EQUILIBRIUM OF ROTATING MASSES OF FLUID. 
425 
W = - 
47 rM 3 /a\ 3 2 ft + 1 fA\ 3 /a\ 2*-2 1 
3c *\cj 2k-2\c 
P ' , 
i — I 
ft + »! 
l c \ i\ ^ iH,c \fi I ^ 
Wlc \ 2 
+ 
lc + t! 
/ <> 2 / 
i- 2 ! ft + 2 ! 
P -,pic { - 
WJc -if- 2 
r^lc 
dm 
}■ 
Then, effecting the integrations, and putting in the , we get 
'« = 
47r\ 2 M 3 a 3 
3 
c 2 vc 
a \3 \3 lc = » t = co pi-1 7 *-l r A; 4- i! 
- 2 
C ) Jc = 2 i = 2 i — 1 ft — 1 L ft • i ! 
~rr NjHi + 
ft + 2! 
l + i\ 
2. ft -2 \i- 2! A, + 2! 
o ! 
}• (vi.) 
This involves the two figures symmetrically. 
5th. e 5 is the loss of energy in the yielding of the figure a to centrifugal force. To find it, we 
multiply the potential due to rotation by the mass of each element of the layer a, and integrate. 
By (35) and (52) we know that the rotation potential is 
A,.,2„3 (-V 
rOJ-a- 
J ZC 3 x fVWq 
(r 2 2 r 2 J 
As this only involves harmonics of the second order, we may neglect in the layer a all terms except 
those of the second order. Thus we get 
= 4a> 2 a 2 41 — J a 3 
A\ 3 
Wc, 
= 4?r • tV" 3 ** 5 (7] {% + 2“2’o!^ 2 } 
/4tt\ 2 A 3 a 3 f 3 w 2 /«\2 "I 
\3 / c { 16 ^( 0 ] ^ 2 + } • ' ’ 
aw 
(vii.) 
The expression for E s may be written down by symmetry. Collecting results from (v.), (vi.), 
and (vii.), we get, for the whole exhaustion of energy, 
E + 
47t\ 2 A 3 a 3 
= \h D if) {“* _ w ~ rvvk**’} + 
A 
3 ,n \ 3 r^- 1 
oj k — 1 
k + 2 ! 
lc + 2! 
N, - W - - 7 .. P^ 2 
+ 
lc + i! 
2 . ft- 2 ! i- 2 ! ft + 
4.ft- 2!' 
2 ! Pi ^} 
16tt [_\c j 
- (n 2 + 6 p 2 ) + 
(IV 3 + 6 P 3 ).(viii.) 
The expression is found without any assumption that the two masses are bounded by level surfaces, 
and therefore in equilibrium. But the condition for equilibrium is that the differential coefficients of 
E with respect to any one and all of the parameters n, p, N, P, shall vanish. If we equate to zero 
dV/dn/i, we get 
1 — n k + | 
a \3 i=co pi-i lc + i\ 
pj t = 2 i — 1 ft! i ! 
If, however, ft = 2 , there is on the left-hand side an additional term 
Ni = 0. 
16 
7 T \ C 
2 W^\ 3 7 = ^M 3 = 
■ " c) 1 8 ttU/ 
r.\3 
A 
The equation of dV/dNi to zero gives a similar equation. 
If we equate to zero dV/dpt, we get 
Pic + f - 
3 pi—1 
ft + i ! 
* - 1 i - 2 ! ft + 2 ! 
3 1 
P; = 0. 
MDCCCIXXXYII. — A. 
