The Virial Theorem: Derivation and Example

Heyo, welcome to AstroNaught! I hope you’re feeling energetic because today is all about the Virial Theorem. The Virial Theorem relates kinetic energy to gravitational potential energy. I’ll

start by giving you the answer so that you see what I mean, and then we’ll go back and prove why this relationship works and discuss when it is relevant. The Virial Therorem states that the kinetic energy of a stationary system is equal to negative ½ of the gravitational potential energy. Since the total energy is equal to the kinetic energy plus potential

energy, this also means that energy equals one half of the potential energy. So how do we get here? Well, let’s start by considering the product of linear momentum and the position vector of a particle. If we take the sum of all the particles in a system, we get The sum of p_i times r_i.

For brevity’s sake, let’s call this quantity Q. Now, I know sometimes it can be confusing to work through a derivation without understanding the physical significance of each step or quantity. Try to think of this purely from a math perspective or, if you really want, you can think of this as the sum of the particles’ angular momentum. However, note that angular

momentum is the cross-product of the position vector and linear momentum, not a dot product, so this is more like angular momentum with respect to the perpendicular component of linear momentum. Now, we can take a time derivative of Q. Applying the product rule for derivatives, this gets

us dp/dt times r plus p times dr/dt on the lefthand side of our equation. Since linear momentum is mass times velocity, we can split Q apart into the product of m times the velocity, or dr/dt, times the position vector r. This is equal to ½ times the second

derivative of the moment of inertia, since I is just the sum of m_i times r_i squared. The ½ comes from the chain rule. This is a bit messier than we would like. We can take the time average over a period 𝜏 to get 1 over the period times the integral of thesecond derivative of I. Note that for

orbital motion, dI/dt at 𝜏 is the same as dI/dt at 0, and so these two terms cancel out. The lefthand side of our equation is now just zero. Yay. Next, let’s look at the righthand side of our equation. According to Newton’s Second

Law, force is equal to mass times acceleration. In other words, force is equal to mass times dv/dt, where mv just equals the linear momentum. So we can rewrite the first term on the righthand side as the sum of F times r.

The total force is equal to the negative gradient of the potential energy, so we can rewrite this as the gradient of U times r, or r times partial U over partial r. For a potential of the form kr^(n+1), this becomes (n+1) times the gravitational potential. By the inverse

square law, n=-2, so our term simplified to U. We can rewrite the second term, p times dr/dt as m times dr/dt times dr/dt, which is just mass times the velocity. You may recognize this as two times the formula for kinetic

energy. By now, you can probably see that the end is near. We have 0=2K+U. We can move the gravitational potential to the other side of our equation and divide by two to get the equation we started

with. This right here is the virial theorem. Cool. Why do we care? Well, the virial theorem allows us to estimate the kinetic energy of a complex system without having to consider every detail about the individual particles in the system. Let’s put the virial theorem into application.

Say we want to calculate the mass of a galaxy. Well, we can approximate its kinetic energy as ½ Mv squared and its potential energy as GM squared divided by its radius. Solving for mass, we get M equals velocity squared times r over the gravitational constant G.

Assuming we know the mean velocity and radius of the galaxy, we can calculate its mass.