I think, there is practically everything given to that: many point-like masses, able to exchange energy pseudo-randomly, and far long enough time to reach a thermodynamical equilibrium.

Of course, the large-scale distribution behavior shouldn't calculated here.

Can we consider the galaxies as rotating gas disks, where the atoms of the gas are stars?

  • $\begingroup$ The system is able to eject stars, so even the conditions for the Virial theorem are not fully met, though you can probably just apply it with a patch-up term of some kind. $\endgroup$ – dmckee --- ex-moderator kitten Jan 9 '15 at 23:08
  • $\begingroup$ Yes. Regarding the fact that the intergalactic space isn't full with rogue stars, this ejection shouldn't happen really big rate. Although it could be considered maybe some like an evaporation mechanism, which cools the system. $\endgroup$ – peterh - Reinstate Monica Jan 9 '15 at 23:11
  • $\begingroup$ An important difference is that in gas-poor ellipsoidal systems, energy partition doesn't happen by collision, but through exchange of gravitational energy. This has the interesting consequence that the "temperature" of the distribution can be different along different axes. Gas-rich disk galaxies are different. $\endgroup$ – Thriveth Jan 10 '15 at 14:40
  • $\begingroup$ @Thriveth Thank you, it is very important information! If you converted it to an answer, I were happy to upvote and maybe accept it. $\endgroup$ – peterh - Reinstate Monica Jan 12 '15 at 4:33
  • $\begingroup$ I have an important deadline, maybe I can find the time to write an answer after that. But it is somewhat buried, have not worked with DM kinematics in quite a while. $\endgroup$ – Thriveth Jan 12 '15 at 11:24

If you're interested in the dynamics of stars in a disk, or any other configuration, the equation you want is the collisionless Boltzmann equation. It also applies to the dynamics of dark matter or other "collisionless fluids". Galaxies typically also have a gas component (which is sometimes in a disk), which should be modelled using the usual hydrodynamics.

A "gas" of stars cannot be modelled as a normal fluid because the cross section for collisions between stars is vanishingly small - for instance if you take two star clusters and put them on a collision course, none of the stars really collide and the two clusters pass through each other, probably deforming a bit in the process. This is in contrast to putting two clouds of gas on a collision course - they will decidedly not pass through each other.

This is the collisionless Boltzmann equation (one of the many possible choices of variables and coordinate systems):

$$\frac{\partial f}{\partial t} + \dot{\mathbf{q}}\cdot\frac{\partial f}{\partial \mathbf{q}} + \dot{\mathbf{p}}\cdot\frac{\partial f}{\partial \mathbf{p}} = 0$$

$f=f(\mathbf{q},\mathbf{p},t)$ is called the distribution function, and describes the probability that a particle (in this context, a star) is found at phase space coordinates $(\mathbf{q},\mathbf{p})$ at time $t$. One interesting property of systems described by this equation is that their phase space density is conserved.

This web page, despite some crappy formatting, gives a nice bit of additional detail.

This book is the canonical reference on this topic in the context of astrophysics.

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  • $\begingroup$ Thank you the answer! Maybe I am misunderstanding something, but the arms of the spiral galaxies are rotating (which is essentially a phase change and differs from the rotation period of the stars around the galactic core, it can be maybe even retrograde), which means $f$ isn't conserved? If $f$ is conserved, why you said $f(\underline{q},\underline{p},t)$ and not $f(\underline{q},\underline{p})$? $\endgroup$ – peterh - Reinstate Monica Jan 13 '15 at 14:34
  • $\begingroup$ Next to that: yes I understand it is collisionless. But why really changes it anything? There is a many-body system, with pseudorandom energy interchange. The Boltzman statistics has a very clear, pure mathematical, combinatorics-based deduction, why it wouldn't stay? $\endgroup$ – peterh - Reinstate Monica Jan 13 '15 at 14:40
  • $\begingroup$ ...I think it is better if I ask this second as a new question. $\endgroup$ – peterh - Reinstate Monica Jan 13 '15 at 14:41
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    $\begingroup$ $f$ is a phase space density, and is conserved in time. The simple analogy I know is this: consider a marathon where each runner runs at constant speed. At the start of the race there is high spatial density, and a large spread in velocities. Later in the race, the spatial density is lower since the runners have spread out, but adjacent runners have similar speeds. The phase space density of runners is constant. In this 1+1 dimensional example this is obvious if you plot position vs speed at different times and look at the density of points. This is harder to do in 6D. $\endgroup$ – Kyle Oman Jan 13 '15 at 16:24
  • $\begingroup$ Also, it is a function of $t$ because it is conserved in the lagrangian sense. The phase space density in the vicinity of a given star particle is constant, but not at a fixed coordinate position. $\endgroup$ – Kyle Oman Jan 13 '15 at 16:25

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