Blog #17, Worksheet #6.1, Question 3, 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. 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 \[U \thickapprox - \frac{GM^2}{R}\] 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.

We can start by relating the potential energy to the work of the system through \[U = - \int_{0}^{R} F dr\] where \[F = \frac{GMm}{r^2}\] and \[M = 4 \pi r^2 \rho dr\ \qquad m = \frac{4}{3} \pi r^3 \rho\]and once we integrate we get \[U = - \frac{16}{15} G \pi^2 \rho^2 R^5\]Finally we replace rho with a general expression for density with \[\rho = \frac{M}{\frac{4}{3} \pi R^3}\]which gives us the desired expression \[U = - \frac{3}{5} \frac{GM^2}{R} \thickapprox - \frac{GM^2}{R}\] 

(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 vi 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. Show that the kinetic energy of the system is: \[K = \frac{3}{2} N m \sigma^2\]The proof is quite straightforward for this part of the theorem. We begin with the general form for kinetic energy where \[K = \frac{1}{2} m v^2 \]However, this only accounts for line of sight, one direction of the 3 possible directions that the system can move so we get \[K = \frac{3}{2} m v^2 \] but we have N stars in our system and our velocity profile, \(v = \sigma \). Substituting into our equation we get our desired final answer \[K = \frac{3}{2} N m \sigma^2\](c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity): 
\[M \thickapprox \frac{\sigma^2 R}{G} \]

We begin with the general Virial Theorem \[K = - \frac{1}{2} U\]We plug in our answers from part a) and b) where \[U = - \frac {GM^2}{R} \qquad K = \frac{3}{2} N m \sigma^2 \]so \[\frac{3}{2} N m \sigma^2 = \frac{1}{2} \frac{GM^2}{R} \]Solving for M we get our final answer of \[M = \frac{3 \sigma^2 R}{G}\] which simplifies to our desired final solution of \[M \thickapprox \frac{\sigma^2 R}{G} \]