184 L. Page — Relativity and the Ether. 



Suppose to be a charged particle. Then (21) to (25) show 

 that when O is at rest in K every point in #'s field will be at 

 rest in K. Hence at every point, c must be parallel to E and 

 C X E must be zero. (25) shows that all points in a plane per- 

 pendicular to the Z axis must have the same acceleration, and 

 (23) gives this acceleration as a function of z. Since <f> becomes 



infinite for z = — — the field comes to an abrupt end at the 



<Po 



& 

 plane z — — . In fact it can easily be shown that a light 



wave travelling in the — z direction must have left the 

 charged particle at a time t = — qo in order to have reached 



any point on the plane z = — — at the time t =■ 0. As we pro- 

 ceed in the + z direction the acceleration of the field at the 

 instant t = becomes less and less, vanishing at infinity. 



Since C and E have the same direction at any point at the 

 time t = 0, it follows that the tubes of strain are circles of the 

 form 



x + ky = 



where p = V * a + y'% and h and k are the parameters. These 

 circles pass through O and have their centers in the plane 



G' 



Z = — — . 

 4>o 



The equipotential surfaces are spheres of the form 



" + /"+[-(»-*)T- IF -^ 



(27) 



where b is the parameter. 



Making use of the fact that in the immediate vicinity of 

 the charged particle equal solid angles must contain equal 

 numbers of tubes of strain, we get the previously derived 

 equations (10) for the strain at any point at the time t = 0. 



Let the dotted circle (fig. 3) be an equipotential surface, 

 with center at Q and radius QP = a. Then from (27) 



OQ=- C \ j/^xrf = » 



r + a * 



