478 PROCEEDIXGS OF THE AMERICAN ACADEMY, 



The value of a is, from (80), 



a= V[l-(wxc)]-[l-(wxc)] = i?V(nxc)-(nxc). (Ill) 



Now the vector 1', being a singular vector continuously distributed, 

 can be treated by the method of § 47 to give at any point a discrete 

 vector of the second order in 1', namely,*^ 



dg= (f/Sxl')*r (112) 



where c?S is the vector volume cut off on any planoid by an infinitesimal 

 tube of singular lines parallel to 1'. If f/s is an infinitesimal portion 

 of the locus of the electron which gives rise to the fields M' and 1', 

 and if we consider the region of the 1' field bounded by the two forward 

 hypercones from the termini of ds, then all the vectors dg belonging 

 to this region can be redistributed continuously on one of the hyper- 

 cones, and just as in § 47 we obtain the vector 



dg ,,1 —vcos(t>., J 

 — = /4 — , 1 ds. 



dS Vl — 1)2 



Now we may substitute the value of V and obtain 



rfg = |, aMr/Sxl)*l, (113) 



dg e- 1 — »cos(/) , . 



y5=5fia-/4 , Ids. (114) 



dS R^ Vl _ t<2 



Before proceeding further with the second of these equations, let 

 us obtain dg in another form. We may first show that 



dg = (f/Sxl')*l' = (f/S*.]V[>M'. (115) 



For M' = uxl' where U is a unit vector perpendicular to 1'. Hence 



(f/S*.M').M'= [f/S*-(uxr)]-(uxl') = [(f/5*-l')u 



-(rfS*-u)l'].(uxl') 



by (34). Applying this rule again and noting that U'U = 1 and 



u-r = 0, 



(r/s*'M')-M'= - (f/s* -or. 



From this, (Ho) follows by (24). Now we have written M' as 



M' = H' + E' = ] Lxe' - e'xk4. 



47 Since 1' involves a and therefore nxc, the vector dg is zero for all points 

 in the line of C, and is a maximum when n is perpendicular to C. 



