292 Free Path Phenomena [CH. xv 



and calculates for A 2 the value 



AI = 1-3682 (695). 



Hence Au 2 = ^ 2mKA s (- 2IT + v^ + w 7 -) ............... (696). 



III. C = 0. 



350. We also require Aw, but this is more easily found by transformation 

 of axes than by direct calculation. 



Let us write Ix + my + nz instead of x, so that we write lu + mv + nw 

 instead of u. The left-hand member of equation (696) becomes 



The bracket on the right-hand may be written 



u 2 + V 2 + w 2 - 3U 2 , 

 and therefore transforms into 



(I 2 + m 2 + n 2 ) {u 2 + v 2 + W 2 - 3 (Zu + mv + ww) 2 }. 

 Equating coefficients of 2,1m, we have at once 



IV (697). 



IV. Q = u( 



351. To evaluate Aw (u 2 + v 2 + w 2 ), we need the complete system of three 

 equations of the type of (683). 



On putting m^ = ra 2 , equation (683) becomes 



u = u + a a' cos (e Wj) ........................ (698), 



in which 



a? _ 



a = (w'-w)cos 2 -, a' = ^VF 2 -(M'-w) 2 sin^, 



3C 



and a>! = 0. The angle o^ has been introduced for the sake of symmetry, for 

 we have now, from equations (679) and (680), 



v = v + b - V cos (e - &> 2 ) ........................ (699), 



w = w + c c' cos (e 3 ) ........................ (700), 



where b, b' and c, c' are obtained from a, a' by replacing u by v and w 

 respectively, and, by equation (681), 



(u - u) (v' -v) . 

 cos o> = - sm 



(u' u)(w'-w) . a , 

 cos o 3 = - A-; sm 2 ^ . 



