Equations in a Homogeneous Isotropic Medium. 31 



A sufficient proof is the satisfrction of the equations (1), 

 (2), and of the two initial conditions. 

 An alternative form of (7) is 



E = 6-p' [cosh qt + ^^^^ (p + P)] Eo, 



Slllll ot 



H = e-p'[cosh^i+^ ^(j9 + /o)]Ho, 



shomng the derivation of E from Eq and ^Eo in precisely the 

 same way as H from Hq and pHo. In this form of solution 

 the initial values of ^Eq and pK^ occur. But they are not 

 arbitrary, being connected by equations (1), (2). The form 

 (7) is much more convenient, involving only Eq and Hq as 

 functions of position, although (7 a) looks simpler. The form 

 (7) is also the more useful for interpretations and derivations. 



If, then, Eq and Hq be given as continuous functions ad- 

 mitting of the performance of the differentiations involved in 

 the functions of q^, (7) will give the required solutions. The 

 original field should therefore be a real one, not involving 

 discontinuities. We shall now consider special cases. 



3. Persistence or Subsidence of Polar Fields. — We see im- 

 mediately by (7) that the E resulting from Hy depends solely 

 upon its curl, or on the initial electric current, and, similarly, 

 that the H due to Eq depends solely upon its curl, or on the 

 magnetic current. Notice also that the displacement due to Hq 

 is related to Hq in the same way as the induction -. — A.tt due 

 to Eq is related to Eq. Or, if it be the electric and magnetic 

 currents that are considered, the displacement due to electric 

 current is related to it in. the same way as the induction 

 -i-47r due to magnetic current is related to it. 



Observe also that in passing from the E due to Eq to the H 

 due to Hq the sign of a is changed. 



By (7) a distribution of Hq which has no curl, or a polar 

 magnetic field, does not, in subsiding, generate electric force ; 

 and, similarl}^, a polar electric field does not, in subsiding, 

 generate magnetic force. Let then Eq and Hq be polar fields, 

 in the first place. Then, by (5), 



q^ = (T^, 



that is, a constant, and using this in (7) we reduce the general 

 solutions to 



E = Eoe-2p>^, H = Hoe-2p2< (8) 



The subsidence of the electric field requires electric conduc- 

 tivity, that of the magnetic field requires magnetic conductivity; 



