4 ms·

Is this right?: * Although you can make the enveloping sphere as large as you want, the (anti-)equilibration process requires a sphere of some finite radius be

Is this right?: * Although you can make the enveloping sphere as large as you want, the (anti-)equilibration process requires a sphere of some finite radius because if you wait long enough a few stars eventually get launched at escape velocity, and if these actually escaped they would effectively cool the remaining stars. * Therefore, the characteristic time scale for this process (i.e., the timescale on which the average kinetic energy rises substantially) gets longer and longer as the sphere gets larger. * In order for the pressure and average speed of the stars to keep rising, the gravitational potential needs to keep falling, so at least some stars need to get and stay very close. In real life, these turn into black holes, which cuts off the process by limiting the amount of gravitational potential energy that can be unlocked in any given volume with a given mass.

pavel_lishin 2y ago
> In real life, these turn into black holes I think this is right, and I think he explicitly calls out that these calculations were done with Newtonian physics modeling point particles - and we know that those two factors severely limit the application of this to the real-world.
- jessriedel 2y ago
  Right, it wasn't criticism, but the point I added (that I think was not explicit in the article) is this: black holes provide a lower bound on the potential energy, not just an indication that the model is breaking down.
- hnuser123456 2y ago
  Yes, so some of the gravitational potential energy gets converted into gravitational waves instead of flinging some stars further and further out, enabling at least something to stay behind.
drudd 2y ago
I was asked this question on my grad candidacy exam (which was quite a while ago, so my memory is hazy), and I believe stars tend to form binary systems which can halt the runaway gravitational collapse (the potential energy in a hard binary can be a very significant fraction of the total energy of a relatively loosely bound globular cluster).
- jessriedel 2y ago
  In real life this can happen, but I think only by flinging out stars on escape trajectories. That’s the thing explicitly prevented by the reflecting sphere in the model discussed in the post. Basically, the issue is that you can’t end up in a stable equilibrium of binaries (and binaries of binaries) in a bounded phase space because the dynamics are time-reversible. The only way you get coarse-grained irreversible behavior is with an unbounded phase space where there are no recurrences.

pavel_lishin 2y ago

> In real life, these turn into black holes I think this is right, and I think he explicitly calls out that these calculations were done with Newtonian physics modeling point particles - and we know that those two factors severely limit the application of this to the real-world.

jessriedel 2y ago
Right, it wasn't criticism, but the point I added (that I think was not explicit in the article) is this: black holes provide a lower bound on the potential energy, not just an indication that the model is breaking down.
hnuser123456 2y ago
Yes, so some of the gravitational potential energy gets converted into gravitational waves instead of flinging some stars further and further out, enabling at least something to stay behind.

drudd 2y ago

I was asked this question on my grad candidacy exam (which was quite a while ago, so my memory is hazy), and I believe stars tend to form binary systems which can halt the runaway gravitational collapse (the potential energy in a hard binary can be a very significant fraction of the total energy of a relatively loosely bound globular cluster).

jessriedel 2y ago
In real life this can happen, but I think only by flinging out stars on escape trajectories. That’s the thing explicitly prevented by the reflecting sphere in the model discussed in the post. Basically, the issue is that you can’t end up in a stable equilibrium of binaries (and binaries of binaries) in a bounded phase space because the dynamics are time-reversible. The only way you get coarse-grained irreversible behavior is with an unbounded phase space where there are no recurrences.