204 
ME. J. H. JEANS ON THE CONFIGUEATIONS 
The equatorial radius is M\> t By a simple integration the volume of this critical 
equipotential is found to be 
„ fl " A fno 2 f ) — 3 
32ttMo)- 2 cos 2 6 sin 6 d6 = 32ttM«- 2 x 0‘0225466. 
Jo 4 cos 2 0—1 
This volume is of course equal to M/p where J> is the mean density, from which 
we find 
ft) 2 /27 rp = 0-360746.(180) 
Since p = 0, the critical value of « is also zero, and the innermost strata differ 
only imperceptibly from spheres. Thus, when y = 1|, equatorial break-up occurs as 
soon as the mass is set into rotation at all, and therefore long before there can be 
any question of the pseudo-ellipsoidal form being attained. 
At the other end of the scale (y = oo) comes the incompressible mass, for which 
the ellipsoidal form is attained long before there is any question of equatorial 
breaking up. 
Thus for some value of y, intermediate between y = IT and y = oo } there must be 
a crossing over from equatorial break-up to fission through pseudo-ellipsoidal and 
pear-shaped figures. For a mass for which y has this critical value, the point of 
bifurcation is reached and the pseudo-ellipsoidal form assumed at the very instant at 
which equatorial break-up is about to begin. 
49. A comparison with the corresponding two-dimensional problem is of interest 
at this stage. Here again a solution in finite terms is only possible for one value of 
y other than y = 2, and here again this value happens to be that one for which the 
total mass is first finite, while the matter extends to infinity. The value in question 
is y = 1 and the solution is 
P = A(l+r 2 /ci 2 )- 2 , .(181) 
which may be compared with the three-dimensional solution (179). Again it is clear 
that, somewhere between y = 1 and y = co 5 there must be a critical value of y at 
which a transition occurs from equatorial break-up to fission into two detached 
masses. The two-dimensional problem can, however, be fully solved, and the 
critical value is found to be y = 2 exactly.* 
Seeing that the two problems run fairly close together in all their essential 
features, we might suspect that the critical value in the three-dimensional problem 
would not be very far from 2. An alternative guess might be y = 2§, this corresponding 
more closely to the two-dimensional value y = 2, since y = 1 J in the three-dimensional 
problem has been found to correspond to y = 1 in the two-dimensional problem. 
50. In the present paper the critical value of y has been shown to be determined 
in the three-dimensional problem by the equation 
l+e[(y-2) — r0509]+ e 2 [i(y — 2) 2 —0-4063 (y —2)—0-0510] + ... = 0, . (182) 
* ‘ Phil. Trans.,’ A, vol. 213, p. 471. 
