Electron necessarily Radiate Energy ? 409 



(3) There is no irreversible radiation of energy from the 

 system. This is evident, for the field at t— +^ is identical 

 throughout all space with that at t— — t x except that the 

 sign of H is reversed. Hence, whatever flux of energy 

 outwards from either charge ma}^ have occurred at t=— £ 1? 

 it will be exactly annulled by an equal inward flux at 

 t= +t 1 < 



(4) It will be useful to express the energy and the 

 momentum of the system. The volume of the orthogonal 

 element comprised between the spheres % and % + <^%, ^ and 

 yjr-\-d\jr, and the planes <f> and + dcf> is 



dY= ¥ ^X d X d ^d& 

 (cosh^— cos^) 3 ' 



and the total energy of the field is 



fi(E» + H')<*V. 



The integration extends from 6 = to 2ir, % — to tt, 

 but with regard to that for i/r, the space occupied by the 

 nucleus or charged surface of the electron must be excluded 

 from the integration. The result will depend on the shape 

 assumed for the nucleus. If we integrate from i|r = to 

 the surface given by 



sinh yjr _1 / 1 — /3 2 \* 

 ~T"~~« U-/3 2 sin 2 x/ ' • ' * ' \*K 



we get for the field energy of either charge in the region 

 external to the surface 



liV-n-'-liV-P*: ■ ■ ■ ^ 



This is identical with the field energy of a Lorentz electron 

 of the same e and a in uniform motion with the same 

 velocity. 



The surface (5) is not precisely the same as the spheroid 

 which forms the surface of the Lorentz electron in a state 

 of uniform motion; but it reduces to it in two cases: when 

 the acceleration is zero, and when the radius of the electron a 

 is indefinitely small. In each of these cases f/sinh ^ (the 

 radius of the -yfr sphere passing through a given point) 

 reduces to identity with the radius vector to the point 

 drawn from the centre of the nucleus. There is a difficulty 

 in dealing with the problem for an electron of finite size, 

 partly mathematical if the Lorentz spheroid is assumed, and 



