434 MATHEMATICAL APPENDICES 40* 



suppose is either in the interior of the space occupied by a gas 

 or on the wall of the containing vessel. We take this surface- 

 element at the origin of a system of rectangular coordinates 

 x, y, z, whose directions we assume to be such that the direction of 

 x is perpendicular to df. Let an element of volume dx dy dz be 

 at a distance r from df, and let the line r make an angle s with 

 the negative direction of x. In the element dx dy dz there are at 

 each moment 



****Pdia dx dy dz 



particles with speed w, of which each collides on an average B 

 times per unit time with other particles and starts a new path. In 

 unit time, therefore, there proceed from the element dx dy dz 



^-^(kmyBe-^^du dx dy dz 

 particles with speed w. Of these the number 



~ 2 cos s dfdx dy dz 



move in such a direction that the surface-element df is met when 

 the length of path is greater than, or at least equal to, the distance 

 r. The number of the particles which traverse this distance 

 without collision is 



N(km/ir)*Be- fr e- km '*u*du r~ 2 cos s dfdx dy dz 

 if /3 is the reciprocal of the mean free path I, or 



/3 = l/Z = Bw. 



To deduce from this number of the particles meeting df the 

 force exerted by them, we must multiply by 



2?7iW COS S, 



if df is part of the fixed wall, and integrate over the whole volume 

 of the gas. 



But if df lies within the gas, we must multiply by mw cos s 

 and integrate, and thus determine the momentum which passes 

 through df in unit time in the direction of increasing x, and sub- 

 tract from this the momentum simultaneously carried over in the 

 reverse direction. 



The integrations are easily performed when the Cartesian 

 coordinates x, y, z are replaced by the polar coordinates r, s, <j> 

 given by 



x = r cos s, y = r sin s cos 0, z = r sin s sin 0. 



