Absorption of Energy by Electrons. 83 



This is the part of the emitted energy which travels within 



dS 

 the solid angle -™- ^ n ^ ne direction 1 and is polarized along L. 



Put dS ,_ 



IP =dn. 



The total emission for all electrons is found by integration 

 of (51), it is 



•or using the same notation as in (46) or (47) 



.*>£££" •W4IT. • • • (52) 



§ 7. The Complete Radiation. 



In complete radiation there must be equality in the emission 

 and absorption of energy in any direction. Where (47) 

 represents the emission (52) shows that 



* 2 + /3 2 = R<9^^m (53) 



The total energy per unit volume in the wave train (25) is, 



And the whole energy per unit volume in the interval dp is, 

 according to (53), 



wj 4 



JP 3 dn or mP^l, 



4t 2 c 3 ire 6 



allowing for two waves polarized at right angles and both 

 travelling along 1. 



Since \= -^- this gives Lorentz's result 

 E x = $irm\-\ 



If (46) gives the absorption instead of (47) we have no 

 longer (53), and instead of Lorentz's formula 



E k f° - -^.{log f(H) }<S>(\)dH = 8tt\" 4 i <!>(\)dH. 

 Jo d±l Jo 



Feb. 2, 19D. 



G2 



