296 BELL SYSTEM TECHNICAL JOURNAL 



That the term in question does actually vanish is easily proved. 

 For on the one hand it follows, from the coupling of / and r into the 

 linear combination (/ — rjc) in the argument of U, that 



and on the other hand it follows, from the facts that the partial deriva- 

 tives of U at any point and moment have the same values as the cor- 

 responding derivatives of 5 at the same point at some other moment, 

 while the derivatives of 5 at every point and moment conform to (6) — 

 from these it follows that 



df \dx^^ a/ ^ dz^ I' ^ ^ 



Therefore the first term of the right-hand member of (15) is zero 

 everywhere, and we have to perform the volume-integration only over 

 the second: 



f,WW,V^f}l{r'Jl-u). ,18) 



Employ spherical coordinates for the integration ; then the element 

 of volume is ^F = r^ sin ddddipdr, and we have: 



fd'ivWdV = fj^sin e f d^ f^''j'r(^TF~ ^) 



(19) 



This signifies that the long narrow volume-element comprised within 

 any elementary solid angle dw = sin dddd^p, and limited at its two ends 

 by the surface of the sphere and the surface S, contributes to the 

 volume-integral the difiference between the values of (r d U/dr — 60 at 

 its two extremities. Completing the integration by considering all of 

 these volume-elements together, we see that the volume-integral there- 

 fore becomes a pair of angle-integrals, those of the function (rdU/dr 

 — U) over the surface 5 and over the sphere. We may transform 

 the first of these into an area-integral by reflecting that the elementary 

 solid angle dw intercepts upon the surface 5 the area-element dS, 

 given by the equation : 



— dS cos {n, r) = rHw. (20) 



in which («, r) stands for the angle between the normal to dS and the 

 radius r drawn to dS from the origin. We must choose positive 



