492 Mr. 0. Heaviside on Electromagnetic Waves, and the 



whilst the curious relation (353) in the problem of a filament 

 of curl e is now replaced by 



A=rfiZ/t, (374) 



where Z is the time-integral of the magnetic force ; so that 

 H.=pZ, and curl Z = cE, . . . . (375) 



Z being merely the vectorized Z. It is the vector-potential 

 of the magnetic current. 



The following reciprocal relation is easily seen by comparing 

 the differential equations of an infinitely fine filament e and 

 a finite filament. The electric current-density at the axis due 

 to a longitudinal cylinder of e (uniform) of radius a is nume- 

 rically identical with the total current through the circle of 

 radius a due to the same total impressed force (that is, 7ra 2 e) 

 concentrated in a filament at the axis, at corresponding 

 moments. 



68. Having got the solutions (372) for a filament e , it 

 might appear that we could employ them to build up the 

 solutions in the case of, for instance, a cylinder of longitu- 

 dinal impressed force of finite radius a. But, according to 

 (372), E would be positive and H negative everywhere and 

 at every moment, in the case of the cylinder, because the 

 elementary parts are all positive or all negative. This is 

 clearly a wrong result. For it is certain that, at the first 

 moment of starting the longitudinal impressed force of inten- 

 sity e in the cylinder, E just outside it is negative ; thus 



E= +%e, in or out, at r=a, £=0 ; 



and that H is positive ; viz. 



1 H. = e/2fjbv at r=a, t = 0. 



We know further that, as E starts negatively just outside the 

 cylinder, E will be always negative at the front of the outward 

 wave, and H positive ; thus 



-E=fjivH.=iex{a/rf, .... (376) 



the variation in intensity inversely as the square root of the 

 distance from the axis being necessitated in order to keep the 

 energy constant at the wave-front. The same formula with 

 + E instead of — E will express the state at the front of the 

 wave running in to the axis. There is thus a momentary 

 infinity of E at the axis, viz. when t = a/v. 



So far we can certainly go. Less securely, we may con- 

 clude that during the recoil, E will be settling down to its 

 steady value e within the cylinder, and therefore the force of 



