Blog #21: White Dwarf Goes Type 1a Supernova

1) White dwarfs are supported internally against the force of gravity by "electron degeneracy" pressure (encountered in Astronomy 16). 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 ~ $12 \times 10^8$ cm.

Given the 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 $k = 1.4 \times 10^{-16}$ erg $K^{-1}$ is the Boltzmann constant. You can also assume the interior of the white dwarf is an ideal gas, and its mass is uniformly distributed.

The internal kinetic energy of a single particle is $\frac{3}{2}kT$, so the kinetic energy of the total system of $N$ particles is going to be:

\begin{align*}
K = \frac{3}{2}NkT
\end{align*}

Since the interior of a star can be considered an ideal gas, we can use the ideal gas relation, $PV = NkT$, where $V = \frac{4}{3}\pi R^3$ is the volume of the star. Using this equation, we can solve for $K$ in terms of pressure and volume as follows:

\begin{align*}
K &= \frac{3}{2}NkT\\
K &= \frac{3}{2}PV\\
K &= \frac{3}{2}P \left(\frac{4}{3}\pi R^3\right)\\
K &= 2\pi PR^3
\end{align*}

We also know that the total potential energy in the system from the previous worksheet is:

\begin{align*}
U = -\frac{3GM^2}{5R}
\end{align*}

Using the Virial Theorem, we can solve for the internal pressure as follows:

\begin{align*}
K &= -\frac{1}{2}U\\
2\pi PR^3 &= -\frac{1}{2}\left(-\frac{3GM^2}{5R}\right)\\
P &= \frac{\frac{3GM^2}{10R}}{2\pi R^3}\\
P &= \frac{3GM^2}{20\pi R^4}
\end{align*}

The internal pressure inside the white dwarf star is: $P = \frac{3GM^2}{20\pi R^4}$.