170 
MR. J. H. JEANS ON THE CONFIGURATIONS 
Then the complete equation (44) becomes, on dividing throughout by irp 0 abc, 
-T-[y,(l)+ e AV i (l) + e^V i (l) + ...- e (E i+ eAE i +...)] 
irClOC 
+ — + - 4 " 7 -- +T:! " + (n + e A n + e 2 Sn+ ...) (:r 2 + y 2 ) 
irpyfXOC 
— cons. — f) 
2 —2 + eP 0 + e"Q 0 + e 3 R 0 + ... ) — Je (y —2) ( 2 — + eP 0 +e 2 Q 0 + ... 
a' 
a~ 
+ i e 2 (y-2)(y-3)(25: + t P, l + ...)-... 
a 
(50) 
an equation in which each side is equal to &/irp 0 abc. 
13. On equating terms which are independent of e, we of course arrive merely at 
equations (38) to (40); these determine a, b, c when the data of the problem are 
given. 
On equating terms in e, we obtain 
(AV f (l)—E,.) + An(x 2 + y 2 ) = -0 P 0 -*(y-2}(s£ 
, 2\2 
a 
(51) 
In evaluating E, we put e = 0, and so may neglect (f> (q). Thus, 
equation (28), 
E. = . 
from 
(52) 
in which, from equation (25), q is given by 
x 
+ 
y 
+ 
a 2 + \ b 2 + \ c 2 + A 
(53) 
Thus, in the notation defined by equation (37), 
E { = — \irabc ( Jaa^ 4 + J B b y x + J CC 2 4 + 2 J BC y 2 z 2 + ... — J ). . 
(54) 
Using this value for E f , it appears that all terms in equation (51) are of degrees 4 
and 2 except P 0 and AY, (l). The value of A V { (1) depends on that of P 0 , and terms 
of degree n in P y give rise to terms of degrees n, n — 2, ... in AV f (l). It is accord¬ 
ingly clear that the solution of equation (51) must be such that P 0 consists of terms 
of degrees 4 and 2; the value of A V i (1) is then also of degrees 4 and 2, and, on 
equating coefficients, we can determine the coefficients in P 0 . 
Following closely the notation previously used,* let us assume for P 0 the value 
P 0 = 
Lx 4 
a 8 
b 8 
/> 4 c 4 
+ 
2 mz 2 x 2 2nx 2 y 2 2px 2 2 qy 2 2rz 2 
A 7 A * 7~~ •” T~l ' A J 
cW 
a 4 6 4 
a 
(55) 
* ‘ Phil. Trans.,’ A, vol. 215, p. 54. The small quantity \e l of that paper is replaced by e in this 
paper, otherwise the figure of that paper reduces to the present figure on taking a, (3, y and s all zero. 
