Note: Some readers may first wish to review this basic physics blog here: http://www.brane-space.blogspot.com/2011/05/basic-physics-thermodynamics-pt-12.html

Besides gas pressure, there is also the radiation pressure:

P

_{R}= aT^{4}/ 3^{-16}Jm

^{-3 }K

^{-4}and T is the absolute temperature (in K degrees). The equation above is valid only if the material is in thermodynamic equilibrium (i.e. the distribution of radiation is that of a black body). This is generally true, but some exceptions do exist.

In dealing with stars, energy considerations also arise. In particular, the thermal and gravitational energy of a star are very closely related. In a perfect gas, the total thermal energy is found by multiplying the number of particles N by the degrees of freedom, f, possessed by each particle. The thermal energy per unit volume is then: ½Nf kT.

The number of degrees of freedom f is also related to the ratio of specific heats (g) of the material by:

g = (f + 2) / f

where g is the ratio of the specific heat at constant pressure to the specific heat at constant volume (if confused, refer to link given above) or:

g =

**C**/_{p}**C**_{v}
Using the equation for a perfect gas and introducing the thermal energy per kg, U, we find:

U = P/ (g - 1) r

Using the previous equation for U in conjunction with the virial theorem, it can be shown that for a star with g = 5/3:

2U + W = 0

where U is the thermal energy for the whole star and W is the gravitational energy such that:

W = - G ò

_{o}^{R }M(r) dM(r) / r = - GM^{2}/ 2R
But the potential energy, V = 2W = - GM

^{2}/ RThis can be put into an even more useful form based on the kinetic theory of gases, for which:

P = r v

^{2}/ 3

And: - W = 3 ò PdV = 3 ò (P/r) dm

Whence:

3 ò (P/r) dm = 3 ò v

^{2}dm/ 3 = m v^{2 }= 2 K
Where K denotes the (gravitational) kinetic energy, which is also expressed:

K = GM

^{2}/R
Then in terms of the gravitational energy, W:

- W = 2K or 2K + W = 0

But: W= - GM

^{2}/ 2R so: K = -W/ 2 = GM^{2}/ R
Which checks out. We are left with these conclusions:

1) 2K + W = 0 applies to any spherical system in equilibrium where K is the gas kinetic energy and also the gravitational kinetic energy. (K = 3/2(g - 1)U). As we saw before:

U = P/ (g - 1) r so: P/r = U (g - 1)

**3 ò (P/r) dm = 3 ò U (5/3 - 1) dm = 2 K**

*Then:*
2) The binding or total energy of a star E

_{T}is then:**E**

_{T}

**K + V = GM**

*=*^{2}/ 2R + (- GM

^{2}/ R) = - GM

^{2}/ 2R

Or: E

_{T}**= W/ 2 = -K**
Thus, the total energy of a star is negative and equal to half the gravitational potential energy or the negative of the gas kinetic (or gravitational kinetic). From the preceding we see that if for some reason E

_{T}decreases, then K increases but W decreases (e.g. the sphere must contract).
Problem for the budding astrophysicist:

For a uniform sphere with a polytropic index n = 0 for uniform density, show that V = -6/5 (GM

^{2}/R). Take the potential to be:
W= 3/ (n – 5) (GM

^{2}/ R )
For any polytropic gas sphere.*

* Polytropic gas spheres are basically mathematical entities used for modeling of actual stars. As usual, some basic assumptions are made (often in terms of temperatures, pressures, potential energies etc.) and these are then used to develop one or more "polytropic" models to test to see if they can work for a given star. Or, more likely, be employed as a guide to model a star.

A primary objective is to develop a basis for a self-gravitating sphere. In the most desirable of cases, one works to attain a simple relationship between the pressure P, and density (r) of a form:

*P = K (r)*^{(1 + 1/n)}where K and n are constants, and n is known as "the

**" and K the "**

*polytropic index***polytropic constant**".

**:**

*The polytropic index n can be defined***n = 1/ (g - 1)**

(where 'g ' is the ratio of specific heats.)

The problem thus asks the reader to regard the hypothetical uniform sphere as having a polytropic index n = 0 associated with a sphere of uniform density, and adopting the given mathematical form for the gravitational energy.

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