## 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.

 Radiusx1 Mass[ - x2 dQ/dx ] x = x1 rcentral/rmean From TABLE 4 of Ch67, Chapter IV, § 5 n = 0 Ö6 2Ö6 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 ¥ Ö3 ¥

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.