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CHAPTER IV 
POLYTROPIC GAS SPHERES 
54. We shall consider the equilibrium of an isolated mass of gas held 
together by its own gravitational attraction. In the absence of rotation 
or other disturbing causes the mass will settle down into a distribution 
with spherical symmetry. In view of the intended application of the 
results, such a gas sphere will be referred to as a “star.” 
At any point in a fluid in equilibrium there is a hydrostatic pressure P, 
the same in all directions. If any closed surface is drawn in the fluid the 
reaction of the fluid outside the surface on the fluid within consists of a 
force P per unit area along the inward normal; and for equilibrium these 
surface forces must balance the body forces such as gravitation acting on 
the interior. 
Let p be the density at any point and g the acceleration of gravity. 
Since there is spherical symmetry P, p and g will depend only on the 
distance r from the centre. 
The gravitational force at r is due entirely to the mass M r interior to 
r, since the symmetrical shell outside r exerts no resultant attraction in its 
interior. Hence 
g = GM T \r 2 (54-1), 
where G is the constant of gravitation 6-66 . 10 -8 in c G.s. units. Also if 
<f) is the gravitational potential, we have by definition 
g = — d(f>ldr (54-2). 
The first condition to be satisfied is the well-known hydrostatic 
equation 
dP = — gpdr (54-3), 
expressing the increase of pressure as we descend in a column of fluid. 
From (54*2) and (54-3) 
dP = pdcf> (54-4). 
A second condition is given by Poisson’s equation in the theory of 
attractions 
VV = - 4t tG p (54-5), 
which for spherical symmetry takes the form 
= _ 4t tG p (54-6). 
dr 2 t dr 
55. We have now two relations (54-4) and (54-6) between the three 
unknown functions of r, namely, P, p, <f>. For further progress a third