The Structure, Stability, and Dynamics
of Self-Gravitating Systems

Joel E. Tohline
tohline@rouge.phys.lsu.edu

Physical Properties of Polytropes

The following text was drawn from the "Polytropes.html" page of Tohline's on-line textbook.


From TABLE 4 of
Ch67, Chapter IV, § 5

Radius
x1
Mass
[ - x2 dQ/dx ] x = x1
rcentral/rmean
n = 0 6 26 1
n = 0.5 2.7528 3.7871 1.8361
n = 1 p p p2/3
n = 1.5 3.65375 2.71406 5.99071
n = 3 6.89685 2.01824 54.1825
n = 5 3


Radius:
Once x1 has been determined for a given polytropic index, it is clear from the definition of x [III.A.19 & III.A.20], that the total radius of a spherical polytrope is,

R = an x1 = {[(n + 1)Kn /(4p G)] (rcentral)(1/n - 1)}1/2 x1,

[Equation III.A.31]
=
Ch67, Chapter IV, Eq. (62)

or, expressing Kn in terms of the model's central (maximum) density and pressure [II.C.2],

R = { Pcentral (4p Gr2central)-1 }1/2 (n + 1)1/2 x1.

[Equation III.A.32]

Now, in Charles Bradley's numerical model for an n = 1 polytrope, we know that x1 = p. Therefore, when he sets R = 1, we know that the Lane-Emden scale factor an = (R/x1) = 1/p. If he sets R = p, then an = 1.


Mass:
The total mass of a spherical polytrope is,

M = 4p an3rcentral [ - x2 dQ/dx ] x = x1.

[Equation III.A.34]

In Charles Bradley's numerical model for an n = 1 polytrope, we know that [ - x2 dQ/dx ] x = x1 = p. Furthermore, he is setting rcentral = 1. Hence,

M = 4p2an3.

Therefore, when he sets R = 1, he should find that M = 4/p = 1.27324; and when he sets R = p, he should find that M = 4p2 = 39.4784.


Central Pressure:
From eq. [III.A.32], above, we see that,

Pcentral = 4pG/(n + 1) (R/x1)2 .

So, for n = 1 when R = p, Pcentral = 2pG = 4.19 x 10-7.