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reducing to n simply when kR is very small. uniform distribution is thus
(sin kR — kR cos
The intensity due to fche
exactly the fi2 term of (28). The distinction between (28) and (32), at least when kR is very great, has its origin in the circumstance that in the first case the n separate centres, however numerous, are discrete and scattered at random, while in the second case the distribution of the same total is uniform, and continuous.
When we examine more attentively the composition of the velocity-potential <£ in (30), we recognize that it may be regarded as originating at the surface of the sphere R. Along any line parallel to z, the phase varies uniformly, so that every complete cycle occupying a length \ contributes nothing. Any contribution which the entire chord may make depends upon the immediate neighbourhood of the ends, where incomplete cycles may stand over. And, since this is true of every chord parallel to z, we may infer that the total depends upon the manner in which the volume terminates, viz. upon the surface. At this rate the ?i2 term in (28) must be regarded as due to the surface of the sphere, and if we limit attention to what originates in the interior this term disappears, and (kR being sufficiently large) (19) reduces
to n.
« When we speak of an effect being due to the surface, we can only mean
the discontinuity of distribution which occurs there, and the best test is the consideration of what happens when the discontinuity is eased off. Let us then in the integration with respect to r in (31) extend the range beyond R to R' with introduction of a factor decreasing" from unity (the value from 0 to R), as we pass outwards from R to R'. The form of the factor is largely a matter of mathematical convenience.
As an example we may take e~h'(~r~:R\ or e~M(r-B>)> -which is equal to unity when r = R and diminishes from R to R. The complete integral (31) is now
4?r T
rS,                        4^ fK
sin kr.rdr + -y-      e-^(r-J2) sin^r.?^?1..........(34)
Jo                   * JE
From the second integral we may extract the constant factor ehkR, and if we then treat sin kr as the imaginary part of eikr, we have to evaluate
Jy supposing UR very small, when the right-hand member reduces •to n2.