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)