64-66] 
Emden's Solutions 
71 
where T is the temperature corresponding to the given values of p and p. 
Comparing equations (651) and (65'2), we see that 
T=®p “- 1 (65*3), 
so that © is the value of the temperature of any element of the star when it is 
compressed to the density p = 1 , and so is the actual temperature inside the 
star at the point at which the density is unity. 
If we now write u for p K-1 and introduce r defined by 
r 2 
2 4>Try (k — 1) mp, 
r m~ K 
(65-4), 
equation (63'3) is found to assume the form 
d*u 2 du 
dr 2 + rdr 
1 (T 
—J.sothatp-ig 
+ u n = 0 
where n stands for 
(65'5), 
For any given value of n, the most direct procedure would be to assume 
an initial value u c at the centre, together with the condition du/dx = 0 at the 
centre and calculate values of u and t by successive stages outward. But it 
is readily seen that the solutions so obtained would fall into homologous 
series, the central density p C} and so also u c , the value of u at the centre, 
having all values from zero to infinity as we pass along any one series. It 
is accordingly sufficient to calculate the solution for any one standard value 
of u c , when the solutions for all other values can be derived immediately by 
homologous contraction or expansion. Emden obtains his standard solution 
by taking u c = 1 , so that the central density p c is also unity. If u lf p lt r x are 
the values of u, p and r for this particular solution, the values corresponding 
to any other central density p c are given by 
l 71-1 
u — u e u 1 =p e nu l \ p = pcpx\ r = ri = pc 2« r x (65’6). 
66 . As a first illustration of the use of this solution, Emden studies the 
internal arrangement of the sun on the improbable supposition that it is 
made of atmospheric air. For this, k, the ratio of the specific heats, is equal 
to T4, so that n — 2 5. 
His numerical solution for the value n — 2’5 shews that the density at the 
centre is 24 - 07 times the mean density. As the mean density of the sun,is 
1'416, the density at the centre on this model must be p c =34 , 02. The 
numerical solution gives the radius of the mass of gas to be x 1 = 5’417, so 
that the last of equations (65'6) gives 
r = 5 , 417p c ~° 3 
The actual radius of the sun is r = 6 ‘95 xl0 10 cms., and on inserting these 
values for r and r in equation (65*4) and putting re — T4, we obtain 
R