424 Mr. 0. Heaviside on the 



~K+$(d/dt), where K is the conductance, or reciprocal of the 

 resistance, of the dielectric across from one conductor to the 

 other. Then both K and S are functions of z. The conduc- 

 tion current is KV, and the displacement current SV, whilst 

 their sum, or S /; Y, is the true current across the dielectric 

 per unit length of line. We have now, by (145), with e = 0, 



--|(VC) = VS // V + CR"G 



= KV a +^4SV 2 + CR"C. . . . (146) 



The additional quantity KV 2 is the dissipativityin the dielectric 

 per unit length, whilst now CR"C includes the whole mag- 

 netic energy increase, and the dissipativity (rate of dissipation 

 of energy) in the conductors. 



Let Vj, Cj, and V 2 , C 2 be two systems satisfying (145) with 

 e=0. Then 



- 1 V 2 C 1 = Y 2 S // V 1 + CxE'C, ; 

 ctz 



from which we see that if the systems be normal, d/dt be- 

 coming p 1 and p 2 respectively, we shall have 



= (p 1 -^){sViV,-^|^0 1 C 2 }, . (147) 



R/' and R 2 " being what R" becomes with p^ and p 2 for djdt. 

 As the quantity in the \ \ is the U 12 -T 12 of Part III., and the 

 first term is U l2 , we see that the mutual magnetic energy is 



T 12 =C 1 2 (Bi w -R/)^(^i-i> 2 ). • • • (148) 



The division by p\—p 2 can be effected, and the right mem- 

 ber of (148) put in the form 



OJtPi) x V 2 f(p 2 ). 

 When this is done, we can find the mutual magnetic energy 

 of any magnetic field (proper to our system) and a normal 

 field, in terms of the total current in the wire and its differ- 

 ential coefficients with respect to t ; so that, in the expansion 

 of an arbitrary initial state, C, C, 0, &c, may be the data of 

 the magnetic energy, instead of the magnetic field itself. 



