58 G. F. Firzczratp—On the Mechanical Theory of Crookes’s Force. 
Following the method adopted by Clausius in his paper already referred to, I 
assume that the mean velocity of a molecule is a function of its direction of motion, 
and that the number of molecules in the unit volume moving in a given direction, 
is also a function of that direction. If, then, we define the direction by means of 
p, the cosine of the angle it makes with a given direction, 9, the angle the plane of 
these two directions make with a fixed plane through the given direction we may 
evidently assume, 
v=0, f(g) n=nE(yd) 
Where v and 7 are the mean velocities, and number of molecules moving in this 
direction, and v, and n, are certain given values of v and n when f and F are unity. 
Now we may evidently in addition take nodule where N is the total number 
of molecules per unit volume, so that we have generally 
N 
n= ZF (uo) dpd$. 
The quantities I intend to calculate are, the number of molecules carried through 
the unit area in any direction, the total momentum carried through the same, and 
the quantity of energy carried through it. ‘The number of molecules going in one 
direction through the unit area, must evidently be equal to that of those going in 
the opposite direction, if there are no gaseous currents going on, and, even if present, 
their existence is evidently beside the question in hand. Hence, if we sum the 
number of molecules passing the unit area, taking those that go in opposite direction 
through it with opposite signs, the sum must vanish. I shall calculate the numbers 
in three cases of unit areas, Ist, perpendicular to the line from which + is measured, 
or X; 2nd, parallel to the plane from which 9 is measured,? ¢., perp. to Y; and 8rd, 
for the case of a unit area perpendicular to these two, «e., perp. to Z. The 
number of molecules going in the direction (+, ¢), that pass through the first of these 
per unit of time, is evidently =” » » and it is likewise evident that the number 
going in the opposite direction will have an opposite sign, so that we have the sum 
of all such zero. Similarly for the other two planes the numbers are 
mW 1 —p*. sin ¢ and nvV 1 — p?. cos ¢ 
So that we get 
0=Invp=Inww 1 = p? sin ¢ = Baw 1 = p2. cos ¢. 
The momentum carried through the 1st of these unit areas per unit of time, by 
molecules moving in the direction, .¢ is = Mzv*y? if M be the mass of each mole- 
cule, and as it does not change sign with », we see that the sum of all such will re- 
present the normal pressure per unit area, at the given place. We can similarly 
get the normal pressures on the other two unit areas, and calling them P,,, P,,, and 
P,,, we obtain 
Pi2=Mnv*p? 
P,,=M2nv?(1 — p?) sin 26 
P2z=M 2nv?(1 — p?) cos. 26 
