An Introduction To The Basic Equations Of Elementary Astrophysics

                     Webb Telescope image containing hundreds of galaxies

 The fundamental equations of astrophysics,  embody its primary principles and are the steppingstone to mastering advanced courses. The equations that comprise the basic inputs into stellar properties are generally covered first. Most students of  physics are already familiar with the gas pressure: P= nkT.  But in astrophysics there is also the even more important  radiation pressure:

P _R = aT⁴/ 3

where a is the radiation density constant which is equal to:  7.55 x 10 ^-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:

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

Bear in mind the material inside a star is highly ionized and in physics a fully ionized gas or plasma is mono-atomic, for which g = 5/3.

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 = - GM²/ 2R

But note the potential energy, V = 2W = - GM²/ R

This can be put into an even more useful form based on the kinetic theory of gases, for which:

P = r v²/ 3

And: -  W  =  3  ò PdV = 3 ò  (P/r) dm

Whence:

   3 ò(P/r) dm    =  3 òv²dm/ 3  =  m v²   = 2 K

Where K denotes the (gravitational) kinetic energy, which is also expressed:

K = GM²/R

Then in terms of the gravitational energy, W:

-  W  =  2K   or  2K + W  = 0

But: W= - GM²/ 2R  so: K = -W/ 2 = GM²/ 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 for the internal energy:

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

Then:

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

2) The binding or total energy of a star E_T is then:

 E_T  = K + V  = GM²/ 2R  + (- GM²/ R) = - GM²/ 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 (GM²/R). 

Take the potential energy to be:

W=  3/ (n – 5) (GM²/ R )