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