How does dark matter collapse?: Entropy considerations 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.
 A: 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.
A: 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. 
A: 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. 
A: 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.
