of Electrons, and the Spectrum of Canal Rays. 665 



electromagnetic momentum of the system (%) in the limit 

 when the surface S is at infinity. Hence the surface integral 

 (1) in the limit represents the reaction of the aether due 

 to radiation, that is, the radiation pressure of Stark. Call 

 it P. 



We shall choose S to coincide with the position at time t of 

 the wave-surface through P. Hence we have 



XX + fiY + vZ^O, X 2 + Y 2 + Z 2 = L 2 + M 2 + N 2 = I/i + JY. (T /2 + P' 2 ), 



while dJJn has the components (Z-f-U/C, Km, kw)/(1 + ZU/C), 

 and d$ = r t2 dco as before. Hence the components [X, Y, Z) 

 of F are given by the equations 



x =- ifj( ;+ S)( i+ ^)( T ' 2+p ' 2 )^ 



r =-i^if m ( i+ ^) (T ' 2+pv ^' 



z =-^jJH 1+/ S) (T ' 2+p ' v% '- 



(9) 



Since the surface force is everywhere normal to the ray EP, 

 F passes through E ; the stress has no couple resultant, but F 

 has a moment about equal to EO . ^/ Y 2 -\-Z 2 , tending to 

 produce rotation about an axis perpendicular to U. 



§ 10. We shall now consider the third problem, to find 

 the energy received by a fixed surface in unit time. 



The element <:ZS, with direc- 

 tion cosines (X. /x, v) receives 

 per second the energy 



(XP + /iQ. + vR)d&, 



<?' Q 



■>-r 



jc Two particular cases are 

 specially important : 



(1) Element d$ along Ox perpendicular to it. 

 Let it be at P. We have 



X = l, /u=;v = 0, 1=1, 



m 



:n = 0, 



r/* 



V l-U/O 



The radiation is 



Si = 



±7Tp 



(S^)> 2+p v- 



(10) 



