1918] RANDOM DISTRIBUTION OF LUMINOUS SOURCES 573 in (19). But inasmuch as the evanescence is but approximate, we have to consider what may happen when n is exceedingly great, The number of terms is of order n2, so that the question arises whether n^ir/kl can be neglected in comparison with n. The ratio is of the order n\/l, and it cannot be neglected unless the mean distance of consecutive sources is much greater than X. It is only under this restriction that we can assert the reduction of the mean intensity to the value n when the initial phases are not at random. The next problem proposed is the application of (19) when the n sources are distributed at random over the volume of a sphere of radius R. In this case the distinction between the mean in one direction and in the mean of all directions disappears. If for the moment we limit our attention to a single pair of sources, the chance of the first source lying in the element of volume d V is dV/V, and similarly of the second source lying in dV is dV'fV, As the individual sources may be interchanged, the chance of the pair occupying the elements dV, dV is ZdVdV/V*, so that from the second part of (19) we get for a single pair the expectation of intensity . rf&mkrdVdV A, I I_______________________ JJ kr V V ' and for the ^n (/? — 1) pairs 2tf. (n — 1) ffsin kr •dVdV......................(26) Here V is the whole volume of the sphere, viz. $7rR3, and r is written in place of Z>. The function of r may be regarded as a kind of potential, so that the integral in (26) represents the work required to separate thoroughly every pair of elements. As in Theory of Sound, § 302, we may estimate this by successive removals to infinity of outer thin shells of thickness dR. The first step is the calculation of the potential at 0, a point on the surface of the sphere. The polar element of volume at P is r2sin0 dwdOdr, where r = OP, 6 = angle OOP. The integration with respect to CD will merely introduce the factor 2?r. For the integration with regard to r, we have r sin kr n kr sin kr — kr cos kr r now standing for OQ. In terms of p (= cos 6), r = 2R[A, and we have next to integrate Avith respect to fjb. We get "lsin kr — kr cos kr ., 1 — cos 2/c.R ~ A;,R sin 2A;JS ks ___ which, multiplied by 2-7J-, now expresses the potential at 0. by supposition a large quantity. write