- a) Mass is conserved

b) The star is in pressure equilibrium (e.g. hydrostatic equilibrium)

c) The equation of state is given by the ideal gas law

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:

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.

So,

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 10^{10} cm then yields a Pressure of
10^{15} dynes cm^{-2} which is the equivalent of
about 10^{9} 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 *dM _{r} *. 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 * dM _{r} *.

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.

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.