168 
SIR G. II. DARWIN ON THE 
Since w is constant, we may write this 
S(F+ico 2 /)-io, 2 ^P- 2 . 
On developing this by Taylor’s theorem, it becomes 
, , 8 , i 2 s 2 , , a 2 
a? +< %07 
The condition for a figure of equilibrium is that the first differentials of the energy 
with respect to the ellipticities shall vanish. If, therefore, e 0 ,f 0 denote the equilibrium 
ellipticities, the equations for finding them are 
+^+f«ZrM v +Wi)- 
de 0 3e 2 J ' > 3edf) 
df +e °dedf 
= o 5 
= 0. 
Multiplying the first of these by e and the second by f and adding them together, 
we find 
(F+|V/) 
8 2 
■^2 
C 
0 2 
e<, »g ? + (^ + e »/)^+Ay5j 
(V+^I) 
+ 7 
'\0e 
On substituting this in the expression for the lost energy, it becomes 
(F+iori) 
0 2 
i e ( e-2<3 °)^2 + ( e /- e /o-G/)v^. + kf(f-Zfo) 
(Jj 
7 
<! ( e - 2 e »)(iy+ 2(e/-ef a -e,f)f e d ^+f(f-2f a ) (|)‘ 
Now let 8e, 8/ be the excesses of e and f above their equilibrium values e 0 , f 0 , so 
that e = e 0 + S e,f — / 0 + S/. Then on substitution in the expression for the lost energy 
it becomes 
^2 
C 
i {(8e) 2 -e 0 2 } ~ + {Se8/-e 0 / 0 } ^j.+ h W) a -fo a } |v 2 
(F+ic o 2 I) 
O) 
7 
^y -*■} ( d 7 f + 2 (|) j 
Since e 0 , f 0 are constants, the portion of this involving e u , 7 explicitly is constant, 
and may be dropped. 
