530 Dr. H. Stanley Allen on 



Also 



I mxv 2 aq> = 1 mr cos A» 2 ad> = mi' 2 1 — V~; r — Z ~T\ 



Jo Jo Vo m<r(l + ecos£) 



M Q l + ecos<£ y J V l + ecosc/>/ ■ 



: -H>-vib] 



= -«'* (8) 



The values o£ these integrals are confirmed by noticing 



that their sum represents S mr 2 <p d<f> and is equal to 2irp 



* o 

 or nh, as it should be. 



9. The advantage of resolving the motion into the com- 

 ponents v 1 and v 2 lies in the fact that each of these is 

 constant in magnitude. The component v 2 is also constant 

 in direction, and corresponding to it there will be a con- 

 stant momentum perpendicular to the major axis of amount 



me 2 

 mv 2 = e. 



" P 



Let us now express this momentum in terms of the mag- 

 netic field produced by the charge e moving with velocity v 2 . 

 The equivalent mass of unit volume at any point in the field 

 of the electron is ^tt/j^T) 2 sin 2 #, where D is the electric 

 polarization at the point, and is the angle between the 

 direction of the electrostatic tube and its velocity. Accom- 

 panying the moving Faraday tubes there is a magnetic field 

 at right angles to their leno-th and to their direction of 

 motion given by H = 47rD sin 6v 2 . Hence D sin <9 = H/47rr 2 , 



and the momentum for unit volume takes the form — x — ■ — . 



2 aH 2 V2 ^ 57r 

 The total momentum is, accordingly, — 2-^ — , the sum- 



V 2 07T 



mation extending over the whole space occupied by the 

 magnetic tubes. But in the steady orbital motion of the 



electron we may regard the sum S^ — as representing 



07T 



the electrokinetic energy associated with the magnetic tubes 

 of induction, N 2 in number, corresponding to the velocity r 2 . 



uH 2 

 Then 2^ — = ^N 2 ez^, the moving electron being regarded as 



07T 



