1
$\begingroup$

I just need to confirm or refute something I'm believing since a very long time, maybe without real evidence.

Can we distinguish stellar clusters by their total mechanical energy ?

Is it true that all "open" star clusters have a positive energy (i.e. more kinetic energy that potential energy), while all globular clusters are "closed" systems with negative total energy (i.e less kinetic energy than total potential energy) ?

So for example, the Peiades M45 are "open" : they have $E > 0$, while M4, M53 and M80 (globulars) have $E < 0$ ? Is this true ?

$\endgroup$
  • 1
    $\begingroup$ The virial state of almost all open clusters remains to be determined because their velocity dispersions are very small. Watch this space for Gaia DR2 results. $\endgroup$ – Rob Jeffries Jun 16 '17 at 21:45
1
$\begingroup$

Stars in an open star cluster are - according to wikipedia - "loosly bound by mutual gravity". Whereas a globular cluster is more dense and will be "less loosly" bound - to use the same terminology.

Since the stars in the open star cluster still are gravitationally bound, the total energy $E$ must be less than zero. This criterium that $E<0$ is always correct for (gravitationally or otherwise) bound systems for exactly the reason you stated, the kinetic energy is 'less positive' than the potential is negative. Otherwise the particles (stars in this case) would break free from the gravitational interactions with the other particles - and that's the complete opposite of a bound system.

However, the fate of most open star clusters is that it will fall apart. This can be due to

  • mass loss of the cluster or,
  • internal processes like a slingshot due to close encounters between constituent stars.

From the get-go these open clusters are quite unstable, so $E$ will be close to zero, but still negative ofcourse because it's a bound system. But it is instable enough for the following to happen. Young, bright stars radiate a lot of energy, and this means that the radiation pressure on the surronding cloud from which these stars are born is high. The light from these young stars can blow large amounts of mass in the form of this gas and dust out of the system - which has an influence on the total energy of the system! Because the mass of the system drops, the total energy rises (read: becomes less negative) - wich makes it easier for the constituent stars to reach escape velocity. If you don't like thinking in terms of energy, the escape velocity is $v_e = \sqrt{2GM/r}$, so $v_e \propto M $. If $M$ drops, so does the escape velocity.

The second bullet speaks for itself, I think.

You can deduce that $E$ has to be negative from a conservation viewpoint. When the stars in the cluster are born from the same cloud, this cloud already is gravitationally bound, so $E<0$, and unless some energy is lost at some point, the cloud will stay bound. But like I said and as you said, most of these clusters don't stay bound for very long, and eventually will evaporate.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Are you sure that stars in an "open" cluster are still bounded ? AFAIK, the stars in M45 are slowly migrating away, like our Sun after its formation, so this implies $K > |U|$ and then $E > 0$. $\endgroup$ – Cham Jun 16 '17 at 18:41
  • $\begingroup$ You are quite right, the Pleiades' eventual fate will be that it will fall apart, like - according to what wikipedia has to say on M45 - most open star clusters. There are several very interesting reasons on why this happens and I'll edit them in my answer so it doens't clutter the comments section. $\endgroup$ – PaleBlueDot Jun 16 '17 at 18:48
  • 1
    $\begingroup$ This is all correct but there is no universally acknowledged definition of a cluster. Many things which are called clusters may not actually be gravitationally bound and are in the process of disintegrating. $\endgroup$ – Rob Jeffries Jul 17 '17 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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