24 The Law of Distribution of Velocities [on. n 



Next put %= u in equation (27). We obtain 



and the right-hand vanishes, since the value of the first integral is zero. 

 Hence Su = 0, or, what is the same thing, 



Thus u is the mean value of w, and is therefore the ^-component of the mean- 

 velocity of the gas. Similarly v 0> W Q are the y and z components of this 

 velocity. 



Lastly, let us put -%= u 2 -f V 2 + w 2 . If we substitute this value in equa- 

 tion (29) we obtain 



2 (u 2 + v 2 + w 2 ) = 4-rrvA I e- hmc *c*dc. 



J o 



3 / 7T 



The value of the integral on the right-hand is known to be -A/ ,-^ -, and 

 this leads to 



/ 7T 3 



2(u 2 + v 2 + w 2 ) = f A A/ ; v, 

 or, if we substitute the value of A from equation (30), 





The kinetic energy per unit volume of the gas is 2^w(w 2 + v 2 + w 2 ), and since 

 2u = 2v = Sw = 0, we have 



2A?tt(w 2 + V 2 + W' 2 ) = i?tt2 ((U + U Y + (V + V ) 2 + (W + W ) 2 ) 



A \ /ft \\ W/ ' \ U/ I \ I'/ / 



1 

 = ^raS (u 2 + v 2 + w 2 4- uf + v 2 + w 2 ) 



= L mv ( 3 M 2 

 - f wz/ ^g^ m o 



= j, 4.10(^24. Vfl 2 + ^^ ^32), 



4<h 



where p is the mass density of the gas, given by equation (1). 



We have now determined five relations between the unknown constants 

 and the density, kinetic energy and momenta of the gas. It appears that for 

 given density, kinetic energy and momenta the values of the constants are 

 unique, h being determined by equation (32), u , v , w by the momenta, and 

 A by equation (SO). Hence there is only one steady state possible for given 

 values of the density, energy and momenta. 





