Blog Post 21, Worksheet 7.1, Problem 1

1. White dwarfs are supported internally against the force of gravity by "electron degeneracy" pressure. The maximum mass that can be supported by this exotic form of pressure is the 1.4 \(M_\odot\) (also known as the Chandrasekhar Mass). The radius of our white dwarf is approximately twice the radius of the Earth or ~ \(12 * 10^8\) cm. 

Given this mass, M, and radius, R, derive an algebraic expression for the internal pressure of a white dwarf with these properties. Start with the Virial theorem, recall that the internal kinetic energy per particle is \(\frac{3}{2}kT\), where k = \(1.4*10^{-16}\) erg \(\text{K}^{-1}\) is the Boltzmann constant. You can also assume the interior of the white dwarf is an ideal gass, and its mass is uniformly distributed. 

Starting with the Virial theorem as suggested, we know that the total kinetic energy of the white dwarf will be equal to the opposite of half of the potential energy: \[K = -\frac{1}{2}U\] We know that the potential energy of the system comes from gravity, so the total (gravitational) potential energy of the system is given by \[U = -\frac{GM^2}{R}\] We are also given that the kinetic energy of each particle in the system is \(\frac{3}{2}kT\). If we say there are N total particles in the system, the total kinetic energy in the system is given by: \[K = N(\frac{3}{2}kT)\] Plugging these values for K and U into the Virial theorem above, we have: \[N(\frac{3}{2}kT) = \frac{1}{2}\frac{GM^2}{R}\] If we rearrange the constants, we get the following relation: \[NkT = \frac{GM^2}{3R}\] Now, the left hand side of the equation should look a little familiar, especially because we are told that we can assume the interior of the white dwarf is an ideal gas. One form of the ideal gas law gives us the following equation: \[PV = NkT,\] where P is the absolute pressure of the gas, N is the number of molecules in the given volume V, k is the Boltzmann constant, and T is the absolute temperature. Because the right hand side of this equation is equal to the left hand side of our previous equation, we can set the remaining two sides equal to one another to get: \[PV = \frac{GM^2}{3R}\] If we solve for V, we get: \[P = \frac{GM^2}{3VR}\] We know that V is the volume of a sphere with a radius of R, or \(V = \frac{4}{3}πR^3\), so we have \[P = \frac{GM^2}{3(\frac{4}{3}πR^3)R}\] \[P = \frac{GM^2}{4πR^4}\]