# Difference between revisions of "Virial Theorem"

### Virial Theorem

\\Start out with hydrostatic equilibrium:

${\displaystyle {\frac {dP}{dr}}=-\rho {\frac {GM_{r}}{r^{2}}}\,\!}$

We want this to have units of energy when we integrate this equation over the entire star. In order to do this, we multiply by the factor ${\displaystyle 4\pi r^{3}dr}$. Now let’s look at the right hand side of this equation.

${\displaystyle -\int _{0}^{R}\rho {\frac {GM_{r}}{r^{2}}}{\dot {4}}\pi r^{3}dr=-\int _{0}^{R}dM_{r}{\frac {GM_{r}}{r}}=U.\,\!}$

So the right side just gives us the gravitational potential energy. Now we look at the left hand side, using the same factor of ${\displaystyle 4\pi r^{3}dr}$ and a little trick. That is, use the fact that

${\displaystyle {\frac {d}{dr}}\left(4\pi r^{3}P\right)=4\pi r^{3}{\frac {dP}{dr}}+3\times 4\pi r^{2}P.\,\!}$

Now we use this on the left hand side of hydrostatic equilibrium to get

${\displaystyle \int _{0}^{R}4\pi r^{3}dr{\frac {dP}{dr}}=\int _{0}^{R}dr{\frac {d}{dr}}\left(4\pi r^{3}P\right)-3\int _{0}^{R}4\pi r^{2}Pdr.\,\!}$

The first factor just goes to zero, since the radius at the center is zero, and the pressure at the boundary is (nearly) zero. Thus we have

${\displaystyle U=-3\int _{0}^{R}4\pi r^{2}Pdr,{\rm {\;or}}\,\!}$
${\displaystyle U=-3

V.\,\!}$

The total energy in the star is

${\displaystyle E_{tot}=K+U=-3P

+K.\,\!}$

Here ${\displaystyle K}$ represents the total kinetic energy in the star. This kinetic energy is dominated by the random motions of particles in the gas, otherwise known as the thermal energy. Ordered motions like convection and rotation are unimportant. For an ideal gas at temperature ${\displaystyle T}$, the energy per particle is given by

${\displaystyle \epsilon ={\frac {3}{2}}k_{B}T,{\rm {\;(non-relativistic)}}\,\!}$
${\displaystyle \epsilon =3k_{B}T{\rm {\;(relativistic).}}\,\!}$

More generally, this is

${\displaystyle \epsilon ={\frac {1}{\gamma -1}}k_{B}T.\,\!}$

Thus the total thermal energy of the a star is given by

${\displaystyle K=\int _{0}^{R}{\frac {1}{\gamma -1}}kTndV.\,\!}$

The integrand is just the energy per volume, which is the same thing as a pressure. Thus the total thermal energy of a star is given by

${\displaystyle K={\frac {1}{\gamma -1}}\int _{0}^{R}PdV={\frac {1}{\gamma -1}}

V.\,\!}$

The gravitational potential energy and kinetic energy are related by the adiabatic index, with

${\displaystyle K={\frac {-U}{3(\gamma -1)}}.\,\!}$

In the non-relativistic case,

${\displaystyle K=-{\frac {U}{2}},\,\!}$

and in the relativistic case

${\displaystyle K=-U.\,\!}$

We can also write the total energy of the star in these terms:

${\displaystyle E_{tot}=K(1-3(\gamma -1)).\,\!}$

This entire derivation assumes the fact that the gas is classical and ideal. Even so, the result holds much more generally, including in white dwarfs (which are degenerate) and in systems where photons dominate the pressure (high mass stars). \\Now we want to find a general relationship between pressure, thermal energy, and the cases where photons or quantum effects are important. We will imagine a system with particles zooming around and bouncing off of a wall, where the wall has area ${\displaystyle A}$. Then the force on the wall from collisions is

${\displaystyle {\frac {\Delta p}{\Delta t}}=PA.\,\!}$

We could also write the change in momentum as the product of the rate at which paticles hit the wall and the change in momentum per collision. The number of particles in a length is ${\displaystyle nA}$, where ${\displaystyle n}$ is the number density of particles. Then the rate is just

${\displaystyle {\rm {rate\;}}={\frac {1}{2}}nAv_{x}.\,\!}$

The change in momentum is use two times the x momentum of the particles, so that

${\displaystyle {\frac {\Delta p}{\Delta t}}=nAv_{x}p_{x}.\,\!}$

If we look at three dimensions rather than one, we have a factor of one third come in, which gives us

${\displaystyle {\frac {\Delta p}{\Delta t}}={\frac {1}{3}}nAvp=PA.\,\!}$

The area cancels out, as it should, and we have

${\displaystyle P={\frac {1}{3}}nvp.\,\!}$

This is a fundamental relationship between the microscopic particles and the macroscopic, thermodynamic properties of that gas. Generally, we have a gas with a range of velocities, so we need to average over the distribution function of particles, meaning the pressure is

${\displaystyle P={\frac {1}{3}}n.\,\!}$

For non-relativistic particles, the momentum and velocity are just related by the mass of the particle, meaning

${\displaystyle P={\frac {2}{3}}n<{\frac {1}{2}}mv^{2}>.\,\!}$

The averaged quantity is just the kinetic energy per particle, so

${\displaystyle P={\frac {2K}{3V}}.\,\!}$

This is exactly what we had before for the specific case of an ideal gas, and gives us the same result for the Virial theorem we had before. We could do the same thing for a photon, or relativistic massive particle (and Eliot did), but the procedure and result are the same. \\Now let’s use the Virial theorem to derive some useful results. We start by imaginging a system in HE that has no internal energy source (like fusion) that is radiating energy with power ${\displaystyle L}$. Thus the total energy, which is already negative, is decreasing, or, in an absolute value sense, the total energy is becoming more negative. That is, ${\displaystyle |E_{tot}|}$ is increasing, ${\displaystyle |U|}$ is increasing, and ${\displaystyle K}$ is increasing. The way that this happens is by contracting, specifically a process called Kelvin-Helmholtz contraction. At any stage in the life of a star, if it does not have an internal energy source, this will happen. We can estimate the timescale of this contraction (meaning the time it takes to change appreciably, say ${\displaystyle R}$ changing of order ${\displaystyle R/2}$) using the total energy and the rate at which energy is being radiatied:

${\displaystyle t_{KH}\sim {\frac {|U|}{L}}.\,\!}$

If we imagine we know nothing about the internal energy source of the Sun, and ask what the Kelvin-Helmholtz timescale is for the Sun, we would find that

${\displaystyle t_{KH}=3\times 10^{7}{\rm {years}}.\,\!}$

This is much less than the age of the Earth or the solar system, but that did not stop Kelvin and Helmholtz from proposing that this is how the energy of the Sun was generated, which caused great debates with biologists and geologists who knew the Earth must be older. What Kelvin and Helmholtz did not know is that fusion kicked in at the center and halted this process, though the Sun did undergo this contraction during its formation. \\Now let’s imagine the opposite situation, where more energy is being generated than is being radiatied. Then the total energy is increasing, but ${\displaystyle |E_{tot}|}$ actually goes down, meaning ${\displaystyle K}$ decreases and ${\displaystyle |U|}$ decreases. This is pretty weird. If you add energy to the star, it gets cooler. From a thermodynamics standpoint, this means stars have a negative heat capacity. In fact, all self gravitating objects have a negative heat capacity. This is unlike anything we experience on Earth, where adding energy to a system causes it to heat up. \\This is the fundamental reason that fusing stars are different from fusion bombs. A bomb is heated, which causes fusion and energy release, which further heats the bomb, leading to a runaway explosion. The star fuses in a controlled manner thanks to gravity, which acts as a safety valve. Any extra fusion in a star causes the star to expand and cool, thereby slowing the rate of fusion until it comes back in to equilibrium. Recreating this on Earth is pretty challenging, and is the reason we do not have fusion energy yet. \\The Virial Theorem can also give us an estimate of the temperature in stars. Consider a star (our Sun) supported by a non-relativistic classical gas. Then

${\displaystyle U=-2K,\,\!}$
${\displaystyle {\frac {GM^{2}}{R}}\sim 3Nk_{B}T\sim 3{\frac {M}{m_{p}}}k_{B}T.\,\!}$

This gives

${\displaystyle 3k_{B}T\sim {\frac {GMm_{p}}{R}}.\,\!}$

For the sun, this is about ${\displaystyle 5\times 10^{6}}$ K, which is much larger than the effective temperature, and a little bit smaller than the central temperature.