188 
VARIABLE STARS 
Eliminating P 1 from (127-51) and (127-52) we have 
•i = 0 (127-6), 
where 
A 1 — 9oPoio/Po • 
The equilibrium values which appear as coefficients in equation (127-6) 
can be tabulated with the help of Table 6*. It is easily shown that p, 
(which is of the dimensions of a pure number) is given by 
the accents denoting differentiation with respect to i 0 . 
128. The equation (127-8) has to be solved numerically. We must 
first decide on a value of a, which depends on the effective ratio of specific 
heats. The maximum value of a is 0-6, corresponding to the maximum 
value y = | for a monatomic gas ; the minimum value of a is 0, correspond 
ing to y = §, since a star is unstable for smaller values (§ 104). 
As an example we shall consider a — 0-2. It is then necessary to try 
various values of to 2 , i.e. try various periods, until we find a solution which 
satisfies the boundary conditions and so represents a possible free oscil 
lation. For the fundamental oscillation the first node (place of constant 
pressure) must fall at the boundary of the star. 
We start from the centre with an arbitrary value of i x (according to 
the amplitude of the pulsation) which is here taken as unity. Evidently 
ii must be taken zero. Proceeding first by a solution in series, and changing 
to quadratures when the series becomes inconvenient, we calculate i x , 
ii, ii" at successive points. Table 26 contains the results of the calculation 
for three values of to 2 , viz. -055, -060, -065. The unit of length here used 
is 1/6-9 of the radius of the star, so that the first column A corresponds to 
2 in Table 6. 
Consider the solution for to 2 = -060. At = 5, i/' has become negative 
and is decreasing very rapidly, so that i x is diminishing and will probably 
become negative before the boundary is reached. The node is given by 
* An auxiliary table giving values of ¡x will be found in Monthly Notices, 79, p. 10. 
^ Z du 
u dz 
also 
where the suffix c denotes the values at the centre of the star. 
Let 
(127-71), 
(127-72). 
a = 3 — 4/y 
Then (127-6) can be written 
(127-8),