I understand that dark matter does not collapse into dense objects like stars apparently because it is non-interacting or radiating and thus cannot lose energy as it collapses. However why then does it form galactic halos? Isn't that also an example of gravitational collapse?
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3$\begingroup$ Thanks for the nice question, it's well timed in that answering it has been useful in my efforts to study for my PhD qualifying/comprehensive/candidacy exam coming up next week :) $\endgroup$– Kyle OmanCommented Apr 8, 2015 at 22:07
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1 Answer
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The answer comes from the virial theorem, which can be derived from the Jeans equations, which are the equivalent of the Euler equations of fluid dynamics for collisionless particles (i.e., dark matter). Incidentally, the virial theorem is also valid for an ideal fluid. For a derivation see Mo, van den Bosch & White 2010 (or I'm sure many other texts). The theorem is:
$$\frac{1}{2}\frac{{\rm d}^2I}{{\rm d}t^2} = 2K + W + \Sigma$$
$I$ is the moment of inertia, $K$ is the kinetic energy of the system, $\Sigma$ is the work done by any external pressure and $W$ is the gravitational energy of the system (if external masses can be ignored in the calculation of the potential).
If $\Sigma$ is negligible (as it is in the collapse of DM haloes), then a system which has $2K < -W$ will have a dynamical evolution that drives an increase in $I$, or in other words the system contracts. Collapse halts and a quasi-stable structure results when $2K\sim-W$.
To sum that up in somewhat less technical terms, the absence of dissipation (e.g. radiative cooling or collisions between particles) does not mean that collapse cannot occur. The dynamics of a collisionless system are described by the Jeans equations, and these equations allow for collapse until virialization occurs.
The difference with gas collapsing into a star is that radiation can carry away energy, so the system can dissipate $K$ and continue to collapse for longer. In the case of a star, collapse continues until pressure support is sufficient to halt it.
answered Apr 8, 2015 at 22:01
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2$\begingroup$ You should know - was that you on arXIv this morning? $\endgroup$– ProfRobCommented Apr 8, 2015 at 22:06
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1$\begingroup$ @RobJeffries Yup, that was me :) It's a small world! My reaction was indeed "oh I should really know this!", though I did end up having to reach for a book. $\endgroup$ Commented Apr 8, 2015 at 22:08
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8$\begingroup$ Nice answer. Just wanted to add that there are additional relaxation processes that help a halo collapse. For instance, some particles are ejected from the system at the expense of reducing the energy of other particles. Also, there's Landau damping, where particles that overtake a density wave (from a disturbance of the system e.g. in case of mergers) with a speed comparable to that of the wave will have a net transfer of energy to the wave. $\endgroup$– pelaCommented Apr 8, 2015 at 22:46
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$\begingroup$ "The dynamics of a collisionless system are described by the Jeans equations, and these equations allow for collapse until virialization occurs." Just to clarify your wording, are you saying here that when $K < -2W$, the system obeys the Jeans equations but does not obey the virial theorem, and that "virialization occurs" when $K = -2W$? Is this why your answer is compatible with John Baez's derivation here, in which he assumes a ball of ideal gas obeys the virial theorem and shows a decrease in volume would always mean a decrease in entropy? $\endgroup$ Commented May 14, 2015 at 19:54
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$\begingroup$ @Hypnosifl I'm not sure what you're asking... I'm arguing that when $K<-2W$ the system obeys both the Jeans equations and the virial theorem, and indeed it is by the virial theorem that we can most easily see that the DM must collapse. I would agree that "virialization occurs" when $K=-2W$, but this is distinct from obeying the virial theorem, which is obeyed also when $K\neq -2W$, and drives the evolution of $I$ until virialization occurs. $\endgroup$ Commented May 15, 2015 at 8:46