100 
RADIATIVE EQUILIBRIUM 
is not in an ideal enclosure with opaque walls at constant temperature; 
but the stellar conditions approach the ideal far more closely than any 
laboratory experiments can do. The material of a star is very opaque to 
the kind of radiation existing there. According to the figures deduced for 
Capella a screen containing 0T gm. per sq. cm. would let through e~ 12 or 
•000006 of the radiation falling on it; at the average density in Capella 
such a screen would be 50 cm. thick. If then we draw a sphere of radius 
50 cm. round a point in Capella the radiation at this point will be practically 
isolated from everything outside the sphere. The temperature gradient in 
Capella is about 1° per kilometre so that the enclosing walls with which 
we have surrounded the point are at constant temperature to within 
0°-001—a uniformity scarcely likely to be attained in laboratory con 
ditions. 
Take the axis of x in the direction of the temperature gradient and 
consider a thin slab of stellar material at right angles to Ox having 1 sq. cm. 
area and thickness dx. Let the temperatures of the two faces be T and 
T + dT. The radiation pressure acting normally on the two faces will 
give forces + p R and — ( p R + dp R ), so that the resultant force in the 
direction Ox is j 
— dp R , 
This resultant gives the amount of momentum which is being acquired 
per second in the region occupied by the slab owing to the flow of radiation. 
Since there is equilibrium this momentum must be got rid of; and it 
can only be got rid of by being first passed on to the matter of the slab. 
(The ultimate fate of the momentum does not concern us here, but it may 
be explained that when handed over to the matter it helps to neutralise 
the momentum communicated to the matter by gravitation.) The 
momentum passes into the matter by the process of absorption (including 
scattering). 
Let k be the mass coefficient of absorption. The definition of k is that 
a thin screen of material of mass w per sq. cm. absorbs the fraction kw 
of the radiation passing through it normally. In our problem w — pdx, 
and accordingly if H ergs of radiation per sq. cm. per sec. are travelling 
along Ox the amount absorbed will be Hkpdx and its ^-momentum will 
be Hkpdx/c per sq. cm. per sec. 
Consider next H ergs per sec. passing obliquely through a sq. cm. of 
the slab at an angle of incidence 9. The distance travelled through the slab 
is now dx sec 9 and the absorption is increased proportionately. The 
momentum absorbed is now Hkpdx sec 9/c; but multiplying by cos 9 
to obtain the ^-component, the absorption of «-momentum is as before 
H kpdxjc 
for any angle of incidence up to 90°. Of course, if the angle is greater than 
90° so that the radiation passes in the other direction through the slab,