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



and the mean value with respect to the time comes to 



en 



Tr^ifoTram cos mz J 1o ) 2 , .... (363) 



if / is the maximum value of/. This may of course be again 

 averaged to get rid of the cosine. 



66. s = 0. Vanishing of external E. — When n=mv, we 

 make s=0, and then (362) reduce to the singular solution 



a 2 



H (in) = i rcpf H (out) = i - cpf, 

 E(in)=/, E (out) = 0, 



(364) 



Observe that the internal longitudinal displacement is pro- 

 duced entirely by the impressed force (if it be internal), though 

 there is radial displacement also on account of the divergence 

 of e (if internal). Outside the cylinder the displacement is 

 entirely perpendicular to it. 



H and F do not alternate along r. This is also true when 

 s 2 is negative, or n lies between and mv. Then, q 2 being 

 positive, we have 



E (out) = ia 2 W — Jiaj[Jorloggr + L 0l .]/, . . (365) 



as the rational form of the equation of the external E when 

 the frequency is too low to produce fluctuations along r. 



The system (364) may be obtained directly from (358) to 

 (360) on the assumption that sjy is zero when s is zero. But 

 (364) appears to require an unbounded medium. Even in 

 the case of the boundary condition E = at r = x, which har- 

 monizes with the vanishing of E externally in (364), there 

 will be the undissipated initial effects continuing. 



If, on the other hand, H x =0, making y=J 0x /G 0x , we shall 

 not only have the undissipated initial effects, but a different 

 form of solution for the forced vibrations. Thus, using this 

 expression for y, and also s=0, in (358) to (360), we obtain 



H <w=l( 1 "5)^5 H « BB= I( 1 -5);w) ,_ 



/ a\ a 2 ( ' ( } 



E (in) = ^1 - - 2 j e ; E (out) = -. ^ e; ) 



representing the forced vibrations. 



67. Effect of suddenly starting a filament of e. — The vibra- 

 tory effects due to a vibrating filament we find by taking a 



