Globular clusters like Omega Centauri certainly don't seem to be very coplanar at all.

In other words, why doesn't the explanation at Why are our planets in the solar system all on the same disc/plane/layer? (quoted below) apply here?

We haven't ironed out all the details about how planets form, but they almost certainly form from a disk of material around a young star. Because the disk lies in a single plane, the planets are broadly in that plane too.

But I'm just deferring the question. Why should a disk form around a young star? While the star is forming, there's a lot of gas and dust falling onto it. This material has angular momentum, so it swirls around the central object (i.e. the star) and the flow collides with itself. The collisions cancel out the angular momentum in what becomes the vertical direction and smear the material out in the horizontal direction, leading to a disk. Eventually, this disk fragments and forms planets. Like I said, the details aren't well understood, but we're pretty sure about the disk part, and that's why the planets are co-planar.


2 Answers 2


Because the stars each form from their own region of the cloud. Each star forming event is separate* from the other stars that are not in the same orbital plane. All the stars are not in a disk like the planets are. This PDF (OBSERVATIONS AND THEORY OF STAR CLUSTER FORMATION) may give you the answers you are looking for.

Furthermore, the Solar System is dominated by one single mass, the Sun, which means that to most planets, the solar system is approximately a two-body problem. This is not the case in a globular cluster, in which the stars are all of more or less comparable size. N-body systems are notoriously chaotic, so even if the stars had started out in ordered motion (which they haven't), their orbits would be highly unstable, and the cluster would quickly be relaxed into a virial system.

*notwithstanding multiple star systems that do form around each other, but this only applies to a limited number of stars compared to the millions in the cluster.

  • 1
    $\begingroup$ @Thriveth Your edit is probably too radical as it changes the essential point of the OP. It would make a good subsidiary answer, though. $\endgroup$ Jun 4, 2013 at 22:40
  • $\begingroup$ @EmilioPisanty OK, maybe I should split it off into a separate answer. Will get back to it. $\endgroup$
    – Thriveth
    Jun 5, 2013 at 9:36

The crucial concept behind the formation of planar orbital systems like spiral galaxies and our solar system is that the transition from a vague cloud with some orbital motion to a planar system dissipates some of the kinetic energy in the orbital motion (which it can, by turning it into heat) but cannot get rid of the angular momentum.

As a cloud of gas and dust collapses, the gas can "fall" into the plane along the direction of the cloud's angular momentum, and stop its fall by turning its gravitational potential energy into interactions and thus heat in the neighbouring gas and dust; the linear momentum of the fall is cancelled by symmetric contributions from the other side of the plane.

On the other hand, conservation of angular momentum stops gas falling in along the orbital plane, because the particles will either be flung back out after apoapsis, or will fling out other particles when they try to stop via interactions. There is no helping symmetry here (unless $\mathrm L=\mathrm 0$, in which case no disk will form).

SO, to answer your question, a globular cluster is a many-body system but it doesn't have nearly the number of bodies that go, from planets to hydrogen molecules, into creating a planar planetary system. Close encounters are probably quite common, but I would imagine three-way scattering is pretty rare and I would be very surprised if inelastic fly-byes are common. Given that, you're waiting on the stellar motions themselves to thermalize, which I imagine will take ages. I don't know if the stellar kinetics are in thermal equilibrium, but I doubt it; if they are it would make a great answer to this old question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.