astrophysicsstarsthermodynamics
I've just found out that a negative specific heat capacity is possible. But I have been trying to find an explanation for this with no success.
Negative heat capacity would mean that when a system loses energy, its temperature increases. How is that possible in the case of a star?
Musn't there be a source of energy to increase the temperature of any system?
So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.)
Only about 1.5% of the sun's mass is anything other than hydrogen and helium-4. This is true all the way out from the core to the surface. We'll assume that the sun contains only hydrogen and helium-4.
All but the outermost 0.2% of the sun's mass (out to 90% of the sun's radius) is at a temperature $kT>54\,\mathrm{eV}$, which is the energy needed to turn $\mathrm{He^+}$ into $\mathrm{He^{2+}}$. (This energy is four times the Rydberg energy.) So somewhere above 99% of the sun's mass is completely ionized.
The core temperature $kT\approx 1300\,\mathrm{eV}$ is much less than the electron mass, so the matter in the core is not relativistic.
I'm going to assume that the electrons aren't degenerate; this tool (via this question) makes me think that's a pretty safe assumption for matter at the core with density $\rho \approx 150\,\mathrm{g/cm^3}$ and temperature $T \approx 10^7\,\mathrm K$.
In that case we can treat the core of the sun as a mixture of three non-interacting ideal gases, $\mathrm H^+$, $\mathrm{He}^{2+}$, and $\mathrm e^-$. As
George Herold says, each ideal gas particle has mean kinetic energy $\frac32 kT$, so we'll want the number densities. The number density for hydrogen $n_\mathrm{H}$ is
$$
n_\mathrm{H} = \rho f_\mathrm{H}/{\mu_\mathrm{H} }
$$
where $\rho$ is the mass density, $f_\mathrm{H}$ is the hydrogen mass fraction, and $\mu_\mathrm{H} = 1\,\mathrm{gram/mole}$ is the atomic mass of hydrogen. You have a similar expression for helium (with $\mu_\mathrm{He} = 4\,\mathrm{gram/mole}$). The electron number density, thanks to complete ionization, is just
$$
n_\mathrm{e} = n_\mathrm{H} + 2n_\mathrm{He}.
$$
Here's a figure showing temperature, mass density, and composition from my source above and number density as computed here:
Note that the horizontal scale (radius) is mass-weighted: you find about half the mass of the sun between 0.1 and 0.3 solar radii, so that interval takes up about half the horizontal axis. This is purely a visualization technique, so that your eye isn't distracted by the (relatively) cool, diffuse outer layers of the sun.
To find the total thermal energy density, we have to integrate. We find the thermal energy density $$ \epsilon = (n_\mathrm{H} + n_\mathrm{He} + n_\mathrm{e})\frac32 kT $$ and the volume of a thin shell at radius $r$ is $$ dV = 4\pi r^2 dr $$ This integral $\int\epsilon\, dV$ gives me a total stored kinetic energy $E=3.09\times10^{41}\,\mathrm{J}$, of which about 95% is contained within half the sun's radius.
Now, if the sun had uniform density you could estimate its gravitational potential energy, the energy that was released when all the pieces fell together, as $$ U_\text{uniform sphere} = -\frac35 \frac{GM_\text{sphere}^2}{R_\text{sphere}} = - 2.3\times10^{41}\,\mathrm J \text{ (uniformly dense sun)}. $$ That's pretty close to our stored heat! We can do a little bit better since we actually know the density profile of the sun, by finding the potential energy released as you lay down each spherical shell, $$ U = - \int_0^{M_\text{sun}} \frac{G M_\text{enclosed}(r)}{r} dM = -6.15\times10^{41}\,\mathrm{J}. $$ This gravitational self-energy is roughly twice the stored kinetic energy --- which a real astronomer would have predicted as a consequence of the virial theorem.
So, is there any macroscopic physical meaning to this statement that the negative temperature is "hotter"? Doesn't this mean that the negative temperature required for that laser to exist ought to be so "hot" no mere mortal can handle such a device?
If you go far up enough in the atmosphere, you'll reach regions where the temperature is over 3000 K. Yet rockets aren't exploding in flames when they get there, because the atmosphere is very sparse. It will transfer energy to the rocket, because its temperature is very high, but only incredibly slowly.
In general, human beings cannot detect temperature; we can only detect heat transfer. This is why metal surfaces often feel cold; they're the same temperature as everything else, but they conduct better, so you lose heat faster when touching them. Similarly, if you go up really high into the atmosphere, you won't feel like it's hot. Instead, you'll feel nothing at all.
So our intuitive sensory notions of "hot" or "cold" have very little to do with temperature. So it's indeterminate how "hot" a negative temperature system will feel, because that depends on how quickly it transfers heat to your body. All the temperature means is that it will transfer heat.
Best Answer
Consider a satellite in orbit about the Earth and moving at some velocity $v$. The orbital velocity is related to the distance from the centre of the Earth, $r$, by:
$$ v = \sqrt{\frac{GM}{r}} $$
If we take energy away from the satellite then it descends into a lower orbit, so $r$ decreases and therefore it's orbital velocity $v$ increases. Likewise if we add energy to the satellite it ascends into a higher orbit and $v$ decreases.
This is the principle behind the negative heat capacity of stars. Replace the satellite by a hydrogen atom, and replace the Earth by a large ball of hydrogen atoms. If you take energy out then the hydrogen atoms descend into lower orbits and their velocity increases. Since we can relate velocity to temperature using the Maxwell-Boltzmann distribution this means that as we take energy out the temperature rises, and therefore the specific heat must be negative.
This is all a bit of a cheat of course, because you are ignoring the potential energy. The total energy of the system decreases as you take energy out, but the decrease is accomplished by decreasing the potential energy and increasing the kinetic energy. The virial theorem tells us that the decrease of the potential energy is twice as big as the increase in the kinetic energy, so the net change is negative.