The Virial Theorem

Here is one of the strangest facts about stars: when a star loses energy, it gets hotter. The reason is a balance law called the virial theorem, which you have already met in disguise. In orbital energy, a circular orbit has $E = -K$, so a satellite that loses energy to air drag drops lower and speeds up. A star is a swarm of particles held by their own gravity, and the same rule applies.

Why it works

Multiply hydrostatic equilibrium, $dP = -Gm\rho\,dr/r^2$, by $4\pi r^3$ (three times the volume inside $r$) and add it up over the whole star. On the right, with $dm = 4\pi r^2\rho\,dr$, you get $-\int Gm\,dm/r = U$, the star's gravitational energy. On the left, integrating by parts (with $P = 0$ at the surface) gives $-3\int P\,dV$. For an ideal monatomic gas, pressure is two-thirds of the kinetic energy per unit volume, so $3\int P\,dV = 2K$, where $K$ is the total thermal kinetic energy. Therefore $$2K + U = 0.$$

Key idea

For a gas ball in equilibrium held by its own gravity: $2K + U = 0$, so $K = -U/2$ and the total energy is $E = K + U = U/2 = -K$. Radiating energy ($E$ falls) makes $K$ rise: the star contracts and heats up. Half of the gravitational energy released is radiated; the other half heats the gas.

A contracting cloud. As gravity releases energy, half goes into heating the gas and half is radiated away, so the cloud gets hotter while it shines.

Worked examples

Worked example: Energy bookkeeping

A protostar radiates $1.0 \times 10^{38}$ J while staying in equilibrium. How do $K$ and $U$ change?

$\Delta E = -1.0 \times 10^{38}$ J, and $E = -K = U/2$.
$\Delta K = -\Delta E = +1.0 \times 10^{38}$ J; $\Delta U = 2\Delta E = -2.0 \times 10^{38}$ J.

Answer: Gravity releases $2.0 \times 10^{38}$ J: half heats the star, half leaves as light.

Worked example: The Sun's thermal energy

For a uniform sphere, $U = -\tfrac35GM^2/R$. Find $U$ and $K$ for a uniform Sun.

$$U = -\frac{0.6(6.67 \times 10^{-11})(1.99 \times 10^{30})^2}{6.96 \times 10^{8}}$$
$U \approx -2.3 \times 10^{41}$ J.

Answer: $K = -U/2 \approx 1.1 \times 10^{41}$ J of thermal energy. (The real Sun is more centrally concentrated, so both are a few times larger.)

Worked example: An average temperature

With $K = \tfrac32NkT$ and $N = M/(\mu m_p)$, find the mean temperature of a uniform Sun with $\mu = 0.6$.

$\tfrac32\dfrac{M}{\mu m_p}k\bar T = \tfrac{3}{10}\dfrac{GM^2}{R}$ gives $\bar T = \dfrac{GM\mu m_p}{5kR}$.
$\bar T = 0.6 \times 2.31 \times 10^{7}\text{ K}/5$.

Answer: About $2.8 \times 10^{6}$ K: millions of kelvins on average.

Common mistake

Thinking a star that radiates energy must cool down, like a cooling cup of coffee. Coffee is not held together by its own gravity. A star is: losing energy lets it contract, contraction releases gravitational energy, and the virial theorem sends half of that into heat. So stars and protostars heat up as they lose energy, a "negative heat capacity". Check against orbits: a satellite slowed by drag ends up moving faster, for exactly the same reason.

This is how a protostar's core gets hot enough to start fusion, and how a star's core heats up to burn helium after its hydrogen runs out.

The Virial Theorem

Worked examples

Practice

Builds on

Leads to

Sources and further reading