212 
SIR G. H. DARWIN ON THE 
Thus 
jj 1 + A. 1 " + X 2 " a q a 3 
2 (1 + X) 13 (1 + X 1 ' 3 ) r r 3 
I his is the form obtained in the solution of the restricted problem of § 3. 
We conclude that if / (a fr), as expressed in (44), with values derived from any 
solution of the problem of the equilibrium of two ellipsoids unconnected by a pipe, 
is positive, the two figures are too far apart to admit of junction, and vice versd. I 
have in fact always found it positive, although always diminishing as r diminishes, so 
that junction would seem to be always impossible, at least so long as the approxima¬ 
tion retains any validity. This might indicate that there is no figure of equilibrium 
shaped like an hour-glass with a thin neck. However, I return to this subject in 
discussing numerical solutions, and in the summary of results. 
In the case where the two masses are equal, X=l, and the above formula for /(a/r) 
fails by becoming indeterminate. As this is a case of especial interest, it must be 
considered. 
Since the two shapes are now exactly alike, we may take k, y, /3, k to define either 
of them. 
When X = 1, the first term of f (a/r) becomes- - • 
The second term becomes 
_3ai// (X — l) _ 30a (1 + X 1/3 +X 2/3 ) _ 90a 
2(1 + X) 1/3 (1 - X 2/3 ) 2(1+ X) 1/3 (1 + X 1 / 3 ) 4.2 1/3 ' 
All the terms in 1/r 3 are one of the two forms 
Now, 
F 
~c 2 -\C 2 j 
1-X 2/3 
or 
G 
~\c 2 -C 2 1 
1-X 2 ' 3 
r; 2 = a 2 _ x 2/3 n2 _ a 2 1 
(cos /3 cos y) 2/3 (1 + X) 2/3 5 (cos B cos T) 2 ' 3 (1 + X) 2/3 ’ 
In the limit /3 = B, y = r, and X = l ; and we find that the first of these forms 
becomes \Fc 2 , and the second —§ Gc 2 . 
Again, of the terms in 1/r 5 , those in c 4 and (7 4 are of one of the two forms 
F 
rc 4 -x<7 4 1 
C 
1 
i— 
i — ' 
i 
CO 
l_ 
or G 
1 
cc 
1 
1 
H 
_J 
In the limit the first of these reduces to — ^Fc 4 , and the second to —kGc*. 
The last term in 1/r 7 has a common factor 1—X, when y = r, ft = B, and 
_ 1-X _ l+X 1/3 +X 2 ' 3 __ 3 
(1+X) 1 / 3 ( 1 -X 2 ' 3 ) (1 + X) 13 (1 +X 13 ) “ 2.2 1/3 " 
