Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

On John Baez's website, http://math.ucr.edu/home/baez/entropy.html, he discusses the problem of how entropy increases when a cloud of ideal gas collapses gravitationally (no black holes - keeping it simple). But he doesn't give away the answer - instead he implies that, I presume, either gravitons or photons are released during collapse and these increase the overall system entropy.

Can someone please answer which it is, gravitons or photons? If gravitons (or gravity waves), then does this mean that pre-20th century physicists, with no notion of gravity waves, would have concluded that gravitational attraction violates the 2nd law? And if photons, then what about if the gas particles were totally neutral - wouldn't they not emit any EM radiation in that case? Or am I wrong that gravitons or photons are the answer? Thanks!

share|improve this question
I think you just have to assume that some heat/energy is transferred to a bath, but you don't have to be specific about what that bath is, or how the transfer happens. E.g. suppose it goes into a bath whose temperature is constant (it's a bath!) and is the final temperature, find the entropy change due to that, and see that it's greater. –  genneth Feb 3 '11 at 17:33
add comment

4 Answers

up vote 4 down vote accepted

The answer would be actually much simpler if the collapse produced a black hole. It can be easily shown to have entropy vastly exceeding the entropy of any gas of the same mass.

Concerning your main question, the answer is, of course, that any system with many degrees of freedom - both in classical physics as well as quantum physics - always satisfies the second law of thermodynamics. The second law may always be proved - quite generally. The proofs are the proofs of the H-theorem or its generalization for any physical system you consider.

Just think about the "balls" in the phase space - any phase space - how it gets deformed via the time evolution. The "smoothened" version of this evolved "spaghetti" has a higher volume whose logarithm represents the entropy increase.

If you didn't allow the molecules to emit photons when they collide, they wouldn't ever shrink spontaneously by obeying the laws of gravity. The probability that a molecule slows down (or gets closer) under the gravitational influence of the other molecules would be equal to the probability that it speeds up (or gets further) - in average. If you introduce some objects and terms in the Hamiltonian that allow inelastic collisions, these inelastic collisions will selectively slow down the molecules that happened to be closer to each other, which is the mechanism that will be reducing the average distance between the molecules (the actual rate will depend on the gravitational attraction, too).

I wrote photons because, obviously, the probability of the emission of a photon is much higher for real-world gases because most of their interactions are electromagnetic interactions. Because a photon carries as much entropy as a graviton would, but you produce many more photons by random collisions, the entropy increase is stored in the photons. The entropy carried by gravitons is smaller by dozens of orders of magnitude.

share|improve this answer
Thanks very much Lubos. It's interesting to me that E&M interactions are needed for gravitational collapse to happen. So a toy model universe of just massive neutral point particles and gravity will not collapse. I guess this makes sense from the point of view of elastic vs inelastic collisions. I wonder whether there's some kind of 'minimum' dipole moment/charge/interaction strength that's required to allow gravitational collapse. A stretch - but maybe it would be equal to e in some appropriate units?? –  user1743 Feb 3 '11 at 18:11
@brian: the em interactions are only necessary to carry energy away. Imagine if the only interactions were energy conserving, then you've simply got a system of gas with a particular interaction, which would have some definite equation of state. It would simply reach equilibrium. –  genneth Feb 3 '11 at 18:28
Dear @genneth, great, I agree with you. But can you authoritatively say what is the equilibrium state into which the gravitationally bound gas (with no inelastic collisions allowed) will converge? ;-) I personally think that the temperature and the number of molecules dictate a particular size and profile of the distribution of $p,T$ away from the center of a sphere. –  Luboš Motl Feb 3 '11 at 18:53
I believe the only constants of motion are energy and momenta (linear+angular), so the equilibrium state will be the particles infinitely separated :-) This is a subtle issue, essentially covered by Ted's comments on his answer below, which point out that in an infinite volume, evaporation will eventually dominate. Statistical physics is often subtle in that minimal toy models will have unphysical/unrealistic output, even though intuitively it seems like the neglected physics is irrelevant (in the renormalisation sense). –  genneth Feb 4 '11 at 11:17
Right, in the "very infinite" timescale limit, many particles escape while a few of them will be really close, and having a huge negative potential energy. ;-) Still, the time scale at which the escape speed becomes relevant may be made grow exponentially with some parameters of the problem, so there must still exist some meta-equilibrium for the cases when the time needed for the molecules to escape is huge. I think it will be a spherical gas object with a particular radial profile of density and velocity. –  Luboš Motl Feb 4 '11 at 18:00
add comment

The situation Baez describes is of a virialized ball of gas shrinking gravitationally. As he points out (and further emphasizes in his hint), this process doesn't conserve energy. Once you know where the energy is going, you've got the answer to what's happening with the entropy.

Your guesses about what's carrying off the energy (photons and gravitons) are both good ones, and in some contexts "photons" is the answer. (I know of no astrophysical context in which gravitons are responsible for a significant fraction of the energy loss.) In other astrophysical contexts the answer is actually something else: it's just gas particles "evaporating off." The way a cloud of gas often shrinks is that some particles acquire speeds greater than escape velocity and fly off, leaving behind less-energetic particles. A small fraction of the particles may escape, but they carry off enough energy to cause the collapse, and enough entropy to satisfy the second law.

share|improve this answer
Binary neutron stars lose enough energy to be observable to be worth giving out Nobel awards for :-) –  genneth Feb 3 '11 at 18:22
True! I should have been clearer there: I meant that gravitons were never the answer in a system like the ones under consideration, namely many-particle clouds where it seems natural to talk about entropy and temperature. –  Ted Bunn Feb 3 '11 at 18:24
One additional point: one great example of a system to which this sort of analysis applies is globular clusters. In this case, the "gas particles" are actually stars: these clusters "cool" by evaporating off stars. –  Ted Bunn Feb 3 '11 at 18:51
Right, good point. Molecules do sometimes evaporate away from the gravitational field. Well, the temperature must be high enough for the Maxwellian suppression of the high velocities - near the escape speed - to allow escape... However, it's easy to design the situation in which the escaping molecules are totally negligible. Then the energy has to be moved by photons etc. –  Luboš Motl Feb 3 '11 at 18:52
In the situation Baez is describing -- a virialized cloud -- the rms velocity is always of order the escape velocity. That is, a significant fraction of the molecules always have escape velocity. –  Ted Bunn Feb 3 '11 at 22:30
add comment

Thermodynamically, only materials which are totally transparent to electromagnetic radiation will fail to emit photons when heated. The stellar clouds are hardly transparent, especially when dense enough.

share|improve this answer
add comment

The result is generic to matter which is under greater pressure due to gravitational implosion. The hydrostatic equilibrium of a star is the condition where the internal pressure of the material keeps the star in a static configuration against the inward push of gravity $$ \frac{dP}{dr}~=~-\frac{GM\rho}{r^2}, $$ for $P$ the pressure and $\rho$ the density of matter. We may think of an adiabatic situation where a small variation in the stellar radius occurs, where the star shrinks inwards from $R$ to $R~-~\Delta R$. The density of a unit of material is then $\rho~=~3m/(4\pi r^3)$, which we sum up to $M$. We consider the equation of state as the natural gas law $PV~=~NkT$. So the change in the pressure is $$ \Delta P~=~\int_P^{P+\Delta P}dP~=~\int_R^{R~-~\Delta R}\frac{3GM^2}{4\pi r^5}dr $$ $$ =~-\frac{15GM^2}{4\pi r^4}\Big|_R^{R~-~\Delta R}~=~-\frac{15GM^2}{4\pi}\Big(\frac{1}{(R~-~\Delta R)^4}~-~\frac{1}{R^4}\Big)~\simeq~\frac{15GM^2}{\pi R^4}\frac{\Delta R}{R} $$ This means the change in the pressure is positive, as we might expect, and the change in the temperature within a unit volume in the star is similarly positive $\Delta T~=~\frac{V}{Nk}\Delta P$. Then the entropy is $\Delta S~=~c\Delta T/T$.

So we expect the star to increase its luminosity, and the hydrostatic equation of Eddington was derived to compute that. This applies up to the point the implosion increases the gravity field enough to red shift light. The delay coordinate $$ r^*~=~r~+~2GM log\Big(\frac{r~-~2GM}{2GM}\Big) $$ maybe used to compute the red shift. General relativity then predicts a dimming of the star as it collapses.

share|improve this answer
Another quantitatively excellent answer from you @Lawrence +1. This is how physics should be done! –  user346 Feb 4 '11 at 7:24
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.