448 Dr. W. F. G. Swann on the Expression for 



symmetrically situated on the other side of the plane PM 



have been bent the other way, they originally started out at 



a smaller angle to the normal than the angle -v|r; so that,, 



quite apart from the momentum communicated by the fields 



there was originally more momentum in the direction from 



right to left in the set of electrons which are to be found at E 



and which have come from the right of PM than there was 



originally momentum in the direction from left to right in 



the set of electrons which are to be found at E, and which 



have come from the left of PM. The two sets of electrons 



coming from elements situated symmetrically on the two 



sides of the plane PM are not corresponding sets, in view of 



the fact that their initial velocities are in different directions. 



Thus, it becomes necessary to add to the current density 



n £^\X 



— deduced above, the current density due to this cause,. 



mv J 



which we will now proceed to calculate. 



Omitting for the moment the part of the current directly 

 due to the velocity communicated by the field, the electrons 

 specified by (13) constitute a current 8i l such that 



8i 1 = ev(8n) 2 cos #=-^-2 e ~ r ^ cos sin ijrdHdr; 



so that using (7) and replacing by yjr in the term mul- 

 tiplied by - — ^ we have 



&i= 9^ e " rA { 1_ S tan2 * } C0S * Sin ^ dUdr ' ( 17 > 



We must now put # = Rcos ^, then integrate with respect 

 to R from P = to R = r, and then from ?' = to r=co , finally 



integrating with respect to ty from ijr = to i/r = — . Th& 



result is 



«-f{rS} a* 



The corresponding quantity due to the flow from the right- 

 hand side of PM is obtained by replacing X by — X in (18).. 

 and the current resulting from i x and i 2 is thus 



ne 2 \vX. 



I. — 2 2 = - -- . 



omv' 



