Random Distribution of Luminous Sources. 439 



occupying the elements dV, dY' is 2dY dV'jY 2 , so that from 

 the second part of (19) we get for a single pair the expect- 

 ation of intensity 



, CC sin krdVdY' 



4 JJ kr V V 



dV. 



(26) 



and for the \n(n—l) pairs 



2n(n — l) ff sin kr , Tr 



Here V is the whole volume of the sphere, viz. §7rR 3 , and 

 r is written in place of D. 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 L Theory of Sound,' § 302, we may 

 estimate this by successive removals to infinity of outer thin 

 shells of thickness dK. The first step is the calculation of 

 the potential at 0, a point on the surface of the sphere. 



Fiff. 2. 



:■ 



The polar element of volume at P is r 2 sin do) d6 Jr, 

 where r=0P, = angle COP. The integration with respect 

 to w will merely introduce the factor 27r. For the inte- 

 gration with regard to r, we have 



j: 



sinkr 97 sin kr + kr cos kr 

 r z dr = — 



kr 



k* 



r now standing for OQ. In terms of /x (= cos#), r = 2R/o., 

 and we have next to integrate with respect to fi. We get 



11 sin kr — kr cos kr _ 1 — cos 2&R — ffisin 2kR 

 — dfi= 



Jo & " r " k*R 



which, multiplied by 27r, now expresses the potential at 0. 



1 







