Blog 17: Modeling Normal Galaxies Using the Virial Theorem

3) One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler's Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law $(F \propto \frac{1}{r^\alpha})$. It relates the kinetic (or thermal) energy of a system $K$, to the potential energy, $U$, giving: 

\begin{align}
k = -\frac{1}{2}U
\end{align}

(a) Consider a spherical distribution of $N$ particles, each with a mass $m$. The distribution has total mass $M$ and a total radius, $R$. Convince yourself that the total potential energy, $U$, is approximately

\begin{align}
U \approx -\frac{GM^2}{R}
\end{align}

You can derive or look up the actual numerical constant out front. But in general astronomy you don't need this prefactor, which is of order unity. 

The total potential energy of the system can be modeled by the following equation:

\begin{align}
U = -\frac{G(\sum_{i = 1}^{N} m_i) (\sum_{j = 1}^{N} m_j)}{ r_{i,j}}
\end{align}

where $G$ is the gravitational constant, $m$ is the mass of a single particle in the spherical distribution, and $r$ is the radius between two particles, $i, j$ within a distribution of $N$ particles. Since there are $N$ particles, and the radius between two particles can be averaged out to the radius of the entire sphere, $R$, the equation above can be simplified to:

\begin{align}
U &= -\frac{G (N \times m) (N \times m)}{ R}\\
U &= -\frac{G N^2 m^2}{ R}\\
\end{align}

The total mass of the distribution, $M$ is the equivalent of the mass of a single particle, $m$, multiplied by the total number number of particles, $N$, given by the formula, $M = Nm$. We can substitute this into the previous equation to solve for the total potential energy as:

\begin{align}
U &= -\frac{G M^2}{ R}
\end{align}

However, the problem says that there should be a prefactor constant that is not included. Since it is a unity factor, it can be ignored, so the true potential energy can be modeled as:

\begin{align}
U \approx -\frac{G M^2}{ R}
\end{align}

(b) Now let's figure out what $K$ is equal to. Consider a bound spherical distribution of $N$ particles (perhaps stars in a globular cluster), each of mass $m$, and each moving away with a velocity of $v_i$ with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter $\sigma^2$. Show that the kinetic energy of the system is: 

\begin{align}
K = N \frac{3}{2}m \sigma^2
\end{align}

The equation for the total kinetic energy of a systemis:

\begin{align}
K = \frac{1}{2}M v^2
\end{align}

However, since the total mass $M$ is composite of the individual particle mass, $m$, times the total number of particles there are, $N$, $M = Nm$, and the equation for total kinetic energy can be written as:

\begin{align}
K = N \frac{1}{2}m v^2
\end{align}

Since the velocity scattering happens in a three dimensional frame, the velocity-squared of the system can be represented as:

\begin{align}
v^2 = 3 \sigma^2
\end{align}

Substituting this into the previous equation, you get a total kinetic energy of the system as:

\begin{align}
K = N \frac{3}{2}m \sigma^2
\end{align}

(c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius $R$ and stellar velocity dispersion $\sigma$ is (to some prefactor of order unity): 

\begin{align}
M \approx \frac{\sigma^2 R}{G}
\end{align}

Let's keep track of everything we know:

(1) The Virial Theorem:

\begin{align}
K = -\frac{1}{2}U
\end{align}

(2) The total kinetic energy of a system:

\begin{align}
K = N \frac{3}{2}m \sigma^2
\end{align}

(3) The total potential energy of a system:

\begin{align}
U \approx -\frac{G M^2}{ R}
\end{align}

(4) The relationship between the total mass, $M$, of a system, and the mass of a single particle of a system, $m$ for $N$ particles:

\begin{align}
M = Nm
\end{align}

Using this information, we can solve for the mass of the globular cluster, $M$, using the Virial Theorem, as follows:

\begin{align}
K &= -\frac{1}{2}U\\
N \frac{3}{2}m \sigma^2 &\approx -\frac{1}{2}\left(-\frac{G M^2}{ R}\right)\\
3 M \sigma^2 &\approx \frac{G M^2}{ R}\\
3 \sigma^2 &\approx \frac{G M}{ R}\\
\frac{3 \sigma^2 R}{G} &\approx M\\
M &\approx \frac{3 \sigma^2 R}{G}
\end{align}

We have derived an approximation for $M$ in terms of the velocity scatter $\sigma^2$, the radius $R$ of the globular cluster, and the gravitational constant $G$ as was intended. However, there is the coefficient 3 included that signifies the 3 dimensions of space. We can remove the coefficient to show the relationship as:

\begin{align}
M &\approx \frac{\sigma^2 R}{G}
\end{align}