We will consider a star as series of shells of mass, each with mass dMr . Within each mass shell, the density is constant. We will further assume that

Equation of mass continuity:

Since we do not know how the density varies with radius, we can not solve this equation. However, the form of this equation will be useful later on as a means of exchanging density and masses. The density term will arise from the ideal gas law.

Hydrostatic Equiliubrium:

The above figure represents a slab of material, or, in our case, a shell of mass within the star.

There is an upward force of gas pressure from the core temperature of the star on this shell.

There is pressure due to the combined layers of mass above the shell which is pushing the shell down. In addition, the shell has mass which is also producing a downward force.

Remember that Force/area is a pressure. The area of the shell is dA and the pressure of the shell at radius = r is P and the pressure at radius r+dr = P+dp.

The mass of the shell is dm . The density in the shell is assumed constant and the volume of the shell is just drdA.

Equilibrium occurs when the total downward force is balanced by the total upward force.


As an aside, if we consider an atmosphere to be in hydrostatic equilibrium then the pressure must exponentially decrease as a function of height above the surface in order for hydrostatic equilibrium to be maintained.

Now we are in a position of estimating the central pressure of the sun. Lets consider a point which is at 1/2 a solar radius. At this point we will take the density to be the mean density of the sun (e.g. 1.4). We can now integrate over the limits R to R/2 where we take the pressure at r=R (e.g. the surface pressure) to be zero:

Taking R = 7 x 1010 cm then yields a Pressure of 1015 dynes cm-2 which is the equivalent of about 109 atmospheres.

Ideal Gas Law:

In Stellar Astrophysics we will use the ideal gas law as expressed in this form:

Where mu is the mean molecular weight which is the average atomic mass per particle (H = mass of the hydrogen atom). To see how mu differs under different situations, consider the following simple example.

A neutral hydrogen atom has a mass = H and is just one particle.

Ionized hydrogen still has a mass of H (because the mass of a proton is 1850 times that of an electron), but now there are two particles, so the average mass per particle is H/2.

Fully ionized hydrogen therefore has mu = 1/2. (a fully ionized gas of purely heavy elements has mu = 2 so, to within astrophysical accuracy, mu can always taken to be 1).

If we put our derived Pressure into the ideal gas law, we derive that T = 5 million degrees.

Therefore, hydrostatic equilibrium considerations necessarily lead to high temperatures in stellar interiors.

Gravitational Contraction:

Finally, we will show that gravitational contraction of a mass will increase the kinetic energy per particle in that mass (e.g. the Temperature) by an amount equal to 1/2 of the change in the gravitational potential energy of that mass.

What happens to the other 1/2? it gets radiated away as energy.

We start by expression the gravitional potential energy associated with a shell of mass dMr . The total amount of potential energy comes from integrating overall mass shells.

To find an expression for the Total kinetic energy, T , we integrate over the star using the average value of kinetic energy per particle (e.g. 3/2 kT). Note that dN = the number of particles in a thin mass shell of mass dMr .

This will allow us to write the Total kinetic energy in units of radii and density when we also use the equation for hydrostatic equilibrium.

We now want to intergrate equation by parts. If you can't remember integration by parts, the rule is:

The first term is zero P = 0 at the surface and R =0 at the center. We now use the hydrostatic condition to get rid of the dP and then use the continuity of mass equation to get rid of the density term.

You will now note that this expression for the kinetic energy, T is now exactly equal to -1/2 the expression derived earlier for Omega, the total gravitational potential. This is the virial theorem.

2T + W = 0

If we consider the sun as contracting from some initial radius to its current radius, then the change in gravitional potential is

Half of that change is available as luminosity for the sun, which leads to the fourth problem in the first homework.