426 Mr. 0. Heaviside on the 



of the two parts into which we may divide the electromag- 

 netic state represented by a single normal solution. 



Also, by (147), integrating with respect to z from to I, 



J'* C l W\ — U" 2 \u^iv 2 -u 2 w^\ 

 hu 1 u 2 dz— \ w l w 2 dz = L Jp . (156) 

 Jo P\—P2 Pl~P2 



either member of which represents the complete U 12 — T 12 of 

 the line. The negative of this quantity, as in Part III., is 

 the corresponding U 12 — T 12 in the terminal arrangements; 

 so that the value of 2(U — T) in a complete normal system, 

 including the apparatus, is 



2(U-n=^-^^wf Wf , • (157) 



if V/C = Z 1 and Z at z = l and 0, these being functions of p 

 and constants, and iv 1} iv are the values of w at z=l and 0. 

 Or, which is the same, 



^- T H w %(l- z )l • • • < 158 > 



as before used. 



There is naturally some difficulty in expressing the state 

 at time t, thus : — 



due to an arbitrary initial state, on account of the difficulty 

 connected with 



and the unstated form of B/\ But when the initial state is 

 such as can be set up by any steadily-acting distribution of 

 longitudinal impressed force (e an arbitrary function of z\ so 

 that whilst V is arbitrary, C is only in a very limited sense 



arbitrary, and C, 0, &c. are initially zero, and certain definite 

 distributions of electric and magnetic energy in the terminal 

 apparatus are also necessarily involved; in this case we may 

 readily find the full solutions, and therefore also determine 

 the effect of any distribution of e varying anyhow with the 

 time. In fact, by the condenser method of Part III., we 

 shall arrive at the solution (135); we have merely to employ 

 the present u and iv, and let M be the value of the right 

 member of (158). The following establishment, however, is 

 quite direct, and less mixed up with physical considerations. 

 To determine how Y and C rise from zero everywhere to 

 the final state due to a steadily-acting arbitrary distribution 

 of e put on at the time £ = 0. Start with e 2 at z = z 2 and 



