444 Dynamical Problems in the Theory of Gases. 



Again from (58), 



Sr d<f> = — qr + q 2 v 2 /r, 



Jo 



r*2ir 



J( 



j. 



cos (j>Brd<j> = qv, 

 (Sr) 2 d<£ = 2?V, 

 cos£(Sr) 2 d<£ = 0. 



The condition for the stationary state is therefore 



v{F(r){-qr + q 2 v 2 lr)-W(r)q 2 v 2 )-rF(r)qv = y 



or 



¥(r){-2r + qv 2 /r} - F'(r) ^ = 0. 



Thus, on integration, 



r 2 — qv 2 log r + g'v 2 log F(r) = const., . . (60) 

 or 



F(r) = Ar*-*/*" 8 (61) 



The mean value of r 2 , expressed by 



/•» CO f»CO 



I r 3 F(f) ^r-r- 1 rF(?*)dr, 



is qv 2 ; that is, the mean value of the maximum kinetic energy 

 attained during the vibration is equal to the kinetic energy of 

 a projectile. The mean of all the actual kinetic energies of 

 the vibrators is the half of this ; but would rise to equality 

 with the mean energy of the projectiles, if the velocities of 

 the latter, instead of being uniform, as above supposed, were 

 distributed according to the Maxwellian law. 



If we are content to assume the law of distribution, pe~ hp2 , 

 leaving only the constant h to be determined, the investiga- 

 tion may be much simplified. Thus from (57) the gain of 

 energy from the collision is 



1^2__l n 2 = 



p 2 = 2qp cos (j> (v — p cos <f>) + 2q 2 v 2 . 



