Theory of Crookes's Force. 27 



while the pressure sideways is 



so that the Crookes's pressure in this case is 

 K=P— p=p 1 u 2 l + p 2 u%. 



In order that there be no accumulation of gas at either sur- 

 face, we must evidently have 



p 1 u 1 =p 2 u 2 . 

 If V 2 and V 2 be the total mean squares of the velocities of 

 agitation, Y^v^ + u^ Y\ = v\ + u\, and the quantity of heat 

 transferred is 



Q = k( Pl Y*ui-p2Vh 2 ), 

 h being, as before, the coefficient by which the vis viva of 

 translation has to be multiplied in order to get the total energy 

 of the gas. 



From these we easily obtain 



K=p 1 u 1 (u 1 + u 2 ), 



V 2 — V 2 



u x + u 2 



We have besides p 1 + p 2 =p, where p is the density of the 

 gas. Hence there are six equations between the six unknowns, 



Pi, P2, v 1} v 2 , u 1} u 2 ; 

 and in order to eliminate them and obtain an equation between 

 K and Q, it is necessary to make one further assumption. I 

 assume, then, that u 1 = \vi and u 2 =\v 2j so that V 2 =(X 2 + l)u\ 

 and V 2 = (\ 2 + l)w 2 . I assume this because, if the streams 

 did not interfere with one another at all, we should have 



w 2 =- V 2 . 



so that, if \ 2 + l=a 2 , we should have 



a 2 =6 and a = 2*5 q. p. 

 Our equations then become 



Vi-V»=«>?-t$ ; 

 ,\ Q=^K« 2 (w 1 --w 2 ). 

 From these we can eliminate u l7 u 2 , p 1} p 2 ; and putting 



V 2 — V 2 =X 2 

 v x v 2 — -<x , 



