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.

Inspired by this question.

I believe that the usual explanation that preserves the second law of thermodynamics as an astrophysical gas cloud collapses under gravity is that the gas must heat and radiate, and while the entropy of the collapsed gas may be lower than the entropy of the uncollapsed gas, the entropy in the emitted radiation is more than enough to compensate.

However, dark matter is thought to undergo a similar collapse process, and it does not radiate by definition. I recall hearing that there is still no contradiction of the second law here, but I can't recall the explanation. What saves the second law here? Is it simply that some mass must be ejected by the collapsing system, i.e. the collapsing halo is "radiating mass" rather than radiating photons?

share|improve this question

4 Answers 4

I think the assumption that radiation is required for a collapse in general is mistaken. Think about a cloud of gas. If it is going to gravitationally collapse it must have a negative total energy; if it doesn't parts of the gas will fly off.

If it has a negative total energy then there is some finite maximum size for the gas cloud, where it only has potential energy and no kinetic energy. In this state the cloud is at absolute zero, so clearly we can increase the temperature of the cloud by reducing its size slightly so that it has a small, none-zero temperature.

Going to the other extreme, if we compress the gas into a very small volume, then it will have a very large temperature, but we can clearly increase its entropy by increasing its volume. Consequently, we would expect the gas to have its maximum entropy at some volume somewhere in the middle. If you think about where pressure in a gas comes from, this point of maximum entropy has to be the point of hydrostatic equilibrium, where the gas pressure equals the gravitational pressure at all points.

Real stars planets and galaxies do more complicated things than this simple model, which they are able to do because they are not closed systems. It is at this point that you need to take into account radiation.

share|improve this answer
So if I understand this correctly you're arguing that a velocity dispersion supported (this would be the analogue of pressure for dark matter) equilibrium dark matter halo has a lower entropy than the same amount of mass in a diffuse dynamically cold uniform distribution? Any chance you could elaborate on this a little bit/show some semi-quantitative support for your argument? –  Kyle Aug 20 at 17:56

A free dark matter cloud (without the presence of ordinary matter) will simply not "collapse" the same way a radiating gas cloud does. In both cases total momentum, angular momentum and energy are conserved, but in the case of a gas cloud the photons can carry away some of the angular momentum and most of the energy, in case of a dark matter cloud they can't, but a fraction of the dark matter particles still can! So while even a dark matter cloud can "thermalize", both its total energy and angular momentum will be conserved in the dark matter particles alone, which means that it has to shed a non-trivial fraction of its mass to attain a more compact core. This means that the radial velocity distribution and the radial density distributions will be different in the two cases. In neither case will any violation of thermodynamics occur.

share|improve this answer
we are talking of dark "matter", no? it has gravitational interactions. At the moment an effective quantization of gravity is accepted otherwise no BB model. Gravitons will take the place of photons in an eventual collapse in the counting of microstates and angular momenta. Photons beat them by far due to the small coupling constant of gravity in the normal collapse/entropy argument. imo –  anna v Aug 16 at 3:25
@Anna: If we are talking about cold dark matter at the density of the current universe, neither quantum gravity nor gravitational waves will make any measurable difference for any imaginable collapse scenario. Gravitational waves, in particular, will be highly ineffective, because the dark matter distribution will thermalize very quickly into configurations without much quadrupole moment. No quadrupole moment, no gravitational waves. As for the long term collapse due to quantum gravity decay to the ground state... I don't have a verified theory for that. Do you? –  CuriousOne Aug 16 at 3:33
I accept that the time taken may be much longer but your argument then would mean that no gravitational collapse of dark matter will exist either. If we assume, as the question assumes that there will be a gravitational collapse of dark matter, then gravitons will take the role of photons. as long as there exists a gravitational force, gravitons are there by construction –  anna v Aug 16 at 3:38
Now I see what you mean! I have no physical intuition about the quantum case, except, that it should take a very, very long time for galactic size clouds to decay like that. It's an interesting question, though. What do you expect would happen? Bose-Einstein condensation? –  CuriousOne Aug 16 at 3:41
It will depend on what dark matter is. no? fermions? bosons already? –  anna v Aug 16 at 3:44

Dark matter does not radiate photons by definition, but as I said in the comment to CuriousOne, dark matter may not have electromagnetic radiations to first order, but it does have gravitational radiation. The current Big Bang model accepts an effective gravitational interaction and thus the existence of gravitons, i.e. elementary particles of mass zero and spin 2.

Gravitons will take over the role of photons in counting microstates for the entropy increase so the argument would be the same, as they carry off angular momentum etc. The model will be the same except for the time constants which will be much longer since the gravitational coupling constant is so much smaller than the electromagnetic one. This smallness is the reason that gravitons do not appear in the argument of increase of microstates/entropy for the usual collapse explanation.

share|improve this answer
This is fine, but it is well known (from simulations) that dark matter collapses to form a cosmic web like structure in less than a Gyr, even without any baryons present. Gravitons may carry away entropy on long timescales, but on the collapse timescale ($\sim\sqrt{G/\rho}$, which will be quite short) it would appear that entropy is in fact decreasing... and I don't think your answer can help here? Or am I missing something? –  Kyle Aug 20 at 17:52
Well, I do not think that the gravitational field disappears even in very small time scales. It is weak but it is there, otherwise there would be no accumulation of this dark matter. gravitons must be exchanged and radiated analogously to the photons always, no? –  anna v Aug 20 at 18:35
Exchanged yes, but gravitational radiation occurs only when there is a time-varying quadrupole moment in the mass distribution (rather than time-varying dipole in the charge distribution for photons). There will probably be a very weak quadrupole that will vary slowly, so some gravitational radiation, but I think it will be too weak to do what you claim it does? –  Kyle Aug 20 at 18:39
It is my opinion, after all. As gravity is quantized ad hoc at the moment maybe we should wait for the definitive work to decide whether only quadrupoles, which after all are a classical construct and should be emergent behavior in a full quantized gravity, are the only source of graviton radiation. It is probable that a dark matter particle interacting gravitationally with another dark particle will radiate as interaction is change . arxiv.org/abs/hep-th/9909012 . There might be a type of black body radiation of bulk matter from an ensemble of quadrupole elementary particles –  anna v Aug 21 at 4:08

Well i'm not into dark matter, but i am into entropy and stuff, so i will post an answer.

How one is supposed to measure the entropy before and after, by counting micro-configurations, by counting volume/size, all these together?

i suggest one or both of the above may give you an answer as to how the 2nd law may still be valid.

No need for radiation (and photons/gravitons) if the final number of micro-configurations is bigger than before collapse.

Once more i am not into dark matter (at least in these frameworks of Big Bang et al)

share|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.