336 DR. S. CHAPMAN ON THE LAW OF DISTRIBUTION OF MOLECULAR VELOCITIES, 
In finding l)C 2 it is convenient to write 
(243) ‘ f= 
0 / 0 
and to recall (cf. (170)) that 
(244) B 0 = fC 0 . 
Thus we have 
(245) UC S = — |B 0 (2 hm)~ a | ^ ^ 
where we have eliminated /3_ x by means of (148). 
Again, if Q = (u) 2 +(v) 2 + (w) 2 , we have 
R 0 T 
m dx ’ 
Q — u 2 + v 2 -f w 2 + 2 (m u U + v 0 V + iv^N ) + C , 
Q = u 2 + v 2 + iv 2 -\-G 2 , 
(u) Q = u 0 (u 2 + v 2 + w 2 ) + u 0 G 2 + 2 (m o L |2 +uUV + w 0 UW) + UC 2 . 
Hence, putting u 0 = v 0 = w u — 0 except in differential coefficients, we have 
|(,Q) = 3^ ' 
_ 3 3i/ _ 3 / 1 \ 
at 1 , 2 /wn i 2 hrn dt + ' dt\2hmj 
= —3 
RvT fdu n , , 0w o \ „ Rb 3T 
2^_(b(m)Q) - b2(C 2 + 2U 
m \dx + 3?/ ' 02 
0 W, 
+ 3 
_ _ 
" 1 2UV + 2UW 
m dx ' 
div 
dx dx 
5C 3 A +«3U J -C 2attl 
0 X 
o + 2uv hffi + 2UW ~4 +2 4 (bUC 2 ) 
da: dx ox J dx 
5 RbT fdu„ dv 0 dw 0 
m \dx dy dz j 
/x j A -y ( du„ \ 4 / -y M-n \ 9 
- ^4 
P 
dx 
2 X 
0ltn V2 
0 X 
s (tr + ©} +2 IX uci) ' 
m 
0 Q 
. \0 (w) 
= 0 . 
Also, since no energy is gained or lost in molecular encounters, 
