436 Mr. 0. Heaviside on the 



electric displacement across the dielectric, when the ends are 

 short-circuited (Z = 0); so that if, in the first place, the cur- 

 rent had been steady, and there had been no displacement, 

 there would have been none during the subsidence. 



The transition from the combined inertia and elasticity 

 solutions to elasticity alone is very curious. Thus, let Z = 

 at both ends, and R" = R + Bp, where R and L are constants 

 not containing p. The rise of current due to e is shown by 

 ri ri 



1 edz 9 cos mz\ e cos mzdz 



q-V< 1 -'-^) + h rUl p -> • ( 183 > 



the m's in the summation being irjl, 2tt/1, &c; and each 

 having two p's, given by 



The m = part is exhibited separately, and is what the solu- 

 tion would be if e were a constant (owing to the constancy of 

 R). But, whatever e be, as a function of z, the summation 

 comes to nothing initially, on account of the doubleness of 

 the p's, just as in (172 a) the part in the second line vanishes 

 by reason of every p summation vanishing when £ = 0. 



Now, in (183), let L be exceedingly small. The two p's 

 approximate to — w' 2 /RS, the electrostatic one, and to — R/L, 

 the electromagnetic one, which goes up to co , the storehouse 

 for roots. The current then rises thus : 



i edz 2 Cl 



C = ^j- (1-e - R */ L ) + ~ 2 cos mz \ e cos mz dz . (1 - 6~^ L ) 



2 C l 



— ^ 2 cos mz \ e cos mz dz . (1 — e~ m2tlRS 

 m Jo 



But the first line on the right side is equivalent to 



( 6 /R)(l-6-^/ L ), 



and here the exponential term vanishes instantly, on L being 

 made exactly zero, so that (184) becomes 



C= ~ - % t cos mz \ e cos mz dz . (I - e -™W), (185) 



R -RI 



except at the very first moment, when it gives C = ^/R, which 

 is quite wrong, although the preceding formula, giving C = 

 at the first moment, is correct. Or, (185) is equivalent to 



n 1 / dY\ 



