732 BELL SYSTEM TECHNICAL JOURNAL 



function /, and remembering that throughout the spherical layer the 

 combination (^'- + -q'- + f '")''' is confined within a narrow range of 

 values around v we obtain : 



b-a = (v/I)-Hv)- J (r - ^)dT'/V. (101) 



To effect the integration it is expedient to change over to polar 

 coordinates in the velocity-space. Leaving the origin where it was, 

 and directing the polar axis towards C, we make the radial coordi- 

 nates and the colatitude-angle identical with our v and 6, and for the 

 meridian-angle we use the symbol 0. Dividing up the layer into 

 compartments by latitude circles and meridians, we have for any one 



of them : 



dr'/V = (l/47r) sin edddcf). (102) 



Consequently we obtain:-*^ 



b - a = (v/4Tl)g(v)- I dd sin 6 t d<t>{k' ~ 0- (103) 



Jq Jo 



Everything therefore hinges on the evaluation of (|' — ^) — the change 

 in the jc-component of velocity which the electron incurs when it 

 leaps from C to C — as a function of 6 and 0. Now it may be shown 

 without much difficulty,-^ that: 



^' — ^ — — 2v cos ;/' cos w = — 2v sin \d cos co (104) 



wherein yp stands as before for the angle between the line of approach 

 of the electron and the line of centres at the instant of impact, and co 

 stands for the angle between the line of centres and the axis of x. 

 There is also a standard formula " relating cos co to \^, </> and the 



2° Lest someone be disconcerted by the apparent difference between this equation 

 and that given by Lorentz, I remark that I am using d to designate an angle twice 

 as great as the one which he denoted by 6. 



21 Let V, v' represent the vector velocities of the electron before and after impact, 

 Ci the unit vector along the line of centres at the moment of impact. The components 

 of V and v' along the line of centres are equal in magnitude and opposite in direction; 

 the components perpendicular to the line of centres are equal in magnitude and 

 direction. Writing these statements down in vector notation: 



v-C) = v'-Ci] V — (y-ci)ci = v' — iv'-Ci)ci = v' + (y-Ci)ci, hence, 

 v' — V = — 2{v-Ci)ci = — 2v cos \p-Ci; ^' — ^ = — 2v cos i/' cos w. 



2^ Imagine two planes Pi and P2 intersecting along a vertical axis, the dihedral 

 angle between them being 4,. Through a point O on the axis draw a horizontal 

 plane N, and from O draw two lines of unit length ORi and OR2, the former in plane 

 Pi and inclined at /3 to the vertical axis, the latter in plane Po and inclined at \p 

 to the vertical axis; u is the angle between them. The points Pi and P2 are at 

 heights cos /3 and cos \p above the plane H. On the vertical line through P2 put 

 a point P3 at distance cos 13 above H. Expressions for the sides of the right-angled 

 triangle R1R2R3 are easily obtained, and (214) is derived by applying the theorem 

 of Pythagoras to them. 



