 4 kb s ,
then this equation is
(a 4 x + by 4 zf + (a v x 4 6 t y)* + (<yc + 6 2 2/) 2 + (a 3 + 6 3 2/) 2
= (a, b, c, f, g, hja?, y, 0) 2 ,
that is, a, b, c, f, g, h denote as follows
a = a? + a* + a* + a 3 2 , f = 6 4 ,
b = 6 4 2 + ^ 2 +6 2 2 + 6 3 2 , g=a 4 ,
c = 1, h = a 4 6 4 + a A 4 a A + a A
We then have
norm (a w) = (a 4  w;) 2 + a^ 4 2 8 4 a s 2 ,
seal, (a w)(bt) = (a 4  w) (6 4  4 a l b l 4 a 2 6 8 4 a s 6 8 ,
norm (b  t) = (6 4  *) a + 6 : 2 4 6, 2 4 & 3 2 ,
or expressing these in terms of (a, b, c, f, g, h) the foregoing three
equations become
a  2gw 4 cw* = X,
h g fw + ctw = Z\w,
b2ft 4c^ 2 =4M,
where c (introduced only for greater symmetry) is = 1.
Writing moreover A, B, C, F, G, #=bcf 2 , cag 2 , abh 2 ,
gh  af, hf  bg, fg  ch, and K = abc  af 2  bg 2  ch 2 + 2fgh ; also
in place of w, t introducing into the equations n =w g, and
v = t f, the equations become
u* + B=\,
uv H = 2\(u + g),
tf+A= 4X (v 4 f).
ANALYTICAL THEORY. 159
We deduce u* = X  B,
and we thence obtain, to determine X, the cubic equation
(X  B) (4X 2 + 4Xf A)  (2Xg + H) 2 = 0,
viz. this is
4X 3 4 4X 2 (f  a) + X { be + f 2 + 4 (gh  af)}
4 c (abc  af 2  bg 2  ch 2 + 2fgh) = 0,
that is, 4X 3 + 4X 2 (f  a) + X ( A + 4<F) + K = 0,
and, X being determined by this equation, then
and then w = u + g, t = v+f; consequently
2q = (agu) 1 (bfv).
Write for a moment a g u = , then
( + 2u) = (agyu 2 = a z 2ag + &Bu\ =X
(since a = g + ia t +ja 9 + ka v a = g z + a 1 2 + a* + a* and thus the
identical equation for a is a J 2ag + a = 0): that is O 2 +2^@+X = 0,
or X 1 =  ( f 2u) = (ag + u) ; that is 1 , = (a  g  u)~\
= (a g+u) , and the value of q is 2q =  (ag+u)(bfv\
X A
or say it is
where X is determined by the cubic equation, and u is = + J\ B;
we have thus six roots of the given quadric equation q 2 2aq + 6 = 0.
CHAPTER VII.
GEOMETRY OF THE STRAIGHT LINE AND PLANE.
201. HAVING, in the preceding Chapters, given a brief ex
position of the theory and properties of quaternions, we intend to
devote the rest of the work to examples of their practical appli
cation, commencing, of course, with the simplest curve and surface,
the straight line and the plane. In this and the remaining Chapters
of the work a few of the earlier examples will be wrought out in
their fullest detail, with a reference to the previous part of the
book whenever a transformation occurs; but, as each Chapter
proceeds, superfluous steps will be gradually omitted, until in
the later examples the full value of the quaternion processes is
exhibited.
202. Before proceeding to the proper business of the Chapter
we make a digression in order to give a few instances of applica
tions to ordinary plane geometry. These the student may multiply
indefinitely with great ease.
(a) Euclid, I. 5. Let a and /3 be the vector sides of an iso
sceles triangle ; /3 a is the base, and
TOL = TIB.
The proposition will evidently be proved if we shew that
a(a0r l = K0(i3ar (52).
This gives a (a  /3) 1 = (  a) 1 ^,
or (a)a = (a/9),
or  = _*.
(b) Euclid, I. 32. Let ABC be the triangle, and let
2O2.] GEOMETRY OF STRAIGHT LINE AND PLANE. 161
where 7 is a unitvector perpendicular to the plane of the triangle.
If 1= 1, the angle CAB is a right angle ( 74). Hence
A=l7r/2 (74). Let = m7r/2, C = n7r/2. We have
Hence UBA = y m . 7 n . 7 UAB,
Or 1 = cy*++.
That is I f m + n = 2,
or A + B + (7 = TT.
This is, properly speaking, Legendre s proof; and might have been
given in a far shorter form than that above. In fact we have for
any three vectors whatever,
U ?i
/3 7 a
which contains Euclid s proposition as a mere particular case.
(c) Euclid, I. 35. Let /3 be the common vectorbase of the
parallelograms, a. the conterminous vectorside of any one of them.
For any other the vectorside is a + xj3 ( 28), and the proposition
appears as
TV$ (a + x&) = TV/301 ( 96, 98),
which is obviously true.
(d) In the base of a triangle find the point from which lines,
drawn parallel to the sides and limited by them, are equal.
If a, j3 be the sides, any point in the base has the vector
p = (1 x) a
For the required point
(\x)Ta =
which determines x.
Hence the point lies on the line
which bisects the vertical angle of the triangle.
This is not the only solution, for we should have written
instead of the less general form above which tacitly assumes that
1 x and x have the same sign. We leave this to the student.
T.Q.I. 11
162 QUATERNIONS. [ 2 O3
(e) If perpendiculars be erected outwards at the middle
points of the sides of a triangle, each being proportional to the
corresponding side, the mean point of the triangle formed by their
extremities coincides with that of the original triangle. Find the
ratio of each perpendicular to half the corresponding side of the
old triangle that the new triangle may be equilateral.
Let 2a, 2/3, and 2 (a f 0) be the vectorsides of the triangle,
i a unitvector perpendicular to its plane, e the ratio in question.
The vectors of the corners of the new triangle are (taking the
corner opposite to 2/3 as origin)
pt = a + eia,
From these
i (Pi + P* + P.)
which proves the first part of the proposition.
For the second part, we must have
Substituting, expanding, and erasing terms common to all, the
student will easily find
3^ = 1.
Hence, if equilateral triangles be described on the sides of any
triangle, their mean points form an equilateral triangle.
203. Such applications of quaternions as those just made are
of course legitimate, but they are not always profitable. In fact,
when applied to plane problems, quaternions often degenerate into
mere scalars, and become ( 33) Cartesian coordinates of some
kind, so that nothing is gained (though nothing is lost) by their
use. Before leaving this class of questions we take, as an
additional example, the investigation of some properties of the
ellipse.
204. We have already seen (31 (&)) that the equation
p = a cos 6 + /3 sin
represents an ellipse, being a scalar which may have any value.
Hence, for the vector tangent at the extremity of p we have
OT = ~ = a sin 6 f ft cos 6,
2O7] GEOMETRY OF STRAIGHT LINE AND PLANE. 163
which is easily seen to be the value of p when is increased by
7T/2. Thus it appears that any two values of p, for which
differs by vr/2, are conjugate diameters. The area of the
parallelogram circumscribed to the ellipse and touching it at
the extremities of these diameters is, therefore, by 96,
4TVp = 4TV (a cos + ft sin 0) ( a sin d + /3 cos 0)
a constant, as is well known.
205. For equal conjugate diameters we must have
T (a cos 6 + ft sin 0) = T ( a sin 6 + ft cos 0),
or (a 2  ft 2 ) (cos 2  sin 2 0) H 4a/3 cos sin = 0,
The square of the common length of these diameters is of course
because we see at once from 204 that the sum of the squares of
conjugate diameters is constant.
206. The maximum or minimum of p is thus found ;
d0~ TpdB
=  ~ { (a 2  ft 2 ) cos sin + Saft (cos 2  sin 2 0)}.
For a maximum or minimum this must vanish*, hence
and therefore the longest and shortest diameters are equally
inclined to each of the equal conjugate diameters ( 205). Hence,
also, they are at right angles to each other.
207. Suppose for a moment a and ft to be the greatest and
least semidiameters, so that
Saft=0.
* The student must carefully notice that here we put * / = 0, and not ^ = 0.
da do
A little reflection will shew him that the latter equation involves an absurdity.
112
164 QUATERNIONS. [208.
Then the equations of any two tangentlines are
p = a cos + ft sin \ x ( a sin 6 + ft cos 6),
p = a cos X + (3 sin l + X I (GL sin l + /3 cos X ).
If these tangentlines be at right angles to each other
8 ( a sin + cos 0) ( a sin t + cos 0J =
or a 2 sin 6 sin X + /3 2 cos 6 cos t = 0.
Also, for their point of intersection we have, by comparing
coefficients of a, ft in the above values of p,
cos 6 x sin 6 = cos a x l sin O v
sin 6 + a? cos = sin a + ^ cos r
Determining a^ from these equations, we easily find
2V = ( + /S"X
the equation of a circle ; if we take account of the above relation
between and r
Also, as the equations above give x x^ the tangents are
equal multiples of the diameters parallel to them ; so that the line
joining the points of contact is parallel to that joining the
extremities of these diameters.
208. Finally, when the tangents
p = acos9 +/3sin0 +x ( asin0 +$cos0),
p = a cos 0j + j3 sin l + x l (a sin O l + ft cos 0^,
meet in a given point
p = aa + bft,
we have a = cos x sin = cos X ^ sin lf
6 = sin + x cos = sin X + ^ cos r
Hence 2 = a 2 + 6 2  1 = a;*
and a cos + b sin = 1 = a cos t + b sin :
determine the values of and x for the directions and lengths of
the two tangents. The equation of the chord of contact is
p = y (a cos + ft sin 0) + (1  y) (a cos 6 l + ft sin 0J.
If this pass through the point
we have p = y cos f (1 y) cos 1?
# = y sin + (1 y) sin t ,
211.] GEOMETRY OF STRAIGHT LINE AND PLANE. 165
from which, by the equations which determine 6 and V we get
ap + bq = y + ly = l.
Thus if either a and b, or p and q, be given, a linear relation
connects the others. This, by 30, gives all the ordinary properties
of poles and polars.
209. Although, in 28 30, we have already given some of
the equations of the line and plane, these were adduced merely for
their applications to anharmonic coordinates and transversals ;
and not for investigations of a higher order. Now that we are
prepared to determine the lengths and inclinations of lines we
may investigate these and other similar forms anew.
210. The equation of the indefinite line drawn through the
origin 0, of which the vector OA, = a, forms a part, is evidently
p = xa,
or p  a,
or Yap = 0,
or Up = VOL ;
the essential characteristic of these equations being that they are
linear, and involve one indeterminate scalar in the value of p.
We may put this perhaps more clearly if we take any two
vectors, /3, 7, which, along with a, form a noncoplanar system.
Operating with S . Fa/5 and S . Vay upon any of the preceding
equations (except the third, and on it by S . j3 and 8. 7) we get
and
Separately, these are the equations of the planes containing a, ft,
and a, 7 ; together, of course, they denote the line of intersection.
211. Conversely, to solve equations (1), or to find p in terms
of known quantities, we see that they may be written
so that p is perpendicular to Va{3 and Vay, and is therefore
parallel to the vector of their product. That is,
p\\V.VaLJ3VcLy,
Hoflf.fl^y,
or p = XOL.
166 QUATERNIONS. [212.
212. By putting p /3 for p we change the origin to a point
B where OB = ft, or BO = /3 ; so that the equation of a line
parallel to a, and passing through the extremity of a vector /3
drawn from the origin, is
p  /3 = xa.,
or p = /3 + xa.
Of course any two parallel lines may be represented as
p = ft + xu,
or Fa<>/3) = 0,
Va(p^)=0.
213. The equation of a line, drawn through the extremity of ft,
and meeting a perpendicularly, is thus found. Suppose it to be
parallel to 7, its equation is
p = /3 + any.
To determine 7 we know, first, that it is perpendicular to a,
which gives
Say = 0.
Secondly, a, /3, and 7 are in one plane, which gives
S.a/3y = 0.
These two equations give
7lF.aF/3,
whence we have p = /3 + xaVaft.
This might have been obtained in many other ways ; for
instance, we see at once that
/3 = a 1 a/3 = a 1 Sa{3 + a 1 Fa/3.
This shews that a 1 Fa/3 (which is evidently perpendicular to a) is
coplanar with a and /3, and is therefore the direction of the
required line ; so that its equation is
i/
the same as before if we put  ~ z for x.
214. By means of the last investigation we see that
a 1 Fa/3
is the vector perpendicular drawn from the extremity of $ to the
line
p = oca..
2l6.] GEOMETRY OF STRAIGHT LINE AND PLANE. 167
Changing the origin, we see that
of 1 Fa (7)
is the vector perpendicular from the extremity of /3 upon the line
p = 7 + a?o.
215. The vector joining B (where OB /3) with any point in
p = 7 + XVL
is 7 + XQL J3.
Its length is least when
or So. (7 + XOL  /8) = 0,
i.e. when it is perpendicular to a.
The last equation gives
or XOL =  a" 1 $a (7  0).
Hence the vector perpendicular is
or cT l Va (7  /3) =  a 1 Fa (  7),
which agrees with the result of last section.
216. To find the shortest vector distance between two lines in
space
and Pi = A + !!;
we must put dT (p  p : ) = 0,
or S(p P 3(dpd Pl ) = O t
or S o adx  adx = 0.
Since x and x 1 are independent, this breaks up into the two
conditions
proving the wellknown truth that the required line is perpendicular
to each of the given lines.
Hence it is parallel to VOLOL V and therefore we have
p/? 1 = ^ + a?a^ 1 ~a? l o 1 = y.Faa 1 ............ (1).
Operate by 8 . ao^ and we get
168 QUATERNIONS. [217.
This determines y, and the shortest distance required is
[Note. In the two last expressions T before S is inserted simply
to ensure that the length be taken positively. If
8 . attj (/3 ySJ be negative,
then ( 89) 8 . a,a (0  ft) is positive.
If we omit the T, we must use in the text that one of these two
expressions which is positive.]
To find the extremities of this shortest distance, we must operate
on (1) with S.OL and S.a.^ We thus obtain two equations, which
determine x and oc v as y is already known.
A somewhat different mode of treating this problem will be
discussed presently.
217. In a given tetrahedron to find a set of rectangular coordi
nate axes, such that each axis shall pass through a pair of opposite
edges.
Let a, /S, 7 be three (vector) edges of the tetrahedron, one
corner being the origin. Let p be the vector of the origin of the
sought rectangular system, which may be called i, j, k (unknown
vectors). The condition that i, drawn from p, intersects a is
S.ioip = Q ............................... (1).
That it intersects the opposite edge, whose equation is
ts=r + a?087),
the condition is
S.;(/37)(/>/3) = 0, or & {(  7) p  7} = ...(2).
There are two other equations like (1), and two like (2), which can
be at once written down.
Put 7 = a 1 , 7a = ft, a^ = 7 1 ,
a, Fya = ft, Fa/3 = 7,,
and the six become
The two in i give i \\ aSa. 2 p  p (Saa 2 + Sa 3 p).
2IQ.] GEOMETRY OF STRAIGHT LINE AND PLANE. 169
Similarly,
 p (S/3/3, + S{3 3 p), and k \\ ySy 2 p  p (77, + S 7s p).
The conditions of rectangularity, viz.,
at once give three equations of the fourth order, the first of which
s
= SapSajSfo  SapSaj (S/3/3 2
The required origin of the rectangular system is thus given as
the intersection of three surfaces of the fourth order.
218. The equation Sap =
imposes on p the sole condition of being perpendicular to a ; and
therefore, being satisfied by the vector drawn from the origin to
any point in a plane through the origin and perpendicular to a, is
the equation of that plane.
To find this equation by a direct process similar to that usually
employed in coordinate geometry, we may remark that, by 29,
we may write
p = xp + 7/7,
where and 7 are any two vectors perpendicular to a. In this
form the equation contains two indeterminates, and is often useful;
but it is more usual to eliminate them, which may be done at
once by operating by S . a, when we obtain the equation first
written.
It may also be written, by eliminating one of the indeter
minates only, as
where the form of the equation shews that Sa/3 = 0.
Similarly we see that
represents a plane drawn through the extremity of and perpen
dicular to a. This, of course, may, like the last, be put into various
equivalent forms.
219. The line of intersection of the two planes
8. a (p0) =0
and S . a, (p  /3.) = (
170 QUATERNIONS. [22O.
contains all points whose value of p satisfies both conditions. But
we may write ( 92), since a, a 1? and Vaa t are not coplanar,
pS. aa l Vaa l = Vaafi.aaj + V. a^Vaafiap + F. F(aa 1 ) a/Sfc^p,
or, by the given equations,
 P T 2 Vaa^ = F. ^Vaafiap + V. V(aaJ aSa 1 {3 1 + xVaa^ . .(2),
where x, a scalar indeterminate, is put for S . aa^p which may have
any value. In practice, however, the two definite given scalar
equations are generally more useful than the partially indeter
minate vectorform which we have derived from them.
When both planes pass through the origin we have ft = ft l = 0,
and obtain at once
as the equation of the line of intersection.
220. The plane passing through the origin, and through the
line of intersection of the two planes (1), is easily seen to have the
equation
or 8 (aSa^  afiajS) p = Q.
For this is evidently the equation of a plane passing through the
origin. And, if p be such that
Sap = Sa/3,
we also have Sa^ = Safi^
which are equations (1).
Hence we see that the vector
aSa l j3 l  afia(3
is perpendicular to the vectorline of intersection (2) of the two
planes (1), and to every vector joining the origin with a point in
that line.
The student may verify these statements as an exercise.
221. To find the vectorperpendicular from the extremity of @
on the plane
Sap = 0,
we must note that it is necessarily parallel to a, and .hence that the
value of p for its foot is
where xa is the vectorperpendicular in question.
Hence Sa (ft + xa) = 0,
223.] GEOMETRY OF STRAIGBT LINE AND PLANE. 171
which gives xa? = Sa{3,
or oca = of 1 Sa{3.
Similarly the vectorperpendicular from the extremity of /3 on the
plane
may easily be shewn to be
222. The equation of the plane which passes through the ex
tremities of a, /3, y may be thus found. If p be the vector of any
point in it, p a, a /3, and ft y lie in the plane, and therefore
( 101)
or Sp(Va/3+ V/3y + Vya) S. a/3y = 0.
Hence, if B = as( Va/3 + Vj3y + Vya)
be the vectorperpendicular from the origin on the plane containing
the extremities of a, ft, 7, we have
S = ( Fa/3 + Vj3y + Vya.) 1 S . a{3y.
From this formula, whose interpretation is easy, many curious pro
perties of a tetrahedron may be deduced by the reader. Thus, for
instance, if we take the tensor of each side, and remember the
result of 100, we see that
T(Va{3+V/3y+Vya)
is twice the area of the base of the tetrahedron. This may be
more simply proved thus. The vector area of the base is
i F(a  /3) (7  /S) =  i ( Fa/3 + F/3 7 + F 7 a).
Hence the sum of the vector areas of the faces of a tetrahedron,
and therefore of any solid whatever, is zero. This is the hydrostatic
proposition for translational equilibrium of solids immersed in a
fluid subject to no external forces.
223. Taking any two lines whose equations are
p = fi + XOL,
p = ft + a? 1 o l ,
we see that 8 . aa x (p 8) =
is the equation of a plane parallel to both. Which plane, of course,
depends on the value of S.
172 QUATERNIONS. [224.
Now if 8 = /?, the plane contains the first line ; if 8 = ft, the
second.
Hence, if 7/Facq be the shortest vector distance between the
lines, we have
S. oa^/3 ft yFoo^O,
or T (yVaaJ = TS . (/3  ft) UVaa lt
the result of 216.
224. Find the equation of the plane, passing through the origin,
which makes equal angles with three given lines. Also find the angles
in question.
Let a., ft, y be unit vectors in the directions of the lines, and let
the equation of the plane be
SSp = 0.
Then we have evidently
SaS = S/3S $y = x, suppose,
where  ^
is the sine of each of the required angles.
But ( 92) we have
SS . a{By = x ( Va/3 + F/3y + Fya).
Hence 8 . p (Va/3 + F/3y + Fya) =
is the required equation ; and the required sine is
ff.gffy
5 r ( Fa/3 +F/3y+ Fya)
225. Find the locus of the middle points of a series of straight
lines, each parallel to a given plane and having its extremities in two
fixed straight lines.
Let Syp =
be the plane, and
the fixed lines. Also let x and ac l correspond to the extremities of
one of the variable lines, r being the vector of its middle point.
Then, obviously, 2or = /3 + xa + ft + x^.
Also $7 ( _ ft + xa.  x^) = 0.
This gives a linear relation between x and x v so that, if we sub
stitute for x t in the preceding equation, we obtain a result of the
form
OT = + xe,
22;.] GEOMETRY OF STRAIGHT LINE AND PLANE. 173
where 8 and e are known vectors. The required locus is, therefore,
a straight line.
226. Three planes meet in a point, and through the line of
intersection of each pair a plane is drawn perpendicular to the
third ; prove that these planes pass through the same line.
Let the point be taken as origin, and let the equations of the
planes be
Sap = 0, Sj3p = 0, Syp = 0.
The line of intersection of the first two is  Fa/3, and therefore the
normal to the first of the new planes is
Hence the equation of this plane is
or
and those of the other two planes may be easily formed from this
by cyclical permutation of a, /3, 7.
We see at once that any two of these equations give the third
by addition or subtraction, which is the proof of the theorem.
227. Given any number of points A, B, C, &c., whose vectors
(from the origin) are a l} 2 , 3 , <&c., find the plane through the origin
for which the sum of the squares of the perpendiculars let fall upon
it from these points is a maximum or minimum.
Let Svp =
be the required equation, with the condition (evidently allowable)
2V = 1.
The perpendiculars are ( 221) ^~ l S^oi v &c.
Hence 2$Va
is a maximum. This gives
2 . SvrCLScLdvr = ;
and the condition that vr is a unitvector gives
Svrdvr = 0.
Hence, as d^ may have any of an infinite number of values,
these equations cannot be consistent unless
where a? is a scalar.
174 QUATERNIONS. [228
The values of a are known, so that if we put
cf) is a given selfconjugate linear and vector function, and therefore
x has three values (g v g 2 , g s , 175) which correspond to three
mutually perpendicular values of or. For one of these there is a
maximum, for another a minimum, for the third a maximum
minimum, in the most general case when g v g 2 , g 3 are all different.
228. The following beautiful problem is due to Maccullagh.
Of a system of three rectangular vectors, passing through the origin,
two lie on given planes, find the locus of the third.
Let the rectangular vectors be w, p, a. Then by the conditions
of the problem
Svrp = Spcr = $CTCT = 0,
and San = 0, S/3p = 0.
The solution depends on the elimination of p and OT among these
five equations. [This would, in general, be impossible, as p and &
between them involve six unknown scalars ; but, as the tensors are
(by the very form of the equations) not involved, the five given
equations are necessary and sufficient to eliminate the four unknown
scalars which are really involved. Formally to complete the requisite
number of equations we might write
Tvr = a, Tp = b,
but a and 6 may have any values whatever.]
From Soivr = 0, W = 0,
we have OT = xVacr.
Similarly, from 8/3p = 0, Sop = 0,
we have py V/3<r.
Substitute in the remaining equation
Svrp = 0,
and we have S . VacrVffcr = 0,
or SaaSP<r<r*Sap = 0,
the required equation. As will be seen in next Chapter, this is a
cone of the second degree whose circular sections are perpendicular
to a and ft. [The disappearance of x and y in the elimination
instructively illustrates the note above.]
GEOMETRY OF STRAIGHT LINE AND PLANE. 175
EXAMPLES TO CHAPTER VII.
1. What propositions of Euclid are proved by the mere form
of the equation
P = (1X) OL + XP,
which denotes the line joining any two points in space ?
2. Shew that the chord of contact, of tangents to a parabola
which meet at right angles, passes through a fixed point.
3. Prove the chief properties of the circle (as in Euclid, III.)
from the equation
p = a cos 6 + /3 sin 6 ;
where Ta = T0, and Sa/3 = 0.
4. What locus is represented by the equation
where Ta = 1 ?
5. What is the condition that the lines
intersect ? If this is not satisfied, what is the shortest distance
between them ?
6. Find the equation of the plane which contains the two
parallel lines
7. Find the equation of the plane which contains
F0>/3) = 0,
and is perpendicular to Sjp = 0.
8. Find the equation of a straight line passing through a given
point, and making a given angle with a given plane.
Hence form the general equation of a right cone.
9. What conditions must be satisfied with regard to a number
of given lines in space that it may be possible to draw through
each of them a plane in such a way that these planes may intersect
in a common line ?
10. Find the equation of the locus of a point the sum of the
squares of whose distances from a number of given planes is
constant.
176 QUATERNIONS.
11. Substitute "lines" for " planes" in (10).
12. Find the equation of the plane which bisects, at right
angles, the shortest distance between two given lines.
Find the locus of a point in this plane which is equidistant from
the given lines.
13. Find the conditions that the simultaneous equations
Sap = a, 8/3 p = b } Syp = c,
may represent a line, and not a point.
14. What is represented by the equations
where a, ft, 7 are any three vectors ?
15. Find the equation of the plane which passes through two
given points and makes a given angle with a given plane.
16. Find the area of the triangle whose corners have the
vectors a, ft, 7.
Hence form the equation of a circular cylinder whose axis and
radius are given.
17. (Hamilton, Bishop Law s Premium Ex., 1858.)
(a) Assign some of the transformations of the expression
Fa/3
0a
where a and ft are the vectors of two given points A and B.
(b) The expression represents the vector 7, or OG, of a point C
in the straight line AB.
(c) Assign the position of this point G.
18. (Ibid.)
(a) If a, ft, 7, 8 be the vectors of four points, A, B, G, D, what
is the condition for those points being in one plane ?
(b) When these four vectors from one origin do not thus
terminate upon one plane, what is the expression for the volume
of the pyramid, of which the four points are the corners ?
(c) Express the perpendicular S let fall from the origin on
the plane ABG, in terms of a, ft, 7.
GEOMETRY OF STRAIGHT LINE AND PLANE. 177
19. Find the locus of a point equidistant from the three
planes
Sap = 0, S/3p = 0, Syp = 0.
20. If three mutually perpendicular vectors be drawn from a
point to a plane, the sum of the reciprocals of the squares of their
lengths is independent of their directions.
21. Find the general form of the equation of a plane from the
condition (which is to be assumed as a definition) that any two
planes intersect in a single straight line.
22. Prove that the sum of the vector areas of the faces of any
polyhedron is zero.
T. Q. I. 12
CHAPTER VIII.
THE SPHERE AND CYCLIC CONE.
229. AFTER that of the plane the equations next in order of
simplicity are those of the sphere, and of the cone of the second
order, To these we devote a short Chapter as a valuable prepara
tion for the study of surfaces of the second order in general.
230. The equation
Tp = Ta,
or p 2 = a 2 ,
denotes that the length of p is the same as that of a given vector a,
and therefore belongs to a sphere of radius To, whose centre is the
origin. In 107 several transformations of this equation were ob
tained, some of which we will repeat here with their interpretations.
Thus
shews that the chords drawn from any point on the sphere to the
extremities of a diameter (whose vectors are a and ct) are at right
angles to each other.
shews that the rectangle under these chords is four times the area
of the triangle two of whose sides are a and p.
p = (p + a)" 1 a (p + a) (see 105)
shews that the angle at the centre in any circle is double that at
the circumference standing on the same arc. All these are easy
consequences of the processes already explained for the interpreta
tion of quaternion expressions,
232.] THE SPHERE AND CYCLIC CONE. 179
231. If the centre of a sphere be at the extremity of a, the
equation may be written
which is the most general form.
If Tct = T/3,
or 2 = /3 2 ,
in which case the origin is a point on the surface of the sphere, this
becomes
f  ZSctp = 0.
From this, in the form
8p (p  2a) =
another proof that the angle in a semicircle is a right angle is
derived at once.
232. The converse problem is Find the locus of the feet of
perpendiculars let fall from a given point, p = /3, on planes passing
through the origin.
Let Sap =
be one of the planes, then ( 221) the vectorperpendicular is
and, for the locus of its foot,
= a 1 Fa/3.
[This is an example of a peculiar form in which quaternions some
times give us the equation of a surface. The equation is a vector
one, or equivalent to three scalar equations ; but it involves the
undetermined vector a in such a way as to be equivalent to only
two indeterminates (as the tensor of a is evidently not involved).
To put the equation in a more immediately interpretable form, a
must be eliminated, and the remarks just made shew this to be
possible.]
Now (pp)* = a*S*aP,
and (operating by S . /3 on the value of p above)
Adding these equations, we get
or
V 2/ 2
122
180 QUATERNIONS. [233.
so that, as is evident, the locus is the sphere of which /3 is a
diameter.
233. To find the intersection of the two spheres
and
square the equations, and subtract, and we have
J  1 p = <xa I / 1)
which is the equation of a plane, perpendicular to a a,, the vector
joining the centres of the spheres. This is always a real plane
whether the spheres intersect or not. It is, in fact, what is called
their Radical Plane.
234. Find the locus of a point the ratio of whose distances from
two given points is constant.
Let the given points be and A, the extremities of the vector a.
Also let P be the required point in any of its positions, and OP = p.
Then, at once, if n be the ratio of the lengths of the two lines,
This gives p 2  2Sap + cr = ?i 2 /a 2 ,
or, by an easy transformation,
noL
Thus the locus is a sphere whose radius is T (= J > an d whose
centre is at B, where OB =   ^ a definite point in the line OA.
235. If in any line, OP, drawn from the origin to a given plane,
OQ be taken such that OQ . OP is constant, find the locus of Q.
Let Sap = a 2
be the equation of the plane, OT a vector of the required surface.
Then, by the conditions,
T&Tp = constant = b 2 (suppose),
and Uvr = Up.
VU b 2
From these p = =  ^ .
_/ w t&
Substituting in the equation of the plane, we have
a V 2 + tfScLv = 0,
238.] THE SPHERE AND CYCLIC CONE. 181
which shews that the locus is a sphere, the origin being situated
on it at the point farthest from the given plane.
236. Find the locus of points the sum of the squares of whose
distances from a set of given points is a constant quantity. Find
also the least value of this constant, and the corresponding locus.
Let the vectors from the origin to the given points be ct v 2 ,
...... a n , and to the sought point p, then
/ 2a\ 2 c 2 + 2(a 2 ) (2a) 2
Otherwise (p  ) =  ^^ +  ~ ,
\ r n J n n 2
the equation of a sphere the vector of whose centre is   , i.e.
whose centre is the mean of the system of given points.
Suppose the origin to be placed at the mean point, the equation
becomes
p* =  2 + ^ (a2) (because Sa = 0, 31 (e)).
The righthand side is negative, and therefore the equation denotes
a real surface, if
c 2 >22V,
as might have been expected. When these quantities are equal,
the locus becomes a point, viz. the new origin, or the mean point
of the system.
237. If we differentiate the equation
we get Spdp = 0.
Hence ( 144), p is normal to the surface at its extremity, a well
known property of the sphere.
If or be any point in the plane which touches the sphere at the
extremity of p, w p is a line in the tangent plane, and therefore
perpendicular to p. So that
or
is the equation of the tangent plane.
238. If this plane pass through a given point B, whose vector
is ft we have
182 QUATERNIONS. [239.
This is the equation of a plane, perpendicular to /3, and cutting
from it a portion whose length is
To?
Tp
If this plane pass through a fixed point whose vector is 7 we must
have
so that the locus of j3 is a plane. These results contain all the
ordinary properties of poles and polars with regard to a sphere.
239. A line drawn parallel to 7, from the extremity of /3, has
the equation
p = ft + x<y.
This meets the sphere
in points for which x has the values given by the equation
The values of x are imaginary, that is, there is no intersection, if
The values are equal, or the line touches the sphere, if
or
This is the equation of a cone similar and similarly situated to the
cone of tangentlines drawn to the sphere, but its vertex is at the
centre. That the equation represents a cone is obvious from the
fact that it is homogeneous in Ty, i.e. that it is independent of the
length of the vector 7.
[It may be remarked that from the form of the above equation
we see that, if x and x be its roots, we have
which is Euclid, III. 35, 36, extended to a sphere.]
240. Find the locus of the foot of the perpendicular let fall from
a given point of a sphere on any tangentplane.
Taking the centre as origin, the equation of any tangentplane
may be written
The perpendicular must be parallel to p, so that, if we suppose it
241.] THE SPHERE AND CYCLIC CONE. 183
drawn from the extremity of a. (which is a point on the sphere) we
have as one value of TX
r = a + scp.
From these equations, with the help of that of the sphere
P 2 = a 2 ,
we must eliminate p arid x.
We have by operating on the vector equation by S . vr
cr 2 = Savr + xStxp
= Saw + xof.
a a 2 (W a)
Hence p =
x
si 2 Savr
Taking the tensors, we have
O 2  Saw) = a 2 O  a) 2 ,
the required equation. It may be put in the form
S 2 ^U(^a) = oL\
and the interpretation of this (viz. that the projection of vr on
(w a) is of constant length) gives at once a characteristic
property of the surface formed by the rotation of the Cardioid
about its axis of symmetry.
If the perpendiculars be let fall from a point, ft, not on the
sphere, it is easy to see that the equation of the locus is
> v!7() = a B ,
whose interpretation is equally easy.
241. We have seen that a sphere, referred to any point what
ever as origin, has the equation
Hence, to find the rectangle under the segments of a chord drawn
through any point, we may put
P = ^\
where 7 is any unitvector whatever. This gives
0V  2a$e*y + a 2 = /3 2 ,
and the product of the two values of x is
This is positive, or the vectorchords are drawn in the same direc
tion, if
T0 < Ta,
i.e. if the origin is outside the sphere.
184 QUATERNIONS. [242.
242. A t B are fixed points ; and, being the origin and P a
point in space,
AP li + BP* = OP 2 ;
find the locus of P, and explain the result when Z A OB is a right, or
an obtuse, angle.
Let OA=a,Ofi = P, OP = p, then
or
or
While Sa/3 is negative, that is, while Z AOB is acute, the locus
is a sphere whose centre has the vector a + /:?. If $a/3 = 0, or
Z^.0.5 = 7r/2, the locus is reduced to the point
p = a + 0.
If Z AOB> r jrj^ there is no point which satisfies the conditions.
243. Describe a sphere, with its centre in a given line, so as to
pass through a given point and touch a given plane.
Let xa., where x is an undetermined scalar, be the vector of
the centre, r the radius of the sphere, ft the vector of the given
point, and
Syp = a
the equation of the given plane.
The vectorperpendicular from the point xa. on the given plane
is ( 221)
(a xSya) y l .
Hence, to determine x and r we have the equations
T. (a  xSya) y 1 = T(xafy = r,
so that there are, in general, two solutions. It will be a good
exercise for the student to find from these equations the condition
that there may be no solution, or two coincident ones.
244. Describe a sphere whose centre is in a given line, and
which passes through two given points.
Let the vector of the centre be xa, as in last section, and let
the vectors of the points be ft and 7. Then, at once,
Here there is but one sphere, except in the particular case when
we have
Ty = T0, and Say =
in which case there is an infinite number.
245] THE SPHERE AND CYCLIC CONE. 185
The student should carefully compare the results of this
section with those of the last, so as to discover why in general two
solutions are indicated as possible in the one problem, and only
one in the other.
245. A sphere touches each of two straight lines, which do not
meet : find the locus of its centre.
We may take the origin at the middle point of the shortest
distance ( 216) between the given lines, and their equations will
then be
p = a + xft,
where we have, of course,
Let a be the vector of the centre, p that of any point, of one
of the spheres, and r its radius ; its equation is
Since the two given lines are tangents, the following equations in
x and # t must have pairs of equal roots,
The equality of the roots in each gives us the conditions
p {(**?+,*},
Eliminating r we obtain
which is the equation of the required locus.
[As we have not, so far, entered on the consideration of the
quaternion form of the equations of the various surfaces of the
second order, we may translate this into Cartesian coordinates to
find its meaning. If we take coordinate axes of x, y, z respectively
parallel to /3, /3 V a, it becomes at once
(x + myf (y + mxf pz,
where m and p are constants ; and shews that the locus is a
hyperbolic paraboloid. Such transformations, which are exceed
ingly simple in all cases, will be of frequent use to the student
who is proficient in Cartesian geometry, in the early stages of his
study of quaternions. As he acquires a practical knowledge of
186 QUATERNIONS. [246.
the new calculus, the need of such assistance will gradually cease
to be felt]
Simple as the above solution is, quaternions enable us to give one
vastly simpler. For the problem may be thus stated Find the
locus of the point whose distances from two given lines are equal.
And, with the above notation, the equality of the perpendiculars is
expressed ( 214) by
which is easily seen to be equivalent to the equation obtained above.
246. Two spheres being given, shew that spheres which cut them
at given angles cut at right angles another fixed sphere.
If c be the distance between the centres of two spheres whose
radii are a and 6, the cosine of the angle of intersection is evidently
a 2 + b 2  c 2
2ab
Hence, if a, a v and p be the vectors of the centres, and a, a v r the
radii, of the two fixed, and of one of the variable, spheres ; A and
A l the angles of intersection, we have
(p a) 2 + a 2 + r 2 = 2ar cos A,
(p  aj 2 + a x 2 + r 2 = 2^9" cos A t .
Eliminating the first power of r, we evidently must obtain a result
such as
where (by what precedes) e is the vector of the centre, and e the
radius, of a fixed sphere
(p  e ) 2 + e 2 = 0,
which is cut at right angles by all the varying spheres. By effect
ing the elimination exactly we easily find e and e in terms of
given quantities.
247. If two vectors divide one another into parts a and ea,
p and p, respectively, the lines joining the free ends meet in
(l + e)p 2ea
1  e
We may write this as
1 + e 2ea.
lf e = a=
247] THE SPHERE AND CYCLIC CONE. 187
For another such pair of vectors, passing through the same
point, we have, say,
/ 1 <r = OT + /3 1 .
Thus the plane quadrilateral, whose diagonals are made up
respectively of the complete a and /3 vectors, will be projected (by
lines from the point r) as a square on the plane of p, cr, provided
Tp=T<7, and Sp<r = 0.
That is, provided
.................. (1),
(2).
These equations obviously belong to spheres which intersect
one another at right angles. For the centres are at
Thus the distance between the centres is
T e * + f* ( a ft}
"2(/i O (l ft)
and this is obviously less than the sum, and greater than the
difference, of the radii
Z\^L( ai ft), and r.ifoft).
J\ ~ e i
And because its square is equal to the sum of their squares, the
spheres intersect at right angles. Hence
The locus of the points, from which a plane (uncrossed) quadri
lateral can be projected as a square, is a circle whose centre is in
the plane of the quadrilateral, and whose plane is perpendicular to
that plane.
To find the points (if any) of this circle from which the quadri
lateral is seen as a square, we must introduce the additional
conditions
Svrp = 0, >SW = 0,
or Sw (r + OJ = 0, flr(isr + ft)=0 ............ (3).
Hence the points lie on each of two spheres which pass through
the origin : i.e. the intersection of the diagonals.
[If we eliminate r among the four equations (1), (2), (3), we
find the condition
S(,/8 I )(/,X + e, 1 A) = 0
This we leave to the student.]
188 QUATERNIONS. [248.
Another mode of solving the last problem, viz. to find the points
from which a given plane quadrilateral is seen as a square, consists
in expressing that the four portions of the diagonals subtend equal
angles, and that the planes containing them are at right angles to
one another.
The first condition gives, with the notation of the beginning of
this section,
S . VF U (ar  a) = S . w U (w + ea)
= 8.vU(v0)=8.vU(w+fl3) ...... (4).
The second condition is
or w^Sfa  oter/&j = ..................... (5),
the cone whose cyclic normals are a, ft.
[It will be excellent practice for the student to shew that (4)
and (5) are equivalent to (1} ; (2), (3). Thus, in particular, the
first equality in (4)
S.vrU (sr a) = 8 .vfU(af + ea),
is equivalent to the first of (3), viz.
S . w (j + aj = 0.]
It is obvious that the solution of the first problem in this
section gives at once the means of solving the problem of projecting
an ellipse into a circle, so that any given (internal) point may be
projected as the centre of the circle. And numerous other con
sequences follow, which may be left to the reader.
248. To inscribe in a given sphere a closed polygon, plane or
gauche, whose sides shall be parallel respectively to each of a series
of given vectors.
Let Tp = l
be the sphere, a, ft, 7, ...... , tj, 6 the vectors, n in number, and let
p lt p z , ...... p n , be the vectorradii drawn to the angles of the polygon.
Then p 2 p l = XJL, &c., &c.
From this, by operating by 8.(p 1t + p l ), we get
Also = Vap 2 Va Pl .
Adding, we get = a/? 2 4 Kap^ = ap 2 + p^a.
Hence p z =  oL^pfr
or, if we please, p 2 = a/OjGf 1 .
249] THE SPHERE AND CYCLIC CONE. 189
[This might have been written clown at once from the result of
105.]
Similarly p 3 =  /3T l pJ3 = ft ^p^ft, &c.
Thus, finally, since the polygon is closed,
P m /,() 0*1 * ...... Fa pflfi ...... ri0.
We may suppose the tensors of a, ft ...... r), 6 to be each unity.
Hence, if
a = aft ...... rjO,
l ~ l
we have a~ l = 6~rj
which is a known quaternion ; and thus our condition becomes
This divides itself into two cases, according as n is an even or an
odd number.
If n be even, we have
a Pl = p,a.
Removing the common part p^Sa, we have
7/0,7(1 = 0.
This gives one determinate direction, + Fa, for p l ; and shews that
there are two, and only two, solutions.
If n be odd, we have
a Pi = ~ PA
which (operating, for instance, by S . /oj requires that we have
Sa = Q,
i.e. that there may be a solution, a must be a vector.
Hence Sap^ = 0,
and therefore p l may be drawn to any point in the great circle of
the unitsphere whose pole is on the vector a.
249. To illustrate these results, let us take first the case of
n = 3. We must have
or the three given vectors must (as is obvious on other grounds) be
parallel to one plane. Here afty, which lies in this plane, is ( 106)
the vectortangent at the first corner of each of the inscribed tri
angles ; and is obviously perpendicular to the vector drawn from
the centre to that corner.
If n = 4t } we have
as might have been at once seen from 106.
190 QUATERNIONS. [250.
250. Hamilton has given (Lectures, p. 674 and Appendix C.),
an ingenious process by which the above investigation is rendered
applicable to the more difficult problem in which each side of the
inscribed polygon is to pass through a given point instead of being
parallel to a given line. His process, which (see his Life, Vol. in.,
pp. 88, 426) he evidently considered as a specially tough piece of
analysis, depends upon the integration of a linear equation in
finite differences.
The gist of Hamilton s method is (briefly) as follows (Lectures,
676):
Let the (unit) vectors to the corners of the polygon be, as
above, p v p# ...... p n . Also let ct v 2 , ...... a n be the points through
which the successive sides are to pass. The sides are respectively
parallel to the vectors
which correspond to a, /3, ...... 0, of 248. Hence, if we write
we have (as in that section), since the expressions are independent
of the tensors of the qs,
P* = ?*& *> &*>
These give, generally, (with the condition p? 1)
Where r m = V,nl + *,!.
[We may easily eliminate s, by the use of the separable symbol D
or 1 + A, but this leads to a troublesome species of equation of
second differences. Hamilton ingeniously avoids this by the use
of biquaternions.]
Putting i for the algebraic J I, we have
r m + t* m = ( m + (r m _,  is m _,\
(where, as usual, we have a second equation by changing through
out the sign of i).
The complete solution of this equation is, of course, obtained
250.] THE SPHERE AND CYCLIC CONE. 191
at once in the form of a finite product. But it is sufficient to
know some of its characteristic properties only.
The squared tensor is
so that, by equating real and imaginary parts, we have
But, by the value of q l above, we have r, = a. v s l = l, so that
TV.  T\ = (TV.  1) (ra m _,  1) ...... (2a,  1),
Thus it appears that we may write
?. = 6 + /9 + (l)"( + 7)ft .................. (2),
with the condition
bc = S/3y ............................. (3).
But, if we write, putting i instead of p l in q n ,
we have ~ Q = \ + /", suppose,
where X and //, are real vectors whose values can be calculated
from the data. And we now have
(1 4 X
= /_ 1\
i
When w, is odd, this gives at once
which, since pj does not vanish, leads to the two equations
S\ Pl = 0,
These planes intersect in a line, whose intersections with the unit
sphere give the possible extremities of the required first radius.
When n is even, we have
V\ Pl = fi + p l 8fj,p l = V. piVppv
or V. Pl (\ + V^ Pl ) = 0. (199.)
With the notation of (2) the condition (3) becomes
 c
192 QUATERNIONS. [250*.
For further details, see especially Appendices B and C to the
Lectures.
By an immediate application of the linear and vector function
of Chapter V., the above solutions may be at once extended to any
central surface of the second order.
250*. The quaternions which Hamilton employed (as above)
were such as change the radius to one corner of the polygon into
that to the next by a conical rotation. It may be interesting and
useful to the student to compare with Hamilton s solution the
following, which employs the quaternions which directly turn one
side of the polygon to lie along the next. The successive sides are
expressed as ratios of one of these quaternions to the next.
Let p^ p z , &c., p n be (unit) vectors drawn from the centre of
the sphere to the corners of the polygon; a l5 2 ,...a n , the points
through which the successive sides are to pass. Then (by Euclid)
we have
(P* ~ (Pi  i) = 1 + , 2 = A> suppose.
&c. &c.
These equations ensure that if the tensor of any one of the ps be
unit, those of all the others shall also be units. Thus we have
merely to eliminate 2 , . .., p n \ and then remark that (for the
closure of the polygon) we must have
Pn+l = Pi
That this elimination is possible we see from the fact already
mentioned, which shews that the unknowns are virtually mere
unitvectors ; while each separate equation contains coplanar
vectors only. In other words, when p m and cc m are given, p m+l is
determinate without ambiguity.
We may now write the first of the equations thus :
(ft  <Pi ~ i) = A + (i ~ (ft ~ = ?i suppose.
Thus the angle of q^ is the angle of the polygon itself, and in the
same plane. By the help of the second of the above equations
this becomes
^2(fti)=
whence
& = ^ 2 (Pi  a i) + K
250 ". THE SPHEEE AND CYCLIC CONE. 193
By the third, this becomes
(ft ~ ) 9 2 = A 3 ^ 5
whence
(P4 ~ 4> & = A & + (a ~ 4 ) ? 2 = &
The law of formation is now obvious; and, if we write
</o = />,,, ft =!*> ft = 2  &C.,
we have
&c.,
We have also, generally,
or
Pm = ff^i^aa = ^g^. m ,9,,, 2 _ P,,, 2 > _
!* S ma ^m2
From (1), and the value of q , we see that all the values of q
are linear functions of p l of the form
(3).
.
Similarly ^^  ^, t _ 2  m ^_ 2 j
But the first equations in (1) give at once
P = + ^l whence ^
^o=,+ pj
and ^ 1= a 2  a i+( 1
p, = + q lPl
This suggests that
By (4) we have
^l = ^ 2 +n^
^i = ^,n 2 a m ^_ 2
Let ?>i be odd, then we should have by (5)
PM = * + %,
9m = BA Pt ;
T. Q. I.
194 QUATERNIONS. [ 2 5*
whence p m _ l = B Ap l + a m (A 4 BpJ,
or
These agree with (5), because m  1 is even. And similarly we
may prove the proposition when m is even.
If now, in (2), we put n + 1 for m, we have
f ,
p, U
CD Pl . f ,
= =r ^p if n be odd,
# + />!
G and Z) being quaternions to be calculated (as above) from the
data. The two cases require to be developed separately.
Take first the odd polygon :
then piD + pflp^CDpv
or Pl (d + 8) + Pl (c + 7) Pl = c + 7  (d + 3) Pl ,
if we exhibit the scalar and vector parts of the quaternions C and
D. Cutting out the parts which cancel one another, and dividing
by 2, this becomes
which, as p is finite, divides itself at once into the two equations
 c = 0.
These planes intersect in a line which, by its intersections (if real)
with the sphere, gives two possible positions of the first corner of
the polygon.
For the even polygon we have
or Vp$  7  p l Sjp 1 = ;
which may be written
This equation gives, as in 199 above,
where x is to be found from
253] THE SPHERE AND CYCLIC CONE. 195
The two values of a? have opposite signs. Hence there are two
real values of x, equal and with opposite signs, giving two real
points on the sphere. Thus this case of the problem is always
possible.
251. To find the equation of a cone of revolution, whose vertex
is the origin.
Suppose a, where Ta= 1, to be its axis, and e the cosine of its
semivertical angle ; then, if p be the vector of any point in the
cone,
or S ap =  ey.
252. Change the origin to the point in the axis whose vector
is XOL, and the equation becomes
 x
Let the radius of the section of the cone made by
retain a constant value b, while x changes ; this necessitates
x
so that when x is infinite, e is unity. In this case the equation
becomes
which must therefore be the equation of a circular cylinder of
radius 6, whose axis is the vector a. To verify this we have only
to notice that if TO be the vector of a point of such a cylinder we
must ( 214) have
= b,
which is the same equation as that above.
253. To find, generally, the equation of a cone which has a
circular section :
Take the origin as vertex, and let the circular section be the
intersection of the plane
with the sphere (passing through the origin)
132
196 QUATERNIONS. [ 2 54
These equations may be written thus,
~,
Hence, eliminating Tp by multiplying the right, and left, members
together, we find the following equation which Up must satisfy
SaUpS{3Up = I,
or f  SapSj3p = 0,
which is therefore the required equation of the cone.
As a. and /3 are similarly involved, the mere form of this
equation proves the existence of the subcontrary section dis
covered by Apollonius.
254. The equation just obtained may be written
or, since a and (3 are perpendicular to the cyclic planes ( 59*),
sinp sin / = constant,
where p and p are arcs drawn from any point of a spherical conic
perpendicular to the cyclic arcs. This is a wellknown property of
such curves.
255. If we cut the cyclic cone by any plane passing through
the origin, as
then Vary and V/3y are the traces on the cyclic planes, so that
p = scUVa.y+yUVpy (24).
Substitute in the equation of the cone, and we get
 a?  y* + Pxy = 0,
where P is a known scalar. Hence the values of x and y are the
same pair of numbers. This is a very elementary proof of the
proposition in 59*, that PL = MQ (in the last figure of that
section).
256. When x and y are equal, the transversal arc becomes a
tangent to the spherical conic, and is evidently bisected at the
point of contact. Here we have
260.] THE SPHERE AND CYCLIC CONE. 197
This is the equation of the cone whose sides are perpendiculars
(through the origin) to the planes which touch the cyclic cone,
and from this property the same equation may readily be deduced.
257. It may be well to observe that the property of the
Stereographic projection of the sphere, viz. that the projection of
a circle is a circle, is an immediate consequence of the above form
of the equation of a cyclic cone.
258. That 253 gives the most general form of the equation
of a cone of the second degree, when the vertex is taken as origin,
follows from the early results of next Chapter. For it is shewn
in 263 that the equation of a cone of the second degree can
always be put in the form
This may be written Sptyp 0,
where </> is the selfconjugate linear and vector function
<f) P = 2 V. ap/3 + (A + ZScijS) p.
By 180 this may be transformed to
< t ) P=PP+ V V/^>
and the general equation of the cone becomes
( p  SXfju) f + 2S\p Sf^p = 0,
which is the form obtained in 253.
259. Taking the form
Sp(f)p =
as the simplest, we find by differentiation
Sdpfyp + Spd^p = 0,
or ZSdpfo = 0.
Hence $p is perpendicular to the tangentplane at the extremity
of p. The equation of this plane is therefore (w being the vector
of any point in it)
Sj>p (*rp) = 0,
or, by the equation of the cone,
Ssr$p = 0.
260. The equation of the cone of normals to the tangentplanes
of a given cone can be easily formed from that of the cone itself.
For we may write the equation of the cone in the form
198 QUATERNIONS. [261.
and if we put tfjp = a, a vector of the new cone, the equation
becomes
Str4>~ l tr = 0.
Numerous curious properties of these connected cones, and of the
corresponding spherical conies, follow at once from these equations.
But we must leave them to the reader.
261. As a final example, let us find the equation of a cyclic
cone when five of its vectorsides are given i.e. find the cone of the
second degree whose vertex is the origin, and on whose surface lie
the vectors a, ft, 7, S, e.
If we write, after Hamilton,
Q = S.V(Va/3V8e)V(Vl3vVep)V(VvSVpoL) ......... (1)
we have the equation of a cone whose vertex is the origin for the
equation is not altered by putting xp for p. Also it is the equation
of a cone of the second degree, since p occurs only twice. Moreover
the vectors a, ft, 7, 8, e are sides of the cone, because if any one
of them be put for p the equation is satisfied. Thus if we put
ft for p the equation becomes
= S .V (VaftVZe] [VftaS.V^Vft^Veft VySS .Vfta VftyVeft J.
The first term vanishes because
and the second because
since the three vectors Vfta., V/3y, Fe/3, being each at right angles
to ft, must be in one plane.
As is remarked by Hamilton, this is a very simple proof of
Pascal s Theorem for (1) is the condition that the intersections of
the planes of a, ft and 8, e ; ft, 7 and e, p ; 7, S and p, a ; shall lie
in one plane ; or, making the statement for any plane section of
the cone : In order that the points of intersection of the three pairs
of opposite sides, of a hexagon inscribed in a curve, may always lie
in one straight line, the curve must be a conic section.
THE SPHERE AND CYCLIC CONE. 199
EXAMPLES TO CHAPTER VIII.
1. On the vector of a point P in the plane
a point Q is taken, such that QO . OP is constant ; find the equation
of the locus of Q.
2. What spheres cut the loci of P and Q in (1) so that both
circles of intersection lie on a cone whose vertex is ?
3. A sphere touches a fixed plane, and cuts a fixed sphere.
If the point of contact with the plane be given, the plane of the
intersection of the spheres contains a fixed line.
Find the locus of the centre of the variable sphere, if the plane
of its intersection with the fixed sphere passes through a given
point.
4. Find the radii of the spheres which touch, simultaneously,
the four given planes
[What is the volume of the tetrahedron enclosed by these planes ?]
5. If a moveable line, passing through the origin, make with
any number of fixed lines angles 6, 6 V # 2 , &c., such that
a cos 6 + a v cos 6 l + ...... = constant,
where a, a v ...... are constant scalars, the line describes a right cone.
6. Determine the conditions that
Spcf)p =
may represent a right cone, </> being as in 258.
7. What property of a cone (or of a spherical conic) is given
directly by the following particular form of its equation,
S . ipicp = ?
8. What are the conditions that the surfaces represented by
Sp(j>p = 0, and S . cpfcp = 0,
may degenerate into pairs of planes ?
200 QUATERNIONS.
9. If arcs of great circles, drawn from any given point of a
sphere to a fixed great circle, be bisected, find the locus of these
middle points ; and shew that the arcs drawn from the pole of this
fixed great circle, to that of which the given point is pole, are also
bisected by the same locus.
10. Find the locus of the vertices of all right cones which
have a common ellipse as base.
11. Two right circular cones have their axes parallel. Find
the orthogonal projection of their curve of intersection on the
plane containing their axes.
12. Two spheres being given in magnitude arid position, every
sphere which intersects them in given angles will touch two other
fixed spheres and cut a third at right angles.
13. If a sphere be placed on a table, the breadth of the
elliptic shadow formed by rays diverging from a fixed point is
independent of the position of the sphere.
14. Form the equation of the cylinder which has a given
circular section, and a given axis. Find the direction of the
normal to the subcontrary section.
15. Given the base of a spherical triangle, and the product of
the cosines of the sides, the locus of the vertex is a spherical conic,
the poles of whose cyclic arcs are the extremities of the given
base.
16. (Hamilton, Bishop Laws Premium Ex., 1858.)
(a) What property of a spheroconic is most immediately
indicated by the equation
(b) The equation
also represents a cone of the second order ; X is a focal line, and
//, is perpendicular to the directorplane corresponding.
(c) What property of a spheroconic does the equation most
immediately indicate ?
17. Shew that the areas of all triangles, bounded by a tangent
to a spherical conic and by the cyclic arcs, are equal.
THE SPHERE AND CYCLIC CONE. 201
18. Shew that the locus of a point, the sum of whose arcual
distances from two given points on a sphere is constant, is a
spherical conic.
19. If two tangent planes be drawn to a cyclic cone, the four
lines in which they intersect the cyclic planes are sides of a right
cone.
20. Find the equation of the cone whose sides are the
intersections of pairs of mutually perpendicular tangent planes
to a given cyclic cone.
21. Find the condition that five given points may lie on a
sphere.
22. What is the surface denoted by the equation
p 2 = xoc + yp 4 ^
where p = xa. + yf$ + zy,
a, ft, 7 being given vectors, and x, y, z variable scalars ?
Express the equation of the surface in terms of p, a, ft, 7 alone.
23. Find the equation of the cone whose sides bisect the
angles between a fixed line, and any line in a given plane, which
meets the fixed line.
What property of a spherical conic is most directly given
by this result ?
CHAPTER IX.
SURFACES OF THE SECOND DEGREE.
262. THE general scalar equation of the second degree in a
vector p must evidently contain a term independent of p, terms of
the form S. apb involving p to the first degree, and others of the
form S.apbpc involving p to the second degree, a, b, c, &c. being
constant quaternions. Now the term S.apb may be written as
S P V(ba),
or as
S.(Sa+ Va) p (Sb + Vb) = SaSpVb + SbSpVa + 8. pVbVa,
each of which may evidently be put in the form Syp, where 7 is
a known vector.
Similarly* the term S . apbpc may be reduced to a set of terms,
each of which has one of the forms
the second being merely a particular case of the third. Thus (the
numerical factors 2 being introduced for convenience) we may
write the general scalar equation of the second degree as follows:
22 . SapSffp + Ap* + 2%> = G ............... (1).
263. Change the origin to D where OD = S, then p becomes
p + S, and the equation takes the form
22 . SapSQp + Ap 9 + 22 (SapSpS + SppSaS) + ZAS&p + 2Syp
* For S . apbpc = S . capbp = S . a pbp = (2Sa Sb  Sa b) p 2 + ZSa pSbp ; and in parti
cular cases we may have Va Vb.
264.] SURFACES OF THE SECOND DEGREE. 203
from which the first power of p disappears, that is the surface is
referred to its centre , if
2 (aSfiS + /3Sa8) + AS + y = (2),
a vector equation of the first degree, which in general gives
a single definite value for 8, by the processes of Chapter V. [It
would lead us beyond the limits of an elementary treatise to
consider the special cases in which (2) represents a line, or a plane,
any point of which is a centre of the surface. The processes to be
employed in such special cases have been amply illustrated in the
Chapter referred to.]
With this value of S, and putting
D = G 2SyS  AV  22 . SaS3/3S,
the equation becomes
22 . SapS/3p + Ap 2 = D.
If D = 0, the surface is conical (a case treated in last Chapter) ;
if not, it is an ellipsoid or hyperboloid. Unless expressly stated
not to be, the surface will, when D is not zero, be considered an
ellipsoid. By this we avoid for the time some rather delicate
considerations.
By dividing by D, and thus altering only the tensors of the
constants, we see that the equation of central surfaces of the
second degree, referred to the centre, is (excluding cones)
^(S a pS^p)+gp"=l (3).
[It is convenient to use the negative sign in the righthand
member, as this ensures that the important vector <frp (which we
must soon introduce) shall make an acute angle with p ; i.e. be
drawn, on the whole, towards the same parts.]
264. Differentiating, we obtain
22 {SadpS/3p + SapS/Bdp] + ZgSpdp = 0,
or S.dp {2 (aS/3p + %>) + gp] = 0,
and therefore, by 144, the tangent plane is
Sfrp) {2 (aSp P + 0S*p) + gp} = 0,
i.e. 8 . {2 (aS/Bp + {3ScL P ) +gp} = l,\)y (3).
Hence, if v=*2(aS/3p + l3S*p) + gp (4),
the tangent plane is Sv& = 1,
and the surface itself is Svp 1,
204 QUATERNIONS. [265.
And, as v~ l (being perpendicular to the tangent plane, and
satisfying its equation) is evidently the vectorperpendicular from
the origin on the tangent plane, v is called the vector of proximity .
265. Hamilton uses for v, which is obviously a linear and
vector function of p, the notation (f>p, $ expressing a functional
operator, as in Chapter V. But, for the sake of clearness, we will
go over part of the ground again, especially in the interests of
students who have mastered only the more elementary parts of
that Chapter.
We have, then, </o = 2 (aSflp + /3Sap) + gp.
With this definition of <, it is easy to see that
(a) </> (p + <r) = </>/> + $&, &c., for any two or more vectors.
(b) </> (asp) = xfyp, a particular case of (a), x being a scalar.
(c) d$p = 4>(dp}.
(d) 3o<f)p = S (ScL(rS/3p + S/3aSap) + gSp<r = Spfa,
or <f) is, in this case, selfconjugate.
This last property is of great importance in what follows.
266. Thus the general equation of central surfaces of the
second degree (excluding cones) may now be written
Sp4>p= i .......................... (i).
Differentiating, Sdp(f>p f Spd<j)p = 0,
which, by applying (c) and then (d) to the last term on the left,
gives
= 0,
and therefore, as in 264, though now much more simply, the
tangent plane at the extremity of p is
S (DT p) <pp = 0,
or Stztyp = Sptyp = 1.
If this pass through A (OA = a), we have
Sa<l>p =  1,
or, by (d), Spfa =  1,
for all possible points of contact.
This is therefore the equation of the plane of contact of tangent
planes drawn from A.
268.] SURFACES OF THE SECOND DEGREE. 205
207. To find the enveloping cone whose vertex is A, notice that
(Sp<t>p + 1) +p (Sp<f>CL + I) 2 = 0,
where p is any scalar, is the equation of a surface of the second
degree touching the ellipsoid along its intersection with the plane.
If this pass through A we have
(8a<j)OL + 1) + p (Safa + I) 2 = 0,
and p is found. Then our equation becomes
(Sp<l)p + I)(Sai<l>oL + l)(Sp<l>a + l) 9 = Q (1),
which is the cone required. To assure ourselves of this, transfer
the origin to A, by putting p + a for p. The result is, using (a)
and (d),
(Spfo + 28pcf)CL + Safa + 1) (Sa<t>a + 1)  (SpQa + SaQoL + I) 2  0,
or Sp^p (Safct + 1)  (Sp<t>a.y = 0,
which is homogeneous in Tp, and is therefore the equation of a
cone.
[In the special case when A lies on the surface, we have
$</> + 1 = 0,
and the value of p is infinite. But this is not a case of failure, for
the enveloping cone degenerates into the tangent plane
/>< + 1 = 0.]
Suppose A infinitely distant, then we may put in (1) oca. for a,
where x is infinitely great, and, omitting all but the higher terms,
the equation of the cylinder formed by tangent lines parallel to a is
(Sp<j>p + 1) SOL^OL  (Spfa) 9 = 0.
See, on this matter, Ex. 21 at end of Chapter.
268. To study the nature of the surface more closely, let us
find the locus of the middle points of a system of parallel chords.
Let them be parallel to a, then, if & be the vector of the middle
point of one of them, 57 + xa. and r xa. are values of p which
ought simultaneously to satisfy (1) of 266.
That is S .(or xa)<f>(iffa;a) = I.
Hence, by (a) and (d), as before,
+ afScvpa = 1,
(1).
206 QUATERNIONS. [269.
The latter equation shews that the locus of the extremity of w,
the middle point of a chord parallel to a, is a plane through the
centre, whose normal is (/> ; that is, a plane parallel to the tangent
plane at the point where OA cuts the surface. And (d) shews that
this relation is reciprocal so that if ft be any value of r, i.e. be
any vector in the plane (1), a will be a vector in a diametral plane
which bisects all chords parallel to ft. The equations of these
planes are
= 0,
= 0,
so that if V. (fracfrft = 7 (suppose) is their line of intersection, we have
= = $a<^7 }
and (1) gives Sfttya = 0= Satyft)
Hence there is an infinite number of sets of three vectors a, ft, 7,
such that all chords parallel to any one are bisected by the diametral
plane containing the other two.
269. It is evident from 23 that any vector may be expressed
as a linear function of any three others not in the same plane ; let
then
where, by last section,
Sa^ft = Sft$a = 0,
$#(7 = S<y(f)ct = 0,
And let Sa<j>a =  1
Sft<f>ft = l\,
so that a, ft, and 7 are vector conjugate semidiameters of the
surface we are engaged on.
Substituting the above value of p in the equation of the surface,
and attending to the equations in a, ft, 7 and to (a), (6), and (d),
we have
To transform this equation to Cartesian coordinates, we notice that
x is the ratio which the projection of p on a bears to a itself, &c.
271.] SURFACES OF THE SECOND DEGREE. 207
If therefore we take the conjugate diameters as axes of f, 77, ?, and
their lengths as a, b, c, the above equation becomes at once
the ordinary equation of the ellipsoid referred to conjugate
diameters.
270. If we write ^ instead of <, these equations assume an
interesting form. We take for granted, what we shall afterwards
prove, that this extraction of the square root of the vector function
is lawful, and that the new linear and vector function has the same
properties (a), (b), (c), (d) ( 265) as the old. The equation of the
surface now becomes
or Styptyp = 1,
or, finally, T*fyp = 1.
If we compare this with the equation of the unitsphere
r/>=i,
we see at once the analogy between the two surfaces. The sphere
can be changed into the ellipsoid, or vice versa, by a linear deforma
tion of each vector, the operator being the function ty or its inverse.
See the Chapter on Kinematics.
271. Equations (2) 268, by 270 become
#a^ a /3 = = <Styra^v8, &c ...................... (1),
so that i/ra, \/r/3, ^7, the vectors of the unitsphere which correspond
to semiconjugate diameters of the ellipsoid, form a rectangular
system.
We may remark here, that, as the equation of the ellipsoid
referred to its principal axes is a case of 269, we may now suppose
i,j, and k to have these directions, and the equation is
which, in quaternions, is
($jp
We here tacitly assume the existence of such axes, but in all cases,
by the help of Hamilton s method, developed in Chapter V., we at
once arrive at the cubic equation which gives them.
208 QUATERNIONS. [272.
It is evident from the lastwritten equation that
fiSip jSjp kSkp
and Jro^z  r + L^r +   ,
V a 6 c
which latter may be easily proved by shewing that
And this expression enables us to verify the assertion of last section
about the properties of ty.
As Sip x, &c., a?, y, being the Cartesian coordinates
referred to the principal axes, we have now the means of at once
transforming any quaternion result connected with the ellipsoid
into the ordinary one.
272. Before proceeding to other forms of the equation of the
ellipsoid, we may use those already given in solving a few problems.
Find the locus of a point when the perpendicular from the centre
on its polar plane is of constant length.
If or be the vector of the point, the polar plane is
= 1,
and the length of the perpendicular from is ( 264).
(pOT
Hence the required locus is
T$w = C,
or &sr<V   C 2 ,
a concentric ellipsoid, with its axes in the same directions as those
of the first. By 271 its Cartesian equation is
273. Find the locus of a point whose distance from a given point
is always in a given ratio to its distance from a given line
Let p = xft be the given line, and A (OA = a) the given point,
and choose the origin so that Saft = 0. Then for any one of the
required points
This is the equation of a surface of the second degree, which may
be written
274] SURFACES OF THE SECOND DEGREE. 209
Let the centre be at 8, and make it the origin, then
and, that the first power of p may disappear,
a linear equation for S. To solve it, note that Sa/3 = ; operate by
S . 0, and we get
(1  e 2 /3 2 + e 2 /3 a ) S/3B = S/38 = 0.
Hence 8  a =  e*3 2 3
Referred to this point as origin the equation becomes
=o,
which shews that it belongs to a surface of revolution (of the
second degree) whose axis is parallel to ft, since its intersection
with a plane Sftp = a, perpendicular to that axis, lies also on the
sphere
eV
I 1 I yi /Q2 /
In fact, if the point be the focus of any meridian section of an
oblate spheroid, the line is the directrix of the same.
274. A sphere, passing through the centre of an ellipsoid, is cut
by a series of spheres whose centres are on the ellipsoid and which
pass through the centre thereof; find the envelop of the planes of
intersection.
Let (p a) 2 = a 2 be the first sphere, i.e.
One of the others is p 2 2Svrp = 0,
where 8^^ =  1.
The plane of intersection is
S (to  a.) p = 0.
Hence, for the envelop (see next Chapter),
jSfww =
, ^r, where tzr = cfo,
=
or <t>is = xp, {Vx = 0},
i.e. BT = x<f> l p.
T. Q. I, 14
210 QUATERNIONS. [ 2 75
Hence a?Sp(f)~ l p = 1 )
and aSpcfr^p = Sap)
and, eliminating x,
a cone of the second degree.
275. From a point in the outer of two concentric ellipsoids a
tangent cone is drawn to the inner, find the envelop of the plane of
contact.
If $ti7( G7 = 1 be the outer, and Spifrp = 1 be the inner, <
and ty being any two selfconjugate linear and vector functions,
the plane of contact is
Stz ^rp = 1.
Hence, for the envelop, S^ ^p = 0)
Str tytJ = Oj
therefore </>^ = octyp,
or t*7 = xtpT^p.
This gives xS .
and xrS .
and therefore, eliminating x,
S.^rp^~ l ^p= 1,
or 8 . piWtyp = 1 ,
another concentric ellipsoid, as \r^)~ 1 i^ is a linear and vector function
% suppose ; so that the equation may be written
276. Find the locus of intersection of tangent planes at the
extremities of conjugate diameters.
If a, /3, 7 be the vector semidiameters, the planes are
with the conditions 271.
Hence  ^S . ^a^fi^y = ^ = ^a + ^/3 + ^7> b y 92
therefore Tirr = V3,
since v/ra, ^^3, ^7 form a rectangular system of unitvectors.
This may also evidently be written
278.] SURFACES OF THE SECOND DEGREE. 211
shewing that the locus is similar and similarly situated to the given
ellipsoid, but larger in the ratio *J3 : 1.
277. Find the locus of intersection of three tangent planes
mutually at right angles.
If p be the point of contact,
Svnpp =  1
is the equation of the tangent plane.
The vector perpendicular from the origin is ( 264)
1
where a is a unitvector. This gives
whence, by the equation of the ellipsoid,
Thus the perpendicular is
a 7^
The sum of the squares of these, corresponding to a rectangular
unitsystem, is
 2 $a^~ 2 = m 2 ,
by 185. See also 279.
278. Find the locus of the intersection of three spheres ivhose
diameters are semiconjugate diameters of an ellipsoid.
If a be one of the semiconjugate diameters
And the corresponding sphere is
with similar equations in @ and 7. Hence, by 92,
^~ 1 P S . f a ^^r 7 = _
and, taking tensors, Tifr~ l p =
or, finally,
This is Fresnel s Surface of Elasticity, Chap. XII.
142
212
QUATERNIONS.
[279.
279. Before going farther we may prove some useful properties
of the function < in the form we are at present using viz.
fiSip jSjp kSkp\
<PP = ~ 1 ~~a~ H 12 H 2
V a b c 2 /
We have p = ^ p JSjp kSkp,
and it is evident that
Hence
Also cf> 1 p =
and so on.
Again, if a, ft, y be any rectangular unitvectors
fe) 2 , (Sja? , (Sk*T
But as
we have
Similarly
&c. = &c.
(%) 2 f (%) 2 + (Skp)* =  p\
OL +
V + P + ^
Lv (/ O
Again,
$ . (bci(bft<}>v =  S . ( TT f ... I I 5 + ...)(
/ \ a
Sia, SJOL, Ska.
Sift, Sjft, Skft
Siy, Sjy, Sky
And so on. These elementary investigations are given here for
the benefit of those who have not read Chapter V. The student
may easily obtain all such results in a far more simple manner by
means of the formulae of that Chapter.
280. Find the locus of intersection of a rectangular system of
three tangents to an ellipsoid.
If w be the vector of the point of intersection, a, ft, y the
282.] SURFACES OF THE SECOND DEGREE. 213
tangents, then, since CT 4 xa, must give equal values of x when
substituted in the equation of the surface, so that
S (OT 4 xa) O 4 xa) =  1,
we have (Svr^a) 2 = Sacfra (S^rfa 41).
Adding this to the two similar equations in /3 and 7, we have
i),
or ((POT) = I 2 4~ T2 H 2 1 (fa GHjyGr 4~ 1)>
1
2 +  2 j ($ 3T<) S7 4~1),
1
an ellipsoid concentric with the first.
281. If a rectangular system of chords be drawn through any
point within an ellipsoid, the sum of the reciprocals of the rectangles
under the segments into which they are divided is constant.
With the notation of the solution of the preceding problem, nr
giving the intersection of the vectors, it is evident that the
product of the values of x is one of the rectangles in question
taken negatively.
Hence the required sum is
111
This evidently depends on S^^^ only and not on the particular
directions of a, (3, 7 : and is therefore unaltered if or be the vector
of any point of an ellipsoid similar, and similarly situated, to the
given one. [The expression is interpretable even if the point be
exterior to the ellipsoid.]
282. Shew that if any rectangular system of three vectors be
drawn from a point of an ellipsoid, the plane containing their other
extremities passes through a fixed point. Find the locus of the
latter point as the former varies.
With the same notation as before, we have
mzr^xsr = 1,
and S(& + xa) </> (w 4 xa) 1 ;
therefore # =
214 QUATERNIONS.
Hence the required plane passes through the extremity of
 a  ,
$(/>
and those of two other vectors similarly determined. It therefore
(see 30) passes through the point whose vector is
* ~
or
Thus the first part of the proposition is proved.
But we have also tzr = 2 (< ^j 0,
whence by the equation of the ellipsoid we obtain
the equation of a concentric ellipsoid.
283. Find the directions of the three vectors which are parallel
to a set of conjugate diameters in each of two central surfaces of the
second degree.
Transferring the centres of both to the origin, let their equations
be
=I or
and Sp^p =  1 or
If a, /3, 7 be vectors in the required directions, we must have ( 268)
Sa(t>/3 = 0, Sofy/3 = j
$807=0, /S^7=OV .................. (2).
$y(/>a = 0, Sry^ct = J
From these equations (/>  V{3y  tya, &c.
Hence the three required directions are the roots of
V.<j>p^p = ........................... (3).
This is evident on other grounds, for it means that if one of the
surfaces expand or contract uniformly till it meets the oilier, it will
touch it successively at points on the three sought vectors.
We may put (3) in either of the following forms
or
V.
284.] SURFACES OF THE SECOND DEGREE. 215
and, as $ and i/r are given functions, we find the solutions (when
they are all real, so that the problem is possible) by the processes
of Chapter V.
[Note. As <f>~ l ifr and ty~ l <j> are not, in general, selfconjugate
functions, equations (4) do not signify that a, /3, 7 are vectors
parallel to the principal axes of the surfaces
S . p(j>~ l ^rp =  1, S. p*^~ l <t>p =  1.
In these equations it does not matter whether c^rS/r is selfconjugate
or not ; but it does most particularly matter when, as in (4), they
are involved in such a manner that their nonconjugate parts do
not vanish.]
Given two surfaces of the second degree, which have parallel
conjugate diameters, every surface of the second degree passing
through their intersection has conjugate diameters parallel to these.
For any surface of the second degree through the intersection
of
Spfa =  1 and 8 (p  a) i/r (p  a) =  e,
is fSp<j>p S(pa.)^(pa) = ef,
where e and f are scalars, of which f is variable.
The axes of this depend only on the term
Hence the set of conjugate diameters which are the same in all
are parallel to the roots of
or
as we might have seen without analysis.
The locus of the centres is given by the equation
where / is a scalar variable.
284. Find the equation of the ellipsoid of which three conjugate
semidiameters are given.
Let the vector semidiameters be a, /3, 7, and let
Spfo = l
be the equation of the ellipsoid. Then ( 269) we have
=  1, Sot<f>{3 = 0,
= ;
216 QUATERNIONS. [ 2 ^5
the six scalar conditions requisite (178) for the determination of
the selfconjugate linear and vector function </>.
They give a  VtfrPQy,
or sea. = c/T 1 V{3y.
Hence x xSa$a = S . afiy,
and similarly for the other combinations. Thus, as we have
pS . afty = aS . @yp + ftS . yap + <yS . afip,
we find at once
 <f>pS 2 . a{3y = VftyS . ftyp + VyaS . yap + VOL ft 8 . a/3p ;
and the required equation may be put in the form
S 2 . a{3y = S 2 .a{3p + S 2 .(3yp + S* . yap.
The immediate interpretation is that if four tetrahedra be formed
by grouping, three and three, a set of semiconjugate vector axes of
an ellipsoid and any fourth vector of the surface, the sum of the
squares of the volumes of three of tliese tetrahedra is equal to the
square of the volume of the fourth.
285. A line moves with three of its points in given planes, find
the locus of any fourth point.
Let a, b, c be the distances of the three points from the fourth,
a, ft 7 unitvectors perpendicular to the planes respectively. If p
be the vector of the fourth point, referred to the point of inter
section of the planes, and a a unitvector parallel to the line, we
have at once
Thus
Sap H j Sfip
The condition Tcr = 1 gives the equation of an ellipsoid referred to
its centre.
We may write the equation in the form
pSafty = aVpySao + bVyaSffo + cVa/3Sya
= (pa . Sa(3y, suppose,
and from this we find at once for the volume of the ellipsoid
j Q
. afty
287.] SURFACES OF THE SECOND DEGREE. 217
altogether independent of the relative inclinations of the three
planes. This curious extension of a theorem of Monge is due to
Booth.
286. We see from 270 that (as in 31 (m)) we can write the
equation of an ellipsoid in the elegant form
where (/> is a selfconjugate linear and vector function, and we
impose the condition
Te = l.
Hence, when the same ellipsoid is displaced by translation and
rotation, by 119 we may write its equation as
p = 8 + q<t>eq~\
with the condition that e is still a unitvector.
Where it touches a plane perpendicular to i, we must have,
simultaneously,
and = See.
Hence e = U(f) (q~ l iq]
at the point of contact ; and, if the plane touched be that of jk,
= SiSS. q~ l iq<l> U<l> (q~ l iq\
or = SiS + T(j> (q~ l iq).
Thus, if we write
q*iq = a, q~ l jq = ft q~ l kq = 7,
we have
which gives the possible positions of the centre of a given ellipsoid
when it is made to touch the fixed coordinate planes.
We see by 279 that T8 is constant. And it forms an
interesting, though very simple, problem, to find the region of
this spherical surface to which the position of the centre of the
ellipsoid is confined. This, of course, involves giving to a, ft 7
all possible values as a rectangular unit system.
287. For an investigation of the regions, on each of the
coordinate planes, within which the point of contact is confined,
see a quaternion paper by Plarr (Trans. R. 8. E. 1887). The
difficulty of this question lies almost entirely in the eliminations,
which are of a very formidable character. Subjoined is a mere
sketch of one mode of solution, based on the preceding section.
218 QUATERNIONS.
The value of p, at the point of contact with
Sip = 0,
is
or + S
or, finally,
Hence, in ordinary polar coordinates, the point of contact with
the plane ofjk is
To find the boundary of the region within which the point of
contact must lie, we must make r a maximum or minimum, 9
being constant, and a, /3, 7 being connected by the relations
= Ty =1
Differentiating (1) and (2), with the conditions
d0 = 0,
(because 6 is taken as constant) and
dr =
as the criterion of the maximum, we have eight equations which
are linear and homogeneous in da, dj3, dy. Eliminating the two
latter among the seven equations which contain them, we have
= Sd* [(PSafr + &) tf/3 7l + (yScL 7l + 72 ) Sy0J ...... (3)
where ft = <j> ( U$OL  U(j)/3),
7. 
But we have also, as yet unemployed,
Sada = ........................... (4).
Conditions (3) and (4) give the two scalar equations
= (S/3 2J 8  800) Sj,/3 + Sy^S&T
288.] SURFACES OF THE SECOND DEGREE. 219
Theoretically, the ten equations (1), (2), and (5), suffice to
eliminate the nine scalars involved in a, /3, 7 : and thus to leave
a single equation in r, 0, which is that of the boundary of the
region in question.
The student will find it a useful exercise to work out fully the
steps required for the deduction of (5).
288. When the equation of a surface of the second order can
be put in the form
8p^p = l ........................ (1),
where (0 g)(<l> g,) ($  g 9 ) = 0,
we know that g, g v g z are the squares of the principal semi
diameters. Hence, if we put </> + h for <f> we have a second surface,
the differences of the squares of whose principal semi axes are the
same as for the first. That is,
Sp(<j> + h) l p = l ..................... (2),
is a surface confocal with (1). From this simple modification of
the equation all the properties of a series of confocal surfaces may
easily be deduced. We give a couple of examples.
Any two confocal surfaces of the second degree, which meet,
intersect at right angles.
For the normal to (2) is evidently parallel to
and that to another of the series, if it passes through the common
point whose vector is p, is parallel to
But a.Q + kr
and this evidently vanishes if h and \ are. different, as they must
be unless the surfaces are identical.
To find the locus of the points of contact of a series of confocal
surfaces with a series of parallel planes.
Here the direction of the normal at the point, p, of contact is
(*+Arv
and is parallel to the common normal of the planes, say a,
Thus
220 QUATERNIONS. [289.
Thus the locus has the equation of a plane curve
p = XOL + y$cL,
and the relation between x and y is, by the general equation of
the confocals,
1 + y 2 Sct(j)CL + xyo? = 0.
Hence the locus is a hyperbola.
289. To find the conditions of similarity of two central surfaces
of the second degree.
Referring them to their centres, let their equations be
...a).
Now the obvious conditions are that the axes of the one are
proportional to those of the other. Hence, if
9*  2 <7 2 +m l g m =0} ,
fm\f + m dm = v]
be the equations for determining the squares of the reciprocals of
the semiaxes, we must have
m m 2 m , ,
=P, =AC, =/* .................. (3),
m 2 m 1 m
where /A is an undetermined scalar. Thus it appears that there
are but two scalar conditions necessary. Eliminating ft we have
( .
m *
which are equivalent to the ordinary conditions.
290. Find the greatest and least semidiameters of a central
plane section of an ellipsoid.
Here Sp(f>p = 1 j ... ,
Sap= OJ"
together represent the elliptic section; and our additional condition
is that Tp is a maximum or minimum.
Differentiating the equations of the ellipse, we have
8<j>pdp = 0,
Sadp = 0,
and the maximum condition gives
dTp = 0,
or Spdp = 0.
2 9 1.]  SURFACES OF THE SECOND DEGREE. 221
Eliminating the indeterminate vector dp we have
........................... (2).
This shews that the maximum or minimum vector, the normal at
its extremity, and the perpendicular to the plane of section, lie in
one plane. It also shews that there are but two vectordirections
which satisfy the conditions, and that they are perpendicular to
each other, for (2) is unchanged if ap be substituted for p.
We have now to solve the three equations (1) and (2), to find
the vectors of the two (four) points in which the ellipse (1) inter
sects the cone (2). We obtain at once
fa = xV . ((/T 1 a) Vap.
Operating by S . p we have
1 = Xp*Sa.(f)~ l OL.
Sp6~ l a
Hence _ = _
from which 8 . a(l +/>~ 1 a  ..................... (4);
a quadratic equation in p 2 , from which the lengths of the maximum
and minimum vectors are to be determined. By 184 it may be
written
wip Sa^ a + p*S.a (m >2  0) a + a 2  ......... (5).
[If we had operated by $ . (f>~ 1 a. or by S . <f)~ l p, instead of by
S . p, we should have obtained an equation apparently different
from this, but easily reducible to it. To prove their identity is a
good exercise for the student.]
Substituting the values of p 2 given by (5) in (3) we obtain the
vectors of the required diameters. [The student may easily prove
directly that
(l+p^ a and (I + pft) 1 a
are necessarily perpendicular to each other, if both be perpen
dicular to a, and if p* and p 2 2 be different. See 288.]
291. By (5) of last section we see that
2 2 _
Pi T2 ~~
Hence the area of the ellipse (1) is
J
222 QUATERNIONS. [292.
Also the locus of central normals to all diametral sections of an
ellipsoid, whose areas are equal, is the cone
SoLf 1 a = COL*.
When the roots of (5) are equal, i.e. when
(ra 2 a 2  a0a) 2  4ma a a<Jf J a (6),
the section is a circle. It is riot difficult to prove that this
equation is satisfied by only two values of Ua, but another
quaternion form of the equation gives the solution of this and
similar problems by inspection. (See 292 below.)
292. By 180 we may write the equation
Spfp = I
in the new form S . \pf^p + pp 2 = 1,
where p is a known scalar, and X and //, are definitely known (with
the exception of their tensors, whose product alone is given) in
terms of the constants involved in </>. [The reader is referred
again also to 128, 129.] This may be written
ZSXpSpp + (p  S\/JL) p" =  1 (1).
From this form it is obvious that the surface is cut by any plane
perpendicular to X or /^ in a circle. For, if we put
S\p = a,
we have ^aSfip + (p S\fi) p 2 = 1,
the equation of a sphere which passes through the plane curve of
intersection.
Hence X and fju of 180 are the values of a in equation (6) of
the preceding section.
293. Any two circular sections of a central surface of the
second degree, whose planes are not parallel, lie on a sphere.
For the equation
(S\p  a} (Sfjip  b) = 0,
where a and b are any scalar constants whatever, is that of a
system of two nonparallel planes, cutting the surface in circles.
Eliminating the product SXpSpp between this and equation (1) of
last section, there remains the equation of a sphere.
294. To find the generating lines of a central surface of the
second degree.
Let the equation be
2Q5] SURFACES OF THE SECOND DEGREE. 223
then, if a be the vector of any point on the surface, and ta a vector
parallel to a generating line, we must have
p = a + xty
for all values of the scalar x.
Hence S(a + XTZ} $ (a + mar) = 1
gives the two equations
=
=
The first is the equation of a plane through the origin parallel
to the tangent plane at the extremity of a, the second is the
equation of the asymptotic cone. The generating lines are there
fore parallel to the intersections of these two surfaces, as is well
known.
From these equations we have
2/^DT Fate
where y is a scalar to be determined. Operating on this by S./3
and S . 7, where ft and 7 are any two vectors not coplanar with a,
we have
S<sr (yQft+Vaft) = 0, Sw (y<fa  F 7 a) = (1).
Hence S . <a (y<j>ft + Fa/3) (y<f>v  Fya) = 0,
or my*S . afty SafaS . a/3y = 0.
Thus we have the two values
a / 1
\f m
belonging to the two generating lines. That they may be real
it is clear that m must be negative : i.e. the surface must be the
onesheeted hyperboloid.
295. But by equations (1) we have
z*r = V. (y<j>ft + Fa/3) (y$y  F 7 a)
which, according to the sign of y, gives one or other generating
line.
Here F/3 7 may be any vector whatever, provided it is not
perpendicular to a (a condition assumed in last section), and we
may write for it 0.
Substituting the value of y before found, we have
!_
m
224 QUATERNIONS. [ 2 95
= V. c^aFac^ 1 0J. F6tya,
Y^ Wls
or, as we may evidently write it,
= <t> l (V.a.V<l>oL0) J  ? * ."P0*a ............ (2).
Put r =
and we have
ztx = 6~ l VOLT + A /  . T,
 v
with the condition STC^OL = 0.
[Any one of these sets of values forms the complete solution of the
problem ; but more than one have been given, on account of their
singular nature and the many properties of surfaces of the second
degree which immediately follow from them. It will be excellent
practice for the student to shew that
is an invariant. This may most easily be done by proving that
F. ^0^0, = identically.]
Perhaps, however, it is simpler to write a. for V/3y, and we thus
obtain
ZTZ d>~ 1 V. a Vad)OL + * /  . Facia.
V m
[The reader need hardly be reminded that we are dealing with the
general equation of the central surfaces of the second degree the
centre being origin.]
EXAMPLES TO CHAPTER IX.
1. Find the locus of points on the surface
Spfo =  1
where the generating lines are at right angles to one another.
2. Find the equation of the surface described by a straight
line which revolves about an axis, which it does not meet, but
with which it is rigidly connected.
SURFACES OF THE SECOND DEGREE. 225
3. Find the conditions that
may be a surface of revolution, with axis parallel to a given vector.
4. Find the equations of the right cylinders which circum
scribe a given ellipsoid.
5. Find the equation of the locus of the extremities of per
pendiculars to central plane sections of an ellipsoid, erected at the
centre, their lengths being the principal semiaxes of the sections.
[Fresnel s WaveSurface. See Chap. XII]
6. The cone touching central plane sections of an ellipsoid,
which are of equal area, is asymptotic to a confocal hyperboloid.
7. Find the envelop of all noncentral plane . sections of an
ellipsoid when their area is constant.
8. Find the locus of the intersection of three planes, perpen
dicular to each other, and touching, respectively, each of three
confocal surfaces of the second degree.
9. Find the locus of the foot of the perpendicular from the
centre of an ellipsoid upon the plane passing through the extremi
ties of a set of conjugate diameters.
10. Find the points in an ellipsoid where the inclination
of the normal to the radiusvector is greatest.
11. If four similar and similarly situated surfaces of the
second degree intersect, the planes of intersection of each pair pass
through a common point.
12. If a parallelepiped be inscribed in a central surface of the
second degree its edges are parallel to a system of conjugate
diameters.
13. Shew that there is an infinite number of sets of axes for
which the Cartesian equation of an ellipsoid becomes
14. Find the equation of the surface of the second degree
which circumscribes a given tetrahedron so that the tangent plane
at each angular point is parallel to the opposite face ; and shew
that its centre is the mean point of the tetrahedron.
T.Q.I, 15
226 QUATERNIONS.
15. Two similar and similarly situated surfaces of the second
degree intersect in a plane curve, whose plane is conjugate to the
vector joining their centres.
16. Find the locus of all points on
where the normals meet the normal at a given point.
Also the locus of points on the surface, the normals at which
meet a given line in space.
17. Normals drawn at points situated on a generating line
are parallel to a fixed plane.
18. Find the envelop of the planes of contact of tangent
planes drawn to an ellipsoid from points of a concentric sphere.
Find the locus of the point from which the tangent planes are
drawn if the envelop of the planes of contact is a sphere.
19. The sum of the reciprocals of the squares of the perpen
diculars from the centre upon three conjugate tangent planes is
constant.
20. Cones are drawn, touching an ellipsoid, from any two
points of a similar, similarly situated, and concentric ellipsoid.
Shew that they intersect in two plane curves.
Find the locus of the vertices of the cones when these plane
sections are at right angles to one another.
21. Any two tangent cylinders to an ellipsoid intersect in two
plane ellipses, and no other tangent cylinder can be drawn through
either of these.
Find the locus of these ellipses :
(a) When the axes of the two cylinders are conjugate to each
other, and to a given diameter.
(6) When they are conjugate to each other, and to diameters
lying in one plane.
(c) When they are conjugate to each other, and to any
diameter whatever.
22. If a, j3, 7 be unit vectors parallel to conjugate semi
diameters of an ellipsoid, what is the vector
/ ^ ~ / ~~ .  / ^
v
and what the locus of its extremity ?
SURFACES OF THE SECOND DEGREE. 227
23. Find the locus of the points of contact of tangent planes
which are equidistant from the centre of a surface of the second
degree.
24. From a fixed point A, on the surface of a given sphere,
draw any chord AD ; let D be the second point of intersection of
the sphere with the secant BD drawn from any point B ; and
take a radius vector AE, equal in length to BD , and in direction
either coincident with, or opposite to, the chord AD: the locus
of E is an ellipsoid, whose centre is A, and which passes through
B. (Hamilton, Elements, p. 227.)
25. Shew that the equation
l z (e* l)(e + ) = (Sap)*  ZeSapSa p + (Sap)* + (1  e*) p\
where e is a variable (scalar) parameter, and a, a unit vectors,
represents a, system of confocal surfaces. (Ibid. p. 644.)
26. Shew that the locus of the diameters of
which are parallel to the chords bisected by the tangent planes to
the cone
Sp^p = 0,
is the cone S . <~ 1 <> = 0.
27. Find the equation of a cone, whose vertex is one summit
of a given tetrahedron, and which passes through the circle
circumscribing the opposite side.
28. Shew that the locus of points on the. surface
the normals at which meet that drawn at the point p = r, is on
the cone
= 0.
29. Find the equation of the locus of a point the square
of whose distance from a given line is proportional to its distance
from a given plane.
30. Shew that the locus of the pole of the plane
with respect to the surface
is a sphere, if a be subject to the condition
2 a = C.
152
228 QUATERNIONS.
31. Shew that the equation of the surface generated by lines
drawn through the origin parallel to the normals to
Sp(j) l p =  1
along its lines of intersection with
is ^ 2  hSw + h l vr = 0.
32. Common tangent planes are drawn to
2/Skp%> + (jpfiV)p* = l, and Tp = h,
find the value of h that the lines of contact with the former surface
may be plane curves. What are they, in this case, on the sphere ?
Discuss the case of
p*  S V = 0.
33. If tangent cones be drawn to
from every point of
the envelop of their planes of contact is
34. Tangent cones are drawn from every point of
S(pa)<t>(pcL) = n*,
to the similar and similarly situated surface
fy<fr> = l,
shew that their planes of contact envelop the surface
35. Find the envelop of planes which touch the parabolas
p = a? + fit, p = au* + 7M,
where a, /3, 7 form a rectangular system, and t and u are scalars.
36. Find the equation of the surface on which lie the lines of
contact of tangent cones drawn from a fixed point to a series of
similar, similarly situated, and concentric ellipsoids.
37. Discuss the surfaces whose equations are
SapSj3p = Syp,
and
38. Shew that the locus of the vertices of the right cones
which touch an ellipsoid is a hyperbola.
SURFACES OF THE SECOND DEGREE. 229
39. If a,, a 2 , a 3 be vector conjugate diameters of
where < 3 m$? 4 m^ m = 0,
shew that
m 1 2 m
and 2 (</>a) 2 =  ra 2 .
40. Find the locus of the lines of contact of tangent planes
from a given point to a series of spheres, whose centres are in one
line and which pass through a given point in that line.
41. Find the locus of a circle of variable radius, whose plane
is always parallel to a given plane, and which passes through each
of three given lines in space.
CHAPTEE X.
GEOMETRY OF CURVES AND SURFACES.
296. WE have already seen ( 31 (I)) that the equations
p = $t = 2 . af(t\
and p = (f>(t,u) = 2,. af(t, u),
where a represents one of a set of given vectors, and f a scalar
function of scalars t and u, represent respectively a curve and a
surface. We commence the present too brief Chapter with a few
of the immediate deductions from these forms of expression. We
shall then give a number of examples, with little attempt at
systematic development or even arrangement.
297. What may be denoted by t and u in these equations is,
of course, quite immaterial : but in the case of curves, considered
geometrically, t is most conveniently taken as the length, s, of the
curve, measured from some fixed point. In the Kinematical
investigations of the next Chapter t may, with great convenience,
be employed to denote time.
298. Thus we may write the equation of any curve in
space as
p = (f)S,
where is a vector function of the length, s, of the curve. Of
course it is a linear function when, and only when, the equation
(as in 31 (I)) represents a straight line.
If be a periodic function, such that
the curve is a reentrant one, generally a Knot in space.
301.] GEOMETRY OF CURVES AND SURFACES. 231
[In 306 it is shewn that when s is the arc certain forms of c/>
are not admissible.]
299. We have also seen ( 38, 39) that
dp d , ,,
f = j 6s = 6 s
ds ds^
is a vector of unit length in the direction of the tangent at the
extremity of p.
At the proximate point, denoted by s 4 &s, this unit tangent
vector becomes
</> s + fi sSs + &c.
But, because T<j> s 1,
we have S . fisfi s = 0.
Hence fi s, which is a vector in the osculating plane of the curve,
is also perpendicular to the tangent.
Also, if 80 be the angle between the successive tangents <j) s
and fis + <f>"s&s + ...... , we have
so that the tensor of <j>"s is the reciprocal of the radius of absolute
curvature at the point s.
300. Thus, if dP = <t>s be the vector of any point P of the
curve, and if C be the centre of curvature at P, we have
PC  
fV
and thus 00 = 65777
(/> S
is the equation of the locus of the centre of curvature.
Hence also V. <f> s<j>"s or </> s<j>"s
is a vector perpendicular to the osculating plane ; and therefore
is the tortuosity of the given curve, or the rate of rotation of
its osculating plane per unit of length.
301. As an example of the use of these expressions let us find
the curve whose curvature and tortuosity are both constant.
We have curvature = Tfi s = Tp" = c.
232 QUATERNIONS. [3OI.
Hence fisfi s = p p" = ca,
where a is a unit vector perpendicular to the osculating plane.
This gives
if c l represent the tortuosity.
Integrating we get
p> // = c/ + /3 .......................... (1),
where /3 is a constant vector. Squaring both sides of this equation,
we get
c 2 = c, 2  /3 2  ZcfiPp
(for by operating with 8 . p upon (1) we get f c, = $//),
or
Multiply (1) by /o , remembering that
2V 1,
and we obtain p" = c t + p fi,
or, by integration, p = CjS p/3 + a ..................... (2),
where a is a constant quaternion. Eliminating p , we have
of which the vector part is
p" p& =  erf  Fa/3.
The complete integral of this equation is evidently
p = ? cos. 8 T0 + 7]sm.sT/3~(c lS /3+ Fa/3) ...... (3),
f and ?; being any two constant vectors. We have also by (2),
SPp = CjS + Sa,
which requires that Sftj; = 0, Spy = 0.
The farther test, that Tp f = 1, gives us
2 .
C ~T~ C.
This requires, of course,
= 0, Tt=Tr, = ,
so that (3) becomes the general equation of a helix traced on a
right cylinder. (Compare 31 (m).)
303.] GEOMETRY OF CURVES AND SURFACES. 233
302. The vector perpendicular from the origin on the tangent
to the curve
p = $8
is, of course, Vpp, or p Vpp
(since p is a unit vector).
To find a common property of curves whose tangents are all
equidistant from the origin.
Here
which may be written p 2 S*pp = c 2 ..................... (i).
This equation shews that, as is otherwise evident, every curve
on a sphere whose centre is the origin satisfies the condition. For
obviously
p z = c 2 gives Spp = 0,
and these satisfy (1).
If Spp does not vanish, the integral of (1) is
c = s ........................... (2),
an arbitrary constant not being necessary, as we may measure s
from any point of the curve. The equation of an involute which
commences at this assumed point is
or = p sp.
This gives 2V = Tp* + s* + ZsSpp
= Tp* + s 2  2 5 jTp*  c 2 , by (1),
= c 2 , by (2).
This includes all curves whose involutes lie on a sphere about the
origin.
303. Find the locus of the foot of the perpendicular drawn to a
tangent to a right helix from a point in the axis.
The equation of the helix is
o o
p = a cos  + j3 sin  + ys,
a a
where the vectors a, ft, 7 are at right angles to each other, and
Ta=T@ = b, while
(The latter condition is from Tp = 1.)
234 QUATERNIONS. [34
The equation of the required locus is, by last section,
f s a*b 2 s\ O f . s d*b* s\ b*
= a ( cos  H  s sm  + p sm  3 scos  + y 2 s.
\ a a aj \ a a 3 aj a
This curve lies on the hyperboloid whose equation is
as the reader may easily prove for himself.
304. To find the least distance between consecutive tangents to a
tortuous curve.
Let one tangent be or = p + xp ,
then a consecutive one, at a distance s along the curve, is
The magnitude of the least distance between these lines is, by
216, 223,
TVp p"$s
if we neglect terms of higher orders.
It may be written, since p p" is a vector, and Tp = 1,
But ( 140 (2)) j^fr t = V^ffT & = ^ p S . p p p".
Hence ^Ls.Up"Vp p"
is the small angle, S<, between the two successive positions of the
osculating plane. [See also 300.]
Thus the shortest distance between two consecutive tangents is
expressed by the formula
12r
where r, = m , is the radius of absolute curvature of the tortuous
curve.
306.] GEOMETRY OF CURVES AND SURFACES. 235
305. Let us recur for a moment to the equation of the parabola
Here p = (
whence, if we assume $a/3 = 0,
from which the length of the arc of the curve can be derived in
terms of t by integration.
Again, p" = (a + ft
w/o
<M = d L J_ ,dt8 P(+AQ
ds 2 ds T~(a~+~ftt) ds T(a + fttf
Hence p" =
and therefore, for the vector of the centre of curvature we have
( 300),
OT = a t + ^  (a 8 + fftj ( /3<x
= / Y + "
which is the quaternion equation of the evolute.
306. One of the simplest forms of the equation of a tortuous
curve is
where a, ft, 7 are any three noncoplanar vectors, and the numerical
factors are introduced for convenience. This curve lies on a para
bolic cylinder whose generating lines are parallel to 7; and also on
cylinders whose bases are a cubical and a semicubical parabola,
their generating lines being parallel to ft and a respectively. We
have by the equation of the curve
Pa + ftt += ) y,
2 J ds
from which, by Tp = 1, the length of the curve can be found in
terms of t ; and
236 QUATERNIONS. [307.
from which p" can be expressed in terms of s. The investigation
of various properties of this curve is very easy, and will be of great
use to the student.
[Note. It is to be observed that in this equation t cannot stand
for s, the length of the curve. It is a good exercise for the student
to shew that such an equation as
p = as + 0s 2 + 75 3 ,
or even the simpler form
p = as + /3s 2 ,
involves an absurdity.]
307. The equation p = </>*e,
where (/> is a given selfconjugate linear and vector function, t a
scalar variable, and e an arbitrary vector constant, belongs to a
curious class of curves.
We have at once = ( fi ^# ^ >
where log </> is another selfconjugate linear and vector function,
which we may denote by ^. These functions are obviously com
mutative, as they have the same principal set of rectangular vectors,
hence we may write
dp
which of course gives 7^ = ^ 2 p, &c.,
dt
since ^ does not involve t.
As a verification, we should have
where e is the base of Napier s Logarithms.
This is obviously true if ^ = e stx ,
or </> = e*,
or log < = x,
which is our assumption. See 337, below,
[The above process is, at first sight, rather startling, but the
310.] GEOMETRY OF CURVES AND SURFACES. 237
student may easily verify it by writing, in accordance with the
results of Chapter V,
<e =  g.aSae  g^S/Be  gjySyc,
whence <f> f = g\ a.Scx.6 g\/3S/3e g\ySye.
He will find at once
X =  log g l aSae  log gftSpe
and the results just given follow immediately.]
308. That the equation
represents a surface is obvious from the fact that it becomes the
equation of a definite curve whenever either t or u has a particular
value assigned to it. Hence the equation at once furnishes us with
two systems of curves, lying wholly on the surface, and such that
one of each system can, in general, be drawn through any assigned
point on the surface. Tangents drawn to these curves a,t their
point of intersection must, of course, lie in the tangent plane, whose
equation we have thus the means of forming. [Of course, there
may occasionally be cases of indeterminateness, as when the curves
happen to touch one another. But the general consideration of
singular points on surfaces is beyond the scope of this work.]
309. By the equation we have
where the brackets are inserted to indicate, partial differential
coefficients. If we write this as
the normal to the tangent plane is evidently
F
and the equation of that plane
310. Thus, as a simple example, let
p = ta + u(3 + tuy.
This surface is evidently to be constructed by drawing through each
point to., of the line a, a line parallel to /3 + ty ; or through u/3, a
line parallel to a + uy.
238 QUATERNIONS. [3 11 
We may easily eliminate t and u, and obtain
8 . @ypS .jap = S. a/3yS . a/3p ;
and the methods of last chapter enable us to recognise a hyperbolic
paraboloid.
Again, suppose a straight line to move along a fixed straight
line, remaining always perpendicular to it, while rotating about it
through an angle proportional to the space it has advanced ; the
equation of the ruled surface described will evidently be
p = at + u ((3 cos t + 7 sin t) .................. (1 ) ,
where a, /3, 7 are rectangular vectors, and
frj r\ _ fjj
This surface evidently intersects the right cylinder
p = a (j3 cos t + 7 sin t) + VOL,
in a helix ( 31 (m), 301) whose equation is
p = at + a(/3 cos t + 7 sin t).
These equations illustrate very well the remarks made in 31 (I),
308, as to the curves or surfaces represented by a vector equation
according as it contains one or two scalar variables.
From (1) we have
dp = [a u (j3 sin t 7 cos t)] dt + (/3 cos t + 7 sin t) du,
so that the normal at the extremity of p is
Ta (7 cost/3 sin t)u T/3*Ua.
Hence, as we proceed along a generating line of the surface, for
which t is constant, we see that the direction of the normal changes.
This, of course, proves that the surface is not developable.
311. Hence the criterion for a developable surface is that if it
be expressed by an equation of the form
where </> and tyt are vector functions, we must have the direction
of the normal
independent of u.
This requires either Vtytty t 0,
which would reduce the surface to a cylinder, all the generating
lines being parallel to each other ; or
312.] GEOMETRY OF CURVES AND SURFACES. 239
This is the criterion we seek, and it shews that we may write, for a
developable surface in general, the equation
p = <l>t+ ufit (1).
Evidently p = <f)t
is a curve (generally tortuous) and <fit is a tangent vector. Hence
a developable surface is the locus of all tangent lines to a tortuous
curve.
Of course the tangent plane to the surface is the osculating
plane at the corresponding point of the curve; and this is indicated
by the fact that the normal to (1) is parallel to
Vtftf t. (See 300.)
To find the form of the section of the surface made by a normal
plane through a point in the curve.
The equation of the surface in the neighbourhood of the
extremity of p is approximately
s 2
07 = p + Sp +  p" + &C. + X (p + Sp" + &C.).
The part of *& p which is parallel to p is
therefore OT  p = Ap + (  + xs ) p"  ( S  + ~ ) pVp p " + ... .
\z / \o z /
And, when ^4 = 0, i.e. in the normal section, we have approximately
x = s,
s z s 3
so that OT p =  p" + ^ p Vp p".
Hence the curve has an equation of the form
a = S 2 a + s 3 /3,
a semicubical parabola.
312. A Geodetic line is a curve drawn on a surface so that its
osculating plane at any point contains the normal to the surface.
Hence, if v be the normal at the extremity of p, p and p the first
and second differentials of the vector of the geodetic,
S . vp p" = 0,
which may be easily transformed into
V.vdU P =0.
240 QUATERNIONS.
313. In the sphere Tp = avte have
HI*
hence S . pp p" = 0,
which shews of course that p is confined to a plane passing through
the origin, the centre of the sphere.
For a formal proof, we may proceed as follows
The above equation is equivalent to the three
from which we see at once that 6 is a constant vector, and
therefore the first expression, which includes the others, is the
complete integral.
Or we may proceed thus
Q =  P S. pp p" + p"S . p*p = F. Vpp Vpp" = V. Vpp d Vpp,
whence by 140 (2) we have at once
UVpp = const. = suppose,
which gives the same results as before.
314. In any cone, when the vertex is taken as origin, we
have, of course,
since p lies in the tangent plane. But we have also
Hence, by the general equation of 312, eliminating v we get
= 8. pp Vp p" = SpdUp by 140 (2).
Integrating C = 8pUp  j SdpUp = SpUp + j Tdp.
The interpretation of this is, that the length of any arc of the
geodetic is equal to the projection of the side of the cone (drawn
to its extremity) upon the tangent to the geodetic. In other
words, when the cone is developed on a plane the geodetic becomes a
straight line. A similar result may easily be obtained for the
geodetic lines on any developable surface whatever.
315. To find the shortest line connecting two points on a given
surface.
Here I Tdp is to be a minimum, subject to the condition that
dp lies in the given surface. [We employ 8, though (in the
3 I 7] GEOMETRY OF CURVES AND SURFACES. 241
notation we employ) it would naturally denote a vector, as the
symbol of variation.]
Now SJTdp=f STdp =  j ^ p = ~JS. UdpdSp
where the term in brackets vanishes at the limits, as the extreme
points are fixed, and therefore Sp = at each.
Hence our only conditions are
I
S.Spd Udp = 0, and SvSp = 0, giving
V.vdUdp = 0, as in 31 2.
If the extremities of the curve are not given, but are to lie on
given curves, we must refer to the integrated portion of the
expression for the variation of the length of the arc. And its
form
S.UdpSp
shews that the shortest line cuts each of the given curves at right
angles.
316. The osculating plane of the curve
is 8 . $t$"t O  p) = ...................... (1),
and is, of course, the tangent plane to the surface
p = ft + u<j> t ........................... (2).
Let us attempt the converse of the process we have, so far,
pursued, and endeavour to find (2) as the envelop of the variable
plane (1).
Differentiating (1) with respect to t only, we have
8. < </> " (rp) = 0.
By this equation, combined with (1), we have
or 5T = p + uty = cj) + u$ ,
which is equation (2).
317. This leads us to the consideration of envelops generally,
and the process just employed may easily be extended to the
problem of finding the envelop of a series of surfaces whose
equation contains one scalar parameter.
T. Q. I. 16
242 QUATERNIONS. [318.
When the given equation is a scalar one, the process of finding
the envelop is precisely the same as that employed in ordinary
Cartesian geometry, though the work is often shorter and simpler.
If the equation be given in the form
p = Tr, U, v,
where ^r is a vector function, t and u the scalar variables for any
one surface, v the scalar parameter, we have for a proximate surface
u l9 v = p r t v u u v v.
Hence at all points on the intersection of two successive surfaces
of the series we have
y t St + i/r M Su + Tjr . Sv = 0,
which is equivalent to the following scalar equation connecting the
quantities t, u and v ;
s. * ,* * .=<>.
This equation, along with
p=ty(t t u, v),
enables us to eliminate t, u, v, and the resulting scalar equation is
that of the required envelop.
318. As an example, let us find the envelop of the osculating
plane of a tortuous curve. Here the equation of the plane is
( 316)
S.(np) $t$ t = 0,
or OT = <f)t + x(f) t + y<j>"t = ty (cc, y, t),
if p = fa
be the equation of the curve.
Our condition is, by last section,
fl f.fft .O,
or S . fit fi t [fit + xfi t + yfi"t] = 0,
or yS.fitfi tfi"t = 0.
Now the second factor cannot vanish, unless the given curve
be plane, so that we must have
2/ = 0,
and the envelop is r = fit + xfit
the developable surface, of which the given curve is the edge of
regression, as in S 316.
32O.] GEOMETRY OF CURVES AND SURFACES. 243
319. When the equation contains two scalar parameters, its
differential coefficients with respect to them must vanish, and we
have thus three equations from which to eliminate two numerical
quantities.
A very common form in which these two scalar parameters
appear in quaternions is that of an arbitrary unitvector. In this
case the problem may be thus stated :
Find the envelop of the surface whose scalar equation is
where a is subject to the one condition
Ta=l.
Differentiating with respect to a alone, we have
Svda = 0, Sada = 0,
where v is a known vector function of p and a. Since da may have
any of an infinite number of values, these equations shew that
VOLV = 0.
This is equivalent to two scalar conditions only, and these, in
addition to the two given scalar equations, enable us to eliminate a.
With the brief explanation we have given, and the examples
which follow, the student will easily see how to treat any other set
of data he may meet with in a question of envelops.
320. Find the envelop of a plane whose distance from the
origin is constant.
Here Sap =  c,
with the condition Ta = 1.
Hence, by last section, Vpa = 0,
and therefore p = ca,
or Tp = c,
the sphere of radius c, as was to be expected.
If we seek the envelop of those only of the planes which are
parallel to a given vector /3, we have the additional relation
SOL& = 0.
In this case the three differentiated equations are
Spda=Q, Sada = 0, S{3da = 0,
and they give S. afip = 0.
162
244 QUATERNIONS. [3 2 I .
Hence a=U./3V/3p,
and the envelop is TV/3p = cT0,
the circular cylinder of radius c and axis coinciding with /3.
By putting Saj3 = e, where e is a constant different from zero,
we pick out all the planes of the series which have a definite
inclination to /3, and of course get as their envelop a right cone.
321. The equation S"ap + 2S.a/3p = b
represents a parabolic cylinder, whose generating lines are parallel
to the vector aVa(3. For the equation is of the second degree, and
is riot altered by increasing p by the vector ocaVaff , also the
surface cuts planes perpendicular to a in one line, and planes
perpendicular to Va/3 in two parallel lines. Its form and position
of course depend upon the values of a, /3, and b. It is required to
find its envelop if /3 and b be constant, and a be subject to the one
scalar condition
r = i.
The process of 319 gives, by inspection,
pSap f Vftp = xa.
Operating by 8 . a, we get
which gives S . a{3p = x + b.
But, by operating successively by S . F/3p and by 8 . p, we have
and (p 2 x) Sap = 0.
Omitting, for the present, the factor Sap, these three equations
give, by elimination of x and a,
which is the equation of the envelop required.
This is evidently a surface of revolution of the fourth degree
whose axis is ft ; but, to get a clearer idea of its nature, put
c p 1 = *r,
and the equation becomes
which is obviously a surface of revolution of the second degree,
referred to its centre. Hence the required envelop is the reciprocal
of such a surface, in the sense that the rectangle under the lengths
323.] GEOMETRY OF CURVES AND SURFACES. 245
of condirectional radii of the two is constant : i.e. it is the Electric
Image.
We have a curious particular case if the constants are so
related that
b + /3 2 = 0,
for then the envelop breaks up into the two equal spheres,
touching each other at the origin,
P 2 =S/3 P ,
while the corresponding surface of the second order becomes the
two parallel planes
Sfa = c 2 .
322. The particular solution above met with, viz.
Sap = 0,
limits the original problem, which now becomes one of finding the
envelop of a line instead of a surface. In fact this equation, taken
in conjunction with that of the parabolic cylinder, belongs to that
generating line of the cylinder which is the locus of the vertices
of the principal parabolic sections.
Our equations become
2S.a/3p = b,
Sap = 0,
Ta = l
whence V/3p = xa ;
giving x =  S . a(3p =  ^ ,
and thence TV/3p = ^ ;
so that the envelop is a circular cylinder whose axis is {3. [It is
to be remarked that the equations above require that
Sap = 0,
so that the problem now solved is merely that of the envelop of a
parabolic cylinder which rotates about its focal line. This discussion
has been entered into merely for the sake of explaining a peculiarity
in a former result, because of course the present results can be
obtained immediately by an exceedingly simple process.]
323. The equation SapS . a/3p = a 2 ,
with the condition Toe 1,
246 QUATERNIONS. [324.
represents a series of hyperbolic cylinders. It is required to find
their envelop.
As before, we have
pS . aftp + VjSpSap = xa,
which by operating by $ . a, 8 . p, and S .VjBp, gives
W = x y
p*S . afip = xSap,
(Vpp)*Sap = xS.oLpp.
Eliminating a and x we have, as the equation of the envelop,
p Wpftof.
Comparing this with the equations
f = ~ 2a 2 ,
and (V/3p) 2 = 2a\
which represent a sphere and one of its circumscribing cylinders,
we see that, if condirectional radii of the three surfaces be drawn
from the origin, that of the new surface is a geometric mean
between those of the two others.
324. Find the envelop of all spheres which touch one given line
and have their centres in another.
Let p = {3 + yry
be the line touched by all the spheres, and let XOL be the vector of
the centre of any one of them, the equation is (by 213, or 214)
" . 7 .
or, putting for simplicity, but without loss of generality,
Ty = 1, Sa/3 = 0, Spy = 0,
so that @ is the least vector distance between the given lines,
(p  xa? = (/3 xaf + x 2 S 2 ay,
and, finally, p 2  /3 2  ZxSap = x 2 S 2 aj.
Hence, by 317,  2Sap = 2xS 2 ay.
[This gives no definite envelop, except the point p = /3, if
Say = 0,
i.e. if the line of centres is perpendicular to the line touched by
all the spheres.]
Eliminating x, we have for the equation of the envelop
S*ap + S 9 *v (p 2  *)  0,
326.] GEOMETRY OF CURVES AND SURFACES. 247
which denotes a surface of revolution of the second degree, whose
axis is a.
Since, from the form of the equation, Tp may have any
magnitude not less than T/3, and since the section by the plane
Sap =
is a real circle, on the sphere
P P = 0,
the surface is a hyperboloid of one sheet.
[It will be instructive to the student to find the signs of the
values of g v g v g 3 as in 177, and thence to prove the above
conclusion.]
325. As a final example of this kind let us find the envelop
of the hyperbolic cylinder
SapS/3pc = 0,
where the vectors a and (B are subject to the conditions
Tct = T{3 = l,
Say = 0, S/3S = 0,
y and 8 being given vectors.
[It will be easily seen that two of the six scalars involved
in a, still remain as variable parameters.]
We have Sada = 0, Syda = 0,
so that da = x Vay.
Similarly d/3 = yV/38.
But, by the equation of the cylinders,
SapSpd/3 + SpdaSj3p = 0,
or ySapS . /38p + asS . aypS/3p = 0.
Now by the nature of the given equation, neither Sap nor Sftp can
vanish, so that the independence of da and dj3 requires
Hence a=U.yVyp, = U.S VSp,
and the envelop is T.VypVSp  cTy& = 0,
a surface of the fourth degree, which may be constructed by laying
off mean proportionals between the lengths of condirectional radii
of two equal right cylinders whose axes meet in the origin.
326. We may now easily see the truth of the following
general statement.
248 QUATERNIONS. [3 2 6.
Suppose the given equation of the series of surfaces, \vhose
envelop is required, to contain m vector, and n scalar, parameters ;
and that these are subject to p vector, and q scalar, conditions.
In all there are 3m + n scalar parameters, subject to 3p + q
scalar conditions.
That there may be an envelop we must therefore in general
have
l y or = 2.
In the former case the enveloping surface is given as the locus of
a series of curves, in the latter of a series of points.
Differentiation of the equations gives us 3p + q + 1 equations,
linear arid homogeneous in the 3m + n differentials of the scalar
parameters, so that by the elimination of these we have one final
scalar equation in the first case, two in the second ; and thus in
each case we have just equations enough to eliminate all the
arbitrary parameters.
Sometimes a very simple consideration renders laborious cal
culation unnecessary. Thus a rectangular system turns about the
centre of an ellipsoid. Find the envelop of the plane which passes
through the three points of intersection.
If a, /3, 7 be the rectangular unitsystem, the points of
intersection with
Spfo =  1
are at the extremities of
 SOLQOL J &&<!># Syfa
And if these lie in the plane
*/.! . ................. (1),
we must have e = Sa\/ ScL<f>a. ........................ (2).
It would be troublesome to work out the envelop of (1), with
(2) and the conditions of a rectangular unitsystem as the data,
but we may proceed as follows.
The length of the perpendicular from the centre on the
plane (1) is
 (Sa^a + S/3(f>l3 +
a constant, by 185. Hence the envelop is a sphere of which this
is the radius ; and it has the same property for all the ellipsoids
328.] GEOMETRY OF CURVES AND SURFACES. 249
which, having their axes in the same lines as the first, intersect it
at the point on
i+j + k.
We may obtain the result in another way. By 281 the sum
of the reciprocals of the squares of three rectangular central vectors
of an ellipsoid is constant ; while it is easily shewn (see Ex. 20 to
Chap. VII.) that the same sum with regard to a plane is the
reciprocal of the square of its distance from the origin.
327. To find the locus of the foot of the perpendicular drawn
from the origin to a tangent plane to any surface.
If Svdp =
be the differentiated equation of the surface, the equation of the
tangent plane is
S(<SFp)v = 0.
We may introduce the condition
Svp =  1,
which in general alters the tensor of v, so that v~ l becomes the
required vector perpendicular, as it satisfies the equation
Svrv =  1.
It remains that we eliminate p between the equation of the
given surface, and the vector equation
37 = V~ l .
The result is the scalar equation (in BT) required.
For example, if the given surface be the ellipsoid
we have trr* 1 = v =
so that the required equation is
Or
which is Fresnel s Surface of Elasticity. ( 278.)
It is well to remark that this equation is derived from that of
the reciprocal ellipsoid
Sp^p = l
by putting iar~ l for p.
328. To find the reciprocal of a given surface with respect to
the unit sphere whose centre is the origin.
With the condition Spv 1,
250 QUATERNIONS. [S 2 9
of last section, we see that v is the vector of the pole of the
tangent plane
S(vrp)v = Q.
Hence we must put BT = v,
and eliminate p by the help of the equation of the given surface.
Take the ellipsoid of last section, and we have
VT = <f)p,
so that the reciprocal surface is represented by
8vp*v = l.
It is obvious that the former ellipsoid can be produced from
this by a second application of the process.
And the property is general, for
gives, by differentiation, and attention to the condition
Svdp =
the new relation Spdv = 0,
so that p and v are corresponding vectors of the two surfaces :
either being that of the pole of a tangent plane drawn at the
extremity of the other.
329. If the given surface be a cone with its vertex at the
origin, we have a peculiar case. For here every tangent plane
passes through the origin, and therefore the required locus is
wholly at an infinite distance. The difficulty consists in Spv
becoming in this case a numerical multiple of the quantity which
is equated to zero in the equation of the cone, so that of course
we cannot put as above
330. The properties of the normal vector v enable us to write
the partial differential equations of families of surfaces in a very
simple form.
Thus the distinguishing property of Cylinders is that all their
generating lines are parallel. Hence all positions of v must be
parallel to a given plane or
Sav = 0,
which is the quaternion form of the wellknown equation
. dF dF dF
I = f ra j t n , 0.
dx ay dz
332.] GEOMETRY OF CURVES AND SURFACES. 251
To integrate it, remember that we have always
Svdp = 0,
and that as v is perpendicular to a it may be expressed in terms
of any two vectors, fi and 7, each perpendicular to a.
Hence v = x/3 + yy,
and xS{3dp + ySydp = 0.
This shews that S0p and Syp are together constant or together
variable, so that
S0p=f(Syp),
where /is any scalar function whatever.
331. In Surfaces of Revolution the normal intersects the axis.
Hence, taking the origin in the axis a, we have
S.apv = 0,
or v = xa + yp.
Hence xSadp + ySpdp = 0,
whence the integral Tp =f(Sap).
The more common form, which is easily derived from that just
written, is
In Cones we have Svp = 0,
and therefore
Svdp = S.v(TpdUp+ UpdTp) = TpSvd Up.
Hence SvdUp = 0,
so that v must be a function of Up, and therefore the integral is
which simply expresses the fact that the equation does not involve
the tensor of p, i.e. that in Cartesian coordinates it is homogeneous.
332. If equal lengths be laid off on the normals drawn to any
surface, the new surface formed by their extremities is normal to the
same lines.
For we have BT = ^ + a Uv,
and Svdv = Svdp + aSvd Uv = 0,
which proves the proposition.
Take, for example, the surface
S P<I>P =  1 5
252 QUATERNIONS. [333
the above equation becomes
ad>p
= f >+ >
so that a = I =r +1 OT
i
and the equation of the new surface is to be found by eliminating
mT (written x) between the equations
and
333. It appears from last section that if one orthogonal
surface can be drawn cutting a given system of straight lines, an
indefinitely great number may be drawn : and that the portions of
these lines intercepted between any two selected surfaces of the
series are all equal.
Let p = a f XT,
where a and r are vector functions of p, and x is any scalar, be
the general equation of a system of lines : we have
Srdp = = S(po)dp
as the differentiated equation of the series of orthogonal surfaces,
if it exist. Hence the following problem.
334. It is required to find the criterion of integrability of the
equation
Svdp = .............................. (1)
as the complete differential of the equation of a series of surfaces.
Hamilton has given (Elements, p. 702) an extremely elegant
solution of this problem, by means of the properties of linear and
vector functions. We adopt a different and somewhat less rapid
process, on account of some results it offers which will be useful to
us later ; and also because it will shew the student the connection
of our methods with those of ordinary differential equations.
If we assume Fp = C
to be the integral, we have by 144,
8dpVF=0.
Comparing with the given equation, (1), we see that the latter
represents a series of surfaces if v, or a scalar multiple of it, can be
expressed as VF.
335] GEOMETRY OF CURVES AND SURFACES. 253
If v = VF,
_ _ fd*F d*F d*F\
we have Vz; = V 2 F =  ^~ 2 4 , + jy ,
V dx dy dz* /
and the lastwritten quantities are necessarily scalars, so that the
only requisite condition of the integrability of (1) is
FVi/ = 0....: ......................... (2).
If v do not satisfy this criterion, it may when multiplied by a
scalar. Hence the farther condition
7V (wv) = 0,
which may be written
VrfwiuWv = b ...................... (3).
This requires that
If then (2) be not satisfied, we must try (4). If (4) be satisfied w
will be found from (3) ; and in either case (1) is at once integrable.
[If we put dv = (frdp,
where is a linear and vector function, not necessarily self
conjugate, we have
by 185. Thus, if (/> be selfconjugate, e = 0, and the criterion (2)
is satisfied. If <f> be not selfconjugate we have by (4) for the
criterion
Sev = 0.
These results accord with Hamilton s, lately referred to, but the
mode of obtaining them is quite different from his.]
335. As a simple example let us first take lines diverging
from a point. Here v  p, and we see that if v = p
Vy = 3,
so that (2) is satisfied. And the equation is
Spdp = 0,
whose integral Tp 2 = C
gives a series of concentric spheres.
Lines perpendicular to, and intersecting, a fixed line.
If a be the fixed line, (3 any of the others, we have
0, S/3dp = 0.
254 QUATERNIONS. [335.
Here v  a. Vap,
and therefore equal to it, because (2) is satisfied.
Hence S . dp a. Vap = 0,
or S. VapVadp = 0,
whose integral is the equation of a series of right cylinders
To find the orthogonal trajectories of a series of circles whose
centres are in, and their planes perpendicular to, a given line.
Let a be a unit vector in the direction of the line, then one of
the circles has the equations
Tp = C]
Sap = C \
where C and C are any constant scalars whatever.
Hence, for the required surfaces
v 1 d lP 1 Vap,
where d^ is an element of one of the circles, v the normal to the
orthogonal surface. Now let dp be an element of a tangent to
the orthogonal surface, and we have
Svdp = S . apdp = 0.
This shews that dp is in the same plane as a and p, i.e. that the
orthogonal surfaces are planes passing through the common ax
[To integrate the equation
S.apdp = Q
evidently requires, by 334, the introduction of a factor. For
=  aSVp + SaV . p ( 90, (1))
so that the first criterion is not satisfied. But
8. VapV.VVap=2S.aVcLp = 0,
so that the second criterion holds. It gives, by (3) of 334,
 V . VapVw + 2wa = 0,
or pSaVw aSpVw + 2wa = 0.
That is ^ , .
= 2w
These equations are satisfied by
1
CLp
33^.] GEOMETRY OF CURVES AND SURFACES. 255
But a simpler mode of integration is easily seen. Our equation
may be written
. ,
P Up /3
which is immediately integrable, (3 being an arbitrary but constant
vector.
As we have not introduced into this work the logarithms of
versors, nor the corresponding angles of quaternions, we must refer
to Hamilton s Elements for a further development of this point.]
336. As another example, let us find series of surfaces wJ tick,
together, divide space into cubes.
If p be the vector of one series which has the required property,
a that of a second, it is clear that (u being a scalar)
da = uq~ l dpq ........................... (1),
where u and q are functions of p. For, to values of dp belonging
to edges of one cube correspond values of dcr belonging to edges of
another. Operate by S . a, where a is any constant vector, then
Sada = iiS . qaq l dp.
As the lefthand member is a complete differential, we have by
334
FV (uqaq~ l ) = 0.
This is easily put in the form {Chap. IV., Ex. (4), and 140 (8)}
VM
V. qaq 1 =  2qaq~ l S . Vqq~ l + 2S (qaq~ l V) q.q 1 .. .(2).
u
Multiply by qaq~ l , and add together three equations of the resulting
form, in which the values of a. form a rectangular unit system.
Then
V?/
+ 2  = + QS .Vq q 1  2V q q 1 .
u
This shews that
Take account of this result in (2), and put dp for qaq~ l , which
may be any vector. Thus
u
From this we see at once that
q = a Up,
256 QUATERNIONS. [33^
where a is any constant versor. Then (3) gives
Vu__2Up du_ ZSpdp
V = "Yp 1T = ^y~
C
so that u = ffj z .
1 P
Thus, from (1), we have
a = a  a~ l + 8.
P
This gives the Electric Image transformation, with any subsequent
rotation, followed (or, as is easily seen, preceded) by a translation.
Hence the only series of surfaces which satisfy the question, are
mutually perpendicular planes ; and their images, which are series
of spheres, passing through a common point and having their
centres on three rectangular lines passing through that point.
[For another mode of solution see Proc. R. 8. E., Dec. 1877.]
In some respects analogous to this is the celebrated physical
problem of finding series of Orthogonal Isothermal Surfaces. We
give a slight sketch of it here.
If three such series of surfaces be denoted by their temperatures
thus :
F^T,, F,= T t , F S = T S ,
the conditions of orthogonality are fully expressed by putting for
the respective values of the flux of heat in each series
where a, /3, 7 form a constant rectangular unit vector system.
But the isothermal conditions are simply
v*F t = v*F 9 = v*F 3 = o.
Hence we have three simultaneous equations, of which the
first is
= ........................... (a).
[In the previous problem a might be any vector whatever, the
values of u were equal, and the vector part only of the lefthand
member was equated to zero. These conditions led to an unique
form of solution. Nothing of the kind is to be expected here.]
From equation (a) we have at once
fi
where v l has been put for logu v and a for qaq~\
337] GEOMETRY OF CURVES AND SURFACES. 257
From the group of six equations, of which (6) gives two, we
have
with three of the type
S.* S(a V)qq* = 0.
We also obtain without difficulty
2Vv + 4tfqq l = 0,
which may be put in the form
V.(u l uji s )*q = 0.
But when we attempt to find the value of dq we are led to
expressions such as
27. a S (a dp) Vw, + Zdqq 1 = 0,
which, in consequence of the three different values of v, are
comparatively unmanageable. (See Ex. 24 at end of Chapter.)
They become, however, comparatively simple when one of the
three families is assumed. The student will find it useful to work
out the problem when F l represents a series of parallel planes, so
that the others are cylinders; or when ^represents planes passing
through a line, the others being surfaces of revolution ; &c.
337. To find the orthogonal trajectories of a given series of
surfaces.
If the equation Fp = C
give Svdp = 0,
the equation of the orthogonal curves is
This is equivalent to two scalar differential equations ( 210),
which, when the problem is possible, belong to surfaces on each of
which the required lines lie. The finding of the requisite criterion
we leave to the student. [He has only to operate on the last
written equation by 8. a, where a is any constant vector ; and,
bearing this italicized word in mind, proceed as in 334.]
Let the surfaces be concentric spheres.
Here p* = C,
and therefore Vpdp = 0.
Hence Tp*d Up = Up Vpdp = 0,
and the integral is Up = constant,
denoting straight lines through the origin.
T. Q. I. 17
258 QUATERNIONS. [338.
Let the surfaces be spheres touching each other at a common
point. The equation is ( 235)
whence V. pctpdp = 0.
The integrals may be written
the first (/3 being any vector) is a plane through the common
diameter ; the second represents a series of rings or tores ( 340)
formed by the revolution, about a, of circles touching that line
at the point common to the spheres.
Let the surfaces be similar, similarly situated, and concentric,
surfaces of the second degree.
Here &PXP ~ @,
therefore V^pdp 0.
But, by  307, the integral of this equation is
where and % are related to each other, as in 307 ; and e is any
constant vector.
338. To integrate the linear partial differential equation of a
family of surfaces.
The equation (see 330)
P du + du + R du = Q
dx dy dz
may be put in the very simple form
8(aV)u = Q ........................... (1),
if we write a = iP +jQ + kR,
. d . d , d
and V =i = + i J +K j .
dx J dy dz
[From this we see that the meaning of the differential
equation is that at every point of the surface
u = const.
the corresponding vector, <r, is a tangent line. Thus we have a
suggestion of the ordinary method of solving such equations.]
(1) gives, at once, Vu
338] GEOMETRY OF CURVES AND SURFACES. 259
where m is a scalar and 6 a vector (in whose tensor m might have
been included, but is kept separate for a special purpose). Hence
= mS . Ocrdp
if we put dr = m Vadp
so that m is an integrating factor of V . crdp. If a value of m can
be found, it is obvious, from the form of the above equation, that
6 must be a function of r alone ; and the integral is therefore
u = F (r) = const.
where F is an arbitrary scalar function.
Thus the differential equation of Cylinders is
S(aV)u = 0,
where a is a constant vector. Here m= 1, and
u = F( Vap) const.
That of Cones referred to the vertex is
S(pV)u=0.
Here the expression to be made integrable is
Vpdp.
But Hamilton long ago shewed that ( 140 (2))
d Up _ ydp _ Vpdp
~~
which indicates the value of m, and gives
u = F( Up) = const.
It is obvious that the above is only one of a great number of
different processes which may be applied to integrate the differen
tial equation. It is quite easy, for instance, to pass from it to the
assumption of a vector integrating factor instead of the scalar m,
and to derive the usual criterion of integrability. There is no
difficulty in modifying the process to suit the case when the right
hand member is a multiple of u. In fact it seems to throw a very
clear light upon the whole subject of the integration of partial
differential equations. If, instead of 8 (o"V), we employ other
operators as 8 (crV) 8 (TV), 8 . crVrV, &c. (where V may or may not
operate on u alone), we can pass to linear partial differential
172
260 QUATERNIONS. [339
equations of the second and higher orders. Similar theorems can
be obtained from vector operators, as F(crV)*.
339. Find the general equation of surf aces described by a line
which always meets, at right angles, a fixed line.
If a be the fixed line, /3 and 7 forming with it a rectangular
unit system, then
where y may have all values, but x and z are mutually dependent,
is one form of the equation.
Another, expressing the arbitrary relation between x and z, is
But we may also write
as it obviously expresses the same conditions.
The simplest case is when F (x) = hx. The surface is one which
cuts, in a right helix, every cylinder which has a. for its axis.
340. The centre of a sphere moves in a given circle, find the
equation of the ring described.
Let a be the unitvector axis of the circle, its centre the origin,
r its radius, a that of the sphere.
Then (pfi)* = a z
is the equation of the sphere in any position, where
These give ( 326) S . a/3p = 0, and /3 must now be eliminated.
The immediate result is that
giving (p 2 r* + a 2 ) a = 4r 2 T 2 Vap,
which is the required equation. It may easily be changed to
(p*a* + f)* = 4ia*p*4?r*S*ap ............... (1),
and in this form it enables us to give a very simple proof of the
singular property of the ring (or tore) discovered by Villarceau.
a/3
(a \
a i 2 2 j = 0,
Tait, Proc. E. S. E., 186970.
341 J GEOMETRY OF CURVES AND SURFACES. 261
which together are represented by
evidently pass through the origin and touch (and cut) the ring.
The latter equation may be written
r*S 2 ap  a* (tfap + S 2 p U0) = 0,
or r*S oLp + a*(p* + S .oipUp) = ................ (2).
The plane intersections of (1) and (2) lie obviously on the new
surface
which consists of two spheres of radius r, as we see by writing its
separate factors in the form
341. It may be instructive to work out this problem from a
different point of view, especially as it affords excellent practice in
transformations.
A circle revolves about an axis passing within it, the perpen
dicular from the centre on the axis lying in the plane of the circle:
shew that, for a certain position of the axis, the same solid may be
traced out by a circle revolving about an external axis in its own
plane.
Let a Jb* + c 2 be the radius of the circle, i the vector axis of
rotation, COL (where To. = 1) the vector perpendicular from the
centre on the axis i, and let the vector
bi + da
be perpendicular to the plane of the circle.
The equations of the circle are
Also  /o 2 = S 2 ip + tfctp 4 S 2 . ictp,
= S*ip + S*ctp + ,S*ip
c
by the second of the equations of the circle. But, by the first,
(p 2 + by = 4c 2 S 2 a / D =  4 ( C y + a*S*ip),
which is easily transformed into
or
262 QUATERNIONS. [342.
If we put this in the forms
and (p_ a ) + c = 0,
where (3 is a unitvector perpendicular to i and in the plane of i
and p, we see at once that the surface will be traced out by, a
circle of radius c, revolving about i, an axis in its own plane,
distant a from its centre.
[This problem is not well adapted to shew the gain in brevity
and distinctness which generally attends the use of quaternions ;
as, from its very nature, it hints at the adoption of rectangular
axes and scalar equations for its treatment, so that the solution
we have given is but little different from an ordinary Cartesian
one.]
342. A surface is generated by a straight line which intersects
two fixed straight lines : find the general equation.
If the given lines intersect, there is no surface but the plane
containing them.
Let then their equations be,
p = a + x$, p = a l + xfi r
Hence every point of the surface satisfies the condition, 30,
P = y(* + x&) + (ly) (a, + *A) ............... (1).
Obviously y may have any value whatever : so that to specify a
particular surface we must have a relation between x and # r By
the help of this, x l may be eliminated from (1), which then takes
the usual form of the equation of a surface
p = <l>(x> y\
Or we may operate on (1) by F. (a + x/3 a x xfi^) t so that we
get a vector equation equivalent to two scalar equations ( 98, 123),
and not containing y. From this a? and x l may easily be found in
terms of p, and the general equation of the possible surfaces may
be written
f(x, x^ = 0,
where f is an arbitrary scalar function, and the values of x and x^
are expressed in terms of p.
This process is obviously applicable if we have, instead of two
straight lines, any two given curves through which the line must
pass; and even when the tracing line is itself a given curve,
situated in a given manner. But an example or two will make the
whole process clear.
344] GEOMETRY OF CURVES AND SURFACES. 263
343. Suppose the moveable line to be restricted by the condition
that it is always parallel to a fixed plane.
Then, in addition to (1), we have the condition
y being a vector perpendicular to the fixed plane.
We lose no generality by assuming a and a v which are any
vectors drawn from the origin to the fixed lines, to be each per
pendicular to 7 ; for, if for instance we could not assume Sya. = 0,
it would follow that Syj3 = 0, and the required surface would either
be impossible, or would be a plane, cases which we need not con
sider. Hence
= 0.
Eliminating x v by the help of this equation, from (1) of last section,
we have
= y (a + a/3) + (1  y) , + a/3,
Operating by any three noncoplanar vectors and with the charac
teristic S, we obtain three equations from which to eliminate x and
y. Operating by 8 . y we find
Syp = xSfty.
Eliminating x by means of this, we have finally
which appears to be of the third degree. It is really, however, only
of the second degree : since, in consequence of our assumptions,
we have
and therefore Syp is a spurious factor of the lefthand side.
344. Let the fixed lines be perpendicular to each other, and let
the moveable line pass through the circumference of a circle, whose
centre is in the common perpendicular, and whose plane bisects that
line at right angles.
Here the equations of the fixed lines may be written
where a, /3, y form a rectangular system, and we may assume the
two latter to be unitvectors.
The circle has the equations
p 2 =  a 2 , Sap = 0.
264 QUATERNIONS. [345
Equation (1) of 342 becomes
Hence SoC 1 p = y (ly) = 0, or y = J.
Also f =  a 2 = (2y  1) 2 a 2  *y  < (1  y)\
or 4a 2
so that if we now suppose the tensors of /3 and 7 to be each 2a, we
may put x = cos 0, x^ = sin 0, from which
For this specially simple case the solution is not better than the
ordinary Cartesian one ; but the student will easily see that we
may by very slight changes adapt the above to data far less sym
metrical than those from which we started. Suppose, for instance,
/3 and 7 not to be at right angles to one another ; and suppose the
plane of the circle not to be parallel to their plane, &c., &c. But
farther, operate on every line in space by the linear and vector
function <f>, and we distort the circle into an ellipse, the straight
lines remaining straight. If we choose a form of < whose principal
axes are parallel to a, /3. 7, the data will remain symmetrical, but
not unless. This subject will be considered again in the next
Chapter.
345. To find the curvature of a normal section of a central
surface of the second degree.
In this, and the few similar investigations which follow, it will
be simpler to employ infinitesimals than differentials ; though for
a thorough treatment of the subject the latter method, as it may
be seen in Hamilton s Elements, is preferable.
We have, of course, Sp(f>p = 1,
and, if p + Sp be also a vector of the surface, we have rigorously,
whatever be the tensor of Sp,
Hence ZSSpfo + SSp&p = ..................... (1).
Now (j>p is normal to the tangent plane at the extremity of p,
so that if t denote the distance of the point p + Sp from that plane
346.] GEOMETRY OF CURVES AND SURFACES. 265
and (1) may therefore be written
2tT<l>p  T 2 SpS . U8p<t> USp = 0.
But the curvature of the section is evidently
or, by the last equation,
In the limit, 8p is a vector in the tangent plane ; let is be the
vector semidiameter of the surface which is parallel to it, and the
equation of the surface gives
so that the curvature of the normal section, at the point p, in the
direction of or, is
1
directly as the perpendicular from the centre on the tangent plane,
and inversely as the square of the semidiameter parallel to the
tangent line, a wellknown theorem.
346. By the help of the known properties of the central
section parallel to the tangent plane, this theorem gives us all
the ordinary properties of the directions of maximum and mini
mum curvature, their being at right angles to each other, the
curvature in any normal section in terms of the chief curvatures
and the inclination to their planes, &c., &c., without farther
analysis. And when, in a future section, we shew how to find
an osculating surface of the second degree at any point of a given
surface, the same properties will be at once established for surfaces
in general. Meanwhile we may prove another curious property
of the surfaces of the second degree, which similar reasoning
extends to all surfaces.
The equation of the normal at the point p + Sp on the surface
treated in last section is
(1).
This intersects the normal at p if ( 216, 223)
8 . $p(f)p(f>$p = 0,
that is, by the result of 290, if Bp be parallel to the maximum or
266 QUATERNIONS. [346.
minimum diameter of the central section parallel to the tangent
plane.
Let 0j and cr 2 be those diameters, then we may write in general
where p and q are scalars, infinitely small.
If we draw through a point P in the normal at p a line parallel
to cTj, we may write its equation
tff = p + CKpp + ycr^
The proximate normal (1) passes this line at a distance (see 216)
or, neglecting terms of the second order,
TV^ff) ^ pS ^" 1 ^ " 1 + aq8
The first term in the bracket vanishes because o^ is a principal
vector of the section parallel to the tangent plane, and thus the
expression becomes
Hence, if we take a = To*, the distance of the normal from the
new line is of the second order only. This makes the distance of
P from the point of contact T<f)p Tcr*, i.e. the principal radius of
curvature along the tangent line parallel to cr 2 . That is, the group
of normals drawn near a point of a central surface of the second
degree pass ultimately through two lines each parallel to the tangent
to one principal section, and drawn through the centre of curvature
of the other. The student may form a notion of the nature of this
proposition by considering a small square plate, with normals
drawn at every point, to be slightly bent, but by different amounts,
in planes perpendicular to its edges. The first bending will make
all the normals pass through the axis of the cylinder of which the
plate now forms part; the second bending will not sensibly disturb
this arrangement, except by lengthening or shortening the line in
which the normals meet, but it will make them meet also in the
axis of the new cylinder, at right angles to the first. A small
pencil of light, with its focal lines, presents this appearance, due
to the fact that a series of rays originally normal to a surface
remain normals to a surface after any number of reflections and
(ordinary) refractions. (See 332.)
348.] GEOMETRY OF CURVES AND SURFACES. 267
347. To extend these theorems to surfaces in general, it is
only necessary, as Hamilton has shewn, to prove that if we write
dv = (f)dp,
(f) is a selfconjugate function ; and then the properties of <, as ex
plained in preceding Chapters, are applicable to the question.
As the reader will easily see, this is merely another form of the
investigation contained in 334. But it is again cited here to
shew what a number of very simple demonstrations may be given
of almost all quaternion theorems.
The vector v is defined by an equation of the form
dfp = Svdp,
where / is a scalar function. Operating on this by another inde
pendent symbol of differentiation, 8, we have
Bdfp = SSvdp + SvSdp.
In the same way we have
dSfp = SdvSp + SvdSp.
But, as d and 8 are independent, the lefthand members of these
equations, as well as the second terms on the right (if these exist
at all), are equal, so that we have
SdvSp = SSvdp.
This becomes, putting dv = (frdp,
and therefore $v = <f>Sp,
S&p(f)dp = Sdp<f>Sp,
which proves the proposition.
348. If we write the differential of the equation of a surface
in the form
dfp = 2Svdp,
then it is easy to see that
f(p + dp) =fp + ZSvdp + Sdvdp + &c.,
the remaining terms containing as factors the third and higher
powers of Tdp. To the second order, then, we may write, except
for certain singular points,
and, as before, ( 345), the curvature of the normal section whose
tangent line is dp is
_1_ ydv
TV dp
268 QUATERNIONS. [349
349. The step taken" in last section, although a very simple
one, virtually implies that the first three terms of the expansion of
f(p + dp) are to be formed in accordance with Taylor s Theorem,
whose applicability to the expansion of scalar functions of quater
nions has not been proved in this work (see 142) ; we therefore
give another investigation of the curvature of a normal section,
employing for that purpose the formulae of 299.
We have, treating dp as an element of a curve,
Svdp = 0,
or, making s the independent variable,
Svp = 0.
From this, by a second differentiation,
qdv , ~ lr A
ds p + p
The curvature is, therefore, since v \\ p" and Tp = 1,
m it 1 ci dv ,o 1 a dv , f,
Tp" =  ^ S j p 2 = jfr S y , as before.
TV dp r TV dp
350. Since we have shewn that
dv = <pdp
where $ is a selfconjugate linear and vector function, whose con
stants depend only upon the nature of the surface, and the position
of the point of contact of the tangent plane ; so long as we do not
alter these we must consider < as possessing the properties explained
in Chapter V.
Hence, as the expression for Tp" does not involve the tensor of
dp, we may put for dp any unitvector r, subject of course to the
condition
Svr = Q .............................. (1).
And the curvature of the normal section whose tangent is r is
If we consider the central section of the surface of the second degree
made by the plane Svvr = 0,
we see at once that the curvature of the given surface along the
normal section touched by r is inversely as the square of the
parallel radius in the auxiliary surface. This, of course, includes
Euler s and other wellknown Theorems.
352.] GEOMETRY OF CURVES AND SURFACES. 269
351. To find the directions of maximum and minimum
curvature, we have
ST(J)T = max. or min.
with the conditions, Svr = 0,
Tr= 1.
By differentiation, as in 290, we obtain the farther equation
>Sf.^T</)T = ........................... (1).
If r be one of the two required directions, r =rUv is the other,
for the lastwritten equation may be put in the form
or S . vr <T = 0.
Hence the sections of greatest and least curvature are perpendicular
to one another.
We easily obtain, as in 290, the following equation
S.v(<t> + Srct>T) 1 v = 0,
which gives two values of $T<T, and these divided by TV are
the required curvatures.
352. Before leaving this very brief introduction to a subject,
an exhaustive treatment of which will be found in Hamilton s
ElementSy we may make a remark on equation (1) of last section
S.VT(f)T = )
or, as it may be written, by returning to the notation of 350,
S . vdpdv = 0.
This is the general equation of lines of curvature. For, if we
define a line of curvature on any surface as a line such that
normals drawn at contiguous points in it intersect, then, Bp being
an element of such a line, the normals
tx = p + xv and w = p + Bp + y (v + 8v)
must intersect. This gives, by 216, the condition
as above.
270 QUATERNIONS.
EXAMPLES TO CHAPTER X.
1. Find the length of any arc of a curve drawn on a sphere
so as to make a constant angle with a fixed diameter.
2. Shew that, if the normal plane of a curve always contains
a fixed line, the curve is a circle.
3. Find the radius of spherical curvature of the curve
Also find the equation of the locus of the centre of spherical
curvature.
4. (Hamilton, Bishop Law s Premium Examination, 1854.)
(a) If p be the variable vector of a curve in space, and if the
differential die be treated as = 0, then the equation
dT(ptc) = Q
obliges ic to be the vector of some point in the normal plane to
the curve.
(6) In like manner the system of two equations, where d/c
and d 2 tc are each =0,
dT(p  K) = 0, d 2 T(p  K) = 0,
represents the axis of the element, or the right line drawn through
the centre of the osculating circle, perpendicular to the osculating
plane.
(c) The system of the three equations, in which K is treated
as constant,
dT(p K ) = 0, d*T(pK) = Q, d 5 T(p K ) = 0,
determines the vector K of the centre of the osculating sphere.
(d) For the three last equations we may substitute the
following :
S.(p K )dp = 0,
(e) Hence, generally, whatever the independent and scalar
variable may be, on which the variable vector p of the curve
GEOMETRY OF CURVES AND SURFACES. 271
depends, the vector K of the centre of the osculating sphere admits
of being thus expressed :
3V. dptfpS . dpd*p  dp*V. dpd 5 p
~ p ~ S.dpd*pd 3 p
(/) In general
d(p*V.dpUp) = d( Tp~ z V.pdp)
= Tp~ 5 (3V. pdpS.pdp P *V.pd*p) ;
whence,
and, therefore, the recent expression for K admits of being thus
transformed,
~~
(g) If the length of the element of the curve be constant,
dTdp = 0, this last expression for the vector of the centre of the
osculating sphere to a curve of double curvature becomes, more
simply,
d . d 2 pdp*
~
V.
or K = p +
S.dpd*pd?p
(h) Verify that this expression gives K = 0, for a curve
described on a sphere which has its centre at the origin of vectors ;
or shew that whenever dTp = 0, d?Tp = 0, d?Tp = 0, as well as
dTdp = 0, then
5. Find the curve from every point of which three given
spheres appear of equal magnitude.
6. Shew that the locus of a point, the difference of whose
distances from each two of three given points is constant, is a
plane curve.
7. Find the equation of the curve which cuts at a given angle
all the sides of a cone of the second degree.
Find the length of any arc of this curve in terms of the
distances of its extremities from the vertex.
8. Why is the centre of spherical curvature, of a curve
described on a sphere, not necessarily the centre of the sphere ?
272 QUATERNIONS.
9. Find the equation of the developable surface whose gene
rating lines are the intersections of successive normal planes to a
given tortuous curve.
10. Find the length of an arc of a tortuous curve whose
normal planes are equidistant from the origin.
11. The reciprocals of the perpendiculars from the origin on
the tangent planes to a developable surface are vectors of a
tortuous curve ; from whose osculating planes the cuspedge of
the original surface may be reproduced by the same process.
12. The equation p=Va t {3,
where a is a unitvector not perpendicular to /3, represents an
ellipse. If we put 7 = Fa/3, shew that the equations of the locus
of the centre of curvature are
13. Find the radius of absolute curvature of a spherical
conic.
14. If a cone be cut in a circle by a plane perpendicular to a
side, the axis of the right cone which osculates it, along that side,
passes through the centre of the section.
15. Shew how to find the vector of an umbilicus. Apply
your method to the surfaces whose equations are
and SapSfipSyp = 1.
16. Find the locus of the umbilici of the surfaces represented
by the equation
s P (<f>+hr P =i,
where h is an arbitrary parameter.
17. Shew how to find the equation of a tangent plane which
touches a surface along a line, straight or curved. Find such
planes for the following surfaces
and (p*  a 2 + 6 2 ) 2 + 4 (a> 2 + V&ap) = 0.
GEOMETRY OF CURVES AND SURFACES. 273
18. Find the condition that the equation
(,> + )</> =  1,
where <p is a selfconjugate linear and vector function, may
represent a cone.
19. Shew from the general equation that cones and cylinders
are the only developable surfaces of the second degree.
20. Find the equation of the envelop of planes drawn at each
point of an ellipsoid perpendicular to the radius vector from the
centre.
21. Find the equation of the envelop of spheres whose centres
lie on a given sphere, and which pass through a given point.
22. Find the locus of the foot of the perpendicular from the
centre to the tangent plane of a hyperboloid of one, or of two,
sheets.
23. Hamilton, Bishop Law s Premium Examination, 1852.
(a) If p be the vector of a curve in space, the length of the
element of that curve is Tdp ; and the variation of the length of a
finite arc of the curve is
SfTdp = JSUdpSdp =  ASUdpSp + JSdUdpBp.
(b) Hence, if the curve be a shortest line on a given surface,
for which the normal vector is v t so that SvSp = 0, this shortest or
geodetic curve must satisfy the differential equation,
Also, for the extremities of the arc, we have the limiting
equations,
Interpret these results.
(c) For a spheric surface, Vvp = 0, V.pdUdp = , the inte
grated equation of the geodetics is Vp Udp = r, giving Svrp = Q
(great circle).
For an arbitrary cylindric surface,
the integral shews that the geodetic is generally a helix, making
a constant angle with the generating lines of the cylinder.
(d) For an arbitrary conic surface,
T, Q. I, 18
274 QUATERNIONS.
integrate this differential equation, so as to deduce from it,
TVpUdp = const.
Interpret this result; shew that the perpendicular from the
vertex of the cone on the tangent to a given geodetic line is
constant ; this gives the rectilinear development.
When the cone is of the second degree, the same property is a
particular case of a theorem respecting confocal surfaces.
(e) For a surface of revolution,
S.apv = 0, S . apdUdp = ;
integration gives,
const. = 8 . ap Udp = TVapS U(Vap. dp)
the perpendicular distance of a point on a geodetic line from the
axis of revolution varies inversely as the cosine of the angle under
which the geodetic crosses a parallel (or circle) on the surface.
(/) The differential equation, S . apd Udp = 0, is satisfied not
only by the geodetics, but also by the circles, on a surface of
revolution ; give the explanation of this fact of calculation, and
shew that it arises from the coincidence between the normal plane
to the circle and the plane of the meridian of the surface.
(g) For any arbitrary surface, the equation of the geodetic
may be thus transformed, S.vdpd*p = Q] deduce this form, and
shew that it expresses the normal property of the osculating plane.
(h) If the element of the geodetic be constant, dTdp = 0,
then the general equation formerly assigned may be reduced to
V. vd*p = 0.
Under the same condition, d?p = v~ l Sdvdp.
(i) If the equation of a central surface of the second order be
put under the form fp = 1, where the function / is scalar, and
homogeneous of the second dimension, then the differential of that
function is of the form dfp = 2$ . vdp, where the normal vector,
v = cf)p ) is a distributive function of p (homogeneous of the first
dimension), dv = d<f>p = <f>dp.
This normal vector v may be called the vector of proximity
(namely, of the element of the surface to the centre) ; because its
reciprocal, jT l , represents in length and in direction the perpen
dicular let fall from the centre on the tangent plane to the surface.
(k) If we make Sacfrp =/(&, p) this function f is commu
tative with respect to the two vectors on which it depends,
GEOMETRY OF CURVES AND SURFACES. 275
f(p, o) =f(cr, p) ; it is also connected with the former function /,
of a single vector p, by the relation, /(/o, p) =fp : so that
fp = Spfp.
fdp = Sdpdv ; dfdp = 2S . dvd*p ; for a geodetic, with constant
element,
this equation is immediately integrable, and gives const.
= Tv\/ (fUdp) = reciprocal of Joachimstal s product, PD.
(I) If we give the name of "Didonia" to the curve (discussed
by Delaunay) which, on a given surface and with a given perimeter,
contains the greatest area, then for such a Didonian curve we have
by quaternions the formula,
fS. Uvdp8p + c8fTdp = 0,
where c is an arbitrary constant.
Derive hence the differential equation of the second order,
equivalent (through the constant c) to one of the third order,
C  l dp=V. UvdUdp.
Geodetics are, therefore, that limiting case of Didonias for
which the constant c is infinite.
On a plane, the Didonia is a circle, of which the equation,
obtained by integration from the general form, is
p = TX + c Uvdp,
w being vector of centre, and c being radius of circle.
(m) Operating by 8. Udp, the general differential equation of
the Didonia takes easily the following forms :
c l Tdp =
AJ ~~
ff j
Uvdp
(n) The vector o>, of the centre of the osculating circle to a
curve in space, of which the element Tdp is constant, has for
expression,
182
276 QUATERNIONS.
Hence for the general Didonia,
Tr 7
Uvdp
(o) Hence, the radius of curvature of any one Didonia varies,
in general, proportionally to the cosine of the inclination of the
osculating plane of the curve to the tangent plane of the surface.
And hence, by Meusnier s theorem, the difference of the
squares of the curvatures of curve and surface is constant ; the
curvature of the surface meaning here the reciprocal of the radius
of the sphere which osculates in the direction of the element of
the Didonia.
(p) In general, for any curve on any surface, if f denote the
vector of the intersection of the axis of the element (or the axis of
the circle osculating to the curve) with the tangent plane to the
surface, then
vdp*
Hence, for the general Didonia, with the same signification of the
symbols,
= p c Uvdp ;
and the constant c expresses the length of the interval p ,
intercepted on the tangent plane, between the point of the curve
and the axis of the osculating circle.
(q) If, then, a sphere be described, which shall have its
centre on the tangent plane, and shall contain the osculating
circle, the radius of this sphere shall always be equal to c.
(r) The recent expression for f, combined with the first form
of the general differential equation of the Didonia, gives
(s) Hence, or from the geometrical signification of the con
stant c, the known property may be proved, that if a developable
surface be circumscribed about the arbitrary surface, so as to touch
it along a Didonia, and if this developable be then unfolded into a
plane, the curve will at the same time be flattened (generally)
into a circular arc, with radius = c.
GEOMETRY OF CURVES AND SURFACES. 277
24. Find the condition that the equation
may give three real values of /for any given value of p. If /be a
function of a scalar parameter f, shew how to find the form of this
function in order that we may have
_Vf* + ** + *f_o
* ~ da? c% 2 ^ d* 8
Prove that the following is the relation between /and f,
df
in the notation of 159. (Tait, Trans. R. S. E. 1873.)
25. Shew, after Hamilton, that the proof of Dupin s theorem,
that " each member of one of three series of orthogonal surfaces
cuts each member of each of the other series along its lines of
curvature," may be expressed in quaternion notation as follows :
If Svdp = Q, Sv dp = Q, S.vi/dp = Q
be integrable, and if
Svv = 0, then Vv dp = 0, makes S . vvdv = 0.
Or, as follows,
If 8vVv = Q, &/Vi/ = 0, &/ Vj;" = 0, and
then 8.v"(8v / V)v = O t
where Vt+j +.*.
dx J dy dz
26. Shew that the equation
represents the line of intersection of a cylinder and cone, of the
second order, which have ft as a common generating line.
27. Two spheres are described, with centres at A, B, where
OA = a, OB = @, and radii a, b. Any line, OPQ, drawn from the
origin, cuts them in P, Q respectively. Shew that the equation
of the locus of intersection of AP, BQ has the form
Shew that this involves 8 . a{3p = 0,
and therefore that the left side is a scalar multiple of V . aj3, so
that the locus is a plane curve,
278 QUATERNIONS.
Also shew that in the particular case
the locus is the surface formed by the revolution of a Cartesian
oval about its axis.
28. Integrate the equations
Shew that each represents a series of circles in space. What
is the common property of the circles of each series? [See 140,
(10), (11).]
29. Express the general equation of a knot of any kind, on
an endless cord, in the form
P = (*)
pointing out precisely the nature of the function <.
What are the conditions to which (f> must be subject, when
possible distortions of the knot are to be represented ? What are
the conditions that the string may be capable of being brought
into a mere ring form ?
30. Find the envelop of the planes of equilateral triangles
whose vertices are situated in three given lines in space. What
does it become when two, or all three, of these lines intersect ?
31. Form the equation of the surface described by a circle,
when two given points in its axis are constrained to move on given
straight lines. Also when the constraining lines are two concentric
circles in one plane.
CHAPTER XI.
KINEMATICS.
353. IN the present Chapter it is not proposed to give a
connected account of even the elements of so extensive a subject
as that indicated by the Title. All that is contemplated is to
treat a few branches of the subject in such a way as to shew the
student how to apply the processes of Quaternions.
And, with a view to the next Chapter, the portions selected
for treatment will be those of most direct interest in their physical
applications.
A. Kinematics of a Point.
354. When a point s vector, p, is a function of the time t, we
have seen ( 36) that its vectorvelocity is expressed by fa or, in
Newton s notation, by p.
That is, if p = (f>t
be the equation of an orbit, containing (as the reader may see) not
merely the form of the orbit, but the law of its description also,
then
gives at once the form of the Hodograph and the law of its
description.
This shews immediately that the vectoracceleration of a point s
motion,
tfp
280 QUATERNIONS. [ 3 5 5
is the vectorvelocity in the hodograph. Thus the fundamental pro
perties of the hodograph are proved almost intuitively.
355. Changing the independent variable, we have
dp ds ,
p= cfedi = ^
if we employ the dash, as before, to denote = .
This merely shews, in another form, that p expresses the
velocity in magnitude and direction. But a second differentiation
gives
p = vp + v*p".
This shews that the vectoracceleration can be resolved into two
components, the first, vp, being in the direction of motion and
equal in magnitude to the acceleration of the speed, v or 7 ;
the second, v 2 // , being in the direction of the radius of absolute
curvature, and having for its amount the square of the speed
multiplied by the curvature.
[It is scarcely conceivable that this important fundamental
proposition can be proved more elegantly than by the process
just given.]
356. If the motion be in a plane curve, we may write the
equation as follows, so as to introduce the usual polar coordinates,
r and 0,
p=rJ > "ft,
where a is a unitvector perpendicular to, jB a unitvector in, the
plane of the curve.
Here, of course, r and 6 may be considered as connected by one
scalar equation ; or better, each may be looked on as a function of
t. By differentiation we get
which shews at once that r is the velocity along, r6 that perpen
pendicular to, the radius vector. Again,
p = (r r<9 2 ) ^ $ + (2r<9 + r&) wP v ft
which gives, by inspection, the components of acceleration along,
and perpendicular to, the radius vector.
357. For uniform acceleration in a constant direction, we have
at once
358.] KINEMATICS. 281
Whence p = at f {3,
where ft is the vectorvelocity at epoch. This shews that the
hodograph is a straight line described uniformly.
Also p = h fit,
no constant being added if the origin be assumed to be the position
of the moving point at epoch.
Since the resolved parts of p, parallel to ft and a, vary respect
ively as the first and second powers of t, the curve is evidently a
parabola ( 31 (/)).
But we may easily deduce from the equation the following
result,
T(p + fioL~ l ft) = SUa(p + ^ oT 1 } ,
the equation of a paraboloid of revolution, whose axis is a. Also
8 . ctftp = 0,
and therefore the distance of any point in the path from the
point ^ftoC l ft is equal to its distance from the line whose
equation is
Thus we recognise the focus and directrix property. [The
student should remark here how the distances of the point of
projection (which may, of course, be any point of the path) from
the focus and from the directrix are represented in magnitude and
direction by the two similar but different expressions
 Jcf and  J0V 1 ,
or i(#O and tftfa 1 ).
This is an excellent example of the noncommutative character of
quaternion multiplication.]
358. That the moving point may reach a point 7 (where 7 is,
of course, coplanar with a and /3) we must have, for some real
value of t,
Now suppose T/3, the speed of projection, to be given = v, and,
for shortness, write ts for Up.
Then y = ^^ + vtvf ........................ (a).
282 QUATERNIONS. [359.
Since 2V = 1,
m 2^4
we have   (V  Say) f + T 7 2 = 0.
TP
The values of f are real if
is positive. Now, as TaTy is never less than Say, this condition
evidently requires that v* Say also shall be positive. Hence,
when they are real, both values of f are positive. Thus we have
four values of t which satisfy the conditions, and it is easy to see
that since, disregarding the signs, they are equal two and two,
each pair refer to the same path, but described in opposite direc
tions between the origin and the extremity of 7. There are
therefore, if any, in general two parabolas which satisfy the
conditions. The directions of projection are (of course) given by
the corresponding values of ST. These, in turn, are obtained at
once from (a) in the form
1 t
WJ7J*
where t has one or other of the values previously found.
359. The envelop of all the trajectories possible, with a given
speed, evidently corresponds to
for then y is the vector of intersection of two indefinitely close
paths in the same vertical plane.
Now v 2  Say = TaTy
is evidently the equation of a paraboloid of revolution of which
the origin is the focus, the axis parallel to a, and the directrix
plane at a distance  .
All the ordinary problems connected with parabolic motion are
easily solved by means of the above formulae. Some, however, are
even more easily treated by assuming a horizontal unitvector in
the plane of motion, and expressing /3 in terms of it and of a.
But this must be left to the student.
360. For acceleration directed to or from a fixed point, we
have, taking that point as origin, and putting P for the magnitude
of the central acceleration,
P = PU P .
36 1.] KINEMATICS. 283
From this, at once,
Integrating Vpp = 7 = 3. constant vector.
The interpretation of this simple formula is first, p and p are
in a plane perpendicular to 7, hence the path is in a plane (of
course passing through the origin) ; second, the doubled area of
the triangle, two of whose sides are p and p (that is, the moment
of the velocity) is constant.
[It is scarcely possible to imagine that a more simple proof
than this can be given of the fundamental facts, that a central
orbit is a plane curve, and that equal areas are described by the
radius vector in equal times.]
361. When the law of acceleration to or from the origin is
that of the inverse square of the distance, we have
m_
~T P *
where m is negative if the acceleration be directed to the origin.
mUp
Hence p = ^rr .
The following beautiful method of integration is due to Hamil
ton. (See 140, (2).)
dUp U P . Vpp Up.y
Generally, ^  ^ ^ ,
dUp
therefore py = m ~ ,
and p7 = e m Up,
where e is a constant vector, perpendicular to 7, because
Syp = 0.
Hence, in this case, we have for the hodograph,
p = 67" 1 m Up . 7" 1 .
Of the two parts of this expression, which are both vectors, the
first is constant, and the second is constant in length. Hence the
locus of the extremity of p is a circle in a plane perpendicular to 7
(i.e. parallel to the plane of the orbit), whose radius is T.my 1 ,
and whose centre is at the extremity of the vector ey~ l .
[This equation contains the whole theory of the Circular
Hodograph. Its consequences are developed at length in Hamil
ton s Elements.]
284 QUATERNIONS. [362.
362. We may write the equations of this circle in the form
(a sphere), and Sjp =
(a plane through the origin, and through the centre of the sphere).
The equation of the orbit is found by operating by V.p upon
that of the hodograph. We thus obtain
or ry
or mTp = Se (<fe*  p) ;
in which last form we at once recognise the focus and directrix
property. This is in fact the equation of a conicoid of revolution
about its principal axis (e), and the origin is one of the foci. The
orbit is found by combining it with the equation of its plane,
Syp = 0.
We see at once that 7V 1 is the vector distance of the directrix
from the focus ; and similarly that the excentricity is T. em" 1 , and
2
the major axis 2T 2 .
m 2 + e 2
363. To take a simpler case : let the acceleration vary as the
distance from the origin.
Then p i = + m z p,
the upper or lower sign being used according as the acceleration is
from or to the centre.
This is
Hence p = ae
or p = a. cos mt + ft sin mt,
where a and ft are arbitrary, but constant, vectors ; and e is the
base of Napier s logarithms.
The first is the equation of a hyperbola ( 31, k) of which a
and ft are the directions of the asymptotes ; the second, that of an
ellipse of which a and ft are semiconjugate diameters.
Since p = m{ae mt  fte~ mt ],
or =m { asmmt + @cosmt},
the hodograph is again a hyperbola or ellipse. But in the first
case it is, if we neglect the change of dimensions indicated by the
364.] KINEMATICS. 285
scalar factor m, conjugate to the orbit ; in the case of the ellipse
it is similar and similarly situated.
364 Again, let the acceleration be as the inverse third power of
the distance, we have
.. mUp
P==Ty
Of course, we have, as usual,
Vpp = 7.
Also, operating by S . p,
... mSpp
A 2 r m
p  *,
of which the integral is
the equation of energy.
A a  m
Again, Spp =  t .
Hence Spp + /3 2 = C,
or Spp = Ct,
no constant being added if we reckon the time from the passage
through the apse, where Spp = 0.
We have, therefore, by a second integration,
p * = ce + c .......................... (i).
[To determine C , remark that
pp = Ct + 7,
or p*p*=C*fy*.
But p 2 /3 2 = Cp 2 m (by the equation of energy),
Hence
To complete the solution, we have, by 140 (2),
T7 p dUp, TT ,_! d , Up
= 
where /9 is a unit vector in the plane of the orbit.
But r=2,.
p p
dt
**", s^. 71 f^ .<
v t
286 QUATERNIONS. [365.
The elimination of t between this equation and (1) gives Tp in
terms of Up, or the required equation of the path.
We may remark that if 6 be the ordinary polar angle in the
orbit,
Hence we have = Ty J ^ ^,1
and 7* = (Cf + C ) j
from which the ordinary equations of Cotes spirals can be at once
found. [See Tait and Steele s Dynamics of a Particle, Appendix
(A).]
365. To find the conditions that a given curve may be the hodo
graph corresponding to a central orbit.
If or be its vector, given as a function of the time, fadt is that
of the orbit ; hence the requisite conditions are given by
V<srfadt = y ........................... (1),
where 7 is a constant vector.
We may transform this into other shapes more resembling the
Cartesian ones.
Thus F*r/W<ft = ........................... (2),
and Vtirfadt + F^OT  0.
From (2) fadt = xfr,
and therefore by (1) x Vvriz = 7,
or the curve is plane. And
ajFBJar + FBTBT = ;
or eliminating x, 7 Vizix = ( Vvrdr)*.
Now if v be the velocity in the hodograph, R its radius of curva
ture, p the perpendicular on the tangent ; this equation gives at
once
M = R p",
which agrees with known results.
366. The equation of an epitrochoid or hypotrochoid, referred
to the centre of the fixed circle, is evidently
where a is a unit vector in the plane of the curve and i another
367.] KINEMATICS. 287
perpendicular to it. Here o> and co l are the angular velocities in
the two circles, and t is the time elapsed since the tracing point
and the centres of the two circles were in one straight line.
Hence, for the length of an arc of such a curve,
s=fTpdt = fdt V{o>V + 2o>o> 1 a& cos (to <o^t + to?!?},
= fdt Aua + >,&) 4^06 ^ "^ t\ ,
j )
which is, of course, an elliptic function.
But when the curve becomes an epicycloid or a hypocycloid,
coa T to) = 0, and
which can be expressed in finite terms, as was first shewn by
Newton.
The hodograph is another curve of the same class, whose equa
tion is
and the acceleration is denoted in magnitude and direction by the
vector
Of course the equations of the common Cycloid and Trochoid
may be easily deduced from these forms by making a indefinitely
great and a indefinitely small, but the product aw finite ; and
transferring the origin to the point
= aa.
B. Kinematics of a Rigid System.
367. Let i be the normalvector to any plane.
Let TS and p be the vectors of any two points in a rigid plate
in contact with the plane.
After any small displacement of the rigid plate in its plane, let
fc and dp be the increments of r and p.
Then Sidvr 0, Sidp = ; and, since T (& p) is constant,
And we may evidently assume, consistently with these equations,
dp = a)i (p  r),
dm = a)i (sr T) ;
288 QUATERNIONS. [368.
where of course r is the vector of some point in the plane, to a
rotation co about which the displacement is therefore equivalent.
Eliminating r, we have
. d (OT p)
(01 =  ^ ,
57 p
which gives ew, and thence r is at once found.
For any other point a in the plane figure
Sida = 0,
S (p a) (dp da) = 0. Hence dp da = w l i (p a).
S (a iv) (dm da) = 0. Hence da cZcr = &> 2 % (a or).
From which, at once, a) l = &> 2 = &>, and
da = col (a r),
or this point also is displaced by a rotation co about an axis through
the extremity of r and parallel to i.
368. In the case of a rigid body moving about a fixed point
let OT, p, a denote the vectors of any three points of the body ; the
fixed point being origin.
Then CT 2 , p*, a z are constant, and so are Svrp, Spa, and Savr.
After any small displacement we have, for za and p,
h
........................... (1).
Sndp + Spd = OJ
Now these three equations are satisfied by
dt& Vavr, dp = Fa/o,
where a is any vector whatever. But if d^ and dp are given, then
Vdvrdp = V. VO.TX Vap = ctS . ap&.
Operate by S .Vvrp, and remember (1),
Vd&dp Vdpdtn
Hence a = c , , = aT~ .................. (")
Spdru
Now consider a, Sada =
Spda = Sadp J ,
Svrda = Sad
do = Ya.a satisfies them all, by (2), and we have thus the proposi
tion that any small displacement of a rigid body about a fixed
point is equivalent to a rotation.
3 70.] KINEMATICS. 289
369. To represent the rotation of a rigid body about a given
axis, through a given finite angle. [This is a work of supererogation,
if we consider the results of 119. But it may be interesting to
obtain these results in another manner.]
Let a be a unitvector in the direction of the axis, p the vector
of any point in the body with reference to a fixed point in the axis,
and 6 the angle of rotation.
Then p = of 1 Sap + a~ l Yap,
= aSap a Vap.
The rotation leaves, of course, the first part unaffected, but the
second evidently becomes
or a Vap cos 6 + Vap sin 0.
Hence p becomes
p l = aSap aVap cos 9 + Vap sin 6,
= (cos 6/2 + a sin 0/2) p (cos 0/2  a sin 0/2),
a* pa *.
370. Hence to compound two rotations about axes which meet,
we may evidently write, as the effect of an additional rotation </>
about the unit vector (3,
P^^ pfi* 1 .
Hence p^tf 1 * of 1 pa." 1 * $* *.
If the /3rotation had been first, arid then the arotation, we should
have had
p , = a >lf ft* p^ l " a " ,
and the noncommutative property of quaternion multiplication
shews that we have not, in general,
P* = P*
If a, (3, 7 be radii of the unit sphere to the corners of a spherical
triangle whose angles are 0/2, <j)/2, tfr/2, we know that
yfr/ */ a /ir = _ L (Hamilton, Lectures, p. 2G7.)
Hence /3* /7r a e/7T =  7 "* 7 ,
and we may write p z = y * f * py^,
or, successive rotations about radii to two corners of a spherical
triangle, and through angles double of those of the triangle, are
T. Q. I. 19
290 QUATERNIONS. [371.
equivalent to a single rotation about the radius to the third corner,
and through an angle double of the exterior angle of the triangle.
Thus any number of successive finite rotations of a system, of
which one point is fixed, may be compounded into a single rotation
about a definite axis.
371. When the rotations are indefinitely small, the effect of
one is, by 369,
and for the two, neglecting products of small quantities,
a and b representing the angles of rotation about the unitvectors
a. and /3 respectively.
But this is equivalent to
P, = P + T(aoL + b/3) VU(aoL + b/3) p,
representing a rotation through an angle T (aa + b/3), about the
unitvector U(aa + b/3). Now the latter is the direction, and the
former the length, of the diagonal of the parallelogram whose sides
are aa and b/3.
We may write these results more simply, by putting a for aa,
P for b/3, where a and /3 are now no longer unitvectors, but repre
sent by their versors the axes, and by their tensors the angles
(small), of rotation.
Thus =
372. Given the instantaneous axis in terms of the time, it is
required to find the single rotation which will bring the body from
any initial position to its position at a given time.
If a be the initial vector of any point of the body, w the value
of the same at time t, and q the required quaternion, we have
by 119
53 = qaq l ............................... (1).
Differentiating with respect to t, this gives
tb = qaq~ l qaq~ l qq~ l ,
But tzr = Fecr = F. eqctq~\
373]
KINEMATICS.
291
Hence, as qaq l may be any vector whatever in the displaced
body, we must have
e = 2Vqq* ............................... (2).
This result may be stated in even a simpler form than (2), for
we have always, whatever quaternion q may be,
and, therefore, if we suppose the tensor of q, which, as it is not
involved in q ( ) q~ l , may have any value whatever, to be a
constant (unity, for instance), we may write (2) in the form
7 = 2j ................................. (3).
An immediate consequence, which will be of use to us later, is
q.q~ 1 eq = 2q ............................... (4).
373. To express q in terms of the usual angles fy, 6, ^>.
Here the vectors i, j, k in the original position of the body
correspond to OA, OB, OC, re
spectively, at time t. The trans
position is defined to be effected
by first, a rotation ^r about lc\
second, a rotation about the new
position of the line originally
coinciding with j ; third, a rotation
<f> about the final position of the
line at first coinciding with Jc.
This selection of angles, in
terms of which the quaternion is
to be expressed, is essentially
unsymmetrical, and therefore the results cannot be expected to be
simple.
The rotation ty about k has the operator
This converts^ into T/, where
,7 = A* " 1 jk~* lir =j cos ^  i sin ^.
The body next rotates about rj through an angle 0. This has
the operator
192
292 QUATERNIONS. [374.
It converts k into
OC = f = v e/7T kv~ e/1T = (cos 0/2 + 77 sin (9/2) A; (cos 0/2  77 sin (9/2)
= k cos 4 sin (i cos ^ + j sin ir).
The body now turns through the angle (/> about the operator
being
t,0/7T/ X C.0/7T
Hence, omitting a few reductions, which we leave as excellent
practice for the reader, we find
q = f */V /ff A;* /ir
= (cos 0/2 + ? sin 0/2) (cos 0/2 + r? sin 0/2) (cos ir/2 + A; sin ^r/2)
= cos (0 + ^)/2 . cos 0/2 + i sin (0  ^)/2 . sin 0/2 +
j cos (0  ^)/2 . sin 0/2 + k sin (0 + ^)/2 . cos 0/2,
which is, of course, essentially unsyrnmetrical.
374. To find the usual equations connecting ty, 0, with the
angular velocities about three rectangular axes fixed in the body.
HaviDg the value of q in last section in terms of the three
angles, it may be useful to employ it, in conjunction with equation
(3) of 372, partly as a verification of that equation. Of course,
this is an exceedingly roundabout process, and does not in the .
least resemble the simple one which is immediately suggested by
quaternions.
We have 2q = eq = {co l OA + o> 2 O + o> 3 OC}q,
whence 2q~*q = q~ l {co l OA + o> 2 OB + o> 3 OC} q,
or 2q = q (ico 1 +jo 2 + kco s ).
This breaks up into the four (equivalent to three independent)
equations
2 ~ [cos (0 + ^)/2 . cos 0/2]
= to i sin (</>  ^)/2 . sin 0/2  w 2 cos (0  ^)/2 . sin 0/2
 &) 3 sin (0 + i/r)/2 . cos 0/2,
2 [sin (0^/2. an 0/2]
= &), cos ((/> 4 ^)/2 . cos 0/2  w, sin (</> + ^r)/2 . cos 0/2
+ w 3 cos (c/)  f )/2 . sin 0/2,
375] KINEMATICS. 293
. cos 0/2 + o> 2 cos (< + ^)/2 . cos 0/2
 o> 3 sin (<  i/r)/2 . sin 0/2,
=  o> : cos ((/>  i/r)/2 . sin 0/2 + 2 sin (<f>  ir)/2 . sin 0/2
f o> 8 cos (< + ^)/2 . cos 0/2.
From the second and third, eliminate (j> ^jr, and we get by
inspection
cos 0/2 . = (o) t sin </> f ct> 2 cos <) cos 0/2,
or = co 1 sin < fo> 2 cos $ ...................... (1).
Similarly, by eliminating between the same two equations,
sin 0/2 . ((j> fy = &> s sin 0/2 + co 1 cos <^> cos 0/2 o> 2 sin (/> cos 0/2.
And from the first and last of the group of four
cos 0/2 . ((j> + vfr) = co 3 cos 0/2 w l cos (/> sin 0/2 + o> 2 sin </> sin 0/2.
These last two equations give
< + ^jr cos = w 3 ..................... (2).
cos + ^ = ( o) i cos cf) + o> 2 sin <) sin + &) 3 cos 0.
From the last two we have
T/T sin = co 1 cos </> I o> 2 sin ^> ............ (3).
(1), (2), (3) are the forms in which the equations are usually given.
375. To deduce expressions for the directioncosines of a set of
rectangular axes, in any position, in terms of rational functions of
three quantities only.
Let a, /3, 7 be unitvectors in the directions of these axes. Let q
be, as in 372, the requisite quaternion operator for turning the
coordinate axes into the position of this rectangular system. Then
q = w + xi + yj + zk,
where, as in 372, we may write
1 = w z + x 2 + ?/ 2 + z\
Th en we have q~ l = w xi yj zk,
and therefore
a  qiq 1  (wi  x  yk + zj} (w  xi  yj  zk)
= (w 2 4 a? 2  /  z 2 ) i+2(wz + xy)j 4 2 (xz  wy) k,
294 QUATERNIONS. [3/6.
where the coefficients of i, j, k are the directioncosines of a as
required. A. similar process gives by inspection those of /3 and 7.
As given by Cayley*, after Rodrigues, they have a slightly
different and somewhat less simple form to which, however, they
are easily reduced by putting
w = as/\ = yip = z/v = 1/K*.
The geometrical interpretation of either set is obvious from the
nature of quaternions. For (taking Cayley s notation) if 6 be the
angle of rotation : cosy, cos g, cos h, the directioncosines of the
axis, we have
q = w + ari + yj + zk = cos 6/2 + sin 0/2 . (i cos/+ j cos g + k cos h),
so that w = cos 0/2,
x = sin 0/2 . cos f }
y = sin 0/2 . cos g,
z = sin 0/2 . cos h.
From these we pass at once to Rodrigues subsidiary formulae,
* = l/w 2 = sec 2 0/2,
X = xjw = tan 0/2 . cos /*,
&c. = &c.
C. Kinematics of a Deformable System.
376. By the definition of Homogeneous Strain, it is evident
that if we take any three (nonco planar) unitvectors a, ft, 7 in
an unstrained mass, they become after the strain other vectors,
not necessarily unitvectors, a lf ft v 7 r
Hence any other given vector, which of course may be thus
expressed, p = XOL + yft + zy,
becomes p l = osa 1 + yft t + zy v
and is therefore known if a v ft v y 1 be given.
More precisely
pS.a/By = aS.jSyp + /SS.yap 4 yS.a/3p
becomes
pfi . CL(3y = fpS . OL(3y  o^S . /%> + Pfi . yap + 7^ . aft p.
Thus the properties of <J>, as in Chapter V., enable us to study
with great simplicity homogeneous strains in a solid or liquid.
* Camb. and Dub> Math. Journal Vol. i, (1846).
37&] KINEMATICS. 295
For instance, to find a vector whose direction is unchanged by
the strain, is to solve the equation
Vpcf)p = 0, or ^>p=gp (1),
where g is a scalar unknown.
[This vector equation is equivalent to three scalar equations, and
contains only three unknown quantities ; viz. two for the direction
of p (the tensor does not enter, or, rather, is a factor of each side),
and the scalar g.]
We have seen that every such equation leads to a cubic in g
which may be written
g*  m 2 g 2 + m^g  ra = 0,
where m 2 , m v m are scalars depending in a known manner on the
constant vectors involved in <. This must have one real root, and
may have three.
377. For simplicity let us assume that a, /3, 7 form a rectangular
system, then we may operate on (1) by S . a, S . ft, and $.7; and
thus at once obtain the equation for g, in the form
i =0
W, + <7,
To reduce this we have, for the term independent of g,
8aa lt SOL ft v Say 1 , = $. aftyS . a 1 j3 l y 1 (Ex. 9, Chap. III.),
which, if the mass be rigid, becomes 1.
The coefficient of the first power of g is,
Thus we can at once form the equation; which becomes,
for the special case of a rigid system,
10 (Saa, + Sflfr + SyyJ + g 2 (Saa, + S/3/3, + 8y 7l ) + g* = 0,
or !(
378. If we take Tp = C we consider a portion of the mass
initially spherical. This becomes of course
r*V,0,
an ellipsoid, in the strained state of the body.
296 QUATERNIONS. [379
Or if we consider a portion which is spherical after the strain, i.e.
2>, = C.
its initial form was T(j)p = C,
another ellipsoid. The relation between these ellipsoids is obvious
from their equations. (See 327.)
In either case the axes of the ellipsoid correspond to a rect
angular set of three diameters of the sphere ( 271). But we must
carefully separate the cases in which these corresponding lines in
the two surfaces are, and are not, coincident. For, in the former
case there is pure strain, in the latter the strain is accompanied
by rotation. Here we have at once the distinction pointed out by
Stokes* and Helmholtzf between the cases of fluid motion in
which there is, or is not, a velocitypotential. In ordinary fluid
motion the distortion is of the nature of a pure strain, i.e. is differ
entially nonrotational ; while in vortex motion it is essentially
accompanied by rotation. But the resultant of two pure strains is
generally a strain accompanied by rotation. The question before
us beautifully illustrates the properties of the linear and vector
function.
379. To find the criterion of a pure strain. Take a, /3, 7 now
as unitvectors parallel to the axes of the strainellipsoid, they
become after the strain, ace, 6/3, cy, provided the strain be pure.
Hence p 1 = <f)p = aaSap b{3S/3p cjSyp.
And we have, for the criterion of a pure strain, the property of
the function ^>, that it is selfconjugate, i.e.
Spcfxr = So(j)p.
380. Two pure strains, in succession, generally give a strain
accompanied by rotation. For if </>, ty represent the strains, since
they are pure we have
But for the compound strain we have
Pi = XP
and we have not generally
* Cambridge Phil Trans. 1845.
t Crclle, vol. Iv. 1857. See also PhiL Mag. (Supplement) June 1867.
381.] KINEMATICS. 297
For
by (1), and *//( is not generally the same as <f>*fy. (See Ex. 7 to
Chapter V.)
To find the lines which are most altered in length by the strain.
Here T$p is a maximum or minimum, while Tp is constant ;
so that
Sdp$$p = 0, Spdp = 0.
Hence fy fyp ~ X P>
and the required lines are the principal vectors of <j> (f), which
( 381) is obviously selfconjugate; i.e. denotes a pure strain.
381. The simplicity of this view of the question leads us to
suppose that we may easily separate the pure strain from the
rotation in any case, and exhibit the corresponding functions.
When the linear and vector function expressing a strain is
selfconjugate the strain is pure. When not selfconjugate, it may
be broken up into pure and rotational parts in various ways (ana
logous to the separation of a quaternion into the sum of a scalar
and a vector part, or into the product of a tensor and a versor
part), of which two are particularly noticeable. Denoting by a
bar a selfconjugate function, we have thus either
</> = ^ f or < = w.gr q~,
where e is a vector, and q a quaternion (which may obviously be
regarded as a mere versor). [The student must remark that,
although the same letters have been employed (from habit) in
writing the two last formulae, one is not a transformation of the
other. In the first a pure strain is succeeded by a rotation, in the
second the rotation is followed by the pure strain.]
That this is possible is seen from the fact that (j> involves nine
independent constants, while ty and r each involve six, and e and q
each three. If <j> be the function conjugate to <, we have
f?F.e< ),
so that 2tr = < + < ,
and 2 V. e ( ) = < </> ,
which completely determine the first decomposition. This is, of
course, perfectly well known in quaternions, but it does not seem
to have been noticed as a theorem in the kinematics of strains that
293 QUATERNIONS. [381.
there is always one, and but one, mode of resolving a strain into
the geometrical composition of the separate effects of (1) a pure
strain, and (2) a rotation accompanied by uniform dilatation
perpendicular to its axis, the dilatation being measured by
(sec. 1) where 6 is the angle of rotation.
In the second form (whose solution does not appear to have
been attempted), we have
where the pure strain precedes the rotation, and from this
* =3?.*( )q,
or in the conjugate strain the rotation (reversed) is followed by the
pure strain. From these
and w is to be found by the solution of a biquadratic equation*.
It is evident, indeed, from the identical equation
that the operator </> (/> is selfconjugate.
In the same way
or q 1 (<< q = t? (q l pq) = <j> <f> (q l pq),
* Suppose the cubic in "w to be
Now ro 2 is equal to 0, a known function, which we may call u. Thus
W 2 W,
and therefore ~m and w are commutative in multiplication.
Eliminating 5 between these equations we have, first,
(w  g z ) u + ^W g = Q=i&((a + g l ) g z w  g,
and finally w 3 + (2g^  g^} tf + (g^  2gg z ) u  # 2 = 0.
This must agree with the (known) cubic in w,
co 3  ??i 2 w 2 + mj<ta  m = 0, suppose ;
so that, by comparison of coefficients we have
and thus g is known, and r/ 2 = 
where
The values of the quantities g being found, OT is given in terms of w by the equation
383.] KINEMATICS. 299
which shew the relations between <fxf) f , fy fy, aid q.
To determine q we have
<l>p.q= q^p
whatever be p, so that
S.Vq(<l>v)p = 0,
or S.p((f>  =) Vq = 0,
which gives (</> r) F<? = 0.
The former equation gives evidently
Vq\\ V.(4>v)a(4>S)0
whatever be a and ft ; and the rest of the solution follows at once.
A similar process gives us the solution when the rotation precedes
the pure strain. [Proc. R. S. E. 18701.]
382. In general, if
the angle between any two lines, say p and a, becomes in the
altered state of the body
cosT 1 (8. UfoUjxr).
The plane $(J> = becomes (with the notation of 157)
S^ ^p = 8&*p = 0.
[For if X, IJL be any two vectors in it,
But they become <f>\, fa, and the line perpendicular to both is
Hence the angle between the planes Sp = 0, and Srjp = 0, which
is cos"^ S . UUr)), becomes
cos 1 ( 8 . UQ^UjT 1 *!).
The locus of lines equally elongated is, of course,
or Tfa = eTp,
a cone of the second degree.
383. In the case of a Simple Shear, we have, obviously,
Pi = <t>P = P +
where a is a unit vector, and
300 QUATERNIONS.
The vectors which are unaltered in length are given by
TptTfr
or 2S/3pSoip + /3*S 2 ap = 0,
which breaks up into S . ap = 0,
and
The intersection of this plane with the plane of a, {3 is perpen
dicular to 2/3 + /3 2 a. Let it be a + x/3, then
.e. 2#l=0.
Hence the intersection required is
4
For the axes of the strain, one is of course a/3, and the others
are found by making T(f)Up a maximum and minimum.
Let p = a + x/3,
then p l = (j)p = a + x/B /3,
T Pl
and T=P = max. or mm.,
JP
gives a; 2  ic + g* = 0,
from which the values of x (say oc^ and a? 8 ) are found.
Also, as a verification, we must show that the lines of the
body which become most altered in length are perpendicular to
one another both before and after the shear.
Thus S.(a + xfi) (OL + xfi) =  1 + /3 V? 2 ,
should be = 0. It is so, since, by the equation in x t
1
X \ X * ^2
Again
fif(a + (^l)/9} fa+(^l)/3}=  1 +/3 2 [x^ (^ +a; 9 ) + 1},
ought also to be zero. And, in fact,
by the equation for x ; so that this also is verified.
384. We regret that our limits do not allow us to enter farther
upon this very beautiful application. [The reader is referred to
385.] KINEMATICS. 301
Chapter X. of Kelland and Tail s Introduction to Quaternions ; in
which the treatment of linear and vector equations is based upon
the theory of homogeneous strain ; which, in its turn, is much more
fully developed than in the present work.]
But it may be interesting to consider briefly the effects of any
continuous displacements (of the particles of a body) by the help of
the operator V.
We have seen ( 148) that the effect of the operator SdpV,
upon any scalar function of the vector of a point, is to produce
total differentiation due to the passage from p to p + dp.
Hence if a be the displacement of p, that of p + $p is
Thus the strain of the group of particles near p is such that
T = T(TV)<r.
[Here we virtually assume that a is a continuous function of p.]
But if this correspond to a linear dilatation e, combined with
a rotation whose vectoraxis is e, both being infinitesimal,
< T = T (1 + e ) + Ver.
Thus, for all values of T, each with its proper e,
This gives at once (for instance by putting in succession for T
any three rectangular unit vectors)
from which we conclude as follows :
If a (a continuous function of p) represent the vector displace
ment of a point situated at the extremity of the vector p (drawn
from the origin)
SVo represents the consequent cubical compression of the
group of points in the vicinity of that considered, and
FVcr represents twice the vector axis of rotation of the same
group of points.
385. As an illustration, suppose we fix our attention upon a
group of points which originally filled a small sphere about the
extremity of p as centre, whose equation referred to that point is
Tco = c ................................. (1).
After displacement p becomes p + cr, and, by last section, p + w
becomes p + co f a ($o>V) a. Hence the vector of the new surface
302 QUATERNIONS. [386.
which encloses the group of points (drawn from the extremity of
p + a) is
ft), = ft)(ASfctV) <r (2).
Hence & is a homogeneous linear and vector function of co l ; or
ft) = (f>CO l}
and therefore, by (1), T^co^ = c,
the equation of the new surface, which is evidently a central surface
of the second degree, and therefore, of course, an ellipsoid.
We may solve (2) with great ease by approximation, if we as
sume that TVar is very small, and therefore that in the small term
we may put co l for CD ; i.e. omit squares of small quantities ; thus
&) = co 1 + ($ft> t V) a.
386. If the vector displacement of each point of a medium is in
the direction of, and proportional to, the attraction exerted at that
point by any system of material masses, the displacement is effected
without rotation.
For if Fp = C be the potential surface, we have Sadp a complete
differential ; and, by 334,
FVo = 0.
Conversely, if there be no rotation, the displacements are in the
direction of, and proportional to } the normal vectors to a series of
surfaces.
For =  V. dp FVcj =  (SdpV) a + VSadp,
where, in the last term, V acts on <r alone.
Now, of the two terms on the right, the first is ( 149, (4))
the complete differential da, and therefore the remaining term
must be a complete differential. This, of course, means that
Sadp
is a complete differential.
Thus, in a distorted system, there is no compression if
SVa = 0,
and no rotation if FVcr = ;
and evidently merely transference if <r = a= a constant vector,
which is one case of
V(T=0.
In the important case of a = VFp
there is (as proved already) no rotation, since
387.] KINEMATICS. 303
is evidently a scalar. In this case, then, there are only translation
and compression, and the latter is at each point proportional to the
density of a distribution of matter, which would give the potential
Fp. For if r be such density, we have at once
D. Axes of Inertia.
387. The Moment of Inertia of a body about a unit vector a
as axis is evidently
where p is the vector of the element ra of the mass, and the origin
of p is in the axis. [The letter h has, for an obvious reason, been
put here in place of the k which is usually employed for the radius
of gyration.]
Hence if we put /5 = e^ajh, where e is constant, we have, as
locus of the extremity of /3,
Me =  2m(Vj3p) 2 =  MS/3<t>/3 (suppose),
the wellknown ellipsoid. The linear and vector function, <,
depends only upon the distribution of matter about the (temporary)
origin.
If r be the vector of the centre of inertia, cr the vector of in
with respect to it, we have
p = <& + o,
where (31 (e)) 2m<r = 0;
and therefore Mh* =  2m {( Fern) 2 + ( Few) 2 J
Here fa is the unique value of < which corresponds to the
distribution of matter relative to the centre of inertia. The
equation last written gives the wellknown relation between the
moment of inertia about any line, and that about a parallel line
through the centre of inertia.
Hence, to find the principal axes of inertia at any point (the
origin, whose vector from the centre of inertia is w), note that h
is to be made max., min., or max.min., with the condition
a = l.
Thus we have SOL (crFaor + fact) = 0,
Sa a = 0;
* Proc, R, S, E., 18623.
304 QUATERNIONS.
therefore (/^a + TzVavr pa.= 7i 2 a (by operating by S . a).
Hence (^  tf  w*) a =  vrSam .................. (1),
determines the values of a, k 2 being found from the cubic
Si* (fa  W  w*) 1 <GF =  1 .................. (2).
Now the normal to
S<r(<f> l h* B*) l <r = l .................. (3),
at the point a is parallel to
(*,A rr.
But (3) passes through OT, by (2), and there the normal is
fah* <**)*,
which, by (1), is parallel to one of the required values of a. (3) is,
of course ( 288), one of the surfaces confocal with the ellipsoid
S.o ( j) 1 (T = l.
Thus we prove Binet s theorem that the principal axes at any
point are normals to the three surfaces of the second degree, confocal
with the central ellipsoid, which pass through that point.
EXAMPLES TO CHAPTER XI.
1. Form, from kinematical principles, the equation of the
cycloid ; and employ it to prove the wellknown elementary
properties of the arc, tangent, radius of curvature, and evolute,
of the curve.
2. Interpret, kinematically, the equation
where /3 is a given vector, and a a given scalar.
Shew that it represents a plane curve ; and give it in an
integrated form independent of t.
3. If we write or = (3t p,
the equation in (2) becomes
/3 is = a Uvr.
Interpret this kinematically ; and find an integral.
What is the nature of the step we have taken in transforming
from the equation of (2) to that of the present question ?
EXAMPLES TO CHAPTER XI. 305
4. The motion of a point in a plane being given, refer it to
(a) Fixed rectangular vectors in the plane.
(b) Rectangular vectors in the plane, revolving uniformly
about a fixed point.
(c) Vectors, in the plane, revolving with different, but uni
form, angular velocities.
(d) The vector radius of a fixed circle, drawn to the point of
contact of a tangent from the moving point.
In each case translate the result into Cartesian coordinates.
5. Any point of a line of given length, whose extremities move
in fixed lines in a given plane, describes an ellipse.
Shew how to find the centre, and axes, of this ellipse ; and
the angular velocity, about the centre of the ellipse, of the tracing
point when the describing line rotates uniformly.
Transform this construction so as to shew that the ellipse is a
hypotrochoid.
When the fixed lines are not in one plane, what is the locus ?
6. A point, A, moves uniformly round one circular section of
a cone ; find the angular velocity of the point, a, in which the
generating line passing through A meets a subcontrary section,
about the centre of that section.
7. Solve, generally, the problem of finding the path by which
a point will pass in the least time from one given point to another,
the speed at the point of space whose vector is p being expressed
by the given scalar function
fp
Take also the following particular cases :
( a ) fp = a while Sap > 1,
fp=b while Sap < 1.
(b) fp = TSap.
(c) fp = p 2 . (Tait, Trans. R. S. E. t 1865.)
8. If, in the preceding question,//? be such a function of Tp
that any one swiftest path is a circle, every other such path is a
circle, and all paths diverging from one point converge accurately
in another. (Maxwell, Camb. and Dub. Math. Journal, IX. p. 9.)
T. Q. I. 20
306 QUATERNIONS.
9. Interpret, as results of the composition of successive conical
rotations, the apparent truisms
/,
7#
a K i 8 7 /3
and  z ......  5  = 1.
K L 6 7 p a
(Hamilton, Lectures, p. 334.)
10. Interpret, in the same way, the quaternion operators
Qffl ">
11. Find the axis and angle of rotation by which one given
rectangular set of unitvectors a, /3, 7 is changed into another
given set 15 yS 1? 7 r
12. Shew that, if (f>p = p + Fe/>,
the linear and vector operator $ denotes rotation about the vector
e, together with uniform expansion in all directions perpendicular
to it.
Prove this also by forming the operator which produces the
expansion without the rotation, and that producing the rotation
without the expansion ; and finding their joint effect.
13. Express by quaternions the motion of a side of one right
cone rolling uniformly upon another which is fixed, the vertices of
the two being coincident.
14. Given the simultaneous angular velocities of a body about
the principal axes through its centre of inertia, find the position
of these axes in space at any assigned instant.
15. Find the linear and vector function, and also the quater
nion operator, by which we may pass, in any simple crystal of the
cubical system, from the normal to one given face to that to
another. How can we use them to distinguish a series of faces
belonging to the same zone ?
16. Classify the simple forms of the cubical system by the
properties of the linear and vector function, or of the quaternion
operator, mentioned in (15) above.
EXAMPLES TO CHAPTER XL 307
17. Find the vector normal of a face which truncates symme
trically the edge formed by the intersection of two given faces.
18. Find the normals of a pair of faces symmetrically truncat
ing the given edge.
19. Find the normal of a face which is equally inclined to
three given faces.
20. Shew that the rhombic dodecahedron may be derived from
the cube, or from the octahedron, by truncation of the edges.
21. Find the form whose faces replace, symmetrically, the
edges of the rhombic dodecahedron.
22. Shew how the two kinds of hemihedral forms are indi
cated by the quaternion expressions.
23. Shew that the cube may be produced by truncating the
edges of the regular tetrahedron. If an octahedron be cut from a
cube, and cubes from its tetrahedra, all by truncation of edges, the
two latter cubes coincide.
24. Point out the modifications in the auxiliary vector func
tion required in passing to the pyramidal and prismatic systems
respectively.
25. In the rhombohedral system the auxiliary quaternion
operator assumes a singularly simple form. Give this form, and
point out the results indicated by it.
26. Shew that if the hodograph be a circle, and the accelera
tion be directed to a fixed point ; the orbit must be a conic section,
which is limited to being a circle if the acceleration follow any
other law than that of gravity.
27. In the hodograph corresponding to acceleration f(D )
directed towards a fixed centre, the curvature is inversely as
28. If two circular hodographs, having a common chord, which
passes through, or tends towards, a common centre of force, be cut
by any two common orthogonal s, the sum of the two times of
hodographically describing the two intercepted arcs (small or large)
will be the same for the two hodographs. (Hamilton, Elements,
p. 725 )
202
308 QUATERNIONS.
29. Employ the last theorem to prove, after Lambert, that the
time of describing any arc of an elliptic orbit may be expressed in
terms of the chord of the arc and the extreme radii vectores.
30. If q ( ) q~ l be the operator which turns one set of rect
angular unitvectors a, /3, 7 into another set a 4 , @ lt y lt shew that
there are three equations of the form
31. If a ray, a, fall on a fine, polished, wire 7, shew that on
reflection it forms the surface
a right cone.
32. Find the path of a point, and the manner of its descrip
tion, when
33. In the first problem of 336 shew that
Vq~*q = Vv, or V . uq~ l 0.
Also that (Vv) 2 =  2V 2 <y, or 4tTVw* = 0.
Again, shew that there are three equations of the form
dv _ ._ 2 ^ dv
_ Vv = iV*v + V j.
doc dx
From these last deduce, by a semiCartesian process, the result
as in the text.
34. Give the exact solution of
to, = ft>  Sort . a: ( 385.)
[Note that we may assume, cr being given,
do = <j)dp,
where the constituents of (f> are known functions of p. Thus we
have what is wanted for the problem above : viz.
with various other important results, such as
V<7 = 2^, &c.]
CHAPTEE XII.
PHYSICAL APPLICATIONS.
388. WE propose to conclude the work by giving a few in
stances of the ready applicability of quaternions to questions of
mathematical physics, upon which, even more than on the Geo
metrical or Kinematical applications, the real usefulness of the
Calculus must mainly depend except, of course, in the eyes of
that section of mathematicians for whom Transversals and Anhar
monic Pencils, &c. have a to us incomprehensible charm. Of course
we cannot attempt to give examples in all branches of physics, nor
even to carry very far our investigations in any one branch : this
Chapter is not intended to teach Physics, but merely to shew by
a few examples how expressly and naturally quaternions seem to
be fitted for attacking the problems it presents.
We commence with a few general theorems in Dynamics the
formation of the equations of equilibrium and motion of a rigid
system, some properties of the central axis, and the motion of a
solid about its centre of inertia. The student may profitably
compare, with the processes in the text, those adopted by Hamilton
in his Elements (Book III., Chap. III., Section 8).
A. Statics of a Rigid System.
389. When any forces act on a rigid body, the force y8 at the
point whose vector is a, &c., then, if the body be slightly displaced,
so that a becomes a f Sa, the whole work done against the forces is
SS/SSo.
310 QUATERNIONS. [39
This must vanish if the forces are such as to maintain equilibrium.
Hence the condition of equilibrium of a rigid body is
For a displacement of translation Sa is any constant vector, hence
2/3 = .............................. (1).
For a rotationdisplacement, we have by 371, e being the axis,
and Te being indefinitely small,
SOL = Fea,
and 28.ftVect = 2S.eVaft = S.6$(VcLfi) = 0,
whatever be e, hence 2 . Va.fi = ........................... (2).
These equations, (1) and (2), are equivalent to the ordinary six
equations of equilibrium.
390. In general, for any set of forces, let
2/3 = ft,
2. Fa/3 = a,,
it is required to find the points for which the couple a t has its axis
coincident with the resultant force /3 r Let 7 be the vector of such
a point.
Then for it the axis of the couple is
and by condition xft l = a l
Operate by 8ft l ; therefore
and Vyft, = a,  ft~ l Sa^ = ft, Vafi\
or y = Va 1 ft l + uft v
a straight line (the Central Axis) parallel to the resultant force.
[If the resultant force and couple be replaced by an equivalent
in the form of two forces, ft at a, and ft at of, we have
The volume of the tetrahedron whose opposite edges are @,
(acting as above stated) is as 8 . ftV(a a) ft. But
so that the volume is as 8^ (& + ft) = Sa^ a constant whatever
pair of equivalent forces be taken.]
393] PHYSICAL APPLICATIONS. 311
391. To find the points about which the couple is least.
Here T (a x  Fy^) = minimum.
Therefore 8 . (a,  Fy/^) VPrf = 0,
where 7 is any vector whatever. It is useless to try 7 = ft v but
we may put it in succession equal to OL I and to Vaft^ Thus
and ( FaA) 2  {3*S . 7 FA = 0.
Hence 7 = x VOL ft \ f yfi v
and by operating with S . Va.J3 v we get
or 7 = 1 /
the same locus as in last section.
392. The couple vanishes if
This necessitates $,, = 0,
or the force must be in the plane of the couple. If this be the
case,
7 = 1 /3," + */3 I >
still the central axis.
To assign the values of forces f, f t , to act at e, e p and be
equivalent to the given system.
Hence Fef + Fe, (/9,  f) = a,,
and f = (6  e,) 1 (a,  F^/3,) + x (e  6 t ).
Similarly for f r The indefinite terms may be omitted, as they
must evidently be equal and opposite. In fact the}^ are any equal
and opposite forces whatever acting in the line joining the given
points.
393. If a system of parallel forces act on a rigid body, say
x/3 at a, &c.
they have the single resultant /3S (a), at a, such that
312 QUATERNIONS. [393
Hence, whatever be the common direction of the forces, the
resultant passes through
_ 2 pa)
~
If 5 (#) = 0, the resultant is simply the couple
By the help of these expressions for systems of parallel forces
we can easily proceed to the case of forces generally.
Thus if any system of forces, /3, act at points, a, of a rigid body ;
and if i, j, k be a system of rectangular unit vectors such that
the resultant force is
,
bk acting at , or
as we may write it. Take this as origin, then <& = 0.
The resultant couple, in the same way, is
or V(i(f)i+j(f)j).
Now </>i, $7, <A? are invariants, in the sense that they retain
the same values however the forces and (with them) the system
i, j, k be made to rotate : provided they preserve their mutual
inclinations, and the forces their points of application. For the
as are constant, and quantities of the form S/3i, S/3j t or S/3k are
not altered by the rotation.
We may select the positions of i and j so that (f>i and <fyj
shall be perpendicular to one another. For this requires only
8 . ty <f>j = 0, or S.ipthj = 0.
But ( 381) </> (/> is a selfconjugate function; and, by our change of
origin, k is parallel to one of its chief vectors. The desired result
is secured if we take i, j as the two others.
With these preliminaries we may easily prove Minding s
Theorem :
If a system of forces, applied at given points of a rigid body,
have their directions changed in any way consistent with the
preservation of their mutual inclinations, they have in an infinite
number of positions a single force as resultant. The lines of
action of all such single forces intersect each of two curves fixed
in space.
393] PHYSICAL APPLICATIONS. 313
The condition for the resultant s being a single force in the
line whose vector is p is
bVkp=V(i(j)i+j(f>j\
which may be written as
bp = xk J4>i + i<f)j.
That the two last terms, together, form a vector is seen by
operating on the former equation by S.k , for we thus have
We may write these equations for convenience as
bp = xk jo. + i/3 ....................... ,(1),
SjaSij3 = () ....................... (2).
[The student must carefully observe that a and /3 are now
used in a sense totally different from that in which they first
appeared, but for which they are no longer required. If this
should puzzle him, he may change a into 7, and (3 into 8, in the last
two equations and throughout the remainder of this section.]
Our object now must be to express i and j in terms of the single
variable k, which is afterwards to be eliminated for the final
result.
From (2) we find at once
whence we easily arrive at either of the following
or _ y = a * + + (Ska)* + (Sk/3) z + 28 . kcfi]
Substituting for i and^ in (1) their values (3), we have
= (vrz)ka/3 ........................... (5),
where w t which is now used for a linear and vector function, is
defined by the equation
srp = aSap ftSffp.
Obviously r (a/3) = 0,
so that (w  z)~ l (a/3) =   a/3.
z
Thus yb(^2) l p = k+ ..................... (6).
314 QUATERNIONS. [394
Multiply together the respective members of (5) and (6), and take
the scalar, and we have
or, by (4), = f + z + a* + /3 2 +
z
= /+
z
which, for z a 2 , or z =  /3 2 , gives as the required curves the
focal conies of the system
394. The preceding investigation was based on the properties
of a system of parallel forces ; and thus has a somewhat composite,
semiCartesian, character.
That which follows is much more purely quaternionic. It is
taken from the Trans. R. S. E. 1880.
When any number of forces act on a rigid system ; (3 l at the
point GL V /3 2 at 2 , &c., their resultant consists of the single force
/8 = 2/3
acting at the origin, and the couple
* = 2F/3a ........................... (1).
If these can be reduced to a single force, the equation of the
line in which that force acts is evidently
F/8p = 2F/3a .............................. (2).
Now suppose the system of forces to turn about, preserving
their magnitudes, their points of application, and their mutual
inclinations, and let us find the fixed curves in space, each of
which is intersected by the line (2) in every one of the infinite
number of its positions.
Operating on (2) by F. /3, it becomes
with the notation of Chap. V. Now, however the forces may
turn, 0/3 = 2a/3/3
is an absolute constant ; for each scalar factor as S/3J3 is unaltered
by rotation. Let us therefore change the origin, i.e. the value of
each a, so as to make
= ........................ (3).
395] PHYSICAL APPLICATIONS. 315
This shews that / is one of the three principal vectors of $, and
we see in consequence that $ may be expressed in the form
where 7 and S are unit vectors, forming with JB a rectangular
system. They may obviously be so chosen that 7 and 8 shall be
at right angles to one another, but these (though constants) are
not necessarily unit vectors.
Equation (2) is now
&F/3 P =F77 +FSS ..................... (2 ),
where b is the tensor, and ft the versor, of ft.
The condition that the force shall lie in the plane of the couple
is, of course, included in this, and is found by operating by S .0.
Thus 8 (&/  78 ) = ........................... (4).
395. We have here all the data of the problem, and solutions
can only differ from one another in the mode of attacking
and (4).
Writing (4) in the form
7 (8
we have at once ty = F/3S + /3 Vfiy, } ,.,.
whence tS =  V@y + ft V/3& j"
where t is an undetermined scalar.
By means of these we may put (2 ) in the form
where * f = ySy ( )S SS ( ).
Let the tensors of 7 and 8" be e v e 2 respectively, and let be a
unit vector perpendicular to them, then we may write
btp^xpe.efl + isp ........................ (5).
Operating by (or + a;)" 1 , and noting that
we have bt (cr + = /3   ................. (5 ).
M?
Taking the scalar of the product of (5) and (5 ) we have
+ x)~ l p =  (sc/3  e&p
316 QUATERNIONS. [39$.
But by (4 ) we have
(6),
so that, finally,
(7).
Equation (7), in which f is given by (6) in terms of ft, is true
for every point of every single resultant. But we get an immense
simplification by assuming for x either of the particular values
e* or e*. For then the righthand side of (7) is reduced
to unity, and the equation represents one or other of the focal
conies of the system of confocal surfaces
a point of each of which must therefore lie on the line (5).
396. A singular form, in which Minding s Theorem can be
expressed, appears at once from equation (2 ). For that equation
is obviously the condition that the linear and vector function
bpS0( ) + y8y( ) + m( )
shall denote a pure strain.
Hence the following problem : Given a set of rectangular unit
vectors, which may take any initial position : let two of them, after
a homogeneous strain, become given vectors at right angles to one
another, find what the third must become that the strain may be
pure. The locus of the extremity of the third is, for every initial
position, one of the single resultants of Minding s system ; and
therefore passes through each of the fixed conies.
Thus we see another very remarkable analogy between strains
and couples, which is in fact suggested at once by the general
expression for the impure part of a linear and vector function.
397. The scalar t, which was introduced in equations (4 ), is
shewn by (6) to be a function of ft alone. In this connection it
is interesting to study the surface of the fourth degree
 (e* + e*) r 2  2e 1 e 2 2Wr = 1,
where r =  ft.
L
But this may be left as an exercise.
Another form of t (by 4 ) is $77 + SSB .
399] PHYSICAL APPLICATIONS. 317
Meanwhile (6) shews that for any assumed value of ft there
are but two corresponding Minding lines. If, on the other hand,
p be given there are in general four values of /3.
398. For variety, and with the view of further exploring this
very interesting question, we may take a different mode of
attacking equations (4) and (2 ), which contain the whole matter.
In what follows b will be merged in p, so that the scale of the
result will be altered.
Operating by V. ft we transform (2 ) into
p + pS0p=(<ySyi3 + 8SVl3) (2").
Squaring both sides we have
p* + S*/3p = S/3/3 (8).
Since /3 is a unit vector, this may be taken as the equation of a
cyclic cone; and every central axis through the point p lies upon it.
For we have not yet taken account of (4), which is the condition
that there shall be no couple.
To introduce (4), operate on (2") by S.y and by S.8 . We
thus have, by a double employment of (4),
Next, multiplying (8) by Sftvrfi, and adding to it the squares of
(9), we have
p*Sfa/3  2S0pSfap  Spvp =  Sfa P (10).
This is a second cyclic cone, intersecting (8) in the four directions
/3. Of course it is obvious that (8) and (10) are unaltered by the
substitution of p + y/3 for p.
If we look on ft as given, while p is to be found, (8) is the
equation of a right cylinder, and (10) that of a central surface of
the second degree.
399. A curious transformation of these equations may be made
by assuming p 1 to be any other point on one of the Minding lines
represented by (8) and (10). Introducing the factor /3 2 ( = 1) in
the terms where @ does not appear, and then putting throughout
II PiP (11)
(8) becomes  p 2 Pl 2 + S*pp 1 = S ( Pl  p) <& ( Pl  p) (8 ).
As this is symmetrical in p, p v we should obtain only the same
318 QUATERNIONS. [400.
result by putting p l for p in (8), and substituting again for /3 as
before.
From (10) we obtain the corresponding symmetrical result
(p 2  SppJ Sp l ^p l + (p, 2  SppJ Spvp =  Sppfi (p,  p) VF (p l  p)
S( pi p)^( Pl p) ...... (10 ).
These equations become very much simplified if we assume p and p l
to lie respectively in any two conjugate planes ; specially in the
planes of the focal conies, so that SS p = 0, and Sy f p l = 0.
For if the planes be conjugate we have
i = 0,
and if, besides, they be those of the focal conies,
Bph SffpBffto
8pvr 2 p = e*Spwp, &c.,
and the equations are
P 2 p 2
and p*Sp l &p l + p*Spwp = 8p 1 vr 3 p l Spiv 2 p ......... (1 0").
From these we have at once the equations of the two Minding
curves in a variety of different ways. Thus, for instance, let
p t =p*
and eliminate p between the equations. We get the focal conic in
the plane of ft , 7 . In this way we see that Minding lines pass
through each point of each of the two curves ; and by a similar
process that every line joining two points, one on the one curve,
the other on the other, is a Minding line.
400. Another process is more instructive. Note that, by the
equations of condition above, we have
Then our equations become
+ Shmpt = 0>
and (p 2 + e 2 ) Sp^p, + (p* + e 2 ) Spvp = 0.
If we eliminate p 2 or p 2 from these equations, the resultant
obviously becomes divisible by Spivp or Sp^p v and we at once
obtain the equation of one of the focal conies.
403.] PHYSICAL APPLICATIONS. 319
401. In passing it may be well to notice that equation (10)
may be written in the simpler form
8 . p/3pvr/3 + Spvp = Sj3<n*l3.
Also it is easy to see that if we put
we have (8) in the form S/36 = 0,
and by the help of this (10) becomes
This gives another elegant mode of attacking the problem.
402. Another valuable transformation of (2") is obtained by
considering the linear and vector function, x suppose, by which fi,
7, S are derived from the system ft , Uy, US . For then we have
obviously
P = *>XP + X**XP ..................... (2 ")
This represents any central axis, and the corresponding form of the
Minding condition is
S.Jxvr*V = S.&x SF ~*y .................. (4").
Most of the preceding formulae may be looked upon as results of
the elimination of the function x from these equations. This forms
probably the most important feature of such investigations, so far
at least as the quaternion calculus is concerned.
403. It is evident from (2 ") that the vectorperpendicular
from the origin on the central axis parallel to xfi is expressed by
But there is an infinite number of values of % for which Ur is a
given versor. Hence the problem ; to find the maximum and
minimum values of Tr, when Ur is given i.e. to find the surface
bounding the region which is filled with the feet of perpendiculars on
central axes.
We have 2V =  S .
= TrS.x/3 Ur.
Hence = S .
But as Tj3 is constant = 8.
320 QUATERNIONS. [404.
These three equations give at sight
where u, u are unknown scalars. Operate by S . %/3 and we have
 T 2 ru=0,
so that ST (sr + r 2 ) 1 r = 0.
This differs from the equation of Fresnel s wavesurface only in
having w + r 2 instead of TZ + r~ 2 , and denotes therefore the reciprocal
of that surface. In the statical problem, however, we have izft = 0,
and thus the corresponding wavesurface has zero for one of its
parameters. [See 435.]
[If this restriction be not imposed, the locus of the point
where (/> is now any given linear and vector function whatever,
will be found, by a process precisely similar to that just given, to be
8.(r f ) (f < + T") (T  f /3 ) = 0,
where < is the conjugate of </>.]
B. Kinetics of a Rigid System.
404. For the motion of a rigid system, we have of course
by the general equation of Lagrange.
Suppose the displacements &a to correspond to a mere transla
tion, then So. is any constant vector, hence
2 (ma  /3) = 0,
or, if a x be the vector of the centre of inertia, and therefore
we have at once a^m 2/3 = 0,
and the centre of inertia moves as if the whole mass were
concentrated in it, and acted upon by all the applied forces.
405. Again, let the displacements Sa correspond to a rotation
about an axis e, passing through the origin, then
it being assumed that Te is indefinitely small.
406.] PHYSICAL APPLICATIONS. 321
Hence 28 . e Vet (ma  ) = 0,
for all values of e, and therefore
2 . Fa (ma  /9) = 0,
which contains the three remaining ordinary equations of motion.
Transfer the origin to the centre of inertia, i.e. put a = ^ + CT,
then our equation becomes
2F(a, + tsr) (ma, + WOT  ft) = ;
or, since 2mar = 0,
2 Far (m#  ) + Fa, (a\2m  2y3) = 0.
But ( 404) 3,2m  2)9 = 0,
hence our equation is simply
Now 2 Fsr/3 is the couple, about the centre of inertia, produced
by the applied forces ; call it f, then
? ........................... (1).
406. Integrating once,
2roFM!r = 7 + /f<ft ..................... (2).
Again, as the motion considered is relative to the centre of
inertia, it must be of the nature of rotation about some axis, in
general variable. Let e denote at once the direction of, and the
angular velocity about, this axis. Then, evidently,
CT = Fesr.
Hence, the last equation may be written
2mcr Fear = 7 + fdt.
Operating by S . e, we get
2m(Vei*y = Sev + Sft;dt .................. (3).
But, by operating directly by 2fSedt upon the equation (1), we
get
2m(FS7) 2 = /^ 2 + 2/>Sfe^ ............... (4).
Equations (2) and (4) contain the usual four integrals of the first
order, h being here an arbitrary constant, whose value depends
upon the initial kinetic energy of the system. By 387 we see
how the principal moments of inertia are involved in the left
member.
T. Q. I. 21
322 QUATERNIONS. [407.
407. When no forces act on the body, we have f=0, and
therefore
SmcrFe sr = <y ........................ (5),
2mr 8 = 2m(F6w) 8 = ^ .................. (6),
and, from (5) and (6), Se 7 = /i 2 ........................... (7).
One interpretation of (6) is, that the kinetic energy of rotation
remains unchanged : another is, that the vector e terminates in an
ellipsoid whose centre is the origin, and which therefore assigns
the angular velocity when the direction of the axis is given ;
(7) shews that the extremity of the instantaneous axis is always
in a plane fixed in space.
Also, by (5), (7) is the equation of the tangent plane to (6) at
the extremity of the vector e. Hence the ellipsoid (6) rolls on the
plane (7).
From (5) and (6), we have at once, as an equation which e
must satisfy,
7 2 2 . m (Fe^r) 2 =  h* (2 . mw Fesr) 2 .
This belongs to a cone of the second degree fixed in the body.
Thus all the results of Poinsot regarding the motion of a rigid body
under the action of no forces, the centre of inertia being fixed, are
deduced almost intuitively: and the only difficulties to be met
with in more complex properties of such motion are those of
integration, which are inherent to the subject, and appear what
ever analytical method is employed. (Hamilton, Proc. R. /. A.
1848.)
If we write (5) as
+" 1 6 = 7 ........................... (5),
the special notation <}> indicating that this linear and vector
function is related to the principal axes of the body, and not to
lines fixed in space, and consider the ellipsoid (of which e is a
sernidiameter)
Se+^e^h* ........................ (6),
we may write the equation of a confocal ellipsoid (also fixed in the
body) as
Any tangent plane to this is
S.<r(4, + P r P = h > ..................... (9).
4O8.J PHYSICAL APPLICATIONS. 323
If this plane be perpendicular to 7, we may write
so that, by (5)
(10).
The plane (9) intercepts on 7 a quantity h*/xTy, which is constant
by (8) and (10).
The vector velocity of the point p is
Vep px Vey = p Vyp
(by two applications of (10)). Hence the point of contact, p,
revolves about 7 with angular velocity pTy. That is, if the
plane (9) be rough, and can turn about 7 as an axis, the ellipsoid (8)
instead of sliding upon it, will make it rotate with uniform angular
velocity. This is a very simple mode of obtaining one of Sylvester s
remarkable results. (Phil. Trans. 1866.)
408. For a more formal treatment of the problem of the
rotation of a rigid body, we may proceed as follows :
Let ct be the initial position of r, q the quaternion by which
the body can be at one step transferred from its initial position to
its position at time t. Then
w = qctq 1
and Hamilton s equation (5) of last section becomes
S . mqaq" 1 V . eqaq~ l = 7,
or S . mq {aS . aq~ l eq q~ l eqc?} q~ l = 7.
The vector 7 is now written for 7 + f%dt of 406, as f is required
for a new purpose. Thus 7 represents the resultant moment of
momentum, and will be constant only if there is no applied couple.
Let (f>p = S m (aSap 2 /o) .................. (1),
where $ (compare 387) is a selfconjugate linear and vector
function, whose constituent vectors are fixed in the body in its
initial position. Then the previous equation may be written
or
For simplicity let us write
<T ( 1 =
q~ l yq =
Then Hamilton s dynamical equation becomes simply
0i? = ? (3).
212
324 QUATERNIONS. [409.
409. It is easy to see what the new vectors rj and f represent.
For we may write (2) in the form
from which it is obvious that 77 is that vector in the initial position
of the body which, at time t, becomes the instantaneous axis in the
moving body. When no forces act, 7 is constant, and f is the
initial position of the vector which, at time t, is perpendicular to
the invariable plane.
410. The complete statement of the problem is contained in
equations (2), (3) above, and (4) of 372*. Writing them again,
we have
7? = <Z? .............................. (2),
*? = ? .............................. (3)
We have only to eliminate f and 77, and we get
2<7=</<r(<r 7<?) ........................ (5),
in which q is now the only unknown; 7, if variable, being supposed
given in terms of q and t.
It is hardly conceivable that any simpler, or more easily
interpretable, expression for the motion of a rigid body can be
presented until symbols are devised far more comprehensive in
their meaning than any we yet have.
411. Before entering into considerations as to the integration
of this equation, we may investigate some other consequences of
the group of equations in 410. Thus, for instance, differentiating
(2), we have
and, eliminating q by means of (4),
* To these it is unnecessary to add
Tq = constant,
as this constancy of Tq is proved by the form of (4). For, had Tq been variable,
there must have been a quaternion in the place of the vector 17. In fact,
l
412.] PHYSICAL APPLICATIONS. 325
whence, eliminating y by the help of (2),
which gives, in the case when no forces act, the forms
= W? .............................. (0).
and (as f =(/>??)
(/>7) = F.77c/>?7 ........................ (7).
To each of these the term q~ lf yq must be added on the right, if
forces act,
412. It is now desirable to examine the formation of the func
tion c/>. By its definition 408, (1), we have
<frp = 2 . m (aSap  a. 2 p),
= S. maVap.
Hence  Sp<j>p = 2.m (TVap}\
so that Sp<j)p is the moment of inertia of the body about the
vector p, multiplied by the square of the tensor of p. Compare
387. Thus the equation
evidently belongs to an ellipsoid, of which the radiivectores are
inversely as the square roots of the moments of inertia about them;
so that, if i, j, k be taken as unitvectors in the directions of its
axes respectively, we have
(8),
A, B, G, being the principal moments of inertia. Consequently
<j>p = {AiSip + BjSjp + CkSkp] ..................... (9).
Thus the equation (7) for 77 breaks up, if we put
it] = ia) l \j(o 2 + &o) 3 ,
into the three following scalar equations
which are the same as those of Euler. Only, it is to be understood
that the equations just written are not primarily to be considered
as equations of rotation. They rather express, with reference to
326 QUATERNIONS. [413
fixed axes in the initial position of the body, the motion of the
extremity, a) lt o> 2 , &> 3 , of the vector corresponding to the instan
taneous axis in the moving body. If, however, we consider c^, o^,
o> 3 as standing for their values in terms of w t x, y, z ( 416 below),
or any other coordinates employed to refer the body to fixed axes,
they are the equations of motion.
Similar remarks apply to the equation which determines , for
if we put
(6) may be reduced to three scalar equations of the form
i_i\
in I 23
413. Euler s equations in their usual form are easily deduced
from what precedes. For, let
whatever be p ; that is, let <}> represent with reference to the moving
principal axes what < represents with reference to the principal
axes in the initial position of the body, and we have
c>e
= V.
which is the required expression.
But perhaps the simplest mode of obtaining this equation is to
start with Hamilton s unintegrated equation, which for the case
of no forces is simply
S . m Vvriir = 0.
But from TS
we deduce OT
= sre 2 eSetz +
so that S . m ( Vevr Set? etrr 2 + TV Set?) = 0.
If we look at equation (1), and remember that <j> differs from </>
simply in having trr substituted for a, we see that this may bo
written
<j>e = 0,
4I5J PHYSICAL APPLICATIONS. 327
the equation before obtained. The first mode of arriving at it has
been given because it leads to an interesting set of transformations,
for which reason we append other two.
By (2)
7 = qq \
therefore = qq~\ qgq 1 + qgq 1  qq~ l qq~ l ,
or ^
But, by the beginning of this section, and by (5) of 407, this
is again the equation lately proved.
Perhaps, however, the following is neater. It occurs in Hamil
ton s Elements.
By (5) of 407 <>e = 7.
Hence <{>e = <j>e = 2 . m (&
= V. e^L .
414. However they are obtained, such equations as those of
412 were shewn long ago by Euler to be integrable as follows,
Putting Zfco^^co^dt = s,
we have Aco* = AQ* + (BC) s,
with other two equations of the same form. Hence
9/7*= ^f
. CA \*/^ AB
so that t is known in terms of s by an elliptic integral. Thus,
finally, rj or f may be expressed in terms of t ; and in some of the
succeeding investigations for q we shall suppose this to have been
done. It is with this integration, or an equivalent one, that most
writers on the farther development of the subject have commenced
their investigations.
415. By 406, 7 is evidently the vector moment of momen
tum of the rigid body ; and the kinetic energy is
 2 . mtsr* = 
But #67 = S . q l eqq~ l y
328 QUATERNIONS. [4 1 6.
so that when no forces act
But, by (2), we have also
T=T % or
so that we have, for the equations of the cones described in the
initial position of the body by rj and f, that is, for the cones de
scribed in the moving body by the instantaneous axis and by the
perpendicular to the invariable plane,
This is on the supposition that 7 and h are constants. If forces act,
these quantities are functions of t, and the equations of the cones
then described in the body must be found by eliminating t between
the respective equations. The final results to which such a process
will lead must, of course, depend entirely upon the way in which t
is involved in these equations, and therefore no general statement
on the subject can be made.
416. Recurring to our equations for the determination of q, and
taking first the case of no forces, we see that, if we assume rj to
have been found (as in 414) by means of elliptic integrals, we
have to solve the equation
* To get an idea of the nature of this equation, let us integrate it on the suppo
sition that ?j is a constant vector. By differentiation and substitution, we get
Hence q = Q l cos  t + Q. 2 sin ~ t.
Substituting in the given equation we have
Tn (Q lB in ^t + Q. 2 cos^t^ = (Q lC os T J t + Q.sm^ t) r,.
Hence ^ . Q 2 = Q^,
T*.Qi=Qrfl,
which are virtually the same equation, and thus
tTr,
)* 
And the interpretation of q ( ) q~ l will obviously then be a rotation about tj
through the angle tTrj, together with any other arbitrary rotation whatever. Thus
any position whatever may be taken as the initial one of the body, and Q 1 ( ) Q^ 1
brings it to its required position at time i = 0.
4 1 7] PHYSICAL APPLICATIONS. 329
that is, we have to integrate a system of four other differential
equations harder than the first,
Putting, as in 412,
where o> lt co l2) w 3 are supposed to be known functions of t, and
q = w + ix +jy + kz,
1 ,. dw dx dii dz
this system is 9 W = ~X = ~Y = ~7 y
where W = w^o w^y &>/,
X ft)jW + <DJ ft)./,
Y =
Z =
or, as suggested by Cayley to bring out the skew symmetry,
X = . ft) 3 7/ ft)/ + W V W,
Y = ft> 3 . + ft)/ + (0> 2 W,
Z = ft) 2 wjj . 4 o) 3 w,
W = a)^ a) 2 y  &)/
Here, of course, one integral is
w 2 + x z + y* + z 2 = constant.
It may suffice thus to have alluded to a possible mode of
solution, which, except for very simple values of rj, involves very
great difficulties. The quaternion solution, when TJ is of constant
length and revolves uniformly in a right cone, will be given
later.
417. If, on the other hand, we eliminate TJ, we have to
integrate
q^(q l yq) = 2q,
so that one integration theoretically suffices. But, in consequence
of the present imperfect development of the quaternion calculus,
the only known method of effecting this is to reduce the quaternion
equation to a set of four ordinary differential equations of the first
order. It may be interesting to form these equations.
Put q = w + ix +jy + kz,
7 ia +jb +kc,
330 QUATERNIONS. [4 1 ?
then, by ordinary quaternion multiplication, we easily reduce the
given equation to the following set :
dt _ dw _ dx _ dy _ dz
2~~ W~X = ~~ Y ~T
where
W = x&y3$z<& or X = .
X= W + y(z3$ Y=X<&
Y= iM + z&x<$, Z=
and
<& = 1 [ a (w*  x *f /) + 2x(ax + by + cz) + 2w (bz  cy)\
3$ = n[b(w 2 x 2 y z  z 2 } + 2y(ax + by+ cz) + 2w (ex  at)],
= I [ c (w 2  a? f z 2 ) + 2z(ax + by+ cz) + 2w (aybx~)].
W, X, Y, Z are thus homogeneous functions of w, x, y, z of the third
degree.
Perhaps the simplest way of obtaining these equations is to
translate the group of 410 into w, x, y, z at once, instead of
using the equation from which f and r\ are eliminated.
We thus see that
One obvious integral of these equations ought to be
w z + x 2 4 y 2 + z* = constant,
which has been assumed all along. In fact, we see at once that
identically, which leads to the above integral.
These equations appear to be worthy of attention, partly
because of the homogeneity of the denominators W, X, Y, Z, but
particularly as they afford (what does not appear to have been
sought) the means of solving this celebrated problem at one step,
that is, without the previous integration of Euler s equations
( 2).
A set of equations identical with these, but not in a homo
geneous form (being expressed, in fact, in terms of K, \, p, v of
375, instead of w, x, y, z), is given by Cayley (Camb. and Dub.
419] PHYSICAL APPLICATIONS. 331
Math. Journal, vol. i. 1846), and completely integrated (in the
sense of being reduced to quadratures) by assuming Euler s
equations to have been previously integrated. (Compare 416.)
Cayley s method may be even more easily applied to the above
equations than to his own ; and I therefore leave this part of the
development to the reader, who will at once see (as in 416) that
&, 23, ( correspond to <D V o> 2 , o) 3 of the 77 type, 412.
418. It may be well to notice, in connection with the formulae
for direction cosines in 375 above, that we may write
= \[a (w*+ x 2 f  * 2 ) + 26 (xy + wz) + 2c (xz  wy}\
3$ = [2a (xy wz) + b (w* x* + f z 2 ) + 2c (yz + wx)\
= i [2a (xz + wy) + 26 (yz wx) + c (w*  x* y* + z*)].
(j
These expressions may be considerably simplified by the usual
assumption, that one of the fixed unitvectors (i suppose) is
perpendicular to the invariable plane, which amounts to assigning
definitely the initial position of one line in the body ; and which
gives the relations
60, c = 0.
419. When forces act, 7 is variable, and the quantities a, 6, c
will in general involve all the variables w, x, y y z, t, so that the
equations of last section become much more complicated. The
type, however, remains the same if 7 involves t only ; if it involve
q we must differentiate the equation, put in the form
7 =
and we thus easily obtain the differential equation of the second
order
^ = 4F. & (q q) 3 + 2q<t> ( V. q^j) g ;
if we recollect that, because q~ l q is a vector, we have
Though the above formula is remarkably simple, it must, in the
present state of the development of quaternions, be looked on as
intractable, except in certain very particular cases.
332 QUATERNIONS. [420.
420. Another mode of attacking the problem, at first sight
entirely different from that in 408, but in reality identical with
it, is to seek the linear and vector function which expresses the
Homogeneous Strain which the body must undergo to pass from its
initial position to its position at time t.
Let w = x&,
a being (as in 408) the initial position of a vector of the body,
OT its position at time t. In this case % is a linear and vector
function. (See 376.)
Then, obviously, we have, ^ l being the vector of some other
point, which had initially the value a t ,
$57^7 1 = S .
(a particular case of which is
and FOTOT I = V.
These are necessary properties of the strainfunction x> depending
on the fact that in the present application the system is rigid.
421. The kinernatical equation
tzr = Veiff
becomes %a = V. e^a
(the function % being formed from % by the differentiation of its
constituents with respect to t).
Hamilton s kinetic equation
2 . miff Vein = 7,
becomes 2 . vti^p. V. e^a = 7.
This may be written
2 . m (xaS . e^a  ea") = 7,
or 2 . m (aS . a% e % 1 e . a 2 ) = %~ J 7>
where % is the conjugate of %.
But, because S . x a X a i = ^ aa i>
we have $, = S . % %!>
whatever be a and a 1? so that
x =x"
Hence 2 . m (aS . a%~ J e X~ IG a ") = X J 7>
or, by 408 ^" ^X V
424.] PHYSICAL APPLICATIONS. 333
422. Thus we have, as the analogues of the equations in
408, 409,
and the former result %a = F.
becomes %a = F. X^IX  = %F??a.
This is our equation to determine %, 77 being supposed known.
To find rj we may remark that
** = fc
and ?=X~V
But %%~ IQC = a >
so that %x~ a + XX~ 1 a = 
Hence t = ~ X~ l %X~ l V
or (>?) =
These are the equations we obtained before. Having found
from the last we have to find from the condition
423. We might, however, have eliminated 77 so as to obtain
an equation containing ^ alone, and corresponding to that of
410. For this purpose we have
so that, finally, % 1 % a = ^ < ~ 1 X~ 1 7 a
or ^a=^.%"W 1 %
which may easily be formed from the preceding equation by
putting % IOL f r a > an d attending to the value of ^ given in last
section.
424. We have given this process, though really a disguised
form of that in 408, 410, and though the final equations to
which it leads are not quite so easily attacked in the way of
integration as those there arrived at, mainly to shew how free a
use we can make of symbolic functional operators in quaternions
334 QUATERNIONS. [4 2 5
without risk of error. It would be very interesting, however,
to have the problem worked out afresh from this point of view by
the help of the old analytical methods : as several new forms of
longknown equations, and some useful transformations, would
certainly be obtained.
425. As a verification, let us now try to pass from the final
equation, in ^ alone, of 423 to that of 410 in q alone.
We have, obviously,
OT = qaq~ l = ^a,
which gives the relation between q and ^.
[It shews, for instance, that, as
while S./3xoL = S. jSqaq 1 = S . aq~ l /3q,
we have % /3 = q^ffq,
and therefore that XX $ = # ( ( f l @<l) tf" 1 = &
or x = X~ l > as above.]
Differentiating, we have
qaq 1  qaq~ l qq~ l = tfa.
Hence % 1 % a = <f l <i a "~ a f l q
Also c/r 1 x~ l y = F 1 (f l
so that the equation of 423 becomes
or, as a may have any value whatever,
which, if we put Tq = constant
as was originally assumed, may be written
2q = q^ 1 (q l <yq),
as in 410.
C. Special Kinetic Problems.
426. To form the equation for Precession and Nutation. Let
cr be the vector, from the centre of inertia of the earth, to a particle
m of its mass: and let p be the vector of the disturbing body, whose
mass is M. The vectorcouple produced is evidently
426.] PHYSICAL APPLICATIONS. 335
mVcrp
m
no farther terms being necessary, since ^ is always small in the
actual cases presented in nature. But, because a is measured from
the centre of inertia,
S . ma = 0.
Also, as in 408, </>/> = S . m (aSap <r 2 p).
Thus the vectorcouple required is
L r
Referred to coordinates moving with the body, $ becomes <j> as in
413, and 413 gives
Simplifying the value of <J> by assuming that the earth has two
principal axes of equal moment of inertia, we have
Be  (A  B) aSae = vectorconstant + 3M (A  B) dt.
This gives Sae = const. = H,
whence e = Ha + ad,
so that, finally,
BVc&AOa = ^(AB) VotpSap.
The most striking peculiarity of this equation is that the form
of the solution is entirely changed, not modified as in ordinary cases
of disturbed motion, according to the nature of the value of p.
Thus, when the righthand side vanishes, we have an equation
which, in the case of the earth, would represent the rolling of a
cone fixed in the earth on one fixed in space, the angles of both
being exceedingly small.
336 QUATERNIONS. [4 2 7
If p be finite, but constant, we have a case nearly the same as
that of a top, the axis on the whole revolving conically about p.
But if we assume the expression
pr(j cos mt + k sin mi),
(which represents a circular orbit described with uniform speed,)
a revolves on the whole conically about the vector i, perpendicular
to the plane in which p lies. ( 408426, Trans. R 8. E.,
18G8 9.)
427. To form the equation of motion of a simple pendulum,
taking account of the earth s rotation. Let a be the vector (from
the earth s centre) of the point of suspension, X its inclination to
the plane of the equator, a the earth s radius drawn to that point;
and let the unit vectors i,j,kbe fixed in space, so that i is parallel
to the earth s axis of rotation ; then, if co be the angular velocity
of that rotation
a = a [i sin X 4 (j cos a)t + k sin at) cos X] ......... (1).
This gives a. = an (j sin cot + k cos cot) cos X
= coVia .......................................... (2).
Similarly a = CD Via = &> 2 (a ai sin X) .................. (3).
428. Let p be the vector of the bob m referred to the point of
suspension, R the tension of the string, then if a t be the direction
of pure gravity
m(a + p) = mg UOL^ R Up .................. (4),
which may be written
Vw ............... (5).
To this must be added, since r (the length of the string) is
constant,
Tp = r .............................. (6),
and the equations of motion are complete.
429. These two equations (5) and (6) contain every possible case
of the motion, from the most infinitesimal oscillations to the most
rapid rotation about the point of suspension, so that it is necessary
to adapt different processes for their solution in different cases.
We take here only the ordinary Foucault case, to the degree of
approximation usually given.
430.] PHYSICAL APPLICATIONS. 337
430. Here we neglect terms involving o> 2 . Thus we write
8 = 0,
and we write a for a t , as the difference depends upon the ellipticity
of the earth. Also, attending to this, we have
(7),
where (by (6)) &m = .............................. (8),
and terms of the order TX* are neglected.
With (7), (5) becomes
so that, if we write  = n z .............................. (9),
we have Fa(w + ?iV) = ..................... (10).
Now, the two vectors ai a sin X and Via.
have, as is easily seen, equal tensors ; the first is parallel to the line
drawn horizontally northwards from the point of suspension, the
second horizontally eastwards.
Let, therefore, w = x (ai a sin X) + y Via ............ (11),
which (x and y being very small) is consistent with (6).
From this we have (employing (2) and (3), and omitting &> 2 )
CT = x (ai a sin X) + y Via. xw sin X Via. yco (a ai sin X),
Hs x (ai OL sin X) + y Via. 2xa) sin X Via %yw (a ai sin X).
With this (10) becomes
Va \x (ai a sin X) + y Via 2xo) sin X Via 2?/w (a ai sin X)
+ n*x (ai a sin X) + tfyVia] = 0,
or, if we note that V.aVia = a (ai a sin X),
(as Zi/Q) sin X n z x] aVia + (y 2xco sin X +n z y)a(ai asinX) = 0.
This gives at once x + n 2 x + 2&>?/ sin X = 0]
(12),
y + tfy 2co% sin X = OJ
which are the equations usually obtained; and of which the solution
is as follows :
If we transform to a set of axes revolving in the horizontal plane
at the point of suspension, the direction of motion being from the
T. Q. I. 22
338 QUATERNIONS. [43 1.
positive (northward) axis of as to the positive (eastward) axis of y,
with angular velocity fl, so that
x = f cos fit r) sin fit] ._ _.
y = (f sin 1U + ?? cos fltfj
and omit the terms in H 2 and in wO (a process justified by the
results, see equation (15)), we have
( + ri*t; ) cos fit  (77 + ft 2 ??) sin fit  2y (1  w sin X) =
...(14).
sin &+(*? + ^) cos HZ + 2#(H  G) sin X) =
So that, if we put H = o> sin X (15),
we have simply f + n z j; =
(16)
1 + 11*11 = OJ
the usual equations of elliptic motion about a centre of force in the
centre of the ellipse. (Proc. R. 8. E., 1869.)
D. Geometrical and Physical Optics.
431. To construct a reflecting surface from which rays, emitted
from a point, shall after reflection diverge uniformly, but horizontally.
Using the ordinary property of a reflecting surface, we easily
obtain the equation
(j3 + oLVap\%
P ~ P =
By Hamilton s grand Theory of Systems of Rays, we at once
write down the second form
TpT(/3 + a Vap) = constant.
The connection between these is easily shewn thus. Let TX and
r be any two vectors whose tensors are equal, then
l (1 + SW 1 ) (Chapter III. Ex. 2),
whence, to a scalar factor pres, we have
(^\2 T + VT
T) T
Hence, putting vr= U(fi + aVcLp) and T= ?7/?, we have from the
first equation above
8. dp [Up +U(/3 + aVap)] = 0.
432.] PHYSICAL APPLICATIONS. 339
But d(/3 + a Vap) = a Vadp = dp aSadp,
and S.
so that we have finally
which is the differential of the second equation above. A curious
particular case is a parabolic cylinder, as may be easily seen
geometrically. The general surface has a parabolic section in the
plane of a, /3 ; and a hyperbolic section in the plane of ft, aft.
It is easy to see that this is but a single case of a large class of
integrable scalar functions, whose general type is
$ . dp (
V P
the equation of the reflecting surface ; while
is the equation of the surface of the reflected wave : the integral
of the former being, by the help of the latter, at once obtained in
the form
TpT(trp) = constant*.
432. We next take Fresnel s Theory of Double Refraction, but
merely for the purpose of shewing how quaternions simplify the
processes required, and in no way to discuss the plausibility of the
physical assumptions.
Let far be the vector displacement of a portion of the ether,
with the condition
^ 2 =i a),
the force of restitution, on Fresnel s assumption, is
t (cfiSi + VjSjiff + c Mtar) = t<jysr t
using the notation of Chapter V. Here the function </> is the
negative of that of Chapter IX. (the force of restitution and the
displacement being on the whole towards opposite parts), and it is
clearly selfconjugate, a 2 , b* t c 2 are optical constants depending on
the crystalline medium, and on the wavelength of the light, and
may be considered as given.
Fresnel s second assumption is that the ether is incompressible,
or that vibrations normal to a wave front are inadmissible. If,
* Proc. E. S. E., 187071.
222
340 QUATERNIONS. [433
then, a be the unit normal to a plane wave in the crystal, we have
of course
2 = l .............................. (2),
and >SW = ............................... (3);
but, and in addition, we have
or $ . acr^cr = ........................... (4).
This equation (4) is the embodiment of Fresnel s second assump
tion, but it may evidently be read as meaning, the normal to the
front, the direction of vibration, and that of the force of restitution
are in one plane.
433. Equations (3) and (4), if satisfied by OT, are also satisfied
by CTQC, so that the plane (3) intersects the cone (4) in two lines
at right angles to each other. That is, for any given wave front
there are two directions of vibration, and they are perpendicular to
each other.
434. The square of the normal speed of propagation of a
plane wave is proportional to the ratio of the resolved part of the
force of restitution in the direction of vibration, to the amount of
displacement, hence
v 2 = Stff^vf.
Hence Fresnel s Wavesurface is the envelop of the plane
Sap = Svr<l>vr ........................ (5),
with the conditions or 2 = 1 .............................. (1),
2 = l .............................. (2),
Sav = Q ................................. (3),
S.aw<w = ....................... . ......... (4).
Formidable as this problem appears, it is easy enough. From (3)
and (4) we get at once,
Hence, operating by S . r,
X =
Therefore (< + v*) & =
and ^f.a(0 + z; 5 ) 1 a = ..................... (C).
435] PHYSICAL APPLICATIONS. 341
In passing, we may remark that this equation gives the normal
speeds of the two rays whose fronts are perpendicular to a. In
Cartesian coordinates it is the wellknown equation
70 9 9
p +_?!!_ +^!_ = o.
By this elimination of cr, our equations are reduced to
= ........................ (6),
v = Sp ........................... (5),
2 = l .............................. (2).
They give at once, by 326,
(< + V^OL + vpSa (0 + v*) 2 a = ha.
Operating by S . a we have
Substituting for h, and remarking that
Sa(<l> + v 2 ra = 
because (/> is selfconjugate, we have
This gives at once, by rearrangement,
v (t + vT 1 * = (<# pTp
Hence >> =
Operating by S .p on this equation we have
SpQffp l ..................... (7),
which is the required equation.
[It will be a good exercise for the student to translate the last
ten formulae into Cartesian coordinates. He will thus reproduce
almost exactly the steps by which Archibald Smith* first arrived
at a simple and symmetrical mode of effecting the elimination.
Yet, as we shall presently see, the above process is far from being
the shortest and easiest to which quaternions conduct us.]
435. The Cartesian form of the equation (7) is not the usual
one. It is, of course,
* Cambridge Phil. Trans., 1835.
342 QUATERNIONS.
But write (7) in the form
and we have the usual expression
oV by cV_
^
The lastwritten quaternion equation can also be put into either
of the new forms
or
436. By applying the results of 183, 184 we may introduce
a multitude of new forms. We must confine ourselves to the most
simple ; but the student may easily investigate others by a process
precisely similar to that which follows.
Writing the equation of the wave as
where we have g = p~*,
we see that it may be changed to
s p (^ 1 + hr l p = o,
if mSp<j>p = glip* = h.
Thus the new form is
Sp((F l mSp<l>prp = ..................... (1).
Here m = ,,, Sp^p = aV + Vtf + cV,
Ct C
and the equation of the wave in Cartesian coordinates is, putting
a? y* z*
^ +  * + ** 
c 
437. By means of equation (1) of last section we may easily
prove Pliicker s Theorem :
The Wave Surf ace is its own reciprocal with respect to the
ellipsoid whose equation is
438.] PHYSICAL APPLICATIONS. 343
The equation of the plane of contact of tangents to this surface
from the point whose vector is p is
The reciprocal of this plane, with respect to the unitsphere
about the origin, has therefore a vector cr where
Hence p =
p.
~
and when this is substituted in the equation of the wave we have
for the reciprocal (with respect to the unitsphere) of the reciprocal
of the wave with respect to the above ellipsoid,
1 \ 1
O(J(p (T 1 (T == 0.
in /
This differs from the equation (1) of last section solely in having
0" 1 instead of (/>, and (consistently with this) l/m instead of m.
Hence it represents the indexsurface. The required reciprocal
of the wave with reference to the ellipsoid is therefore the wave
itself.
438. Hamilton has given a remarkably simple investigation
of the form of the equation of the wavesurface, in his Elements,
p. 736, which the reader may consult with advantage. The
following is essentially the same, but several steps of the process,
which a skilled analyst would not require to write down, are
retained for the benefit of the learner.
Let %> = ! (1)
be the equation of any tangent plane to the wave, i.e. of any wave
front. Then //, is the vector of waveslowness, and the normal
velocity of propagation is therefore 1/Tfj,. Hence, if TZ be the vector
direction of displacement, /J,~*TZ is the effective component of the
force of restitution. Hence, (/>OT denoting the whole force of
restitution, we have
and, as CT is in the plane of the wavefront,
= 0,
" 2 )~ l p = (2).
344 QUATERNIONS. [439.
This is, in reality, equation (6) of 434. It appears here,
however, as the equation of the Index Surf ace, the polar reciprocal
of the wave with respect to a unitsphere about the origin. Of
course the optical part of the problem is now solved, all that
remains being the geometrical process of 328.
439. Equation (2) of last section may be at once transformed,
by the process of 435, into
Let us employ an auxiliary vector
T=G* 2 .>,
whence ^ = (/**  (/T^r ........................ (1).
The equation now becomes
SHT = ! .............................. (2),
or,by(l), ^Sr^r = l ........................ (3).
Differentiating (3), subtract its half from the result obtained by
operating with S . r on the differential of (1). The remainder is
But we have also ( 328)
Spdfi = 0,
and therefore, since dp has an infinite number of values,
Xp = yL6T 2 T,
where x is a scalar.
This equation, with (2), shews that
Srp = () .............................. (4).
Hence, operating on it by 8 . p, we have by (1) of last section
V = T*
and therefore p 1 = //, + r" 1 .
This gives p~ 2 = ^  T " 2 .
Substituting from these equations in (1) above, it becomes
T ^^+T 2 <>,
or T($*p*)V.
Finally, we have for the required equation, by (4),
BpyrpTpio,
or, by a transformation already employed,
441]
PHYSICAL APPLICATIONS.
345
440. It may assist the student in the practice of quaternion
analysis, which is our main object, if we give a few of these
investigations by a somewhat varied process.
Thus, in 432, let us write as in 180,
cfiSiv + tfjSjw + tfkSkv = \ Sp!^ + p S^v? p r ix.
We have, by the same processes as in 432,
8 . uroWSp v + S . isroLfi SXfw = 0.
This may be written, 50 far as the generating lines we require are
concerned,
S.vaV. \ V/JL = = 8.
since ora is a vector.
Or we may write
S.fi V. OTX Bra = = 8.
Equations (1) denote two cones of the second order which pass
through the intersections of (3) and (4) of 432. Hence their
intersections are the directions of vibration.
441. By (1) we have
.(1).
Hence rXV, a, fjf are coplanar ; and, as w is perpendicular to a,
it is equally inclined to FX a and F//a.
For, if L, M, A be the projections of X , ///, a on the unit
sphere, BG the great circle whose
pole is A, we are to find for the ^
projections of the values of OT on
the sphere points P and P , such
that if LP be produced till
Q may lie on the great circle AM.
Hence, evidently,
and
which proves the proposition, since
the projections of FX a and F//a on the sphere are points b and
c in BC, distant by quadrants from G and B respectively.
346 QUATERNIONS. [44
442. Or thus, Si* a. = 0,
S . vrV . aX vT/jf = 0,
therefore XTZ = V. aF.
=  F. XV// 
Hence (x>  a?) w = (V + aaX ) / + (/
Operate by $ . X , and we have
(x + SXaSfi a) SxV = [X V  SVa
Hence by symmetry,
J _
~
and as S&OL = 0,
/ a UVpfa).
443. The optical interpretation of the common result of the
last two sections is that the planes of polarization of the two rays
whose wavefronts are parallel, bisect the angles contained by planes
passing through the normal to the wavefront and the vectors (optic
axes) X , fju.
444. As in 434, the normal speed is given by
[This transformation, effected by means of the value of r in
442, is left to the reader.]
Hence, if v iy v 2 be the velocities of the two waves whose
normal is a,
oc sin X a sin /u/a.
That is, the difference of the squares of the speeds of the two
waves varies as the product of the sines of the angles between the
normal to the wavefront and the optic axes (X , //).
447] PHYSICAL APPLICATIONS. 347
445. We have, obviously,
(T 2  S 2 ) . V\ OL Vp a =TV. V\ OL VH OL = S 2 . \
Hence v* =p + (T S) . VXaVpa.
The equation of the index surface, for which
is therefore 1 =  p p* + (T S) . VX p Vp p.
This will, of course, become the equation of the reciprocal of the
indexsurface, i.e. the wavesurface, if we put for the function </>
its reciprocal : i.e. if in the values of X , //, , p we put I/a, 1/6, 1/c
for a, 6, c respectively. We have then, and indeed it might have
been deduced even more simply as a transformation of 434 (7),
as another form of the equation of Fresnel s wave.
If we employ the i, K transformation of 128, this may be
written, as the student may easily prove, in the form
(V  ij = S*(i K )p + (TVi P
446. We may now, in furtherance of our object, which is to
give varied examples of quaternions, not complete treatment of any
one subject, proceed to deduce some of the properties of the wave
surface from the different forms of its equation which we have
given.
447. Fresnel s construction of the wave by points.
From 290 (4) we see at once that the lengths of the principal
semidiameters of the central section of the ellipsoid
Spj p = 1,
by the plane Sap = 0,
are determined by the equation
3.a(<l> l p*) 1 a = 0.
If these lengths be laid off along a, the central perpendicular to the
cutting plane, their extremities lie on a surface for which a = Up,
and Tp has values determined by the equation.
Hence the equation of the locus is
as in 55 434, 439.
348 QUATERNIONS. [44$.
Of course the indexsurface is derived from the reciprocal ellip
soid
SpQp = 1
by the same construction.
448. Again, in the equation
l=pp*+(TS).V\ P V f j,p,
suppose V\p = 0, or Vjj,p = 0,
we obviously have
,U\ tfyu,
P = 7 r P = , ,
\/p *Jp
and there are therefore four singular points.
To find the nature of the surface near these points put
U\
P = / + ,
Vp
where TVT is very small, and reject terms above the first order in
TOT. The equation of the wave becomes, in the neighbourhood of
the singular point,
2pS\v + S.&V. \V\fji = T. V\iz V\p,
which belongs to a cone of the second order.
449. From the similarity of its equation to that of the wave, it
is obvious that the indexsurface also has four conical cusps. As
an infinite number of tangent planes can be drawn at such a point,
the reciprocal surface must be capable of being touched by a plane
at an infinite number of points ; so that the wavesurface has four
tangent planes which touch it along ridges.
To find their form, let us employ the last form of equation of
the wave in 445. If we put
TK P = TV Kp (1),
we have the equation of a cone of the second degree. It meets the
wave at its intersections with the planes
fl((.*)/> = (" ) (2).
Now the wavesurface is touched by these planes, because we
cannot have the quantity on the first side of this equation greater
in absolute magnitude than that on the second, so long as p
satisfies the equation of the wave.
45 J ] PHYSICAL APPLICATIONS. 349
That the curves of contact are circles appears at once from (1)
and (2), for they give in combination
)p ........................ (3),
the equations of two spheres on which the curves in question are
situated.
The diameter of this circular ridge is
= (a*  V) (b*  <f).
[Simple as these processes are, the student will find on trial
that the equation
V = 0,
gives the results quite as simply. For we have only to examine
the cases in which p~ z has the value of one of the roots of the
symbolical cubic in </>~\ In the present case Tp = b is the only
one which requires to be studied.]
450. By 438, we see that the auxiliary vector of the succeed
ing section, viz.
T =(/*  )> = ( p)V,
is parallel to the direction of the force of restitution, ^>t7. Hence,
as Hamilton has shewn, the equation of the wave, in the form
(4) of 439, indicates that the direction of the force of restitution is
perpendicular to the ray.
Again, as for any one versor of a vector of the wave there are
two values of the tensor, which are found from the equation
we see by 447 that the lines of vibration for a given plane front
are parallel to the axes of any section of the ellipsoid
made by a plane parallel to the front ; or to the tangents to the lines
of curvature at a point where the tangent plane is parallel to the
wavefront.
451. Again, a curve which is drawn on the wavesurface so as
to touch at each point the corresponding line of vibration has
350 QUATERNIONS. [452.
Hence Sfypdp = 0, or Sp(f>p = C,
so that such curves are the intersections of the wave with a series
of ellipsoids concentric with it.
452. For curves cutting at right angles the lines of vibration we
have
Hence Spdp = 0, or Tp = C,
so that the curves in question lie on concentric spheres.
They are also spherical conies, because where
Tp = C
the equation of the wave becomes
s. P (4> + <ryv = o,
the equation of a cyclic cone, whose vertex is at the common centre
of the sphere and the wavesurface, and which cuts them in their
curve of intersection. ( 432 452, Quarterly Math. Journal, 1859.)
The student may profitably compare, with the preceding investi
gations, the (generally) very different processes which Hamilton
(in his Elements) applies to this problem.
E. Electrodynamics.
453. As another example we take the case of the action of
electric currents on one another or on magnets; and the mutual
action of permanent magnets.
A comparison between the processes we employ and those of
Ampere (Theorie des Phenomenes Electrodynamiques) will at once
shew how much is gained in simplicity and directness by the use
of quaternions.
The same gain in simplicity will be noticed in the investiga
tions of the mutual effects of permanent magnets, where the
resultant forces and couples are at once introduced in their most
natural and direct forms.
454. Ampere s experimental laws may be stated as follows :
I. Equal and opposite currents in the same conductor produce
equal and opposite effects on other conductors.
455] PHYSICAL APPLICATIONS. 351
II. The effect of a conductor bent or twisted in any manner is
equivalent to that of a straight one, provided that the two are
traversed by equal currents, and the former nearly coincides with
the latter.
III. No closed circuit can set in motion an element of a
circular conductor about an axis through the centre of the circle
and perpendicular to its plane.
IV. In similar systems traversed by equal currents the forces
are equal.
To these we add the assumption that the action between two
elements of currents is in the straight line joining them. [In a
later section ( 473) other assumptions will be made in place of
this.] We also take for granted that the effect of any element of
a current on another is directly as the product of the strengths of
the currents, and of the lengths of the elements.
455. Let there be two closed currents whose strengths are
a and a l ; let a, a x be elements of these, a being the vector
joining their middle points. Then the effect of a on a l must,
when resolved along a lf be a complete differential with respect
to a (i.e. with respect to the three independent variables involved
in a), since the total resolved effect of the closed circuit of which
a is an element is zero by III.
Also by I, II, the effect is a function of Ta, Sa.a, Saa.^ and
SCL CL^ since these are sufficient to resolve a and a l into elements
parallel and perpendicular to each other and to a. Hence the
mutual effect is
a^Uaf^a, Saa , SOL\),
and the resolved effect parallel to a l is
aa^SUa^Uaf.
Also, that action and reaction may be equal in absolute magnitude,
/ must be symmetrical in 8a.a and $aa r Again, a (as differential
of a) can enter only to the first power, and must appear in each
term of f.
Hence /= A8a\ + BScm Sou^.
But, by IV, this must be independent of the dimensions of the
system. Hence A is of 2 and B of 4 dimensions in To..
Therefore
 [ASaa.Safa, + BSaa &aaJ
352 QUATERNIONS. [45 6.
is a complete differential, with respect to a, if da. = of. Let
G
~Tc?
where G is a constant depending on the units employed, therefore
G B
and the resolved effect
Cad . S O.O..
= To? = * TaTc?
^ +
The factor in brackets is evidently proportional in the ordinary
notation to
sin 6 sin & cos co  cos 6 cos 6 . *
456. Thus the whole force is
Caa^a j $ 2 aa t _ Caa^a. , 8*0.0.
~ l ~ 3 >
as we should expect, d^a. being = a r [This may easily be trans
formed into
which is the quaternion expression for Ampere s wellknown form.]
457. The whole effect on a t of the closed circuit, of which a is
an element, is therefore
Oaa
rv >a J ZV)
between proper limits. As the integrated part is the same at both
limits, the effect is
and depends on the form of the closed circuit.
45^.] PHYSICAL APPLICATIONS. 353
458. This vector 0, which is of great importance in the whole
theory of the effects of closed or indefinitely extended circuits, cor
responds to the line which is called by Ampere " directrice de
fraction electrodynamique." It has a definite value at each point
of space, independent of the existence of any other current.
Consider the circuit a polygon whose sides are indefinitely
small ; join its angular points with any assumed point, erect at
the latter, perpendicular to the plane of each elementary triangle
so formed, a vector whose length is a>/r, where &> is the vertical
angle of the triangle and r the length of one of the containing
sides ; the sum of such vectors is the " directrice " at the assumed
point.
[We may anticipate here so far as to give another expression
for this important vector in terms of processes to be explained
later.
We have, by the formula (for a closed curve) of 497 below,
(where ds is an element of any surface bounded by the circuit, Uv
its unit normal)
But the last integral is obviously the whole spherical angle, II
suppose, subtended by the circuit at the origin, and (unless Tp = 0)
we have ( 145)
Hence, generally,
/e=vfl.
Thus II may be considered as representing a potential, for which
ft is the corresponding force.
This is a " many valued " function, altering by 4Tr whenever we
pass through a surface closing the circuit. For if o be the vector
of a closed curve, the work done against /3 during the circuit is
fSjBdo =  J SdoVQ, = fdtt.
The last term is zero if the curve is not linked with the
circuit, but increases by + 4n for each linkage with the circuit.]
T. Q. I. 23
354 QUATERNIONS. [459
459. The mere/orm of the result of 457 shews at once that
if the element a x be turned about its middle point, the direction of the
resultant action is confined to the plane whose normal is {3.
Suppose that the element a t is forced to remain perpendicular
to some given vector 8, we have
Saf = 0,
and the whole action in its plane of motion is proportional to
TV.BVctfi.
But V . SVafi =  afilSS.
Hence the action is evidently constant for all possible positions
of otj ; or
The effect of any system of closed currents on an element of a
conductor which is restricted to a given plane is (in that plane)
independent of the direction of the element.
460. Let the closed current be plane and very small. Let e
(where Te = 1) be its normal, and let 7 be the vector of any point
within it (as the centre of inertia of its area) ; the middle point of
4 being the origin of vectors.
Let a. = 7 + p ; therefore a = p,
to a sufficient approximation.
Now (between limits) jVpp =
where A is the area of the closed circuit.
Also generally
fVyp Syp = J (%>F 7 p + y 7.
= (between limits) A 7 Vye.
Hence for this case
463] PHYSICAL APPLICATIONS. 355
461. If, instead of one small plane closed current, there be a
series of such, of equal area, disposed regularly in a tubular form,
let x be the distance between two consecutive currents measured
along the axis of the tube ; then, putting y = xe, we have for the
whole effect of such a set of currents on a
CAaa f/y
2x 1 A?V + Ty 5
CAaa. Va..y ,, ,. .
^ ^ V (between proper limits).
If the axis of the tubular arrangement be a closed curve this will
evidently vanish. Hence a closed solenoid exerts no influence on an
element of a conductor. The same is evidently true if the solenoid
be indefinite in both directions.
It the axis extend to infinity in one direction, and y n be the
vector of the other extremity, the effect is
and is therefore perpendicular to the element and to the line joining
it with the extremity of the solenoid. It is evidently inversely as
Ty* and directly as the sine of the angle contained between the
direction of the element and that of the line joining the latter with
the extremity of the solenoid. It is also inversely as x, and there
fore directly as the number of currents in a unit of the axis of the
solenoid.
462. To find the effect of the whole circuit whose element is
otj on the extremity of the solenoid, we must change the sign of
the above and put a l = y , therefore the effect is
CAaa
l f V % %
] Ty*
2x Ty*
an integral of the species considered in 458, whose value is easily
assigned in particular cases.
463. Suppose the conductor to be straight) and indefinitely
extended in both directions.
Let hO be the vector perpendicular to it from the extremity of
the solenoid, and let the conductor be  77, where T6 = Trj = 1.
Therefore y = h& + yrj (where y is a scalar),
232
356
QUATERNIONS.
[464.
and the integral in 462 is
Jr*
The whole effect is therefore
_CAaa J
xh
and is thus perpendicular to the plane passing through the conductor
and the extremity of the solenoid, and varies inversely as the distance
of the latter from the conductor.
This is exactly the observed effect of an indefinite straight
current on a magnetic pole, or particle of free magnetism.
464. Suppose the conductor to be circular, and the pole nearly
in its axis. [This is not a proper subject for Quaternions.]
Let EPD be the conductor, AB its axis, and G the pole ; BG
perpendicular to AB, and small in comparison with AE = h the
radius of the circle.
Let A B be aj, BC=bk, AP = h(jx + ley)
where \ = \ . \EAP=\ . \ B.
y] l sm j l sm j
Then CP = y = a^i \bkh (jx + ky).
And the effect on G x lfJj t
where the integral extends to the whole circuit.
465.] PHYSICAL APPLICATIONS. 357
465. Suppose in particular C to be one pole of a small magnet
or solenoid CO whose length is 21, and whose middle point is at G
and distant a from the centre of the conductor.
Let /.CGB = A. Then evidently
a x = a 4 1 cos A,
b = I sin A.
Also the effect on C becomes, if a/ 2 f b 2 + h 2 = A 2 ,
h
1 f/r
[ij
J 3 I ^ I V * . //" < " 1 <~/ I w,.i/it/ i t"I
since for the whole circuit
If we suppose the centre of the magnet fixed, the vector axis
of the couple produced by the action of the current on C is
IV. (i cos A 4 & sin A) I ^^
sin A . L 36 2 . 15 W 3a,6 cos A
A 3 A* 2
If A } &c. be now developed in powers of , this at once becomes
vh*l sin A . f _ 6aZ cos A 15aT cos 2 A
/ I 9 . T >
(a 2 f ^) "I a 2 + W (a 2 + A 2 ) 2 a 2 + /i 2
_3^sin 2 A 15 /iT sin 2 A (a + Zcos A)^cos A > 5a^cosA\)
~ aM A 2 h (a 2 + A 2 ) 2 ~ a* + h* ( ~?"+l ~ / j
Putting ^ for Z and changing the sign of the whole to get that
for pole C , we have for the vector axis of the complete couple
A .
which is almost exactly proportional to sin A, if 2a be nearly
equal to h and I be small. (See Ex. 15 at end of Chapter.)
On this depends a modification of the tangent galvanometer.
(Bravais, Ann. de Ghimie, xxxviii. 309.)
358 QUATERNIONS. [466.
466. As before, the effect of an indefinite solenoid on a t is
CAaa 1 VOL*/
~^T~T^
Now suppose a t to be an element of a small plane circuit, 8 the
vector of the centre of inertia of its area, the pole of the solenoid
being origin.
Let 7 = 8 f p, then a x = p.
The whole effect is therefore
CAaa 1
GAAaa
where A l and e t are, for the new circuit, what A and e were for
the former.
Let the new circuit also belong to an indefinite solenoid, and
let S be the vector joining the poles of the two solenoids. Then
the mutual effect is
GAA
aa
_CAA l aa l S UB
~
which is exactly the mutual effect of two magnetic poles. Two
finite solenoids, therefore, act on each other exactly as two magnets,
and the pole of an indefinite solenoid acts as a particle of free
magnetism.
467. The mutual attraction of two indefinitely small plane
closed circuits, whose normals are e and e lt may evidently be
deduced by twice differentiating the expression U8/T& for the
mutual action of the poles of two indefinite solenoids, making dS
in one differentiation [ e and in the other  e lt
But it may also be calculated directly by a process which will
give us in addition the couple impressed on one of the circuits by
the other, supposing for simplicity the first to be circular. [In
the sketch we are supposed to be looking at the faces turned
towards one another.]
4 6 7] PHYSICAL APPLICATIONS. 359
Let A and B be the centres of inertia of the areas of A and B,
e and e, vectors normal to their planes, cr any vector radius of B
=0.
Then whole effect on cr , by 457, 460,
A V A
* V7e 
But between proper limits,
for generally fV(r rjS0a =%( VrjaSOo +V.<rjV. 6/Vaa).
Hence, after a reduction or two, we find that the whole force
exerted by A on the centre of inertia of the area of B
reh
This, as already observed, may be at once found by twice
TTO
differentiating 7*7  In the same way the vector moment, due to
A, about the centre of inertia of B,
C  777
These expressions for the whole force of one small magnet on
the centre of inertia of another, and the couple about the latter,
seem to be the simplest that can be given. It is easy to deduce
360 QUATERNIONS. [468.
from them the ordinary forms. For instance, the whole resultant
couple on the second magnet
may easily be shewn to coincide with that given by Ellis (Gamb.
Math. Journal, iv. 95), though it seems to lose in simplicity and
capability of interpretation by such modifications.
468. The above formulae shew that the whole force exerted
by one small magnet M, on the centre of inertia of another m
consists of four terms which are, in order,
1st. In the line joining the magnets, and proportional to the
cosine of their mutual inclination.
2nd. In the same line, and proportional to five times the product
of the cosines of their respective inclinations to this line.
3rd and 4th. Parallel to ^ and proportional to the cosine of
the inclination of \ ! to the joining line.
All these forces are, in addition, inversely as the fourth power
of the distance between the magnets.
For the couples about the centre of inertia of m we have
1st. A couple whose axis is perpendicular to each magnet, and
which is as the sine of their mutual inclination.
2nd A couple whose axis is perpendicular to m and to the line
joining the magnets, and whose moment is as three times the product
of the sine of the inclination of m, and the cosine of the inclination
of M, to the joining line.
In addition these couples vary inversely as the third power of
the distance between the magnets.
[These results afford a good example of what has been called
the internal nature of the methods of quaternions, reducing, as
they do at once, the forces and couples to others independent of
any lines of reference, other than those necessarily belonging to
the system under consideration. To shew their ready applicability,
let us take a Theorem due to Gauss.]
47 *] PHYSICAL APPLICATIONS. 361
469. If two small magnets be at right angles to each other, the
moment of rotation of the first is approximately twice as great when
the axis of the second passes through the centre of the first, as when
the axis of the first passes through the centre of the second.
In the first case e II $ _!_ e t ;
C 2C"
therefore moment = T (ee l  3e^) = ^ ^ee r
In the second e x  @ J_ e ;
G
therefore moment = ^ Tee Hence the theorem.
470. Again, we may easily reproduce the results of 467, if
for the two small circuits we suppose two small magnets perpen
dicular to their planes to be substituted. /3 is then the vector
joining the middle points of these magnets, and by changing the
tensors we may take 2e and 2e x as the vector lengths of the
magnets.
Hence evidently the mutual effect
oc
which is easily reducible to
as before, if smaller terms be omitted.
If we operate with F. e x on the two first terms of the unreduced
expression, and take the difference between this result and the
same with the sign of e : changed, we have the whole vector axis of
the couple on the magnet 2e v which is therefore, as before, seen to
be proportional to
471. Let F (7) be the potential of any system upon a unit
particle at the extremity of 7. Then we have
Svdy = 0,
where v is a vector normal giving the force in direction and
magnitude ( 148).
362 QUATERNIONS. [47 2.
Now by 460 we have for the vector force exerted by a small
plane closed circuit on a particle of free magnetism the expression
A / SySye\
T*)
merging in A the factors depending on the strength of the current
and the strength of magnetism of the particle.
Hence the potential is
ASey
Zy
area of circuit projected perpendicular to 7
^ 7p 2
Trf
x spherical opening subtended by circuit.
The constant is omitted in the integration, as the potential must
evidently vanish for infinite values of Ty.
By means of Ampere s idea of breaking up a finite circuit into
an indefinite number of indefinitely small ones, it is evident that
the above result may be at once ex
tended to the case of such a finite closed
circuit.
472. Quaternions give a simple
method of deducing the wellknown
property of the Magnetic Curves.
Let A, A be two equal magnetic
poles, whose vector distance, 2a, is bi A~~ o ~A f
sected in 0, QQ an indefinitely small
magnet whose length is 2/o , where p = OP. Then evidently,
taking moments,
a)j , V(p*)p
*
where the upper or lower sign is to be taken according as the poles
are like or unlike.
[This equation may also be obtained at once by differentiating
the equation of the equipotential surface,
+ T, X = COnSt "
T (p + a) T (p  a
and taking p parallel to its normal ( 148).]
473] PHYSICAL APPLICATIONS. 363
Operate by 8 . Vap,
Sap (p + a) 2 So. (p + a) So (p + a) ..,
~.n/ T 3  = same with  a},
L (p + a)
or <S.aFf J CT(/o + a)= {same with a},
i.e. SadU(p + a) = + $acZ U(p  a),
a {#"(/> + a) + tT(p  a)} = const.,
or cos /. OAP cos OA P = const,
the property referred to.
If the poles be unequal, one of the terms to the left must be
multiplied by the ratio of their strengths.
( 453472, Quarterly Math. Journal, 1860.)
F. General Expressions for the Action between Linear
Elements.
473. The following general investigation of different possible
expressions for the mutual action between elements of linear
conductors is taken from Proc. R. S. E. 1873 4.
Ampere s data for closed currents are briefly given in 454
above, and are here referred to as I, II, III, IV, respectively.
(a) First, let us investigate the expression for t\\e force exerted
by one element on another.
Let a be the vector joining the elements a p a , of two circuits ;
then, by I, II, the action of a. v on a is linear in each of a t , a ,
and may, therefore, be expressed as
<K
where </> is a linear and vector function, into each of whose con
stituents otj enters linearly.
The resolved part of this along a is
8. Z7a <K
and, by III, this must be a complete differential as regards the
circuit of which a, is an element. Hence,
</>a = (8. a,V) ^a + Fa x P
where \r and ^ are linear and vector functions whose constituents
involve a only. That this is the case follows from the fact that
(f>a is homogeneous and linear in each of a l5 a . It farther
364 QUATERNIONS. [473
follows, from IV, that the part of </>a which does not disappear
after integration round each of the closed circuits is of no dimen
sions in To., To. , Ta^ Hence % is of 2 dimensions in To., and
thus
i  ~2V " 5V ~W~
where jo, q, r are numbers.
Hence we have
ofaSaoL, qV*^ rV.a Vaa,
Change the sign of a in this, and interchange of and a,, and we
get the action of a! on a r This, with a and a t again interchanged,
and the sign of the whole changed, should reproduce the original
expression since the effect depends on the relative, not the abso
lute, positions of a, a. v a . This gives at once,
p = 0, q = 0,
and
r V. a Fact
with the condition that the first term changes its sign with a, and
thus that
ira = aSaa F (To) + a!
which, by change of F, may be written
where /and F are any scalar functions whatever.
Hence
<j>a = S (cqV) [aS (a V)f(Ta) + a F (To)] + ~^ 1
which is the general expression required.
(6) The simplest possible form for the action of one current
element on another is, therefore,
2V
Here it is to be observed that Ampere s directrice for the circuit
a. is
/Too,
"JTW
the integral extending round the circuit ; so that, finally,
473] PHYSICAL APPLICATIONS. 365
(c) We may obtain from the general expression above the
absolutely symmetrical form,
if we assume
/(2k) = const,
Here the action of a on a t is parallel and equal to that of a t on a.
The forces, in fact, form a couple, for a is to be taken negatively
for the second and their common direction is the vector drawn
to the corner a of a spherical triangle abc, whose sides ab, be, ca in
order are bisected by the extremities of the vectors Ua, Ua, Ua^
Compare Hamilton s Lectures on Quaternions, 223 227.
(d) To obtain Ampere s form for the effect of one element on
another write, in the general formula above,
and we have
!.  a.* [_s*r[^r.*v mi
To
V. a Faa,
* / 20 3 o o "\
g I Ct oCtjtt Q ottft oOCOCj 1 ,
I \ ^ /
To 5
a_J^/$.F^jrvfj
which are the usual forms.
(e) The remainder of the expression, containing the arbitrary
terms, is of course still of the form
 S (a,V, [aS (a V)/(Ta) + a F (2V.)].
In the ordinary notation this expresses a force whose com
ponents are proportional to
(Note that, in this expression, r is the distance between the
elements.)
(2) Parallel to a ^ .
366 QUATERNIONS. [473.
(3) Parallel to  ^ .
as
If we assume /= F= Q, we obtain the result given by Clerk
Maxwell (Electricity and Magnetism, 525), which differs from the
above only because he assumes that the force exerted by one
element on another, when the first is parallel and the second per
pendicular to the line joining them, is equal to that exerted when
the first is perpendicular and the second parallel to that line.
(/) What precedes is, of course, only a particular case of the
following interesting problem :
Required the most general expression for the mutual action of
two rectilinear elements, each of which has dipolar symmetry in the
direction of its length, and which may be resolved and compounded
according to the usual kinematical law.
The data involved in this statement are equivalent to I and
II of Ampere s data above quoted. Hence, keeping the same
notation as in (a) above, the force exerted by ctj on a must be
expressible as
(j)OL
where (f> is a linear and vector function, whose constituents are
linear and homogeneous in a t ; and, besides, involve only a.
By interchanging a t and a , and changing the sign of a, we get
the force exerted by a on a r If in this we again interchange a t
and a, and change the sign of the whole, we must obviously repro
duce </> . Hence we must have $ changing its sign with a, or
(/> = PaSa/ + QaSaaficta. + Rafiaa! + RaSctct v
where P, Q, R, R are functions of Ta only.
(g) The vector couple exerted by a t on a must obviously be
expressible in the form
V , a waj,
where or is a new linear and vector function depending on a alone.
Hence its most general form is
where P and Q are functions of To. only. The form of these func
tions, whether in the expression for the force or for the couple,
depends on the special data for each particular case. Symmetry
shews that there is no term such as
473] PHYSICAL APPLICATIONS. 367
(ti) As an example, let a l and a be elements of solenoids or of
uniformly and linearly magnetised wires, it is obvious that, as a
closed solenoid or ringmagnet exerts no external action,
Thus we have introduced a different datum in place of Ampere s
No. III. But in the case of solenoids the Third Law of Newton
holds hence
where % is a linear and vector function, and can therefore be of no
other form than
Now two solenoids, each extended to infinity in one direction, act
on one another like two magnetic poles, so that (this being our
equivalent for Ampere s datum No. TV.)
Hence the vector force exerted by one small magnet on another is
.
(i) For the couple exerted by one element of a solenoid, or of
a uniformly and longitudinally magnetised wire, on another, we
have of course the expression
V. * **
where & is some linear and vector function.
Here, in the first place, it is obvious that
for the couple vanishes for a closed circuit of which a t is an
element, and the integral of wo^ must be a linear and vector
function of a alone. It is easy to see that in this case
F(Ta) oc (Taf.
(j) If, again, a t be an element of a solenoid, and a an element
of current, the force is
< = _ Sa^V . Tfra t
where
^a = Pa + QaSacL + RVoaf.
368 QUATERNIONS. [473.
But no portion of a solenoid can produce a force on an element of
current in the direction of the element, so that
whence P = 0, Q = 0,
and we have < = Sa t V (RVaa. ).
This must be of 1 linear dimensions when we integrate for the
effect of one pole of a solenoid, so that
7? P
** = 7/r~3
If the current be straight and infinite each way, its equation being
where Ty = l and Sfty = 0,
we have, for the whole force exerted on it by the pole of a solenoid,
the expression
/+ 00 dx
Pftv I ^ = ~ 2 PP 7i
which agrees with known facts.
(&) Similarly, for the couple produced by an element of a
solenoid on an element of a current we have
where
and it is easily seen that
ra.
(I) In the case first treated, the couple exerted by one current
element on another is, by (g),
V. CL K^,
where, of course, ^^ are the vector forces applied at either end
of a . Hence the work done when a changes its direction is
with the condition S . a So! = 0.
So far, therefore, as change of direction of a alone is concerned,
the mutual potential energy of the two elements is of the form
8 .
473] PHYSICAL APPLICATIONS. 369
This gives, by the expression for OT in (g), the following value
Hence, integrating round the circuit of which cq is an element, we
have ( 495 below)
f(PSa\ + QSaa SaaJ =ffdsfi. Up,V (Pa! + QctSaa),
P
where < = _ + Q.
Integrating this round the other circuit we have for the mutual
potential energy of the two, so far as it depends on the expression
above, the value
ffdsfi. UpJVota. 3>
= ffd Sl S. UpJJds V.
UP (24> + TOL&) + Sa Uv Sa UP.
But, by Ampere s result, that two closed circuits act on one another
as two magnetic shells, it should be
ffds, ffds S . Uv.VS . Uv V ~
= ffds, ffds S . Uv, UP + 3^a UP SOL UP,
Comparing, we have
gving <J) =  * = &>
which are consistent with one another, and which lead to
Hence, if we put
1 n
weget p
T. Q. I. 24
370 QUATERNIONS. [474.
and the mutual potential of two elements is of the form
, Sao, , . Sao. Seta.
which is the expression employed by Helmholtz in his paper Ueber
die Bewegungsgleichungen der Electricitdt, Crelle, 1870, p. 76.
G. Application of V to certain Physical Analogies.
474. The chief elementary results into which V enters, in
connection with displacements, are given in 384 above. The
following are direct applications.
Thus, if o be the vectordisplacement of that point of a homo
geneous elastic solid whose vector is p, we have, p being the
consequent pressure produced,
whence S&pV 2 a = SpVp = $p, a complete differential (2).
Also, generally, p = cSVa,
and if the solid be incompressible
SVo = Q (3).
Thomson has shewn (Camb. and Dub. Math. Journal, ii. p. 62),
that the forces produced by given distributions of matter, electricity,
magnetism, or galvanic currents, can be represented at every point
by displacements of such a solid producible by external forces. It
may be useful to give his analysis, with some additions, in a
quaternion form, to shew the insight gained by the simplicity of
the present method.
475. Thus, if SaSp = 5 , we may write each equal to
This gives
the vectorforce exerted by one particle of matter or free electricity
on another. This value of cr evidently satisfies (2) and (3).
Again, if S . SpVa = 8 ^ , either is equal to
478.] PHYSICAL APPLICATIONS. 371
Here a particular case is
Vao
~ Tf
which is the vectorforce exerted by an element a of a current upon
a particle of magnetism at p. ( 461.)
476. Also, by 146 (3),
and we see by 460, 461 that this is the vector force exerted by
a small plane current at the origin (its plane being perpendicular
to a) upon a magnetic particle, or pole of a solenoid, at p. This
expression, being a pure vector, denotes an elementary rotation
caused by the distortion of the solid, and it is evident that the
above value of cr satisfies the equations (2), (3), and the distortion
is therefore producible by external forces. Thus the effect of an
element of a current on a magnetic particle is expressed directly
by the displacement, while that of a small closed current or
magnet is represented by the vectoraxis of the rotation caused by
the displacement.
477. Again, let SS P V^ = S^
It is evident that a satisfies (2), and that the righthand side of
the above equation may be written
Hence a particular case is
Tp*
and this satisfies (3) also.
Hence the corresponding displacement is producible by external
forces, and V<r is the rotation axis of the element at p, and is seen
as before to represent the vectorforce exerted on a particle of
magnetism at p by an element a of a current at the origin.
478. It is interesting to observe that a particular value of cr
in this case is
as may easily be proved by substitution.
242
372 QUATERNIONS. [479.
 ~
j~ ,
Again, if S8p<r = 
we have evidently cr = V _ .
Now, as ^~ is the potential of a small magnet a, at the origin,
on a particle of free magnetism at p, cr is the resultant magnetic
force, and represents also a possible distortion of the elastic solid
by external forces, since Vcr = W = 0, and thus (2) and (3) are
both satisfied. (Proc. R S, E. 1862.)
H. Elementary Properties of V.
479. In the next succeeding sections we commence with a
form of definition of the operator V somewhat different from that
of Hamilton ( 145), as we shall thus entirely avoid the use of
Cartesian coordinates. For this purpose we write
where a is any unitvector, the meaning of the righthand operator
(neglecting its sign) being the rate of change of the function to
which it is applied per unit of length in the direction of the
vector a. If a be not a unitvector we may treat it as a vector
velocity, and then the righthand operator means the rate of
change per unit of time due to the change of position.
Let a, /3, 7 be any rectangular system of unitvectors, then by
a fundamental quaternion transformation
V   aSaV  3SSV  ^ V = ad
  aa  //  77 = a a + ft + y y ,
which is identical with Hamilton s form so often given above.
(Lectures, 620.)
480. This mode of viewing the subject enables us to see at
once that V is an Invariant, and that the effect of applying it to
any scalar function of the position of a point is to give its vector of
most rapid increase. Hence, when it is applied to a potential u,
we have the corresponding vectorforce. From a velocitypotential
we obtain the velocity of the fluid element at p ; and from the
temperature of a conducting solid we obtain the temperature
481.] PHYSICAL APPLICATIONS. 373
gradient in the direction of the flux of heat. Finally, whatever
series of surfaces is represented by
u = G,
the vector Vu is the normal at the point p, and its length is
inversely as the normal distance at that point between two con
secutive surfaces of the series.
Hence it is evident that
S . dpVa = du,
or, as it may be written,
the lefthand member therefore expresses total differentiation in
virtue of any arbitrary, but small, displacement dp.
These results have been already given above, but they were
not obtained in such a direct manner.
Many very curious and useful transformations may easily be
derived (see Ex. 34, Chap. XI.) from the assumption
da = (frdp, or </> = SaV . cr,
where the constituents of < are known functions of p. For
instance, if we write
_ . d . d j d
v  3f + *i + *3r
where a = i% + jrj + k,
we find at once V = V ff , or V^ = ^ ^V ;
a formula which contains the whole basis of the theory of the
change of independent variables from x, y, z to f , 77, f, or vice versa.
The reader may easily develop this application. Its primary
interest is, of course, purely mathematical : but it has most
important uses in applied mathematics. Our limits, however,
do not permit us to reach the regions of its special physical
usefulness.
481. To interpret the operator V. aV, let us apply it to a
potential function u. Then we easily see that u may be taken
under the vector sign, and the expression
denotes the vectorcouple due to the force at p about a point whose
relative vector is a.
374 QUATERNIONS. [482.
Again, if cr be any vector function of p, we have by ordinary
quaternion operations
F (aV) . o = 8 . a FVo + aV<r  Va<7.
The meaning of the third term (in which it is of course understood
that V operates on a alone) is obvious from what precedes. The
other terms were explained in 384.
J. Applications of V to Line, Surface, and Volume Integrals.
482. In what follows we have constantly to deal with integrals
extended over a closed surface, compared with others taken through
the space enclosed by such a surface ; or with integrals over a
limited surface, compared with others taken round its bounding
curve. The notation employed is as follows. If Q per unit of
length, of surface, or of volume, at the point p, Q being any
quaternion, be the quantity to be summed, these sums will be
denoted by
ffQds and fflQds,
when comparing integrals over a closed surface with others through
the enclosed space ; and by
ffQds and fQTdp,
when comparing integrals over an unclosed surface with others
round its boundary. No ambiguity is likely to arise from the
double use of
ffQds,
for its meaning in any case will be obvious from the integral with
which it is compared. What follows is mainly from Trans. R. S. E.
186970. See also Proc. R. 8. E. 18623.
483. We have shewn in 384 that, if a be the vector displace
ment of a point originally situated at
then S.Va
.
expresses the increase of density of aggregation of the points of
the system caused by the displacement.
484. Suppose, now, space to be uniformly filled with points,
and a closed surface 5 to be drawn, through which the points can
freely move when displaced.
486.] PHYSICAL APPLICATIONS. 375
Then it is clear that the increase of number of points within
the space S, caused by a displacement, may be obtained by either
of two processes by taking account of the increase of density at
all points within 2, or by estimating the excess of those which
pass inwards through the surface over those which pass outwards.
These are the principles usually employed (for a mere element of
volume) in forming the socalled Equation of Continuity.
Let v be the normal to 2 at the point p, drawn outwards, then
we have at once (by equating the two different expressions of the
same quantity above explained) the equation
which is our fundamental equation so long as we deal with triple
integrals. [It will be shewn later ( 500) that the corresponding
relation between the single and the double integral can be deduced
directly from this.]
As a first and very simple example of its use, let p be written
for o: It becomes
i.e. the volume of any closed space is the sum of the elements of
area of its surface, each multiplied by onethird of the perpendicular
from the origin on its plane.
485. Next, suppose a to represent the vector force exerted
upon a unit particle at p (of ordinary matter, electricity, or
magnetism) by any distribution of attracting matter, electricity,
or magnetism partly outside, partly inside 2. Then, if P be the
potential at p,
and if r be the density of the attracting matter, &c., at p,
by Poisson s extension of Laplace s equation.
Substituting in the fundamental equation, we have
47r ///rcfe = 4rf = ffS . VP Uvds,
where M denotes the whole quantity of matter, &c., inside X.
This is a wellknown theorem.
486. Let P and P x be any scalar functions of p, we can of
course find the distribution of matter, &c., requisite to make either
376 QUATERNIONS. [48 7
of them the potential at p ; for, if the necessary densities be r and
r l respectively, we have as before
Now V (PVPJ = VP VP l
Hence, if in the formula of 484 we put
we obtain
JJJS . VP VP^ 9 =  ///P V 2 P^9 + ffPS . V P l Uvds,
= _ fffP^Pds + JJPfi . VP Uvds,
which are the common forms of Green s Theorem. Sir W. Thomson s
extension of it follows at once from the same proof.
487. If P l be a manyvalued function, but V P 1 singlevalued,
and if 2 be a multiplyconnected* space, the above expressions
require a modification which was first shewn to be necessary by
Helmholtz, and first supplied by Thomson. For simplicity, suppose
2) to be doublyconnected (as a ring or endless rod, whether knotted
or not). Then if it be cut through by a surface s, it will become
simplyconnected, but the surface integrals have to be increased
by terms depending upon the portions just added to the whole
surface. In the first form of Green s Theorem, just given, the
only term altered is the last : and it is obvious that if p l be the
increase of P after a complete circuit of the ring, the portion to be
added to the righthand side of the equation is
pJ/S.VPUvds,
taken over the cutting surface only. A similar modification is
easily seen to be produced by each additional complexity in the
space 2.
488. The immediate consequences of Green s theorem are
well known, so that I take only a few examples.
Let P and P 1 be the potentials of one and the same distribution
of matter, and let none of it be within S. Then we have
2 ds = J/PS . VP Uvds,
* Called by Helmholtz, after Biemann, mehrfach susammenhangend. In trans
lating Helmholtz s paper (Phil. Mag. 1867) I used the above as an English
equivalent. Sir W. Thomson in his great paper on Vortex Motion (Trans. R. S> E.
1868) uses the expression "multiplycontinuous."
489.] PHYSICAL APPLICATIONS. 377
so that if VP is zero all over the surface of 2, it is zero all through
the interior, i.e., the potential is constant inside 2. If P be the
velocitypotential in the irrotational motion of an incompressible
fluid, this equation shews that there can be no such motion of the
fluid unless there is a normal motion at some part of the bounding
surface, so long at least as 2 is simplyconnected.
Again, if 2 is an equipotential surface,
///(VP/ ds = PJfti . V PUvds = P ///V 2 Pd?
by the fundamental theorem. But there is by hypothesis no
matter inside 2, so this shews that the potential is constant
throughout the interior. Thus there can be no equipotential
surface, not including some of the attracting matter, within which
the potential can change. Thus it cannot have a maximum or
minimum value at points unoccupied by matter.
489. Again, in an isotropic body whose thermal conductivity
does not vary with temperature, the equation of heatconduction is
S+W0,
dt c
where (for the moment) k and c represent as usual the conductivity,
and the water equivalent of unit volume.
The surface condition (assuming Newton s Law of Cooling) is
Assuming, after Fourier, that a particular integral is
= e~ mt u,
we have V 2 ww = 0,
*
Let u m be a particular integral of the first of these linear
differential equations. Substitute it for u in the second ; and we
obtain (with the aid of the equation of the bounding surface) a
scalar equation giving the admissible values of m.
Suppose the distribution of temperature when = to be
given ; it may be expressed linearly in terms of the various values
of u, thus
w =
For if u iy u^ be any two of these particular integrals, we have
by Green s Theorem, and the differential equations,
378 QUATERNIONS. [4QO.
Hence, unless m A = m 2 , we have
Ofwk~Q.
Thus we have A m fffu* m d<: = ffjwu m ds.
A m being thus found, we have generally
v = I,A m e""u m .
490. If, in the fundamental theorem, we suppose
o = Vr,
which imposes the condition that
i.e., that the cr displacement is effected without condensation, it
becomes
ffS . Vr Uvds = fffSV 2 rck = 0.
Suppose any closed curve to be traced on the surface X, dividing
it into two parts. This equation shews that the surfaceintegral is
the same for both parts, the difference of sign being due to the
fact that the normal is drawn in opposite directions on the two
parts. Hence we see that, with the above limitation of the value
of cr, the double integral is the same for all surfaces bounded by a
given closed curve. It must therefore be expressible by a single
integral taken round the curve. The value of this integral will
presently be determined ( 495).
491. The theorem of 485 may be written
!ffV 2 Pck =JfSUvVPd8=ffS(UvV) Pds.
From this we conclude at once that if
(which may, of course, represent any vector whatever) we have
or, if
This gives us the means of representing, by a surfaceintegral, a
vectorintegral taken through a definite space. We have already
seen how to do the same for a scalarintegral so that we can now
express in this way, subject, however, to an ambiguity presently
to be mentioned, the general integral
493] PHYSICAL APPLICATIONS. 379
where q is any quaternion whatever. It is evident that it is only
in certain classes of cases that we can expect a perfectly definite
expression of such a volumeintegral in terms of a surfaceintegral.
492. In the above formula for a vectorintegral there may
present itself an ambiguity introduced by the inverse operation
to which we must devote a few words. The assumption
VV = T
is tantamount to saying that, if the constituents of a are the
potentials of certain distributions of matter, &c., those of r are the
corresponding densities each multiplied by 4?r.
If, therefore, r be given throughout the space enclosed by S,
cr is given by this equation so far only as it depends upon the
distribution within 2, and must be completed by an arbitrary vector
depending on three potentials of mutually independent distributions
exterior to .
But, if a be given, r is perfectly definite ; and as
Vo = VV,
the value of V 1 is also completely defined. These remarks must
be carefully attended to in using the theorem above : since they
involve as particular cases of their application many curious
theorems in Fluid Motion, &c.
493. As a very special case, the equation
of course gives Vcr = u, a scalar.
Now, if v be the potential of a distribution whose density is u, we
have
We know that when u is assigned this equation gives one, and but
one, definite value for v. We have in fact, by the definition of a
potential,
where the integration (confined to u^ and p t ) extends to all space
in which u differs from zero.
380 QUATERNIONS. [494.
Thus there is no ambiguity in
and therefore a = Vv
47T
is also determinate.
494. This shews the nature of the arbitrary term which must
be introduced into the solution of the equation
FV<7 = T.
To solve this equation is ( 384) to find the displacement of any
one of a group of points when the consequent rotation is given.
Here SVr = S. VFV<7 = VV = 0;
so that, omitting the arbitrary term ( 493), we have
W = V T ,
and each constituent of cr is, as above, determinate. Compare
503.
Thomson* has put the solution in a form which may be written
if we understand by /( ) dp integrating the term in dx as if y
and z were constants, &c. Bearing this in mind, we have as
verification,
J27i \Vri + fV^ dp
495. We now come to relations between the results of integra
tion extended over a nonclosed surface and round its boundary.
Let cr be any vector function of the position of a point. The
lineintegral whose value we seek as a fundamental theorem is
jSadr,
where T is the vector of any point in a small closed curve, drawn
from a point within it, and in its plane.
* Electrostatics and Magnetism, 521, or Phil. Trans., 1852.
495] PHYSICAL APPLICATIONS. 381
Let (7 be the value of a at the origin of r, then
o = a S(rV)a 0)
so that fSadr =JS.{(r S (rV) a Q ] dr.
But fdr = 0,
because the curve is closed ; and (ante, 467) we have generally
fSrVS(7 dr = JV (T&TOT  <T O fVrdr).
Here the integrated part vanishes for a closed circuit, and
where ds is the area of the small closed curve, and Uv is a unit
vector perpendicular to its plane. Hence
fSo dT = S.V<T Q Uv.ds.
Now, any finite portion of a surface may be broken up into small
elements such as we have just treated, and the sign only of the
integral along each portion of a bounding curve is changed when
we go round it in the opposite direction. Hence, just as Ampere
did with electric currents, substituting for a finite closed circuit
a network of an infinite number of infinitely small ones, in each
contiguous pair of which the common boundary is described by
equal currents in opposite directions, we have for a finite unclosed
surface
There is no difficulty in extending this result to cases in which
the bounding curve consists of detached ovals, or possesses multiple
points.
This theorem seems to have been first given by Stokes (Smith s
Prize Exam. 1854), in the form
rrj (/ (<h dfi\ (da dy\ fd/3 da
= Jjds ll(rr) + m( r j L ) + n(^ :
( \dy dz) \dz dxj \dx dy
It solves the problem suggested by the result of 490 above. It
will be shewn, however, in a later section that the equation above,
though apparently quite different from that of 484, is merely a
particular case of it.
[If we recur to the case of an infinitely small area, it is clear
that
ffS.VrUvds
is a maximum when
F. Z
382 QUATERNIONS. [49 6.
Hence FVcr is, at every point, perpendicular to a small area
for which
fSadp
is a maximum.]
496. If <7 represent the vector force acting on a particle of
matter at p, S . adp represents the work done by it while the
particle is displaced along dp, so that the single integral
JS<rdp
of last section, taken with a negative sign, represents the work
done during a complete cycle. When this integral vanishes it is
evident that, if the path be divided into any two parts, the work
spent during the particle s motion through one part is equal to
that gained in the other. Hence the system of forces must be
conservative, i.e., must do the same amount of work for all paths
having the same extremities.
But the equivalent double integral must also vanish. Hence a
conservative system is such that
whatever be the form of the finite portion of surface of which ds
is an element. Hence, as Ver has a fixed value at each point of
space, while Uv may be altered at will, we must have
or Vo = scalar.
If we call X, Y, Z the component forces parallel to rectangular
axes, this extremely simple equation is equivalent to the well
known conditions
dX dY dY dZ dZ dX _
~7  ~7  ^J ~7  7 ^> ~J  7   
dy dx dz dy ax dz
Returning to the quaternion form, as far less complex, we see
that
Vcr = scalar = 47rr, suppose,
implies that cr = VP,
where P is a scalar such that
that is, P is the potential of a distribution of matter, magnetism,
or statical electricity, of volumedensity r.
PHYSICAL APPLICATIONS. 383
Hence, for a nonclosed path, under conservative forces,
depending solely on the values of P at the extremities of the path.
497. A vector theorem, which is of great use, and which cor
responds to the scalar theorem of 491, may easily be obtained.
Thus, with the notation already employed,
Now F ( F. V F. rdr) o =  S(rV) V. <r dr  S (drV) VT<T O ,
and d {S ( T V) FO O T} 8 (rV) F. a Q dr + S (drV) F<T O T.
Subtracting, and omitting the term which is the same at both
limits, we have
Extended as above to any closed curve, this takes at once the form
JF. adp = ffdsV. (FZTi/V) <r.
Of course, in many cases of the attempted representation of a
quaternion surfaceintegral by another taken round its bounding
curve, we are met by ambiguities as in the case of the space
integral, 492 : but their origin, both analytically and physically,
is in general obvious.
498. The following short investigation gives, in a complete
form, the kernel of the whole of this part of the subject. But
495 7 have still some interest of their own.
If P be any scalar function of p, we have (by the process of
495, above)
fPdr=J{P S(TV)P 9 }dT
= fS.rVP Q .dr.
But F. V F. rdr = drS .rVrS. drV,
and d (rSrV) = drS .rV + rS. drV.
These give
fPdr =  } {rSrV  F. JF(rdr) V} P = dsV. UvVP Q .
Hence, for a closed curve of any form, we have
jPdp=JjdsV.UvVP,
from which the theorems of 495, 497 may easily be deduced.
384 QUATERNIONS. [499
Multiply into any constant vector, and we have, by adding
three such results
fdpa = ffdsV(UvV) <r. [See 497.]
Hence at once (by adding together the corresponding members
of the two last equations, and putting
fdpq=ffdsV(UvV)q,
where q is any quaternion whatever.
[For the reason why we have no corresponding formula, with S
instead of V in the righthand member, see remark in [ ] in
505.]
499. Commencing afresh with the fundamental integral
put o = u&
and we have f/fS{3Vuds = ffuS/3 Uvds ;
from which at once fffVuds = ffuUvds ........................ (1),
or fJfVTd*=ffU V .Td8 ..................... (2),
which gives
JJ/FV Yards = Vf}UvV<rrds = ff(rSUv(r  crSUvr) ds.
Equation (2) gives a remarkable expression for the surface of a
space in terms of a volumeintegral. For take
T =Uv= UVP,
where P = const.
is the scalar equation of the closed bounding surface. Then
_ jfds = fJUv Uvds = ///V U VPds.
(Note that this implies
which in itself is remarkable.)
Thus the surface of an ellipsoid
s
the integration being carried on throughout the enclosed space.
(Compare 485.)
5OO.] PHYSICAL APPLICATIONS. 385
Again, in (2), putting u^r for T, and taking the scalar, we have
ffJSrV Ul + w,fifV T ) d* = f
whence fff[S(rV) a + aSVr\ ds = ffaSrUvds .............. (3).
The sum of (1) and (2) gives, for any quaternion,
The final formulae in this, and in the preceding, section give
expressions in terms of surface integrals for the volume, and the
line, integrals of a quaternion. The latter is perfectly general, but
(for a reason pointed out in 492) the former is definite only when
the quaternion has the form Vq.
500. The fundamental form of the Volume and Surface
Integral is (as in 499 (1))
fffVuds = ffUvuds.
Apply it to a space consisting of a very thin transverse slice of
a cylinder. Let t be the thickness of the slice, A the area of one
end, and a a unitvector perpendicular to the plane of the end.
The above equation gives at once
where dl is the length of an element of the bounding curve of the
section, and the only values of Uv left are parallel to the plane of
the section and normal to the bounding curve. If we now put p
as the vector of a point in that curve, it is plain that
V.aUv=Udp, dl = Tdp,
and the expression becomes
V(oiV)u.A =fudp.
By juxtaposition of an infinite number of these infinitely small
directed elements, a (now to be called Uv) being the normal vector
of the area A (now to be called ds), we have at once
ffV (UvV) uds = fudp,
which is the fundamental form of the Surface and Line Integral,
as given in 498. Hence, as stated in 484, these relations are
not independent.
In fact, as the first of these expressions can be derived at once
from the ordinary equation of "continuity," so the second is merely
the particular case corresponding to displacements confined to a
T. Q. I. 25
386 QUATERNIONS.
given surface. It is left to the student to obtain it, simply and
directly, (in the form of 495) from this consideration.
[Note. A remark of some importance must be made here. It
may be asked : Why not adopt for the proof of the fundamental
theorems of the present subject the obvious Newtonian process (as
applied, for instance, in Thomson and Tait s Natural Philosophy,
194, or in ClerkMaxwell s Electricity, 591) ?
The reply is that, while one great object of the present work is
(as far as possible) to banish artifice, and to shew the "perfect
naturalness of Quaternions/ the chief merit of the beautiful
process alluded to is that it forms one of the most intensely
artificial applications of an essentially artificial system. Cartesian
and SemiCartesian methods may be compared to a primitive
telegraphic code, in which the different signals are assigned to
the various letters at haphazard; Quaternions to the natural
system, in which the simplest signals are reserved for the most
frequently recurring letters. In the former system some one word,
or even sentence, may occasionally be more simply expressed than
in the latter : though there can be no doubt as to which system
is to be preferred. But, even were it not so, the methods we have
adopted in the present case give a truly marvellous insight into
the real meaning and " inner nature " of the formulae obtained.]
501. As another example of the important results derived
from the simple formulae of 499, take the following, viz.:
ffV. V (a Uv) rds = ffaSr Uvds  JfUvSards,
where by (3) and (1) of that section we see that the righthand
member may be written
= /// {# (TV) a + oSVr  VSar] ds
= ///F.F(Vo)T<fc (4).
In this expression the student must remark that V operates on r
as well as on a. Had it operated on r alone, we should have
inverted the order of V and <r, and changed the sign of the whole ;
or we might have had recourse to the notation of the end of 133.
This, and similar formulae, are easily applied to find the
potential and the vectorforce due to various distributions of
magnetism. To shew how they are to be introduced, we briefly
sketch the mode of expressing the potential of a distribution.
503.] PHYSICAL APPLICATIONS. 387
K. Application of the V Integrals to Magnetic &c. Problems.
502. Let cr be the vector expressing the direction and intensity
of magnetisation, per unit of volume, at the element ds. Then if
the magnet be placed in a field of magnetic force whose potential
is u, we have for its potential energy
This shews at once that the magnetism may be resolved into a
volumedensity SVcr, and a surfacedensity  SaUv. Hence, for
a solenoid al distribution,
SVa = 0.
What Thomson has called a lamellar distribution (Phil. Trans.
1852), obviously requires that
be integrable without a factor ; i.e., that
FVo = 0.
A complex lamellar distribution requires that the same expression
be integrable by the aid of a factor. If this be u, we have at once
FV (ua) = 0,
or S . <rVcr = 0.
But we easily see that (4) of 501 may be written
 J/F ( F<7 UP) rds =  J//F. rVVods  JJ/F aVrds + JffSaV . rds.
Now, if T = V (
where r is the distance between any external point and the element
cfc, the last term on the right is the vectorforce exerted by the
magnet on a unitpole placed at the point. The second term on
the right vanishes by Laplace s equation, and the first vanishes as
above if the distribution of magnetism be lamellar, thus giving
Thomson s result in the form of a surface integral.
503. As another instance, let a be the vector of magnetic
induction, /3 the vector potential, at any point. Then we know,
physically, that
fSpdp=ffSUv* t d8.
252
388 QUATERNIONS. [504.
But. by the theorem of S 496, we have
t/ <J
fSI3dp=JfS. UvV/3ds.
Since the boundary and the enclosed surface may be any what
ever, we must have
a=FV;
and, as a consequence,
Va = 0.
Hence V/5 = a + u,
= V~ l a + V" 1 ^ + Vq,
where q is a quaternion satisfying the equations
To interpret the other terms, let
V~ l u = o,
so fchat Wo = ;
and o = Vv,
where v is a scalar such that
V*v = u.
Thus v is the potential of a distribution w/4Tr, and can there
fore be found without ambiguity when u is given. And of course
V" 1 ^ = Vv
is also found without ambiguity.
Again, as SVa. = 0,
we have Va = 7, suppose.
Hence, for any assumed value of 7, we have
so that this term of the above value of /3 is also found without
ambiguity.
The auxiliary quaternion q depends upon potentials of arbitrary
distributions wholly outside the space to which the investigation
may be limited. [Compare this section with 494.]
504. An application may be made of similar transformations
to Ampere s Directrice de I action electrodynamique, which, 458
above, is the vectorintegral
Vpdp
Tp 3
505.] PHYSICAL APPLICATIONS. 389
where dp is an element of a closed circuit, and the integration
extends round the circuit. This may be written
so that its value as a surface integral is
ffs (UvV) V i ds 
Of this the last term vanishes, unless the origin is in, or infinitely
near to, the surface over which the double integration extends.
The value of the first term is seen (by what precedes) to be the
vectorforce due to uniform normal magnetisation of the same
surface. Thus we see the reason why the Directrice can be
expressed in terms of the spherical opening of the circuit, as in
459, 471.
505. The following result is obviously but one of an extensive
class of useful transformations. Since
we obtain at once from 484
a curious expression for the gravitation potential of a homogeneous
body in terms of a surfaceintegral.
[The righthand member may be written as
ffSUvVTpds]
and an examination into its nature shews us why we ought not to
expect to have a general expression for
ffSUvVPds
in terms of a lineintegral. It will be excellent practice for the
student to make this examination himself. Of course, a more
general method presents itself in finding the volumeintegral
which is equivalent to the last written surfaceintegral extended
to the surface of a closed space.]
From this, by differentiation with respect to a, after putting
p + a for p, or by expanding in ascending powers of To. (both of
which tacitly assume that the origin is external to the space
390 QUATERNIONS. [506.
integrated through, i.e., that Tp nowhere vanishes within the
limits), we have
_ f r v. u p v. u.Up
~ ~~ 2
and this, again, involves
"^ = II ^ SUvUpds.
506. The interpretation of these, and of more complex formulae
of a similar kind, leads to many curious theorems in attraction and
in potentials. Thus, from (1) of 499, we have
///**///^*n* <
which gives the attraction of a mass of density t in terms of the
potentials of volume distributions and surface distributions. Putting
this becomes
[[[Vods _ rrr Up . ocfc _[[Uv. <
JJJ Tp ])] Tp*"}) Tp
By putting a = p, and taking the scalar, we recover a formula
given above ; and by taking the vector we have
This may be easily verified from the formula
fPdp=VfSUv.VPds,
by remembering that VTp= Up.
Again if, in the fundamental integral, we put
we have ffj^d<2 fjj%= jf tSUvUpds.
[It is curious how closely, in fact to a numerical factor of one
term pres, this equation resembles what we should get by operating
on (1) by S.p, and supposing that we could put p under the signs
of integration.]
508.] PHYSICAL APPLICATIONS. 391
L. Application of V to the StressFunction.
507. As another application, let us consider briefly the Stress
function in an elastic solid.
At any point of a strained body let X be the vector stress per
unit of area perpendicular to i, ft and v the same for planes per
pendicular to j and k respectively.
Then, by considering an indefinitely small tetrahedron, we
have for the stress per unit of area perpendicular to a unit vector
co the expression
\Sia) + pSjw + vSkw = (pco,
so that the stress across any plane is represented by a linear and
vector function of the unit normal to the plane.
But if we consider the equilibrium, as regards rotation, of an
infinitely small parallelepiped whose edges are parallel to i, j, k
respectively, we have (supposing there are no molecular couples)
or
or, supposing V to apply to p alone,
This shews ( 185) that in the present case <p is selfconjugate, and
thus involves not nine distinct constants but only six.
508. Consider next the equilibrium, as regards translation, of
any portion of the solid filling a simplyconnected closed space.
Let u be the potential of the external forces. Then the condition
is obviously
where v is the normal vector of the element of surface ds. Here
the double integral extends over the whole boundary of the closed
space, and the triple integral throughout the whole interior.
To reduce this to a form to which the method of 485 is
directly applicable, operate by S . a where a is any constant vector
whatever, and we have
JfS . </>a Uvds + {ffdsSaVu =
392 QUATERNIONS. [5o8.
by taking advantage of the selfconjugateness of $. This may be
written (by transforming the surfaceintegral into a volume
integral)
fff<k (S . V</> + S . aVu) = 0,
and, as the limits of integration may be any whatever,
S.V<f>oL + SaVu = ...................... (1).
This is the required equation, the indeterminateness of a rendering
it equivalent to three scalar conditions.
There are various modes of expressing this without the a.
Thus, if A be used for V when the constituents of <p are considered,
we may write
Vu =
It is easy to see that the righthand member may be put in
either of the equivalent forms
or
In integrating this expression through a given space, we must
remark that V and p are merely temporary symbols of construction,
and therefore are not to be looked on as variables in the integral.
Instead of transforming the surfaceintegral, we might have
begun by transforming the volumeintegral. Thus the first equa
tion of this section gives
//((/> + u) Uvds = 0.
From this we have at once
fJS. Uv (< + u) ads = 0.
Thus, by the result of 490, whatever be a we have
.V(( + w)a = 0,
which is the condition obtained by the former process.
As a verification, it may perhaps be well to shew that from this
equation we can get the condition of equilibrium, as regards
rotation, of a simplyconnected portion of the body, which can be
written by inspection as
JJF. p<l> ( Uv) ds + /// VpVuds = 0.
This is easily done as follows : (1) gives
if, and only if, <j satisfy the condition
509.] PHYSICAL APPLICATIONS. 393
Now this condition is satisfied by
a = Yap
where a is any constant vector. For
8. (V) Vap = S.aV(l> (V)p = S. aW^ = 0,
in consequence of the selfconjugateness of (/>. Hence
fffds (S . 7< Vap + 8 . apVu) = 0,
or JfdsS . ap(f> Uv + fffdsS . apVu = 0.
Multiplying by a, and adding the results obtained by making a in
succession each of three rectangular vectors, we obtain the required
equation.
509. To find the stressfunction in terms of the displacement at
each point of an isotropic solid, when the resulting strain is small,
we may conveniently apply the approximate method of 384. As
the displacement is supposed to be continuous, the strain in the
immediate neighbourhood of any point may be treated as homo
geneous. Thus, round each point, there is one series of rectangular
parallelepipeds, each of which remains rectangular after the strain.
Let a, /3, 7 ; a 15 {3 V <y l ; be unit vectors parallel to their edges
before, and after, the strain respectively ; and let e v e 2 , e 3 be the
elongations of unit edges parallel to these lines. We shall not
have occasion to determine these quantities, as they will be
eliminated after having served to form the requisite equations.
Since the solid is isotropic and homogeneous, the stress is
perpendicular to each face in the strained parallelepipeds ; and
its amount (per unit area) can be expressed as
P l = 2we 1 + (c  f n) 2e, &c (1),
where n and c are, respectively, the rigidity and the resistance to
compression.
Next, as in 384, let a be the displacement at p. The strain
function is
i^tzr = OT $33V. a
so that at once
S e = SV<r; (2),
and, if q ( ) q 1 be the operator which turns a into a,, &c v we
have
*r (3).
394 QUATERNIONS. [509.
Thus, if (f> be the stressfunction, we have (as in 507)
<< =  2 . P^&VB (4).
But >/r&> = 2 . (1 + e a ) a
so that \J/ft> = 2 . (1 + e x
and qty wq 1 = 2.(1 +ej afia.^ (5).
By the help of (1), (2), and (5), (4) becomes
00) = Zriqifr toq" 1 2nco (c f ??) coSVa ;
and, to the degree of approximation employed, (3) shews that this
may be written
(j)0) = n (SwV. a + VScoa)  (c  f n) wSVa (G),
which is the required expression, the function c/> being obviously
selfconjugate.
As an example of its use, suppose the strain to be a uniform
dilatation. Here
o = ep,
and (/>&) = Zneay + 3 (c f n) ea) = Scew ;
denoting traction See, uniform in all directions. If e be negative,
there is uniform condensation, and the stress is simply hydrostatic
pressure.
Again, let a eaSap,
which denotes uniform extension in one direction, unaccompanied
by transversal displacement. We have (a being a unit vector)
(f)a) = 2neaS(oa. + (c f n) ew.
Thus along a there is traction
(c + jn)e,
but in all directions perpendicular to a there is also traction
(cn)e.
Finally, take the displacement
a = ea.S/3p.
It gives ^o) =  ne (aSco/3 + fiSua) (c f ?i) ewSaff.
This displacement gives a simple shear if the unit vectors a and ft
are at right angles to one another, and then
(f)a) = ne (a.Sco/3 + fiStoa),
which agrees with the wellknown results. In particular, it shews
that the stress is wholly tangential on planes perpendicular either
511.] PHYSICAL APPLICATIONS. 395
to a. or to {3 ; and wholly normal on planes equally inclined to them
and perpendicular to their plane. The symmetry shews that the
stress will not be affected by interchange of the unit vectors,
a and /3, in the expression for the displacement.
510. The work done by the stress on any simply connected
portion of the solid is obviously
because $ (Uv) is the vector force overcome per unit of area on
the element ds. [The displacement at any moment may be written
XCT ; and, as the stress is always proportional to the strain, the
factor xdx has to be integrated from to 1.] This is easily trans
formed to
511. We may easily obtain the general expression for the
work corresponding to a strain in any elastic solid. The physical
principles on which we proceed are those explained in Appendix
G to Thomson and Tait s Natural Philosophy. The mode in
which they are introduced, however, is entirely different; and a
comparison will shew the superiority of the Quaternion notation,
alike in compactness and in intelligibility and suggestiveness.
If the strain, due to the displacement a, viz.
^T = T SrV . a
be a mere rotation, in which case of course no work is stored up
by the stress, we have at once
S . ^TW^TT= S(i)T
for all values of co and r. We may write this as
S . a) (irS/r  1) r = Sw^r = 0,
where % is ( 380) a selfconjugate linear and vector function, whose
complete value is
X r =  SrV . a  VSro + V^rVSoov
The last term of this may, in many cases, be neglected.
When the strain is very small, the work (per unit volume)
must thus obviously be a homogeneous function, of the second
degree, of the various independent values of the expression
396 QUATERNIONS. [5 I 2.
On account of the selfconjugateness of ^ there are but six
such values : viz.
Their homogeneous products of the second degree are therefore
twentyone in number, and this is the number of elastic coefficients
which must appear in the general expression for the work. In
the most general form of the problem these coefficients are to be
regarded as given functions of p.
At and near any one point of the body, however, we may take
i, j, k as the chief vectors of ^ at that point, and then the work
for a small element is expressible in terms of the six homogeneous
products, of the second degree, of the three quantities
Si W> S JXJ> Sk X k 
This statement will of course extend to a portion of the body of
any size if (whether isotropic or not) it be homogeneous and
homogeneously strained. From this follow at once all the
elementary properties of homogeneous stress.
M. The Hydrokinetic Equations.
512. As another application, let us form the hydrokinetic
equations, on the hypothesis that a perfect fluid is not a molecular
assemblage but a continuous medium.
Let a be the vectorvelocity of a very small part of the fluid at
p ; e the density there, taken to be a function of the pressure, p,
alone ; i.e. supposing that the fluid is homogeneous when the
pressure is the same throughout ; P the potential energy of unit
mass at the point p.
The equation of " continuity" is to be found by expressing the
fact that the increase of mass in a small fixed space is equal
to the excess of the fluid which has entered over that which
has escaped. If we take the volume of this space as unit, the
condition is
.................... (1).
We may put this, if we please, in the form
where 3 expresses total differentiation, or, in other words, that we
follow a definite portion of the fluid in its motion.
513.] PHYSICAL APPLICATIONS. 397
The expression might at once have been written in the form
(2) from the comparison of the results of two different methods of
representing the rate of increase of density of a small portion of
the fluid as it moves along. Both forms reduce to
when there is no change of density ( 384).
Similarly, for the rate of increase of the whole momentum
within the fixed unit space, we have
*&1 =  e vp Vp +SJSUvo . eads ;
where the meanings of the first two terms are obvious, and the
third is the excess of momentum of the fluid which enters, over
that of the fluid which leaves, the unit space.
The value of the double integral is, by 49.9 (3),
(eer) + eSoV . o = o ^ + eSaV . o, by (1).
Thus we have, for the equation of motion,
or, finally
fa i * i ^ / ~~ T ^ ()
This, in its turn, might have been even more easily obtained by
dealing with a small definite portion of the fluid.
It is necessary to observe that in what precedes we have
tacitly assumed that a is continuous throughout the part of the
fluid to which the investigation applies : i.e. that there is neither
rupture nor finite sliding.
513. There are many ways of dealing with the equation (3)
of fluid motion. We select a few of those which, while of historic
interest, best illustrate quaternion methods.
We may write (3) as
Now we have always
F. oFVo  SoV . a  V^cr = SaV . a  JV . a 2 .
398
QUATERNIONS . [513.
Hence, if the motion be irrotational, so that ( 384)
Wo = 0,
the equation becomes
But, if w be the velocitypotential,
0 =
and we obtain (by substituting this in the first term, and operating
on the whole by S . dp) the common form
7 dw , 7 2 ir\
d T. + d . v 2 =  dQ,
where l^ 2 (= o 2 ) is the kinetic energy of unit mass of the fluid.
If the fluid be incompressible, we have Laplace s equation
for w, viz.
V% = SVa = 0.
When there is no velocitypotential, we may adopt Helmholtz
method. But first note the following quaternion transformation
(Proc. R. 8. E. 186970)
[The expression on the right has many remarkable forms, the
finding of which we leave, as an exercise, to the student. For our
present purpose it is sufficient to know that its vector part is
This premised, operate on (3) by V. V, and we have
Hence at once, if the fluid be compressible,
= S. VV . <7 + VVa . SVa = V. V V.
ot
But if the fluid be incompressible
Either form shews that when the vectorrotation vanishes, its
rate of change also vanishes. In other words, those elements of
the fluid, which were originally devoid of rotation, remain so during
the motion.
514.] PHYSICAL APPLICATIONS. 399
Thomson s mode of dealing with (3) is to introduce the
integral
(5)
which he calls the " flow " along the arc from the point a to the
point p ; these being points which move with the fluid.
Operating on (3) with S . dp, we have as above
so that, integrating along any definite line in the fluid from a. to
p, we have
which gives the rate at which the flow along that line increases, as
it swims along with the fluid.
If we integrate round a closed curve, the value of df/dt vanishes,
because Q is essentially a singlevalued function. In this case the
quantity / is called the " circulation," and the result is stated in
the form that the circulation round any definite path in the fluid
retains a constant value.
Since the circulation is expressed by the complete integral
}S<rdp
it can also be expressed by the corresponding double integral
//. UvVads,
so that it is only when there is at least one vortexfilament passing
through the closed circuit that the circulation can have a finite
value.
N. Use of V in connection with Taylor s Theorem.
514. Since the algebraic operator
when applied to any function of x, simply changes x into x + h, it
is obvious that if a be a vector not acted on by
. d . d d
V = ^ + ; ,+&,,
dx ay dz
we have e ~ S(rV f (p) =f (p + <r),
400 QUATERNIONS. [515.
whatever function f may be. From this it is easy to deduce
Taylor s theorem in one important quaternion form.
If A bear to the constituents of or the same relation as V bears
to those of p, and if / and F be any two functions which satisfy
the commutative law in multiplication, this theorem takes the
curious form
of which a particular case is (in Cartesian symbols)
The modifications which the general expression undergoes, when
/ and F are not commutative, are easily seen.
If one of these be an inverse function, such as, for instance,
may occur in the solution of a linear differential equation, these
theorems of course do not give the arbitrary part of the integral,
but they often materially aid in the determination of the rest.
One of the chief uses of operators such as $aV, and various
scalar functions of them, is to derive from 1/Tp the various orders
of Spherical Harmonics. This, however, is a very simple matter.
515. But there are among them results which appear startling
from the excessively free use made of the separation of symbols.
Of these one is quite sufficient to shew their general nature.
Let P be any scalar function of p. It is required to find the
difference between the value of P at p, and its mean value
throughout a very small sphere, of radius r and volume v t which has
the extremity of p as centre. This, of course, can be answered at
once from the formula of 485. But the somewhat prolix method
we are going to adopt is given for its own sake as a singular piece
of analysis, not for the sake of the problem.
From what is said above, it is easy to see that we have the
following expression for the required result :
;///<* *>**
where a is the vector joining the centre of the sphere with the
element of volume cfc, and the integration (which relates to <r
and ds alone) extends through the whole volume of the sphere.
5 1 6.] PHYSICAL APPLICATIONS. 401
Expanding the exponential, we may write this expression in the
form
1 JJJ (SaVf Pd,&c.,
+
higher terms being omitted on account of the smallness of r, the
limit of To.
Now, symmetry shews at once that
JIM? = 0.
Also, whatever constant vector be denoted by a,
/// (Scwr) 1 & =  a 2 /// (OrUay tk.
Since the integration extends throughout a sphere, it is obvious
that the integral on the right is half of what we may call the
moment of inertia of the volume about a diameter. Hence
If we now write V for a, as the integration does not refer to V,
we have by the foregoing results (neglecting higher powers of r)
<**!>** jS^.
which is the expression given by ClerkMaxwell *. Although, for
simplicity, P has here been supposed a scalar, it is obvious that in
the result above it may at once be written as a quaternion.
516. As another illustration, let us apply this process to the
finding of the potential of a surfacedistribution. If p be the
vector of the element ds, where the surface density is fp, the
potential at a is
Jfdsfp.FT(p<r),
F being the potential function, which may have any form whatever.
By .the preceding, 514, this may be transformed into
JJdsfp.^FTp;
or, far more conveniently for the integration, into
* London Math. Soc. Proc., vol. iii, no. 34, 1871.
T. Q. I. 26
402 QUATERNIONS. [5 1 7
where A depends on the constituents of a in the same manner as
V depends on those of p.
A still farther simplification may be introduced by using a
vector CT O , which is finally to be made zero, along with its corre
sponding operator A , for the above expression then becomes
where p appears in a comparatively manageable form. It is obvious
that, so far, our formulae might be made applicable to any dis
tribution. We now restrict them to a superficial one.
517. Integration of this last/orm can always be easily effected
in the case of a surface of revolution, the origin being a point in
the axis. For the expression, so far as the integration is concerned,
can in that case be exhibited as a single integral
where F may be any scalar function, and x depends on the cosine of
the inclination of p to the axis. And
da) a
As the interpretation of the general results is a little trouble
some, let us take the case of a spherical shell, the origin being the
centre and the density unity, which, while simple, sufficiently
illustrates the proposed mode of treating the subject.
We easily see that in the above simple case, a. being any
constant vector whatever, and a being the radius of the sphere,
C+a 977V7
ffds e s *P = 2Tra e xTa dx = ^ (e aTa  e aTa ).
J a L &
Now, it appears that we are at liberty to treat A as a has just
been treated. It is necessary, therefore, to find the effects of such
operators as TA, e arA , &c., which seem to be novel, upon a scalar
function of To ; or , as we may for the present call it.
9F
Now <TA) 2 F =  tfF = F" + ~ ,
whence it is easy to guess at a particular form of TA. To make
sure that it is the only one, assume
5 1 8.] PHYSICAL APPLICATIONS. 403
where f and fare scalar functions of ^ to be found. This gives
= ?F" + (ft + f f + ff ) F + (ff + f 2 ) F.
Comparing, we have
From the first, f = + 1,
whence the second gives f = + ==. ;
the signs of f and f being alike. The third is satisfied identically.
That is, finally, ^ = ^ + 1.
Also, an easy induction shews that
Hence wo have at once
n / d
l 1
J
+&c 
by the help of which we easily arrive at the wellknown results.
This we leave to the student*.
O. Applications of V in connection with Calculus of Variations.
518. We conclude with a few elementary examples of the use
of V in connection with the Calculus of Variations. These depend,
for the most part, on the simple relation
Let us first consider the expression
A=fQTdp,
where Tdp is an element of a finite arc along which the inte
gration extends, and the quaternion Q is a function of p, generally
a scalar.
* Proc.R. S. E. t 18712.
262
404 QUATERNIONS. [5 1 9.
To shew the nature of the enquiry, note that if Q be the
speed of a unit particle, A will be what is called the Action. If
Q be the potential energy per unit length of a chain, A is the
total potential energy. Such quantities are known to assume
minimum, or at least "stationary", values in various physical
processes.
We have for the variation of the above quantity
BA = f (SQTdp + QSTdp)
=f(SQTdpQS.Udpd8p)
=  [QSUdpSp] + f {BQTdp + S.S P d (Q Udp)},
where the portion in square brackets refers to the limits only, and
gives the terminal conditions. The remaining portion may easily
be put in the form
SfSp{d(QUdp)VQ.Tdp}.
If the curve is to be determined by the condition that the varia
tion of A shall vanish, we must have, as Bp may have any direction,
d(QUdp)VQ.
or, with the notation of Chap. X,
This simple equation shews that (when Q is a scalar)
(1) The osculating plane of the sought curve contains the
vector VQ.
(2) The curvature at any point is inversely as Q, and directly
as the component of VQ parallel to the radius of absolute curvature.
519. As a first application, suppose A to represent the Action
of a unit particle moving freely under a system of forces which
have a potential ; so that
Q = Tp,
and p* = 2(PH),
where P is the potential, H the energyconstant.
These give TpVTp = QVQ =  VP,
and Qp = p,
so that the equation above becomes simply
522.] PHYSICAL APPLICATIONS. 405
which is obviously the ordinary equation of motion of a free
particle.
520. If we look to the superior limit only, the first expression
for SA becomes in the present case
If we suppose a variation of the constant H, we get the following
term from the unintegrated part
tSH.
Hence we have at once Hamilton s equations of Varying Action in
the forms
VA=p
dA
and j^. = t.
dJd
The first of these gives, by the help of the condition above,
the wellknown partial differential equation of the first order and
second degree.
521. To shew that, if A be any solution whatever of this
equation, the vector VA represents the velocity in a free path
capable of being described under the action of the given system of
forces, we have
= S(VA.V)VA.
But ~.VA = S(p
Cit
A comparison shews at once that the equality
VA=p
is consistent with each of these vector equations.
522. Again, if 3 refer to the constants only,
d(VA)* = S.VAdVA = dH
by the differential equation.
But we have also ^v = t,
oli
which gives %(dA)=*S (/3V) dA = dff.
Cit
406 QUATERNIONS. [523.
These two expressions for dH again agree in giving
and thus shew that the differential coefficients of A with regard to
the two constants of integration must, themselves, be constants.
We thus have the equations of two surfaces whose intersection
determines the path.
523. Let us suppose next that A represents the Time of
passage, so that the brachistochrone is required. Here we have
the other condition being as in 519, and we have
which may be reduced to the symmetrical form
It is very instructive to compare this equation with that of the
free path as above, 519 ; noting how the force VP is, as it
were, reflected on the tangent of the path ( 105). This is the
wellknown characteristic of such brachistochrones.
The application of Hamilton s method may be easily made, as
in the preceding example. (Tait, Trans. R. S. E., 1805.)
524. As a particular case, let us suppose gravity to be the
only force, then
a constant vector, so that
d ._! ,_ 2
jt p p a=0 
The form of this equation suggests the assumption
where p and q are scalar constants, and
Substituting, we get
pq sec 2 qt + (  /3 2  pV tan 2 qt) = 0,
which gives pq = T*$ = p*Fa.
526.] PHYSICAL APPLICATIONS. 407
Now let p{3~ 1 a. = 7 ;
this must be a unitvector perpendicular to a and /3, so that
/?
p~ l = . (cos at 7 sin otf),
cos g v
whence p = cos qt (cos qt + y sin gtf) /3" 1
(which may be verified at once by multiplication).
Finally, taking the origin so that the constant of integration
may vanish, we have
2/o/S = t + ^ (sin 2qt  7 cos 2q),
which is obviously the equation of a cycloid referred to its vertex.
The tangent at the vertex is parallel to/3, and the axis of symmetry
to a. The equation, it should be noted, gives the law of descrip
tion of the path.
525. In the case of a chain hanging under the action of given
forces we have, as the quantity whose lineintegral is to be a
minimum,
Q = Pr,
where P is the potential, r the mass per unitlength.
Here we have also, of course,
the length of the chain being given.
It is easy to see that this leads, by the method above, to the
equation
where u is a scalar multiplier.
526. As a simple case, suppose the chain to be uniform.
Then we may put ru for u, and divide by r. Suppose farther
that gravity is the only force, then
P = Sap, VP =  a,
and y {(Sap + ?/) p\ + a = 0.
408 QUATERNIONS. [527.
Differentiating, and operating by Sp, we find
or, since p is a unitvector,
0
ds~">
which shews that u is constant, and may therefore be allowed for
by change of origin.
The curve lies obviously in a plane parallel to a, and its equation is
(Sap) 2 + a 2 s 2 = const.,
which is a wellknown form of the equation of the common
catenary.
When the quantity Q of 518 is a vector or a quaternion, we
have simply an equation (like that there given) for each of the
constituents.
527. Suppose P and the constituents of cr to be functions
which vanish at the bounding surface of a simplyconnected space
2, or such at least that either P or the constituents vanish there,
the others (or other) not becoming infinite.
Then, by 484,
fJfdsS . V(Pa) = // ds PScr Uv = 0,
if the integrals be taken through and over 2.
Thus fffdsS.aVP = fffd<;PSV<7.
By the help of this expression we may easily prove a very
remarkable proposition of Thomson (Cam. and Dub. Math. Journal,
Jan. 1848, or Reprint of Papers on Electrostatics, 206).
To shew that there is one, and but one, solution of the equation
8V (e*Vu) = 477T
where r vanishes at an infinite distance, and e is any real scalar
whatever, continuous or discontinuous.
Let v be the potential of a distribution of density r, so that
and consider the integral
527.] PHYSICAL APPLICATIONS. 409
That Q may be a minimum as depending on the value of u (which
is obviously possible since it cannot be negative, and since it may
have any positive value, however large, if only greater than this
minimum) we must have
= JSQ =  fffdiS . (e 2 Vu  Vfl) VBu
By the lemma given above this may be written
Thus any value of u which satisfies the given equation is such as to
make Q a minimum.
But there is only one value of u which makes Q a minimum ;
for, let Q l be the value of Q when
u l = u + w
is substituted for this value of u, and we have
Q, =  jj/* (* v (u+w)^vv y
= Q  2 ///cfc S (e 2 Vu  Vw) Vw  fffds e*(Vw) 2 .
The middle term of this expression may, by the proposition at the
beginning of this section, be written
and therefore vanishes. The last term is essentially positive. Thus
if MJ anywhere differ from u (except, of course, by a constant
quantity) it cannot make Q a minimum: and therefore u is a
unique solution.
MISCELLANEOUS EXAMPLES.
1. The expression
Fa/3 FyS + Fa 7 FS/3 + FaS F/3 7
denotes a vector. What vector ?
2. If two surfaces intersect along a common line of curvature,
they meet at a constant angle.
410 QUATERNIONS.
3. By the help of the quaternion formulae of rotation, translate
into a new form the solution (given in 248) of the problem of
inscribing in a sphere a closed polygon the directions of whose sides
are given.
4. Find the point, the sum of whose distances from any
number of given points is a minimum.
If p be the sought point, a l5 2 , &c. the given points: shew
that
Give a dynamical illustration of this solution.
(Proc. R S. E. 18667.)
5. Shew that
8.
6. Express, in terms of the masses, and geocentric vectors of
the sun and moon, the sun s vector disturbing force on the moon,
and expand it to terms of the second order; pointing out the
magnitudes and directions of the separate components.
(Hamilton, Lectures, p. 615.)
7. lfq = r*, shew that
2 dq = 2dr* = %(dr + Kqdrq 1 ) Sq~ l = %(dr + q^drKq) Sq~ l
= (drq + Kqdr) q~ l (q + Kq)~ l = (drq + Kqdr) (r + Tr)~ l
dr + U^drU 1
q~\qdr+Trdrq~*) _drUq+ Uq l dr _
~ WT Tq(l+ Ur) 1 + Ur
 d.r + V. Vdr ~qq 1 = dr  V
q q q
= drq~ l + V(Vq~\Vdr} (l +
and give geometrical interpretations of these varied expressions
for the same quantity. (Ibid. p. 628.)
8. Shew that the equation of motion of a homogeneous solid
of revolution about a point in its axis, which is not its centre
of gravity, is
where II is a constant. (Trans. R. S. E., 1869.)
MISCELLANEOUS EXAMPLES. 411
9. Find the point P, such that, if A^ A 2 , &c. be any fixed
points in space, and m 1? m 2 , &c. positive numerical quantities,
S.mAP shall be a minimum.
Shew that a closed (gauche) polygon can be constructed whose
sides are parallel to PA^ PA^ &c. while their lengths are as m v m 2 ,
&c., respectively.
If T2, . mAP is to be a minimum, what is the result ?
10. Form the quaternion condition that the lines joining the
middle points of the sides of a closed polygon (plane or gauche)
may form a similar polygon.
When this condition is satisfied, find the quaternion operator
which must be applied to the second polygon to make it similarly
situated with the first.
11. Solve the equations in linear and vector functions; or
being given, while $ and ^ are to be found :
%+ $ =
12. Put the equation of a Minding line ( 394) directly into
the normal form for a line passing through each of two fixed
curves :
p = x $t + (1  x) ^u,
where < and v/r are definite vector functions of the arbitrary
scalars t, and u, respectively.
13. Shew that
are different expressions for the same scalar, and give a number of
other forms. ( 513.) How are the values of these quantities
respectively affected, if the V without suffix acts on a^ as well
as on <7?
14. If (as in 511)
ITT = SCOT
for all values of CD and of r, shew by actual transformation, not by
the obvious geometrical reasoning, that we have also
412 QUATERNIONS.
15. Shew that, to the third order in Tfi, the couple, due to
any closed circuit, on a small magnet, 2/3, whose centre is at the
origin, is proportional to
.p f / sr/8 1 US ! /3
r v ~ 2 v + T
Simplify, by means of this formula, the quasiCartesian investiga
tion of 464.
16. Calculate the value of
[Vpp
J Tp 3
for a circular current, the origin being taken at any point.
17. Find an approximate formula for the potential, near the
centre of the field, when two equal, circular, currents have a
common axis, and are at a distance equal to their common
radius. (This is v. Helmholtz Tangent Galvanometer.)
18. Deduce the various forms of Spherical Harmonics, and
their relations, from the results of the application of scalar
functions of $aV, and similar operators, to l/Tp.
19. Integrate the differential equations :
/ \ dq ,
(a) ? + aq = 6,
(b)
where a and b are given quaternions, and ( and ^ given linear
and vector functions. (Tait, Proc. E. S. E., 18701.)
20. Derive (4) of 92 directly from (3) of 91.
21. Find the successive values of the continued fraction
=(/?)
where i and j have their quaternion significations, and x has the
values 1, 2, 3, &c. (Hamilton, Lectures, p. 645.)
MISCELLANEOUS EXAMPLES. 413
22. If we have
;=(/+)"
where c is a given quaternion, find the successive values.
For what values of c does u become constant ? (Ibid. p. 652.)
23. Prove that the moment of hydrostatic pressures on the
faces of any polyhedron is zero, (a) when the fluid pressure is the
same throughout, (6) when it is due to any set of forces which
have a potential.
24. What vector is given, in terms of two known vectors, by
the relation
p^Ma +r 1 )?
Shew that the origin lies on the circle which passes through the
extremities of these three vectors.
25. Tait, Trans, and Proc. R S. E., 18703.
With the notation of 484, 495, prove
(a) fffS (aV) rd9  ffrSot Uvds.
(b) If 8 (pV) r =  nr,
(n + 3) JJ/rd? =  ffrSp Uvds.
(c) With the additional restriction W = 0,
ffS . UP {2np + (n + 3) /> 2 V j . rds = 0.
(d) Express the value of the last integral over a nonclosed
surface by a lineintegral.
(e) JTdp=ffdsS.UvV<r,
if a = Udp all round the curve.
(/) For any portion of surface whose bounding edge lies
wholly on a sphere with the origin as centre
whatever be the vector a.
(g) JVdpV. a = ffds { UvV 2  S ( UvV)V] <r,
whatever be cr.
26. Tait, Trans. E. S. E., 1873.
Interpret the equation
da = uqdpq~ l ,
414 QUATERNIONS.
and shew that it leads to the following results
VV = qVuq\
V. uq~ l = 0,
V 2 . u* = 0.
Hence shew that the only sets of surfaces which, together, cut
space into cubes are planes and their electric images.
27. What problem has its conditions stated in the following
six equations, from which f, ij, are to be determined as scalar
functions of x } y, z, or of
p = ix +jy + kz ?
__ . d . d 7 d
where V= l j+Jj+kr
dx J dy dz
Shew that they give the farther equations
Shew that (with a change of origin) the general solution of these
equations may be put in the form
where (/> is a selfconjugate linear and vector function, and f, 77, f
are to be found respectively from the three values of / at any
point by relations similar to those in Ex. 24 to Chapter X.
28. Shew that, if p be a planet s radius vector, the potential
P of masses external to the solar system introduces into the
equation of motion a term of the form S (pV)VP.
Shew that this is a selfconjugate linear and vector function
of p, and that it involves only^ye independent constants.
Supposing the undisturbed motion to be circular, find the chief
effects which this disturbance can produce.
29. In 430 above, we have the equations
where &> 2 is neglected. Shew that with the assumptions
= qrrr~ l q~ 1
we have
/3 = 0, T/3 = l, S{3r = 0,
provided w/Sfw  ^ = 0. Hence deduce the behaviour of the
MISCELLANEOUS EXAMPLES. 415
Foucault pendulum without the x, y, and f, rj transformations
in the text.
Apply analogous methods to the problems proposed at the end
of 426 of the text.
30. Hamilton, Bishop Law s Premium Examination, 1862.
(a) If OABP be four points of space, whereof the three first
are given, and not collinear ; if also
OA = a, OB = /3, OP = p ;
and if, in the equation
a a
the characteristic of operation F be replaced by S, the locus of P
is a plane. What plane ?
(6) In the same general equation, if F be replaced by F,
the locus is an indefinite right line. What line ?
(c) If F be changed to K, the locus of P is a point. What
point ?
(d) If F be made = V, the locus is an indefinite halfline,
or ray. What ray ?
(e) If F be replaced by T, the locus is a sphere. What
sphere ?
(/) If F be changed to TV, the locus is a cylinder of
revolution. What cylinder ?
(g) If F be made TVU, the locus is a cone of revolution.
What cone ?
(h) If SU be substituted for F, the locus is one sheet of
such a cone. Of what cone ? and which sheet ?
(i) If F be changed to F7, the locus is a pair of rays.
Which pair ?
31. Hamilton, Bishop Law s Premium Examination, 1863.
(a) The equation
expresses that /> and p are the vectors of two points p and p ,
which are conjugate with respect to the sphere
p 2 + a 2 = 0;
or of which one is on the polar plane of the other.
(b) Prove by quaternions that if the right line PP 7 , connect
ing two such points, intersect the sphere, it is cut harmonically
thereby.
416 QUATERNIONS.
(c) If P be a given external point, the cone of tangents
drawn from it is represented by the equation,
and the orthogonal cone, concentric with the sphere, by
(d) Prove and interpret the equation,
T(npa) = T(pna), if Tp = Ta.
(e) Transform and interpret the equation of the ellipsoid,
(/) The equation
(V  ij = (t 2 + * 2 ) Spp +
expresses that p and p are values of conjugate points, with respect
to the same ellipsoid.
(g) The equation of the ellipsoid may also be thus written,
Svp = 1, if O 2  * 2 )V = (i  K)*P + ZiSicp + 2fc8ip.
(h) The last equation gives also,
(i) With the same signification of v, the differential equa
tions of the ellipsoid and its reciprocal become
Svdp = 0, Spdv = 0.
(j) Eliminate p between the four scalar equations,
Sap = a, Sfip = b, Syp = c, Sep = e.
32. Hamilton, Bishop Law s Premium Examination, 1864.
(a) Let AJ$v Afl^ . . . A n B n be any given system of posited
right lines, the 2n points being all given; and let their vector
sum,
AB = Afr + A& + ...+ A n B n ,
be a line which does not vanish. Then a point H, and a scalar h,
can be determined, which shall satisfy the quaternion equation,
namely by assuming any origin 0, and writing,
nrr v Q^^ l + ..
MISCELLANEOUS EXAMPLES. 417
(b) For any assumed point C, let
Q =CA l .A l B t + ... + CA..AJ.;
then this quaternion sum may be transformed as follows,
Qc=Qn +
and therefore its tensor is
in which AB and CH denote lengths.
(c) The least value of this tensor TQ C is obtained by
placing the point C at H; if then a quaternion be said to be a
minimum when its tensor is such, we may write
min. Q c =Q H =h.AB;
so that this minimum of Q c is a vector.
(d) The equation
TQ C =c = any scalar constant > TQ H
expresses that the locus of the variable point C is a spheric
surface, with its centre at the fixed point H, and with a radius r,
or CH, such that
r.AB=(T<?  T<$$ = (6  h* . AB*)*
so that H, as being thus the common centre of a series of
concentric spheres, determined by the given system of right lines,
may be said to be the Central Point, or simply the Centre, of that
system.
(e) The equation
TVQc = c 1 = any scalar constant > TQ H
represents a right cylinder, of which the radius
divided by AB, and of which the axis of revolution is the line,
wherefore this last right line, as being the common axis of a series
of such right cylinders, may be called the Central Axis of the
system.
(/) The equation
SQc ~ c 2 an y scalar constant
represents a plane ; and all such planes are parallel to the Central
Plane, of which the equation is
SQ C =Q.
T. Q. I. 27
418 QUATERNIONS.
(g) Prove that the central axis intersects the central plane
perpendicularly, in the central point of the system.
(h) When the n given vectors A l B l ,...A n B n are parallel,
and are therefore proportional to n scalars, b v ... b n , the scalar h
and the vector Q H vanish ; and the centre H is then determined
by the equation
or by the expression,
where is again an arbitrary origin.
33. Hamilton, Bishop Laws Premium Examination, 1860.
(a) The normal at the end of the variable vector p, to the
surface of revolution of the sixth dimension, which is represented
by the equation
or by the system of the two equations,
and the tangent to the meridian at that point, are respectively
parallel to the two vectors,
v = 2 (p CL) tp,
and r  2 (1  2$) (pa) + tp ;
so that they intersect the axis a, in points of which the vectors
are, respectively,
AOL i u \ JL 
, and ^
(6) If dp be in the same meridian plane as p, then
, and
dp
(c) Under the same condition,
(d) The vector of the centre of curvature of the meridian,
at the end of the vector p, is, therefore,
3 v 6a4$5
MISCELLANEOUS EXAMPLES. 419
(e) The expressions in (a) give
v 2 = a 2 * 2 (1  t)\ r 2 = aY 5 (1  O 2 (4  1) ;
x 2 9 _ , 2 9a 2 ,, 2
hence (cr p) = 7 a r, and ap = . dt ;
4 4> t
the radius of curvature of the meridian is, therefore,
R = T(ap)= 3 2 tTa;
and the length of an element of arc of that curve is
* ^M.
(/) The same expressions give
thus the auxiliary scalar is confined between the limits and 4,
and we may write t = 2 vers 0, where is a real angle, which varies
continuously from to 2?r ; the recent expression for the element
of arc becomes, therefore,
ds
and gives by integration
if the arc s be measured from the point, say F, for which p a,
and which is common to all the meridians; and the total periphery
of any one such curve is = 127rTa.
(g) The value of cr gives
4 (a 2  a 2 ) = 3a 2 * (4  t\ 16 (Fao) 2 =  aV (4  $)" ;
if, then, we set aside the axis of revolution a, which is crossed by
all the normals to the surface (a), the surface of centres of
curvature which is touched by all those normals is represented by
the equation,
4(c7 2 a 2 ) 3 + 27a 2 (Fa<7) 2 = ........ .......... (b).
(h) The point F is common to the two surfaces (a) and
(b), and is a singular point on each of them, being a triple point
on (a), and a double point on (b) ; there is also at it an infinitely
sharp cusp on (b), which tends to coincide with the axis a, but
a determined tangent plane to (a), which is perpendicular to that
axis, and to that cusp ; and the point, say F t of which the vector
= a, is another and an exactly similar cusp on (b), but does not
belong to (a) .
420 QUATERNIONS.
(i) Besides the three universally coincident intersections of
the surface (a), with any transversal, drawn through its triple
point F, in any given direction ft, there are always three other
real intersections, of which indeed one coincides with F if the
transversal be perpendicular to the axis, and for which the
following is a general formula :
(j) The point, say F, of which the vector is p = 2a, is a
double point of (a), near which that surface has a cusp, which
coincides nearly with its tangent cone at that point; and the
/ IT
semiangle of this cone is = ^ .
Auxiliary Equations :
f Svp =  eft (1 t)(l 20,
\28v (PCL) = a 2 ? (1  1).
f SpT = oi*t*(lt)(4it),
\28(pOL)T = aV (1  t) (4  1).
34. A homogeneous function of p, of the wth degree, is
changed into the same function of the constant vector, a, by the
operator
35. If r be the cube root of the quaternion q, shew that
dr =p + ( V . r 2 + r Vr) Vq 1 (rp  pr),
where p ^r~*dq. (Hamilton, Lectures, p. 629.)
36. Shew from 509 that the term to be added to VQ of
512 (3), in consequence of viscosity, is proportional to
37. Shew, from structural considerations, that
must be a linear and vector function of Fa/3. Also prove, directly,
that its value is
MISCELLANEOUS EXAMPLES. 421
.38. The spherical opening subtended at a by the sphere
,
 ds ^ r0
according as To. a < r.
Hence shew, without further integration, that (with the same
conditions)
f f ds 4Trr 2
1 1 TjT, x = 477T, or ,
whence, of course,
lldsU(p a )_
Oor
&lso that
ds 4Trr 4Trr 2
39. Find also the values of
Uvds Uvds
the integration extending over the surface of the sphere Tp = r.
40. Shew that the potential at /3, due to mass m at a, is to
the potential at /3" 1 , due to mass m at of 1 , as 1 : Tfi, provided
m : m :: 1 : Ta.
Hence, by the results of 38 above, shew that, with a < r,
ds 4?rr 4?rr 2
(p)a) 3 ~ (r>  a?) T (/3  a)
according as T/3 < r, the limits of integration being as before.
Obtain these results directly from Green s Theorem ( 486)
without employing the Electric Image transformation.
41. Shew that the quaternion
[jvS ( UvV) rds  jjjaV *rd<;
(in the notation of 482) is changed into its own conjugate by
interchange of a and r. Express its value (by 133, 486) as a
single volumeintegral.
422 QUATERNIONS.
42. Prove that
jjj(o V V + K Va . V r) d? = jf<r UvVrds,
and thence that
43. Find the distribution of matter on a given closed surface
which will produce, in its interior, the same potential as does
a given distribution of matter outside it.
Hence shew that there is one, and only one, distribution of
matter over a surface, which will produce, at each point of it, any
arbitrarily assigned potential.
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