90 , 91 ] Configurations of Equilibrium 99 
large positive generation of energy in the star’s central regions, and a very 
large negative generation of energy in its outer regions. Since 0 becomes 
very small near the star’s surface, these almost balance out, leaving only a 
small residue of net generation of energy. 
This objection applies to all positive, but not to negative, values of l. Our 
model with l negative will give a true picture of the effect of a deficiency of 
energy-generation at the star’s centre, and since there is continuity in passing 
through 1 = 0, small positive values of l will represent the general tendency 
of a central condensation of the generation of energy. 
Solutions when j = 0 and 1 = 0. 
91. We next consider solutions with j = 1 = 0, representing stars with 
uniform effective molecular weight, generation of energy uniformly distributed 
throughout the mass of the star, and opacity coefficient strictly proportional 
to iipT ~ 3 - 5 
The simplest case occurs when the star’s mass is so small that X may be 
treated as a large quantity. The value of n is then 3 25 and B = 358. 
Equation (89*1) now assumes the form 
X c /z 4 il / 2 = 358 (91*1), 
while equation ( 88 - 6 ) becomes 
T e = 22*4 xlO 6 -^ (91-2). 
These formulae make it very easy to calculate the values of X c and T c for 
stars of given mass and radius, provided only this mass is so small that X c 
comes out to be a fairly large number. 
When the masses are not so small as this, the procedure is more com 
plicated. We start with a series of values of n and calculate the corresponding 
values of X c from equation (85*3). Table X next gives the value of B and 
with these values of B and X c known, equation (89‘1) at once gives the value 
of fi 2 M. Thus for any assigned value of /z, we have a system of values of 
corresponding to different values of n. By interpolation we find the value 
of n, and hence the whole solution, corresponding to any given mass. 
The following table shews the result of such calculations. No special 
value of /z has been assumed, but the table is arranged so as to facilitate 
calculation for the value p = 2'5. In this case the factor (/z/2 - 5) of course 
becomes equal to unity, and the fourth column gives the mass directly. For 
other values of /z it is necessary to calculate M(/x/ 2'5 ) 2 before commencing to 
use the table.