Illustration of the Theory of Gases. 433 



Now 



v— w = -(v — u). v + u = z. — -(v-t-u). 



l — q K y 1 — q K " 



and thus 



£^{(v-u)f(u') + (v + u)f(u")}-2vf(u) = 0. 

 In this 



/(»') -A-)+ 2 4^/'(«)+ ^fe^/>) + • • • 



so that 



g±||{2«/ M+ ^/(,) + ( ^/» } -2 B /(«)=0, 



or, when q is small enough, 



8jt>{/(«) + «/'(u)} + 4 s V/"(u)=0. . . (21) 

 Accordingly 



/(»)+»/» + £<?»V'»=0, . . . (22) 

 or on integration 



It is easy to recognize that the constant C of integration 

 must vanish. On putting u = 0, its value is seen to be 



for/(0) is not infinite. Now f(u) is by its nature an even 

 function of u, so that /'(()) must vanish. We thus obtain the 

 same equation (14) of the first order as by the former process. 



Progress towards the Stationary State. 



Passing from the consideration of the steady state, we will 

 now suppose that the masses are initially at rest, and ex- 

 amine the manner in which they acquire velocity under the 

 impact of the projectiles. In the very early stages of the 

 process the momentum acquired during one collision is prac- 

 tically independent of the existing velocity (u) of a mass, and 

 may be taken to be + 2qv. Moreover, the chance of a 

 collision is at first sensibly independent of u. In the present 

 investigation we are concerned not merely, as in considering 

 the ultimate state, with the mass and velocity of a projectile, 

 but also with the frequency of impact. We will denote by v 

 the whole number of projectiles launched in both directions 



Phil. Mag. S. 5. Vol. 32. No. 198. Nov. 1891. 2 G 



