Theory of ' Crookes's Force. 17 



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 ex- 

 istence 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 : — 1st, perpendicular to the line from 

 which ir is measured, or X ; 2nd, parallel to the plane from 

 which (f> is measured (i. e. perpendicular to Y) ; and, 3rd, for 

 the case of a unit area perpendicular to these two (i. e. per- 

 pendicular to Z). The number of molecules going in the direc- 

 tion (/ub, </>) that pass through the first of these per unit of 

 time is evidently = nv/uu ; 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 



nv\/l — fA 2 . sin (f> and nv\/l — ja 2 . cos <f> ; 



so that we get 



= Smy/, = 2ft u\/ 1 — fM 2 sin <j) = Sft v\/ 1 — y?. cos cf>. 



The momentum carried through the first of these unit areas 

 per unit of time by molecules moving in the direction (/x, cj>) is 

 = Mnv 2 fi 2 , if M be the mass of each molecule ; and as it does not 

 change sign with p, we see that the sum of all such wil] repre- 

 sent 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 



P^=M2ftvy, 



P w = M2tm; 2 (1-/a 2 ) sin 2 <f>, 



F zz =Ml i nv 2 (l-/A 2 ) cos 2 <f>. 



Proceeding similarly, we can get the tangential pressures 

 on these areas ; and we easily see that they are 



V yz =V zy = W2nv 2 (l— fj?) sin <£ cos <f>, 



Y zx = V xz = M.Znv 2 fj,\/l — {j?. cos </>, 



P^ = Y yx = MSftv V \/l — tf. sin (f). 



If now we proceed to calculate the energy carried across 

 these areas per unit of time, we get knv 3 /uu as that carried across 

 the first area by molecules moving in the direction (/^, c£) when 

 Jc is the coefficient by which the energy of translation must 



Phil. Mag. S. 5. Vol. 7. No. 40. Jan. 1879. C 



