294 Free Path Phenomena [OH. xv 



This, then, is the value of Aw(w 2 + v~ + w 2 ). In terms of u, V and w, it 

 becomes 



Aw (u 2 + v* + w 2 ) = 2 A 



(02 + v*+w 2 ) 3 (w a u 2 + v uv + w uw) + 3u (u 2 + v 2 + W 2 )}...(703). 



352. Substituting the value which has been obtained for Aw 2 from 

 equation (696) into equation (661), we have 



- + + ...... (70,, 



Similarly substituting the value obtained for Awv from equation (697) 

 into equation (663) 



(705), 



and lastly, substituting the value of Aw (u 2 + v 2 + w 2 ) just obtained into 

 equation (664) we have, after simplification from equations (704) and (705), 



iU (U 2 + v 2 + W 2 ) = - oq .................. (706). 



Time of relaxation. 



353. We have now obtained a sufficient amount of mathematical working 

 material, and proceed to the discussion of physical phenomena. 



Let us in the first place consider a gas in which the law of distribution 

 is initially some law other than that of Maxwell. Our equations enable 

 us to determine the rate at which the gas approaches the steady state. We 

 take the simplest case, and suppose that there is no mass motion, so that 

 u = V Q = w = ; we also suppose that the law of distribution is the same 

 throughout the gas, so that IP, uv, etc. are constants in space. With these 

 suppositions equation (654) becomes 



" = AQ .................................... (707), 



expressing that the whole change in Q is caused by collisions. If we put 



Q = U * - v 2 = U 2 - V 2 , 



we have from equation (696), 



A (w 2 - v 2 ) = - f i/ 2 \llmK A 2 (w 2 - 1^) 

 giving, upon substitution in equation (707), 



- (a" - v 2 ) = - f v \2mKA 2 (u? - 

 ot 



(708), 



where r) = %\/2mKA 2 .............................. (709). 



