144 
DR. S. CHAPMAN ON THE KINETIC THEORY OF A COMPOSITE 
The general element of V is S mn b mn or S mn b' mn , which are the same except when 
n = ± 0 ; the general elements of these two rows are S^b'^, S m _ Q b' m _ 0 {cf. (5‘38)), the 
general elements of the central row and column respectively are S m0 b' m and S 0n b n , while 
the central element is b. From this we obtain V 0 if we replace all the elements 
of the central row by zero, except the two on either side the centre, which are 
j/j (m = 0) and v 2 (m = — 0). 
Finally, we may prove that 
(5 '40) S(w,+ W-J = W7iHH’ 
where V is defined as usual (± m, ±n ranging from 0 to o°), while in V„ all the 
elements are equal to the corresponding elements of V except in the two central 
rows (n = ± 0). In the row n = 0 all the elements are zero except the central two, 
which are v 1 {m = 0) and v 2 {m = — 0). In the row n = — 0 the general element 
i S C m -0 C m0- 
§ 6. The Complete Solution for Maxwellian Molecules. 
In the case of Maxwellian molecules, i. e. , molecules which are point centres of force 
varying inversely as the fifth power of the distance, the solution reduces to finite 
terms. This arises from the fact {cf. § 9 (C), ‘Phil. Trans.,’ A, vol. 216, p. 323) that 
for such a law of inter-action the functions (p , f, x are independent of y, with the 
result that for all values of t we have {cf. (4'25)) 
( 6 ' 01 ) 
k, = 1, k\, = V 
119 
U _ P 
A/ 12 — A/ 
12 ? 
/A, = F 
22 ) 
and consequently, 
( 6 ' 02 ) 
isC^rWk = £,(W~W = (m 3 +m 2 ) s = i. 
0 0 
From (5‘12)-(5T4) and (4'27), (4'29), with the aid of (6'02), we deduce that 
(6-03) 
o OTliTflo >v \ ~XT f l c\\ 
«oo — = TT T~ MM^oAqK. 12(h)) 
m 1 + m 2 
independently of the value of r or s. 
From (5'2l) and (3‘13) we consequently have 
(6-04) a r =0 (r^0), 
a 0 — a_o 
_L = _a4 
^00 MM 
a 0 = 
/ 
«0 
V 
a-o 
/ 
a _0 
X 2 ' 
Again, from (6‘03) and (5'16), (5'17) we conclude that b r = 0 = b s for all values 
of ±r, + s from 0 to 00 . Hence, by (5'36), or more simply from (5'2l)—since 
in V 0 (b mn ) all the elements of the central column are zero, instead of all except one, 
the central element, as in V—we have 
( 6 ' 05 ) 
& 0 = 0. 
