Blog Post 18, Worksheet 6.1, Problem 3

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 (\(F\propto1/r^\alpha\)). It relates the kinetic (or thermal) energy of a system, K, to the potential energy, U, giving \[K = -\frac{1}{2}U\] (a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately \[U \approx -\frac{GM^2}{R}\] You can derive or look up the actual numerical constant out front. But in general in astronomy, you don't need this prefactor, which is of order unity.

In this spherical distribution, there are N particles, each with a mass m, contributing to a total mass M. Thus, Nm = M. If we think of the gravitational force between a single particle, of mass m, and the rest of the particles in the distribution, of essentially mass M (as m is significantly smaller than M, M-m = M), we can see where this expression for potential energy comes from. If we treat the mass M as a point mass at an approximate distance R away, we have: \[F = -\frac{GMm}{R^2}\] Then, we can use the relationship between potential energy and force to find the gravitational potential energy for this particle: \[F = -\frac{dU}{dx}\] \[-\int F\:dx = U(x)\] \[-\int\frac{-GMm}{R^2}dR = U(R)\] \[U = \frac{-GMm}{R}\] If we add the potential energies for each of the N particles, we get a total potential energy of: \[N(\frac{-GMm}{R}) = -\frac{GM^2}{R}\]
(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 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 of the system along our line of sight, the velocity scatter \(\sigma^2\). Show that the kinetic energy of the system is: \[K = N\frac{3}{2}m\sigma^2\]
The kinetic energy of an individual particle with mass m and velocity v is given by \(K = \frac{1}{2}mv^2\). As we can't measure the individual velocity vector of each particle, we can instead look at the kinetic energy of the entire system, given by \[K = \frac{1}{2}Mv_{\text{mean}}^2\] As \(v_{\text{mean}}\) is given by \(\sigma\), and M = Nm, we can write this as: \[K = N\frac{1}{2}m\sigma^2\] Lastly, as velocity is a vector, this expression actually gives the kinetic energy of the particle for its motion in one dimension/direction. As this is a three-dimensional system, we must add the kinetic energy for motion in each direction to get a total kinetic energy of: \[3(N\frac{1}{2}m\sigma^2) = N\frac{3}{2}m\sigma^2\]
(c) Use the Virial Theorem to showt hat the total mass of, say, a globular cluster of radius R and stellar velocity dispersion \(\sigma\) is (to some prefactor of order unity): \[M \approx \frac{\sigma^2R}{G}\]
The Virial Theorem gives a relationship between the kinetic and potential energies of a system. As we know the total kinetic energy of the system (\(K = N\frac{3}{2}m\sigma^2 = \frac{3}{2}M\sigma^2\)) and the potential energy of the system (\(U \approx -\frac{GM^2}{R}\)), we can plug these values into the Virial Theorem: \[K = -\frac{1}{2}U\] \[\frac{3}{2}M\sigma^2 = -\frac{1}{2}(-\frac{GM^2}{R})\] \[\frac{3}{2}M\sigma^2 = \frac{GM^2}{2R}\] \[GM^2 = 3M\sigma^2\] \[M = \frac{3R\sigma^2}{G}\] \[M \sim \frac{R\sigma^2}{G}\]