142 SOLUTION OF THE EQUATIONS 
since the radiation pressure is £ of the energy-density. Hence by (83-3) 
H = 3 (1 — /3) P±irr*dr 
(103-2) 
since the pressure of a gas is fdf the translatory energy per unit volume. 
Hence by (83-3) and (60-5) 
If y is the ratio of specific heats (averaged if necessary) the whole 
material energy K is eK 1 , where 
by (28-4). Hence 
At high temperatures K — K x will consist solely of energy of ionisation*, 
i.e. energy expended in removing electrons from their orbits in the atom 
and setting them free. 
The whole energy of the star is 
If the molecular weight varies as T s the value n = (3 — s)/(l + s) 
must be used in calculating Q. Thus if p cc T~\ Q. = %GM 2 /R. Otherwise 
the investigation is unaltered, and in particular the result (103-5) holds 
good. 
104. If y < f the whole energy is positive, that is to say, there is 
more energy than if the material were in a state of infinite diffusion at 
zero temperature. Quite apart from the loss by radiation during its past 
life, energy must have been supplied to the star to bring it to its present 
state. The contraction hypothesis which denies any extra supply (sub 
atomic or other) accordingly requires that y > |. 
We now generally agree that there is some extra source of energy; but 
the condition 
is still necessary in order that the star may be stable. For suppose that 
a star with y < f undergoes a slight contraction so that H increases. By 
* It is a matter of definition whether we state energy of excitation separately. 
A nucleus attended by a solitary electron in a 3-quantum orbit may be regarded 
(a) as having lost all electrons except one M electron, or (6) all except one K electron 
which has been excited into an M orbit. 
K x = *j8Q 
(103-3). 
y = 1 + 2/3e 
(103-5).