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?

Keep in mind that dark matter collapse is a well studied problem, and occurs naturally even in the absence of baryonic material (no paywall version), so no coupling to a radiating baryonic component is necessary.

  • $\begingroup$ I'm not an astro-physics type. But why does it have to radiate, lose energy? As it collapses it heats up (average velocity increases.) Till there is some balance between the average kinetic and potential energy. (Viral theorem.) (Aside) I had this idea that dark matter could be neutrino's, but read somewhere that this didn't work because they are too light. Too much KE to collapse. $\endgroup$ – George Herold Oct 28 '14 at 19:16
  • $\begingroup$ @GeorgeHerold As collapse proceeds, the entropy of the collapsing cloud decreases. Either the Second Law is in trouble, or there is entropy leaving the cloud (presumably in the radiation, or for DM I speculate in ejected mass). At least, that's my understanding... $\endgroup$ – Kyle Oman Oct 28 '14 at 19:19
  • $\begingroup$ Doesn't the temperature increase account for the entropy change. (It's been much too long since I had to do any "real" thermodynamics.) $\endgroup$ – George Herold Oct 28 '14 at 19:25
  • $\begingroup$ Thanks to Hypnosifl for linking this article in the comments to one of the answers. This is a nice summary of the gas case - but the DM differs in that it can't radiate (gravity waves are NOT sufficient, if collapsing DM even emits any). And hence my question. $\endgroup$ – Kyle Oman Oct 28 '14 at 20:24
  • $\begingroup$ Great, Thanks for the link to the entropy article. Without really knowing I'm giving @By Symmetry (+1). The assumption of homogeneity must (might) be wrong. Fluctuations (noise) in the density lead to areas (volumes) where collapse is energetically/thermodynamically favorable. (I'm not sure how to figure out the details.) $\endgroup$ – George Herold Oct 29 '14 at 0:28

Note: I think my answer below may be incorrect, at least as far as dark matter is concerned. It seems Kyle Oman has studied the issue further since asking this question and given an answer to a different question about dark matter collapse here, if I'm understanding correctly his answer says that for an ideal fluid with kinetic energy $K$ and gravitational energy $W$, the Jeans equations say that it only "becomes virialized" and stops collapsing when $2K$ becomes approximately equal to $-W$, which meaning it is not virialized (though it does still obey the virial theorem, see Kyle's comment below) when $2K < -W$. And John Baez's derivation did assume the ball of ideal gas is virialized, so his demonstration that collapse of an ideal gas decreases the entropy presumably wouldn't apply to a non-virialized collection of dark matter with $2K < -W$, so I presume this means it could collapse without contradicting the second law, and without the dark matter particles needing to radiate or interact as I suggested in my original answer.

If we want to examine gravitational collapse from a statistical mechanics point of view, we find that there's a tradeoff between the fact that a more spread-out collection of matter has more possible position states, whereas a more concentrated collection has more possible momentum states (because more of the system's potential energy has been converted to kinetic energy and thus the particles have higher average velocity/momentum). And in statistical mechanics, entropy is a function of the total number of states available, with higher entropy = more possible states. It turns out, though, that this tradeoff alone is not enough to explain why gravitational collapse can happen in some systems--the decrease in the number of possible position states when a cloud collapses is actually greater than the increase in the number of momentum states, as derived on this page from physicist John Baez, so if these were the only factors at play the entropy would be lower in the collapsed state than the diffuse state, and gravitational collapses would never occur. However, it turns out that if the collapsing matter can radiate energy away as it collapses, in that case the end state of "more concentrated, hotter matter distribution + outgoing radiation" can have a higher entropy than the initial state of "more spread out matter which hasn't yet radiated", and so this is the key to understanding why gravitational collapse respects the 2nd law of thermodynamics. As explained by Lubos Motl in this answer:

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.

And as explained in this answer by Ted Bunn, this is relevant to why dark matter would "clump" only very weakly (as seen in detailed physical simulations like the ones I have linked to)--dark matter particles would only experience irreversible interactions with other particles very rarely, from either occasional interactions involving the weak nuclear force (which would be infrequent, as with neutrinos which normally pass straight through the Earth, with only about 1 in 10^11 interacting with any of the particles that make up the Earth according to this page) or shedding gravitons:

But it's true that dark matter doesn't seem to have collapsed into very dense structures -- that is, things like stars and planets. Dark matter does cluster, collapsing gravitationally into clumps, but those clumps are much larger and more diffuse than the clumps of ordinary matter we're so familiar with. Why not?

The answer seems to be that dark matter has few ways to dissipate energy. Imagine that you have a diffuse cloud of stuff that starts to collapse under its own weight. If there's no way for it to dissipate its energy, it can't form a stable, dense structure. All the particles will fall in towards the center, but then they'll have so much kinetic energy that they'll pop right back out again. In order to collapse to a dense structure, things need the ability to "cool."

Ordinary atomic matter has various ways of dissipating energy and cooling, such as emitting radiation, which allow it to collapse and not rebound. As far as we can tell, dark matter is weakly interacting: it doesn't emit or absorb radiation, and collisions between dark matter particles are rare. Since it's hard for it to cool, it doesn't form these structures.

Detailed cosmological simulations like the "Illustris simulation" discussed in this article and this one indicate that there is some clustering with dark matter, but it doesn't form very condensed clumps on the scale of stars.

  • $\begingroup$ The densities reached by DM in simulations like Illustris are at least as high as $\sim m_{\rm H}/{\rm cm}^3$, which is denser than much of the interstellar medium of galaxies. Right, it doesn't form planets, but it's still forming structures $>10^6$ times more dense than the background... $\endgroup$ – Kyle Oman Oct 28 '14 at 20:43
  • $\begingroup$ Getting to my point, I'm still left wondering how DM collapsing at all is consistent with the second law, even if it only clusters "weakly". Still a +1 from me, of course. $\endgroup$ – Kyle Oman Oct 28 '14 at 20:44
  • $\begingroup$ @Kyle: Are these cases where it's clustering around galaxies or other baryonic-matter structures? If it's interacting gravitationally with a galaxy, then they're in thermal contact, and the CDM can transfer heat to the baryonic matter. $\endgroup$ – Ben Crowell Oct 28 '14 at 21:03
  • $\begingroup$ @BenCrowell If the CDM model is right, there are many "dark halos" which do not host a galaxy or any substantial mass in gas, but have collapsed. In fact, in the CDM picture, the DM collapses before the gas, and is the reason that the gas collapses at all - it has DM potential wells to collapse into. $\endgroup$ – Kyle Oman Oct 28 '14 at 21:07
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    $\begingroup$ I think I see your point of confusion: virialized and "obeys the virial theorem" are distinct. The virial theorem (the one I'm talking about in my other answer) is $\frac{1}{2}\frac{d^2I}{dt^2}=2K+W+\Sigma$. The statement "$2\langle K\rangle=-\langle W\rangle$" is sometimes also called the virial theorem, but this is in fact a special case of the theorem, when (quasi-)equilibrium is reached. Note that on Baez's page, he assumed virialization throughout (which may be fine for a collisional gas cloud, but is not for collisionless DM). He says he'll check his assumption later, but never does... $\endgroup$ – Kyle Oman May 15 '15 at 9:00

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.

  • $\begingroup$ 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? $\endgroup$ – Kyle Oman Aug 20 '14 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.

  • $\begingroup$ 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 $\endgroup$ – anna v Aug 16 '14 at 3:25
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    $\begingroup$ @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? $\endgroup$ – CuriousOne Aug 16 '14 at 3:33
  • $\begingroup$ 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 $\endgroup$ – anna v Aug 16 '14 at 3:38
  • $\begingroup$ 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? $\endgroup$ – CuriousOne Aug 16 '14 at 3:41
  • $\begingroup$ It will depend on what dark matter is. no? fermions? bosons already? $\endgroup$ – anna v Aug 16 '14 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.

  • $\begingroup$ 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? $\endgroup$ – Kyle Oman Aug 20 '14 at 17:52
  • $\begingroup$ 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? $\endgroup$ – anna v Aug 20 '14 at 18:35
  • $\begingroup$ 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? $\endgroup$ – Kyle Oman Aug 20 '14 at 18:39
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    $\begingroup$ 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 $\endgroup$ – anna v Aug 21 '14 at 4:08
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    $\begingroup$ No, this is wrong. Gravitational radiation is far too weak to be an efficient mechanism for dissipating this energy. $\endgroup$ – Ben Crowell Oct 28 '14 at 19:36

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)

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    $\begingroup$ This isn't an answer. $\endgroup$ – Ben Crowell Oct 28 '14 at 21:01
  • $\begingroup$ @BenCrowell, the point is is it relevant? if your (down)vote refres to that, i would be glad to hear, else it is not accepted $\endgroup$ – Nikos M. Oct 29 '14 at 0:44
  • $\begingroup$ @BenCrowell I flagged it as such and it came back 'disputed' :/ $\endgroup$ – Kyle Oman Oct 29 '14 at 3:42
  • $\begingroup$ @BenCrowell, this is longer than a comment so i added in the form of a post (sth that happens on this site), plus this was posted at a time i could not post comments. In any case either as answer or longer comment the criterion should be the relevance. $\endgroup$ – Nikos M. Oct 29 '14 at 11:05

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