G. F, Firzczratp—On the Mechanical Theory of Crookes’s Force. 65 
so that the resultant unequal pressures in the gas may be represented by a pressure 
2 
pa when R’=X?+Y°+Z? in the direction given by 
piV1—pising : V1 —p? cos ¢ 00 NC BML BH LO UB DEH 
and an equal diminished pressure in every direction at right angles to this line. 
Double this pressure will be the Crookes’ force, which is consequently in this case 
1 1 
K— BP ’ Be +-m?-+-n?) 
and it is in the direction whose direction cosines are proportional to 1; m:n, so that 
if we put | 
l—uyp m=vV1—p' sin n=vV 1p! cos 
il 
R= ppp.” 5 
The direction cosines of the line of transference of heat are evidently a: 6: y and so 
far there is no reason why these two lines should coincide although of course in 
most cases they probably differ but little in direction. 
The only other distribution I shall consider is one suggested by Mr. Stoney’s 
investigation (these Transactions, ante p. 39.) of the nature of the distribu- 
tion of the velocities in the gas between two large parallel surfaces at uniform 
unequal temperatures. He has shown that it tends towards a distribution which 
would be represented by two streams of unpolarised gas moving in opposite 
directions across the layer. Now the actual distribution is never exactly this and 
possibly as he has mentioned departs in various degrees from it as you pass across 
the layer. If, however, we assume the distribution to be the same all the way 
across, and that consequently the mean temperature of each stream is that due to 
the surface it is leaving, we can calculate the resultant pressures. 
If v, and v, be the mean velocities of the molecules in each stream respectively 
relatively to the centres of mass of.the molecules, and if wu, and wu, be the velocities of 
the streams, 7.e. of these centres of mass, and p, and », their densities, the pressure 
upon a fixed plane normal to the direction of the streams is 
ieee 
P= Spry +302¥s tpt + ple 
while the pressure sideways is 
1 1 
P=3A.0 + 3pra 
so that the Crookes’ pressure in this case is 
K=P— P=pity + pte’ 
In order that there be no accumulation of gas at either surface, we must evidently 
have 
PiU =Pals 
