22 The Law of Distribution of Velocities [CH. n 



The Solution for a Steady State. 



24. In order that the gas may be in a steady state, it is necessary, as 

 has been already said, that dH/dt shall be zero. Now equation (20), as we 

 have seen, expresses dHfdt as the sum of a number of contributions, one from 

 every type of collision, and each contribution is either negative or zero. Hence 

 for dH/dt to be zero, each contribution separately must be zero. In other 

 words we must have 



for every type of collision. 



This condition has been seen to be necessary for a steady state. Equation 

 (12) shews that it is also sufficient, for if it is satisfied then df/dt = for every 

 value of u, v and w. The problem of determining the steady state is there- 

 fore reduced to the problem of obtaining the solution of equation (21). We 

 shall find it convenient to take logarithms of both sides, and write the 

 equation in the form 



log/+log/'=log/+log/' ..................... (22). 



25. Let ^ be a function of the velocities u> v, w, such that when two 

 molecules collide, the sum of the ^'s appropriate to the two molecules before 

 impact is equal to the sum of the two %'s after impact. Since ^ is, by 

 hypothesis, a function only of u, v, lu, the value of ^ will remain the same for 

 molecule except when it is altered by collision. We may therefore 

 say that ^ is defined as being capable of exchange between molecules at 

 a collision, but is indestructible ; S^ remains the same throughout the motion, 

 where 2 denotes summation which extends over all the molecules of the gas. 



It is clear that a particular solution of equation (22) is 



Further it will be seen that the difference between ^ and the most general 

 solution of (22) is such as to satisfy the conditions postulated for ^. Let 

 %i> %2> %3--- be independent quantities, each satisfying these conditions, and 

 let it be supposed that there are no other such quantities, then the most 

 general solution of (22) must be 



log/=a 1 % 1 + a 2 X2 + 3% 3 + ........................ (23), 



in which a. lt a^, 3 ... are independent and, so far, arbitrary constants. 



-From the dynamics of a collision we know that there are four quantities 

 which satisfy the condition in question : namely, the energy and the three 

 components of linear momentum. These give four ^ forms for ^ : a fifth is 

 obtained by taking ^ equal to a constant, and it is obvious that there can be 

 no others. For if there were any additional form possible for ^, there would 

 be five equations giving u, v, w, u'~, v', w' in terms of u, v, w, u', v', w', so that 

 u, v, w, u', v', w' would be determined except for one unknown. There must 

 however be two unknowns, as the direction of the line of centres is unknown. 



