196 
VARIABLE STARS 
Dissipation of Energy. 
132. The outward flow of radiation across a sphere of radius | is 
Hence differentiating logarithmically and writing F = F 0 + F 0 F X , etc* 
Fi = - K + 4& + (132-1) 
By (127-21) and (129-75) 
Note that 9 is positive and greater than ^. 
Let dQ/dt be the rate of gain of heat per unit mass in the shell between 
| and £ + d£ owing to the transfer by radiation. Then 
The steady part on the right-hand side must be balanced by the rate of 
liberation of sub-atomic energy e in the shell. Hence 
where e m and p m are mean values interior to so that e m = F 0 /± 7 rp m £ 0 3 . 
133. For numerical discussion of (132-5) we shall use the calculations 
of Table 27, which correspond to a = 0-1, y = 1-380. We take 1 — $ = -385 
corresponding to a Cepheid of period about 4 days. Then 
y' = 1-355, tj = 3-90, 9 = 0-24, T = 1-43. 
By (131-3) n 2 £ 0 /g 0 is 1-71 at the boundary and at other points its value 
is easily found from Table 6 since it is inversely proportional to the mean 
where 
Pi = iT, 
V = y/(y ' - !) 
(132-21), 
(132-22). 
Then 
d (PpTJ _ 1 d (P 0 P 1 ) 
dP 0 7] dP 0 
by (131-1). Hence 
(132-3). 
Also with the absorption law k <x p/T*, 
— Pi ~ yPi — — Qpi 
(132-41), 
(132-42). 
where 
* = *(/- 1) ~ 1 
dQ = _ d (F 0 + FM 
dt 4? Tp 0 £ 0 2 d£ 0 
1 /1 , et \ dF 0 Fq dFj 
47r Po£o 2 1 d£ 0 4tt/) 0 | 0 2 d£ o' 
..(132-5)