Forced Vibrations of Electromagnetic Systems, 491 



infinitely small in (362), that is J la =±sa. To find the wave 

 produced by suddenly starting such a filament, transform 

 equations (358), (359) by means of (311). We get 



E (in) =-(7r/2^)%J 0r W>, 



E ( out)=-i(7r/2^)Vgj lrt )w,, 

 H (in)= -(^/2)^(A J lr )w.'«, 



(367) 



H.( 0Ut ) — 



where W is given by (309) ; the accent means differentiation 

 to r, and the suffix a means the value at r = a. 



In these, let e =ira 2 e, which we may call the strength of the 

 filament, and let a be infinitely small. We then obtain 



(0llt) iE = K ^)%<o. } • • (3b8) 



Now if e is a function of t only, it is clear that there is no 

 scalar electric potential involved. We may therefore advan- 

 tageously employ (and for a reason to be presently seen) the 

 vector-potential A, such that 



E=-pA, orA=- j p- 1 E; and//JI=-^\ (369) 



The equation of A is obviously, by the first of (369) applied 

 to second of (368), 



A=i(iV27™ 3 )*W* (370) 



Comparing this equation with that of H in (345) (problem 

 of a filament of curl of e) , we see that f there becomes e 

 here, and //,H there becomes A here. The solution of (370) 

 may therefore be got at once from the solution of (345), viz. 

 (349). Thus 



A= ^ .; (371) 



27rr(r 2 * 2 -r 2 )* V 



from which, by (369), 



E= e J& — -, H= ^ r , . (372) 



2tt{vH 2 -^Y 27rnv(vH 2 -r 2 )* 



the complete solution. It will be seen that 



A=E* + r/xH ; (373) 



