274 Mr. 0. Heaviside on the 



Corresponding to (49) we shall have 



4rf\ = H i Jo( V) - ( J 1 /K 1 )(sia )K ( Sl r) } A t ; . (50) 



omitting, in both, the 2 and t factors. Now, to obtain the 

 corresponding development of the general equation (22), we 

 have only to change the J (Si%) in it to the quantity in the 

 {} in (50) and the Ji(sia a ) to that in the {} in (49), with 



r = a Y in both cases. 



The method by which (22) was got was the simplest pos- 

 sible, reducing to mere algebra the work that would otherwise 

 involve much thinking out ; and, in particular, avoiding some 

 extremely difficult reasoning relatively to potentials, scalar 

 and vector, that would occur were they considered ah initio. 

 But, having got (22), the interpretation is comparatively easy. 

 Starting with the inner tube, (49) is the general solution of 

 (14), with the limitation H x = at r = a , if, in s, given by 



— s\ — 47r/-6 1 & 1 p + m 2 , 



we let p mean d/dt and m 2 mean —d 2 ldz 2 , instead of the con- 

 stants in a normal system of subsidence, and let A l be an 

 arbitrary function of z and t. Similarly, (50) gives us the 

 connection between Y and A 1# From it we may see what A l 

 means. For, put r = a in (50) ; then, since 



(J K 1 -J 1 K )W = - ;f -', 



we see that A x = — 47ra K 1 (s 1 a )r o , if F is the current-density 

 at r = a . When the tube is solid, A 1 = 4:7rTJsi. But, without 

 knowing A 1? (49) and (50) connect Hj and F Y directly, when 

 A x is eliminated by division. Also H^Ci x (2/r), if Oi be 

 the total longitudinal current from r — a to r; hence 



p_ *1 Jq(^)~(Ji/ K i)(^o) K o(^0 p /r-jx 



connects the current-density and the integral current. 



Now pass to the outer tube. Quite similarly, remembering 

 that H 3 = at r = a 3 , we shall arrive at 



F _ *3 J (V) - (Ji/Kj) (s 3 a 3 )K (s 3 r) p 

 lz ~2^J 1 ...- Ki...^ ' ' (5 ^ 



connecting T 3 , the longitudinal current-density at distance r 

 in the outer tube, with C 3 , the current through the circle of 

 radius r in the plane perpendicular to the axis. 



Next, let there be longitudinal impressed electric forces in 

 the wire and return, of uniform intensities e± and e 2j over the 

 sections of the two conductors. We shall have 



p.T^e. + E,, p 3 T 3 = e 3 + E 3 ; .... (53) 



