62 Mr Pocklington, On the Symbolic Integration of 



Hence r = (% - 2p 2 + 2aV 2 - 6 VHV) \ 



= -(p 2 - aV 2 + b VHV) \+b ViVS . iV\, 

 where \ satisfies (p 2 - aV 2 ) (p 2 - aV 2 + b V 2 iV) \ = 0. 



This value of t does not in general "satisfy SVt = 0. It will 

 do so if SV\ = 0, i.e. if \= Wp,. Substituting this value of \, 

 we find as a completely general solution of our original equations, 

 T = _ <j)2 _ a v 2 + bVHV) VVp, 



+ bViV(V 2 Sip, - SiVSVp.) (2), 



where (p 2 - aV 2 ) (p 2 - aV 2 + b VH V) p, = (3). 



There are two especially simple cases. The first is that in 

 which p. = Pi, where of course P satisfies (3). Then 



T = -(p 2 -aV 2 )V.iVP. 



Writing - (p 2 - aV 2 ) P = P U 

 we have t = V . iVP 1} 

 where P 1 satisfies (p 2 - aV 2 + bVHV)P 1 = (4). 



As a particular case of this we have 



Q-<apR 





V.iV 



R 



where U 2 = -^-t±t (5) 



a a— b 



x, y, z being the Cartesian coordinates corresponding to p, 



i.e. x = — Sip, etc. 



This solution gives the magnetic field in the case of a Hertzian 

 vibrator with its axis parallel to the axis of the medium. 



The second simple case is that in which Sip. = 0, SVp = 0, 

 i.e. when p, = ViV . P. Substituting this value of pu, and putting 

 - (p 2 -aV 2 +b VHV) P = P 2 , we find 



T =V(VV.iVP 2 ), 

 where (p 2 -aV 2 )P 2 = (6). 



0-iapr 



A particular value of P 2 is P 2 = where 



r2 = x 2 + y 2 + z 2 = T 2 p ^ _ 



