388 Lord Rayleigh : Problems in 



This has now to be integrated with respect to 6 from to 

 7T. Since the result must be independent of & , we see by 

 putting 0'=O that, 



f'lo (p sin 6 sin 6') e p cos9c09e ' sin 6 dO 



Jo 



f cosQ smed0 = -(e p 



X' <F™"$medd = -{f -e~^). . . (35) 

 Jo P 



Using the simplified form and putting q — arcdd, where a is 

 the superficial density, we obtain for the complete sphere 



v= ^iWV 4 < - e '-^)> ■ ■ • (36) 



agreeing with (6) when we remember that Q = 4:7rc 2 a. 



We will now consider the problem of an instantaneous 

 source arbitrarily distributed over the surface of the sphere 

 whose radius is c. It suffices, of course, to treat the case of 

 a spherical harmonic distribution ; and we suppose that 

 per unit of area of the spherical surface the rate is S„. 

 Assuming that v is everywhere proportional to S», we know 

 that v satisfies 



1 d / . a dv\ 1 d 2 v 



sin dO \ dQJ snr 6 daj 2 K ' v ' 



0, &) being the usual spherical polar coordinates. Hence 

 irom (1) »asa function of r and t satisfies 



dv d 2 v 2 do n(n + l)v _ n 

 dt ~ dr 2 r dr r 2 



or 



d(rv) d 2 (rv) w(«-f 1) 

 dt dr 2 t 2 



(rv) = Q. , . (38) 



When n = 0j this reduces to the same form as applies in one 

 dimension. For general values of n the required solution 

 appears to be most easily found indirectly. 



Let us suppose that S rt reduces to Legendre's function 

 V n (fi), where //, = cos 0, and let us calculate directly from (2) 

 the value of v at time t and at a point Q distant r from the 

 centre of the sphere along the axis of /jl. The exponential 

 term is 



l*2+c2 rCfJL 



f,[l 



u e m = e ±t $ f . „ . . (39) 



