In many of my astronomy classes, I talk about the scene in Star Trek: Generations where the evil Doctor Soran shoots a missile into a star which causes all fusion in the star to stop, leading the star to immediately collapse. I don’t do this because I’m a fanatical Star Trek fan. As a Trekkie, I’d rather pretend that Star Trek: Generations never happened. It’s because this scene gets the astrophysics wrong in such a wonderfully instructive way.

What would actually happen to a star, say the sun, if all nuclear reactions in it were to suddenly cease? Let’s take this in two passes, one for everyone, and then another for the physics majors. Even physics majors should go through the first pass first.

For everyone

Introductory astronomy books sometimes say or imply that what keeps a star like the sun at a constant size and brightness is a balance between gravity and nuclear forces. This isn’t quite right. In fact, there are two different types of balance going on in the sun.

First, there is a balance of forces. The sun is a ball of gas, and gas likes to expand. This is the force of pressure. Each bit of gas has some mass and so also pulls on each other bit of mass by the attractive force of gravity, which overall tries to squeeze the star. Each part of the sun is in force equilibrium between gas pressure and gravity.

Second, there is the balance of heating and cooling, the thermal balance. The sun’s store of heat energy is depleted by radiation coming off the surface. It is replenished by nuclear fusion reactions in the core. In each part of the sun, heat generation and flow are balanced.

The connection between the two is that the pressure of a gas depends on its temperature. Now, though, you’ll see what the issue is. Even if nuclear reactions stop inside a star, so that no more heat is being released by fusion, the star’s store of heat energy doesn’t immediately disappear. Energy is conserved, so the thermal energy has to go somewhere, and the way stars get rid of energy is by radiation.

Even my non-physics major readers should follow me on an estimate of how long it would take for the sun to radiate away an appreciable amount of its store of thermal energy. How do we do this? It’s just like calculating how long you can spend money before you exhaust your bank account: take the current balance and divide by how much you spend per day. We need two numbers: the thermal energy content of the sun and the rate it is lost by radiation. The second number is easy: it is, by definition, the luminosity of the sun:

\[L_{\odot}=3.8\times 10^{26} {\rm Watts}\]

The second number, the energy in the sun, requires a bit of work. Here’s one way that only involves a little bit of physics, namely that the temperature T of a gas gives the average energy per particle E. The rule is

\[E=\frac{3}{2}k_BT\]

where

\[k_B=1.38\times 10^{-23}{\rm J/K}\]

Temperatures deep in the sun are around 10 million Kelvin. We also need the number of particles in the sun N to go from energy per particle to total energy. Let’s assume the sun is made entirely of ionized hydrogen, so almost all of the mass is in protons. To get the number of protons, take the mass of the sun and divide by the mass of a proton:

\[N=M_{\odot}/m_p = 2\times 10^{30}{\rm kg} / 1.7\times 10^{-27}{\rm kg} = 1.2\times 10^{57}\]

That’s a big number, but in astronomy we’re not afraid of big numbers. Actually, we should multiply by two because there are an equal number of electrons, but I’m going to be throwing away numerical factors like 3/2 or 2 pretty soon. I get a total thermal energy

\[Nk_BT\approx 10^{41} {\rm J}\]

Dividing, I get the time to radiate a significant portion of the thermal energy, the Kelvin-Helmholtz timescale,

\[t_{\rm KH}=Nk_BT/L_{\odot}\approx 10^14 {\rm sec} \sim 10^7 {\rm years}\]

(tens of millions of years)

So the star will slowly change over the course of millions of years. It will indeed shrink, but rather than cooling down, it will actually get hotter.

For physics majors

To understand this, consider the timescale on which force balance is maintained. If pressure were to disappear, the star would collapse on a free-fall timescale: \(t_{\rm ff}=(GM/R^3)^{-1/2}\) which is of order an hour. So the star can respond very quickly, compared to the Kelvin-Helmoholtz (HK) timescale, to any changes in the star’s heat content. It will adjust its size nearly instantaneously (compared to the KH timescale) to maintain force balance. (Furthermore, we might expect that, if pushed a little bit away from equilibrium, it will oscillate adiabatically with periods roughly the free-fall timescale.) This equilibrium is expressed via the virial theorem as an equality (up to factors of order unity) between internal energy and gravitational potential energy. Let us make order of magnitude guesses for the average internal energy per proton and the average gravitational potential energy per proton and then equate them. Let’s be general and say the star has mass M and radius R which are not necessarily the same as the sun’s and will change with time as the star loses energy by radiation. To order of magnitude, an average proton will be at distance R from the center with interior mass of order M.

\[\frac{GMm_p}{R} = k_B T\]

As the star shrinks, the temperature must go up. This makes sense because the star is always in hydrostatic (force balance) equilibrium. As the star shrinks, gravity becomes stronger, so the pressure must also be growing to balance it.

Let’s make a model of what will happen to the star as it shrinks. We will continue to assume that the star is pure hydrogen. We will also assume that thermal energy travels through the star by radiative diffusion, i.e. by photons random walking their way through the hot interior until they find their way to the surface. This will make for a nice simple model.

What’s the time it takes for a photon to random walk its way out of a star from the deep interior? In a very hot, dense, ionized gas, opacity is dominated by electron scattering, which has Thomson cross section

\[\sigma_T=6.7\times 10^{-29}\,{\rm m}^{-2}\]

The distance photons travel between scatters is the mean free path (MFP))

\[\ell = (n\sigma_T)^{-1}\]

where n is the number of density, i.e. electrons per volume. Using the fact that there is one electron per proton, we can estimate

\[n\sim N/V\sim M/(m_pR^3)\]

I’ve taken the volume of the star to be

\[V\approx R^3\]

Given all the approximations we’ll be making, it wouldn’t make sense to keep track of factors of pi and whatnot.

Hopefully you’ve studied random walks in at least one of your classes, because you’ll encounter it in many contexts. If a photon takes steps each the size of the MFP in random directions, the number of steps it takes to travel a distance R is on average

\[(R/\ell)^2\]

The time per step is just the time it takes light to travel the MFP, that time being

\[\ell/c\]

naturally. Thus, the total escape time, the ``diffusion time’’, is

\[t_{\rm diff} \sim \frac{R^2}{\ell c}\]

That’s how long it takes photons to stumble their way out of the star. How much energy do they take with them? Well, what we have is a gas of trapped photons, so they obey blackbody formulae. In particular, the radiation energy density is

\[\epsilon_{\rm rad } = a T^4\]

where T is some average interior temperature. Note that T will probably change as the star shrinks. Note also that T will probably be orders of magnitude higher than the temperature of the ``surface’’ (more precisely from the photosphere from which radiation escapes). Thus, even though we’re being sloppy, we can’t equate the two:

\[T \ne T_{\rm surf}\]

Now, to go from an energy density to an energy, we must multiply by a volume, the volume of the star, which is roughly R cubed. So we have radiation energy

\[E_{\rm rad} \sim \epsilon_{\rm rad}V \sim a T^4 R^3\]

The energy released per time is the luminosity L:

\[L \sim \frac{E_{\rm rad}}{t_{\rm diff}} \sim \frac{a T^4 R^4 c m_p}{M\sigma_T}\]

Now plug in the virial theorem equilibrium condition for T:

\[L \sim \frac{a G^4 m_p^5 c}{\sigma_T k_B^4}M^3\]

Ignore the mess of constants. The main point is that luminosity depends on M cubed, which (assuming the star doesn’t blow off much matter) is a constant, and it doesn’t depend on radius at all. Thus, as the star shrinks, its luminosity stays constant.

(How do we know the star shrinks rather than expands? It’s probably intuitive that this is what will happen, but if you want an argument, you can use energy conservation. As the star radiates, its total energy must go down, so it must sink into its gravitational potential well.)

The photons escape when they reach the surface (i.e. the photosphere). This is the radiation of trapped photons leaking off the surface of a dense, opaque gas, so it will obey the Stefan-Boltzmann law for blackbody emission. The flux (luminosity per surface area) is

\[F=\sigma T_{\rm surf}^4\]

The flux and the luminosity are related by L=FA where A is the surface area (which is roughly R squared). We already know what L is–it’s dictated by the interior physics. For a given radius R, we can solve for the surface temperature:

\[T_{\rm surf} \sim \frac{L}{R^2} \propto R^{-2}\]

The surface heats up as the star shrinks, and (using Wein’s law) the star becomes bluer. On a Hertzprung-Russell diagram (addressing myself now to the astronomy majors), the star would move horizontally to the left.

Come to think of it, something like this might be expected to happen in real life when stars are forming, before their interiors become hot enough for pp-chain nuclear fusion. Sure enough, this horizontal H-R track in pre-main sequence stars sufficiently massive to be radiative is called the Hayney track. (If energy transport is by convection, the star follows a vertical path on the H-R diagram called the Hayashi track.) So something like this happens in real life outside of science fiction. It is stopped precisely by the initiation of nuclear fusion and establishment of thermal equilibrium.