Gross "temperature" of a globular cluster

Question

Globular clusters can be very large, which means we can do statistics about the stars in them. And that means we can try matching their star-as-particle potential/kinetic energy distribution against a Boltzmann distribution, which might mean that a suitable globular cluster has a temperature; however a quick search on the internet seems to evoke only stellar surface temperatures, even though I'm sure the preceding is not at all original.

Can anyone quote off-hand typical cluster temperatures or cite such a calculation in a freely-available paper?

This related question was a simple yes/no, or at least its answers are; here star systems come up, but temperature is never mentioned.

While the concept of temperature might be of questionable usefulness in this case, you might find discussions about so-called "binary heating" to be interesting. — kleingordon, Apr 28 '16 at 07:09

ProfRob · Accepted Answer · 2017-05-10T19:10:09.687

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A typical velocity dispersion in a globular cluster is 10 km/s. For a typical 1 solar mass subgiant in an old globular, then equating the kinetic energy to $3kT/2$, we get $T = 5\times 10^{60}$ K.

Doesn't seem that helpful really...

The concept of temperature is only ever applied in a relative sense - i.e. some component is hotter than another. Can't say I've ever seen absolute temperatures used. An example would be the concept of a negative heat capacity, whereby if energy is lost from a cluster by stellar evaporation (or by the hardening of a binary system), the stars gain kinetic energy. i.e Energy is lost but the "temperature" increases. Exactly the same thing happens in a contracting gas cloud or star.

In response to discussion in the comments - globular clusters are very old, usually many times their two-body relaxation timescales. They come into a (virial) equilibrium, where the speed distribution of the stars should be approximately Maxwellian. Yes, the fast moving tail will escape, just like fast moving molecules escape the Earth's atmosphere. But the rate of escape is roughly 1% per relaxation timescale. There is plenty of time for the pseudo-temperature to gradually adjust to such a slow change, meaning that the concept of a temperature can still be used.

Below I show the 1D radial velocity distributions seen in the core of globular cluster M4 (taken from Sommariva et al. 2009). The data are split into bins according to the position in a colour-magnitude diagram (a proxy for stellar mass). The blue lines are Gaussian fits (what you would expect for a Maxwellian velocity distribution) and they are pretty good.The instrumental resolution is of order 0.2 km/s for these observations, so that is a negligible contributor, but there will be some outliers due to binary motion.

The main effect of stellar escape is that it invalidates the concept of a global temperature (for a similar reason). Clusters cannot be isothermal. The velocity dispersion (and pseudo temperature) must (and is measured to) decrease with radius, otherwise the stars at large radii would escape (see below; the 1D velocity dispersion vs radius for a globular cluster $\omega$ Cen, from Scarpa et al. 2003).

edited May 10 '17 at 19:10

answered Apr 27 '16 at 22:07

ProfRob

130,455

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Not compared to a mesoscopic system, but when compared to other clusters? – CuriousOne Apr 27 '16 at 22:11
@CuriousOne: the only way that it would be useful to compare them is to see clusters collide and have some sort of energy you could get out... Which, come to think of it, might actually be physically relevant. – Zo the Relativist Apr 27 '16 at 22:13
@JerrySchirmer: Can we assign something like a temperature to the cluster's galactic environment? – CuriousOne Apr 27 '16 at 22:34
@CuriousOne: sure, but it doesn't matter if there can't be a heat exchange. – Zo the Relativist Apr 27 '16 at 22:36
@JerrySchirmer: That's the question... you are probably right that it's just not close enough to some sort of equilibrium physics. – CuriousOne Apr 27 '16 at 22:42
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though it seems to me, we really want to count how many 1sol stars are moving through their clusters about 10 km/s vs how many about 10-20 km/s ... – Jesse C. McKeown Apr 27 '16 at 22:59
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@JesseC.McKeown The distribution is approximately Maxwellian. – ProfRob Apr 27 '16 at 23:06
well, that is encouraging – Jesse C. McKeown Apr 28 '16 at 02:17
You have seen absolute temperatures, they're just expressed as velocity dispersions in km/s, instead of temperature in K. – Kyle Oman Apr 28 '16 at 11:48
Because it stays closest to the spirit of the question (not that the others aren't informative and having their own depth) and there's something seductive about all those pictures: Accepted. – Jesse C. McKeown Apr 28 '16 at 13:25
So I asked wolfram alpha what's a solar mass, measured in protons, and it said about 10^57; so 5 x 10^60 would be about 5000K/proton, which is a bit cooler than the surface of the sun. – Jesse C. McKeown Jun 21 '16 at 17:48

score 3 · Answer 2 · 2016-04-28T10:47:43.720

The important point here is that there is no thermodynamic limit for gravitating systems, and thus there is no well-defined temperature.

This is, perhaps, not a completely intuitive result, but it comes from work on the stability of matter. This is not as glamorous as it sounds, but revolves around the need to show that the energy of matter is an extensive quantity - one that scales like the number of particles. It's easy to see that, if it's not, very bad things happen: either you can't make bulk matter at all, or you can but (likely enormous amounts of) energy is released in doing so.

It turns out that stability depends on everything you've got: the inverse-square nature of EM interactions, neutrality, QM and Fermi-Dirac statistics as well as, probably, other things I have forgotten. Even then it's hard to show: I think it was first shown by Dyson but there has been a much less hairy (but still reasonably hairy) derivation of the same result since, perhaps by Lieb & perhaps Thirring (and others?) which is the one to look for.

To demonstrate stability you merely need bounds on the energy above and below by $N$: to demonstrate the thermodynamic limit you need to show that there actually is a limit. That's also been done for normal matter.

Well, gravitating systems are missing a bunch of the requirements for all this: they are classical, gravity is always attractive and so on. And the result of all this is that they are not stable in the sense that the energy of the system is not an extensive quantity. That in turn means that there is no well-defined thermodynamic limit for these systems: temperature makes no sense there.

score 1 · Answer 3 · answered Apr 27 '16 at 21:58

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Temperature is not useful concept for describing clusters of stars or other gravitational systems, because such systems are not in the realm described by thermodynamics. There is no way to set up thermodynamic equilibrium - globular clusters partly evaporate and core implodes. Also the velocity distribution can't be Maxwell-Boltzmannian, because very fast stars would quickly run away from the system, leaving only the slow ones. Such systems tend to evolve into virial quasi-equilibrium, where most speeds are lower than escape speed from the system and rate of loss of stars is low.

answered Apr 27 '16 at 21:58

Ján Lalinský

37,229

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The second point is true of the Earth's atmosphere, though, and one would surely say that it's ridiculous to say that the atmosphere doesn't have a temperature. – Zo the Relativist Apr 27 '16 at 21:59
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While a globular cluster does not have a long term stable configuration, in the short term they do have statistical properties that should make a near-equilibrium a useful physical approximation. – CuriousOne Apr 27 '16 at 22:02
you may well be on to something; while I appreciate Jerry running to the defense of the premise, I also note that Earth's atmosphere is something between equilibrium and adiabatic, depending on Stuff (as is the Sun); "temperature" may have interesting localization properties. – Jesse C. McKeown Apr 27 '16 at 23:01
The ages of globular clusters are of order 10 billion years. During that time they will have lost some fraction (estimated to be around half in most cases) of their stars. Clearly they are in a pseudo-equilibrium state, just like the Earth's atmosphere is. In both cases, the speed distribution is close-to Maxwellian. – ProfRob Apr 28 '16 at 06:36
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The phenomenon of escape explains why a cluster cannot be isothermal. It must get cooler at large radii. – ProfRob Apr 28 '16 at 06:49
GCs are in quasi-equilibrium, so you can define a "temperature", usually expressed as a velocity dispersion. – Kyle Oman Apr 28 '16 at 11:47
Yes, Kyle, that was Rob's own contributed Answer (which is, so far, winning) – Jesse C. McKeown Apr 28 '16 at 13:23
@JerrySchirmer, atmosphere as a whole does not have temperature in the sense of thermodynamics. I think you meant local gas temperature, which can be defined as measure of particle speeds, but that is a different concept. This can indeed be introduced also for stars in a cluster, but the distribution of velocities will be somewhat different from Maxwellian (there won't be high speed tail since in virial equilibrium those stars have already escaped the system). – Ján Lalinský Apr 28 '16 at 22:46

Gross "temperature" of a globular cluster

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