Assuming a relatively isolated cloud of dust and gas with a well-defined total mass and well-defined net angular momentum. Perhaps also assume that the size (or density) of the initial cloud is reasonably well-defined. Friction and collisions within the cloud tend to reduce the relative motions of nearby particles. The end result is presumably that the cloud ends up as a thin oblate ellipsoid. The field in a uniform-density ellipsoid is linear with distance from the center. So for a thin disk with uniform density and a thickness following an elliptical profile (or for a constant thickness and a density following an elliptical profile), the result is that all the particles rotate coherently in circular orbits around the center all with the same angular velocity.
However, for a given mass and angular momentum, such a flat rotating ellipsoidal structure would not seem to minimize overall energy - including gravitational potential energy. It seems like a large central mass with a single smaller distant orbiting mass would do a much better job.
This question came up in the context of the formation and evolution of a protoplanetary disk. I see comments about 'gravitational instabilities'. Does this mean that local concentrations of mass in an otherwise uniform disk are self-reinforcing, leading ultimately to the formation of lumps/planets?
I'm particularly curious if there is a distinct mathematical solution to the least energy configuration that might finally develop for a cloud with given intial mass and angular momentum (and perhaps some measure of the initial size of the cloud)?


2 Answers 2


To my understanding the physics of the process of contraction of a gas/dust cloud to a star is a highly specialized branch of astrophysics.

I assume that the specialists that create simulations for that know what they are doing, but I haven't encountered articles in which the author sets out to present the findings in this branch of physics to a wider audience.

I assume that the only materatial that exists is articles by specialists, written for specialists.

What I can give is general considerations.
In order for a cloud of gas/dust to contract it must shed a lot of energy. The cloud must lose rotational kinetic energy. For comparison, if a spacecraft is in orbit, and you want to descend, then you have to fire retro-rockets; you have to kill orbital velocity.

For a contracting cloud the only way it can lose kinetic energy is as follows: collisions between particles act as friction, that friction heats those particle up. That is: collisions convert kinetic energy to heat. The more kinetic energy the cloud can get rid of, the more it can contract.

This process of contraction will halt when that rotating cloud loses ability to dissipate kinetic energy to heat.

Example: the rings of Saturn. Saturns rings are very close to circular, which means there is a low probability of collision, and any collisions that do occur will be hits with very low relative velocity. Saturn's ring are long time stable because there is practially no opportunity for the rings to lose kinetic energy.

If a pre-stellar cloud happens to end up in a configuration where all the mass is very close to circular orbit, and very close to all mass orbiting the center of rotation with the same angular velocity, then that is where contraction will halt. Coherent rotation means: practically no opportunity to lose kinetic energy.

Contraction all the way to a star means something must be going on that keeps processes of stirring and churning going.

It could be, this is sheer speculation on my part, that a necessary condition is that two clouds with roughly opposite spin must approach each other. Then, maybe, at the intersection of those two clouds there is always trouble: stirring, churning, where the contribution of each cloud tends to kill the rotational kinetic energy of the other cloud.

Another comparison, to illustrate that contraction requires something special:
In a globular galaxy there is no alignment of any of the stars in that galaxy. The stars of a globular galaxy are gravitationally bound to that galaxy, but the orbits of the individual stars are not ordered in any way. That is, the orbits of the stars of a globular galaxy are so unordered that the galaxy as a whole has negligable internal angular momentum.

My understanding is that globular galaxies can only come into being as a product of a merging event of two galaxies. If two galaxies that merge have galactic angular momenta that are roughly aligned the resulting merged galaxy will have galactic angualar momentum. When unaligned the two galaxies entering the merging event will scramble each other. The resulting globular galaxy is long time stable because the distances between the stars are so vast that the probability of two stars coming close to each other, and producing some gravitational slingshot, is low.


You can only have a gravitational collapse if the gas cloud continuously loses energy through inelastic collisions (which is then radiated off). Otherwise the collapse would soon come to a halt as the lost potential energy would be turned into kinetic energy and a steady state would be reached once the kinetic energy is -1/2 x the potential energy (virial theorem). The fact that there are always some radiative losses due to inelastic collisions means that such a steady state is in fact not reached but the gravitational collapse proceeds to form stars. An additional angular momentum of the cloud does in principle not change anything here, it only means that the collapsed system also has an angular momentum (like our solar system).

More details in this respect can be found at https://www.plasmaphysics.org.uk/research/starformation.htm (this is my own page I would like to add).

  • $\begingroup$ Thanks, nice article! Re: "A gas cloud in a stable equilibrium will therefore necessarily be isothermal.". Wouldn't particles in the outer regions of the cloud be cooler having exchanged kinetic energy for potential energy? $\endgroup$
    – Roger Wood
    Commented Jan 1, 2021 at 16:53
  • $\begingroup$ @RogerWood The whole cloud loses kinetic energy due to inelastic collisions. But when it then collapses in its own gravitational field this is more than regained from the change in potential energy. As shown in the reference, if you somehow remove all kinetic energy from a cloud in equilibrium (E_kin=-E_pot/2), it shrinks to half its initial size but has twice the initial temperature. You could repeat this in principle indefinitely, but at some point the density and temperature becomes so high that no individual atoms (and thus no inelastic collisions) can exist anymore. A star has formed then $\endgroup$
    – Thomas
    Commented Jan 2, 2021 at 13:20
  • $\begingroup$ Isn't "E_kin=-E_pot/2" only true outside the cloud? Inside the cloud the fields and orbital velocities drop but the potential keeps getting deeper. I'll think about this a bit more. It's an interesting topic. Can I assume everything is monatomic initially, or do I need to worry about molecular Hydrogen (5/2)kT too? $\endgroup$
    – Roger Wood
    Commented Jan 2, 2021 at 22:24
  • $\begingroup$ @RogerWood E_kin=-E_pot/2 is the long term average for any atom/molecule in the cloud. It holds thus also for the whole cloud in a state of quasi-static equilibrium. For the sake of the argument, you shouldn't worry about molecules. It is commonly claimed that they provide the cooling mechanism (as atomic transitions from the ground state can't be excited at the initial low temperature), but highly excited atomic states could provide an energy loss mechanism as well, and whilst molecules cease to exist at sufficiently high temperature, highly excited atoms would be present all the way. $\endgroup$
    – Thomas
    Commented Jan 3, 2021 at 15:50
  • $\begingroup$ I hadn't come across the 'virial theorem' before. It's a surprising and very useful result. But it refers to an average over the whole cloud. I still think that, in equilibrium, atoms on the edge of the cloud will be slower/cooler than those in the middle because of the difference in potential energy. The difference for a uniform density cloud is -(1/2)GM/R. So the center of the cloud would be 50% hotter than the edge. $\endgroup$
    – Roger Wood
    Commented Jan 4, 2021 at 8:26

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