iii the Electromagnetic Field. 143 



where, for purposes of comparison, I have altered his 

 notation to that of my paper. This expression, so far as it 

 goes, is undoubtedly equivalent to my expression (6), though 

 the last term in the form given by Mr. Livens is not ex- 

 pressed explicitly in terms of the charges and their mutual 

 distances, &c. Mr. Livens then states that in the special 

 case I considered — namely, that of two closed linear circuits — 

 div A must be assumed to be zero at all points of the field, 

 and therefore 



tj (di 



ivA) 2 ^-:0. 



The whole point of my argument was that when this integral 

 was evaluated it rlid not vanish — at least, that is what my 

 argument amounted to. 



To simplify matters for the moment, let us neglect terms 



(\ 2 

 - J . Expression (6) of my paper may then 



contain 

 be written 



T = 



i2 — (u 2 + v 2 + icy*) + 22 ^ ( u r n, + v r v, + w r w s )] 



"r f J 



S ^ («,« + »,» + »/) + i lZ e -^> («,.«,+ v r v s )\ . 



CLr 'I -J 



IV 4 



"6 



Comparing this with Mr. Livens's expression above it is 

 clear that 



1 /"* 2 



f- (div Kfdv = 12 — {n 2 + v 2 + w 2 ) + i 22 — (fi r Ui + v r v s + w r w s ), 



as may easily be verified by direct integration. 



If Mr. Livens is correct in saying that div A must vanish 

 at all points in the case of closed linear circuits, then the 

 expression on the right should vanish either when applied to 

 the two complete circuits or to each complete circuit separately. 



Bv subtraction it follows that ^ 22 -^ (u x m + v 1 v 2 ) should 



r 1 C 



vanish, this representing the u mutual" part of rr— 1 (div A) 2 dv 



for the two circuits. The above condition may be written 



. . . i'fcose , , , . . ffcosajcosag , 



ihh] ) — 7 -ds l ds 2 — ^i l i 2 \ \ ds l ds 2 =0. 



If this were true, then expressions (7) and (8) of my paper 



