340 
THE OUTSIDE OF A STAR 
235. First suppose that e and k'/k are constant so that 
p 2 = const., dr/dr = const. 
Then by (234-6) d 2 J/dr' 2 = 0 so that (234-5) becomes 
..(235-1). 
(235-21), 
The solution is 
J' = J + Ae~ VT ' 
the negative sign being taken because J' must approach the equilibrium 
value J at great depths. Differentiating we have 
Following the first approximation the boundary condition at r = 0 is 
with J = H (2 + 3r) and H = const.* 
The value of H'¡H at r = 0 measures the blackness of the absorption 
line that will be observed. 236 
236. Next suppose that the atoms giving the special absorption do not 
extend to the surface. To illustrate this we suppose that k'/k is constant 
as before when t > r 1} but k' — k for r < r x . 
Then equations (235-21) and (235-22) hold for r > r 1 , and at r x we must 
fit on continuously a solution of the form 
which satisfies (235-1) and the boundary conditions (235-3). In this outer 
region p = V3 and r = r. 
* It can be shown from (234-6) that the equation J = H (2 + 3r), originally 
proved only for the integrated radiation, is valid for the present application. (This 
is, however, a consequence of the initial assumption of equilibrium constitution of 
the radiation, and is not a general theorem.) 
dJ' dJ dr 
— pAe~ VT , 
or by (234-42) 
(235-22). 
J = 2H, J’ = 2/7' 
Hence multiplying (235-21) by f and subtracting (235-22) 
0 = 3/Z (1 — k/k') + A (| + p). 
This determines A and the results become 
(235-3). 
H' = H — 
J' — J — 2B ^cosh t-\/ 3 + ^ sinh T-y/3^ 
H' = H — ^ £ (sinh tV 3+^ cosh tV3 
- 
(t<t 1 ) (236-1),