Good Ol' Virial Theorem: Blog 17, Worksheet 6.1, Problem 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 ( \( F \propto 1/r^α\) ). 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. 


We can think about this sphere of particles with a uniform density in terms of thin shells, each with a width of dR.

We also know that \[ U = - \frac{GMm}{R} \] We can treat the entire sphere as our mass, M, and the shell as our mass, m. Then, we can rewrite each mass in terms of density: \[ \rho = \frac{M}{V} \] Where \( V = \frac{4}{3} \pi R^3\) So \[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \] And we can rewrite the equation for potential energy of one shell as: \[ dU = - \frac{GMdm}{R} = - \frac{G(\frac{4}{3} \pi R^3 \rho) (4 \pi R^2 \rho) dR}{R}= - G \frac{16}{3} \pi^2 R^4 \rho^2 dR \] But now we have dU in terms of dR, so we can integrate to get U: \[ \int_{0}^{R} - \frac{16}{3} G \pi^2 R^4 \rho^2 dR =  - \frac{16}{3} G \pi^2 \rho^2  \int_{0}^{R} R^4 dR = - \frac{16}{3} \frac{R^5}{5} G \pi^2 \rho^2 - 0 \] But this still is not the answer we want. We need U in terms of M, not density so we can reuse  \[ \rho = \frac{M}{ \frac{4}{3} \pi R^3 } \] to get: \[ U = - \frac{16}{3} \frac{R^5}{5}\left( \frac{M}{\frac{4}{3} \pi R^3 } \right)^2 G \pi^2  \] Amazingly enough, this simplifies to the much nicer expression: \[ U = - \frac{3GM^2}{5R} \] This simplifies to our original expression: \[ U \approx - \frac{GM^2}{R} \]

(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 σ 2 . Show that the kinetic energy of the system is: \[K =  N \frac{3 }{2} m σ^2\]

Our traditional, classical mechanics, definition of kinetic energy is \[ K = \frac{1}{2} mv^2 \] where velocity is the total or average velocity of the system. That is very difficult to measure for a distant object here on Earth, because the entire system (e.g. a globular cluster) is moving away from us with the universe's expansion. To account for this we must change our frame of reference to the system.
 

So our definition for kinetic energy \[ K = \frac{1}{2} m \Sigma (v_o + \sigma_{vi} )^2 \] turns into  \[ K = \frac{1}{2} m \Sigma ( \sigma_{vi} )^2 =   \frac{1}{2} m \sigma_{vi}^2\] in the reference frame. However, this is a 1 dimensional representation of velocity scatter, as that is all we can observe - the cluster in one plane. To account for the 3 dimensions of space.

So our final expression would be \[ K =   \frac{3}{2} m \sigma_{vi}^2\]

   (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): \[M = \frac{σ^2R}{ G}\]

Using the Virial Theorem, we can combine our two derived equations to find the answer. \[K = - \frac{1}{ 2} U\] \[\frac{3}{2} m \sigma_{v}^2 = -\frac{1}{2} \left( - \frac{GM^2}{R} \right) \] \[3M\sigma_{v}^2 = \frac{GM^2}{R} \]  \[M = \frac{σ^2R}{ 3G}\]  \[M \approx \frac{σ^2R}{ G}\]