Electromagnetic Waves. 161 



Hence (6'1) is equivalent to the vector with spherical polar 

 components 



(0 — 2-?? 4^ W> > (6'2) 



7^ wtf identify the vector (6-2) with electric force * we see that 

 Lamb's solution is the same as (5*4) if 



W=«S»(*r)T»(0 > rt l 

 which agrees with the form quoted in (6*1) above if A=ik. 



Lamb's solution of the " second class" has an ^-component of 

 electric force equal to 



(» + l)+,. 1 ( l r)|»-4, +1 ( k T)A'« 3 |.( ) l» | ) . (6-3) 



where n is a solid harmonic of order n. 



The expressions (6-3) can be transformed at once by simply 



, ' i. •', ' i /did 1 d\ ■ 



using the spherical polar operators U ? -~^ ^^^ -) * 



Then, since <-/>„ is of the form Br n Y n (6, <p), we have 

 -^- l = n fc — ( <f>n \ _ (n-fl)<^ 



and so (6-3) gives the radial component of electric force 

 X = ^±i) { *.-i(*r)-+(^) a ^ +1 («-)} K 



in 



f ^{8»-i(«r) + S n+1 (Kr)f^ 



r(/.r) 

 n(w+l)(2»+l) 



by using (5-52) and (5'72) above. 



The formula (6*4) agrees with (5*3) if we write 



(2n + l)^ w = K P+:r»Y n (e,^), . . . (6-41) 



which is of the appropriate form since cp n is a solid harmonic of 

 order n. 



Similarly the 0-component of electric force is equal to 



Y = («+l)^_ 1 (^)I^-„^ +1 ( Kr)K V^ 



= ? J+ii( B + 1 )8--i(«-)-"8. + i(<*')^ 



1 d'S)i'd(j) n 



(6-42) 



(Kr)"+ l clr dtf 

 again using (5-52) and (5*72). 



* To identify with magnetic force has merely tl e effect of inter- 

 changing the functions U, V of our general solutions (3*1), (o'2). 



Phil. Mag. S. 6. Vol. 38. No. 223. July 1919. M 



