ME. J. H. JEANS ON THE STABILITY OF A SPHERICAL NEBULA. 
33 
which u maintains a constant value throughout.* Passing on, we notice that the 
stability function now depends solely upon the ratio of the density to the elasticity. 
The different elements of the nebula are attracted towards one another by their 
mutual gravitation, and are kept apart by the elasticity of the gas. For certain 
values of the ratio of these two systems of forces, it will be possible to find displace¬ 
ments in which the work done by one system exactly balances that done against the 
other, and these are the critical vibrations. 
The stability function u m is a function only of the quantities determining the 
equilibrium configuration of the nebula, and its value may therefore be found from 
the equations of equilibrium. We proceed to examine the value. 
Evaluation of the Stability Function. 
General Case of a Nebula at Rest. 
§ 30. We have already quoted Professor Darwin’s result that u x = 1 for an 
isothermal nebula at rest, and the considerations put forward in the last section will 
probably suggest that the result in the more general case will be found to be 
independent of variations in temperature at finite distances, provided only that 
the temperature has a definite limit at infinity. We shall, however, examine the 
question ah initio , using a slight modification of Darwin’s method, and making the 
problem more general by retaining a spherically symmetrical system of external 
forces. 
We shall denote the potential of this system of forces by V', and use V to denote 
the gravitational potential of the nebula itself. The total potential is now V + V', 
so that the equation of equilibrium, equation (11), takes the form 
yieu)-Jqv + v') = o, 
and if M is the mass of the solid core, this can be written 
— — (A.Tp) -f- 477 | pr 3 dr -J- M — r 3 - — = 0 . . . . (107). 
Differentiating with respect to r, 
Write 
A 
dr 
/ r 2 d 
\ p dr 
(XTp) j -j- 47rpr 3 — 
VTp = e?, 
= 0 . 
* G. H. Darwin (l.c. ante, p. 16), “ If we view the nebula from a very great distance, . . . the solution 
of the problem becomes y = log 2a: 2 .” Now u = — d 2 y/dx 2 , so that this solution is equivalent 
to u = 1. This justifies our statement, and shows at the same time that for any nebula at rest and 
in equilibrium u m has the critical value u x = 1, provided it is acted upon by no forces except its own 
gravitation. 
VOL. CXCIX.-A. 
F 
