# How much mass must an open cluster lose to become unbound?

August 27, 2015

Although there is strong evidence that nearly all stars are born in open clusters, it is clear that open clusters do exist for a very long time. After all, most stars are not seen in open clusters so they must spend a small fraction of their lives in them. It’s also possible to measure the ages of open clusters directly by measuring the masses of stars just starting to evolve off the main sequence and calculating how long it takes stars of those masses to reach that point in their evolution. This typically gives ages no larger than a few hundred Myr. While there are a few open clusters that are over a Gyr old, they are very massive and are the exception rather than the rule.

So we know that open clusters dissociate after a short period of time, but why does this happen? The very brief answer is that open clusters are only barely bound — they consist of a loose collection of gas and stars. When the most massive stars in the cluster go supernova, the shocks from the supernovae drive out much of the gas. This removes some of the mass from the cluster, which unbinds it. But how much mass must be removed to unbind the cluster? Is it plausible that enough mass is still in the gas phase at the time of the supernova to unbind the cluster? To answer this we turn to a brief paper by J. G. Hills from 1980.

## Sudden mass loss from a static cluster

We will follow Hills (1980) and start with the simpler problem of sudden mass loss from a cluster which is in equilibrium. In this case the virial theorem applies and so the velocity dispersion of the stars in the cluster is given by

\[v_0^2 = \frac{G M_0}{2 R_0},\]where \(M_0\) and \(R_0\) are the initial mass and effective radius of the cluster, respectively. (If all stars in the cluster have the same mass, then the effective radius is the harmonic mean of all the distances between the stars.)

Immediately after the gas is driven out of the system, the stars will all have the same velocities they did before, but the total mass of the cluster will be reduced to \(M\). The cluster will be out of equilibrium now, and will have energy

\[E = \frac{1}{2} \left( M v_0^2 - \frac{G M^2}{R_0} \right).\]Once virial equilibrium is regained, the radius of the cluster will have changed, but the energy will be the same:

\[E = -\frac{G M^2}{4 R}.\]This then results in an initial-final radius relationship for the cluster:

\[\frac{R}{R_0} = \frac{M_0 - \Delta M}{M_0 - 2 \Delta M}.\]We see that this ratio diverges if half the mass of the cluster is lost. Thus, in a virialized system half the mass needs to be lost to unbind it.

*An aside: adiabatic mass loss*

Hills (1980) next briefly states the case of adiadabatic mass loss. This isn’t particularly relevant to the question at hand but is good to know, so I’ll just state it here.

Adiabatic mass loss occurs whenever the fractional mass lost is small on the dynamical timescale. This means that the system always remains in virial equilibrium. It is easy to show that the initial-final radius relationship of the cluster is in this case

\[\frac{R}{R_0} = \frac{M_0}{M_0 - \Delta M}.\]To show this, take the initial-final radius relationship from the case of instantaneous mass loss above and substitute \(-dm\) for \(\Delta M\) and \((R + dr) / R\) for \(R/R_0\) and integrate.

In this case, the initial-final radius relationship diverges only for \(\Delta M = M_0\). In other words, the cluster always remains bound no matter how much mass you remove.

## Mass loss prior to virialization

A real open cluster is not likely to be in virial equilibrium by the time supernovae from massive stars drive gas out of the cluster. The cluster will still be contracting and this will affect the amount of mass loss that will be necessary to unbind the cluster.

Suppose that a cluster of constant density, \(\rho_0\), starts out at a radius \(R_0\) and then some time later has collapsed to a radius \(R_1\) when the supernovae go off and drive mass out of the system. What is the energy of the cluster at this point? Let’s start by considering the velocity of the outer shell. Initially its energy is all in potential energy, so we have

\[E = - \frac{4}{3} \pi G \rho_0 R_0^2.\]The energy of the outer shell is conserved, so we have at this later time

\[v(R_1)^2 = \frac{8}{3} \pi G \rho_0 R_0 \left( \frac{R_0}{R_1} - 1 \right).\]What about the velocity of some interior shell? Well, since we are assuming that the density is constant, the mass interior to any shell is \(M \sim r^3\), and since the force on any shell is \(F \sim M / r^2\), we have that the acceleration scales as \(a \sim r\). Since the velocity after some time is just \(v = a t\), we have just \(v \sim r\). Thus we can scale our result above to arbitrary radii:

\[v(r)^2 = \frac{8}{3} \pi G \rho_0 R_0 \left( \frac{r}{R_1} \right)^2 \left( \frac{R_0}{R_1} - 1 \right).\]Integrating over the entire shell, we can calculate the kinetic energy of the entire cluster at this later time to be

\[T_0 = \frac{3 G M_0^2}{5 R_0} \left( \frac{R_0}{R_1} - 1 \right).\]The mean-squared velocity of the stars in the cluster at the time of mass loss, \(\left< v^2 \right> = 2T / M_0\) is then

\[\left< v^2 \right> = \frac{6 G M_0}{5 R_0} \left( \frac{R_0}{R_1} - 1 \right).\]After the mass loss takes place, the mean-squared velocity of the stars remains the same, but the kinetic energy is now

\[T = \frac{1}{2} M \left< v_{\infty}^2 \right>,\]and the potential energy is (this can be looked up in a textbook or on Wikipedia):

\[U = - \frac{3 G M^2}{5 R_1}.\]Eventually the cluster will come into virial equilibrium with some effective radius \(R_f\), such that the total energy is half the potential energy:

\[E = - \frac{3 G M^2}{10 R_f}.\]We can now use these equations to relate the final radius, \(R\), with the initial radius, \(R_0\), to find:

\[\frac{R_f}{R_0} = \frac{1}{2} \left[ \frac{M_0 - \Delta M}{M_0 - \Delta M (R_0 / R_1)} \right].\]An unbound cluster has \(R_f \to \infty\), so from this we can easily see how much mass loss is necessary to unbind the cluster just by setting the denominator equal to zero:

\[\frac{\Delta M}{M_0} = \frac{R_1}{R_0}.\]This means that the fractional amount of mass loss necessary to unbind the cluster is exactly equal to the fractional change in radius the cluster has undergone. This then raises the question:

## How much does the cluster radius change?

The basic picture of the formation of a star cluster is that the protocluster begins as a molecular cloud with some radius and average density, \(\left< \rho_0 \right>\), which then shrinks under its own gravitational influence. However, the cloud is not of uniform density — some pockets of the cloud will be somewhat denser than others and will therefore collapse faster. To determine how much the cluster radius changes we must estimate by how much the cluster has collapsed at the time the densest pockets have formed stars (which, for the purposes of this estimate is the time that these dense pockets have collapsed to zero radius).

We’ll begin by writing down the equation of motion for some shell that starts at radius \(r_0\):

\[\frac{ d^2 r}{dt^2} = - \frac{G M(r_0)}{r^2} = - \frac{4 \pi G \left< \rho_0 \right> r_0^3}{3 r^2}.\]If we multiply both sides by \(dr/dt\), we can integrate to find

\[\frac{1}{r_0} \frac{dr}{dt} = - \sqrt{\frac{8}{3} \pi G \left< \rho_0 \right> \left( \frac{r_0}{r} - 1 \right)}.\]This is a tricky differential equation to solve, but if we make the clever substitution \(r/r_0 = \cos^2 \beta\), we can rewrite it as

\[\frac{d \beta}{dt} = \frac{1}{2} \sqrt{ \frac{8}{3} \pi G \left< \rho_0 \right>} \sec^2 \beta,\]which is then easily integrated to yield the collapse equation:

\[\beta + \frac{1}{2} \sin 2 \beta = t \sqrt{ \frac{8}{3} \pi G \left< \rho_0 \right>}.\]Now suppose that some pocket in the protocluster began with a slight overdensity, \(\rho^{\prime}\). The time this pocket to collapse to zero radius can be found by setting \(\beta = \pi / 2\) and solving for \(t\):

\[t_c = \sqrt{ \frac{3 \pi}{32 G \rho^{\prime}}}.\]We can then write the collapse equation in terms of \(t_c\) and radii:

\[\frac{R_1}{R_0} + \sqrt{\frac{R_1}{R_0} - \left( \frac{R_1}{R_0} \right)^3} = \frac{\pi}{2} \sqrt{ \frac{\left< \rho_0 \right>}{\rho^{\prime}}}.\]So in order to estimate the radius of the cluster at the time the most massive stars form (i.e., \(R_1/R_0\)), we must estimate \(\left< \rho_0 \right> / \rho^{\prime}\). If we assume that the molecular cloud is rather homogeneous, and the greatest density fluctuations are only 10%, then we can numerically solve the above equation to find \(R_1/R_0 \approx 0.2\). The more homogeneous the cloud begins, the smaller \(R_1/R_0\) is. It is therefore plausible that a typical protocluster will have \(R_0 / R_1 \sim 10\) or more. If we use this estimate with our result from the previous section, we find that the cluster needs to lose only 10% of its mass to become unbound.

## Why does so little mass need to be lost?

Let us picture the process of virialization. We begin with a cluster of stars, all stationary and at a very large distance. Due to their mutual gravitational attraction, they begin to fall to the center of the cluster and pick up speed. As they pass through the center of the cluster, they have strong gravitational interactions with each other. This process transfers energy among the stars and redirects their trajectories. However, the stars generally have enough velocity that they make it out to nearly the same distance that they started at roughly the same time. Over a long enough time, the strong gravitational encounters at the center of the cluster serve to transfer enough energy between stars that the trajectories of the stars become sufficiently randomized that as some stars are at apocenter, other stars are passing through the center of the cluster and the cluster has become virialized.

Now, if the cluster loses mass at the very beginning of its life when the stars are at a very large distance it is not going to affect this process. The stars will pass through the center with lower speed, but qualitatively nothing else will change. This is because when the mass is lost from the system, it takes the gravitational potential energy it had away with it.

If, however, the mass loss takes place as the stars are passing through the center of the cluster the situation is different. At this point, the gravitational potential energy from the mass that is lost has been transferred to the mass that remains in the form of kinetic energy. In other words, the extra mass has increased the speed of the stars as they pass through the center of the cluster. When the mass is then lost, it is unable to contribute to the gravitational attraction that slows the stars down enough to keep them bound. The closer the stars are to the center of the cluster when mass is lost, the more mass needs to remain to keep the cluster bound since nearly all of the energy is in kinetic energy at this point. Since the cluster will have collapsed by quite a bit by the time the first supernovae go off if it began relatively homogeneous, this means that only a modest amount of mass needs to be lost to unbind the cluster.

## What is the velocity of the unbound stars?

If the mass loss succeeds in unbinding the cluster, how fast do the stars escape? Conservation of energy states that after mass loss we have

\[\frac{1}{2} M \left< v_{\infty}^2 \right> = \frac{1}{2} M \left< v^2 \right> - \frac{3}{5} \frac{G M^2}{R_1},\]which can be rewritten in terms of the velocity dispersion that the cluster would have after virialization if mass loss had not occurred, \(\left< v_c^2 \right>\), as

\[\left< v_{\infty}^2 \right>^{1/2} = \left< v_c^2 \right>^{1/2} \sqrt{ \left( \frac{R_0}{R_1} \right) \left( \frac{ \Delta M}{M_0} \right) - 1}.\]So for reasonable amounts of mass loss the expansion velocity will be only a few times larger than the velocity dispersion of the virialized cluster. This means that for a massive O star with a lifetime of a few tens of millions of years and an escape velocity of a few km / s, the supernova will take place of order 100 pc away from its birthsite!

## A postscript: What about magnetic fields?

We have been working with a very simplified model of a protocluster. A realistic protocluster will have magnetic fields which will affect the evolution of the cluster. In particular, as the cluster collapses, the magnetic fields threading the cluster will become more compact, leading to a higher magnetic energy density. This will result in an additional source of pressure which will serve to resist the collapse of the cluster. Thus, although we estimated that the cluster would have collapsed to some fraction of its radius, \(R_1 / R_0\) at the time of mass loss, the cluster will actually have only collapsed to some larger fraction of its radius \(R^{\prime} / R_0\) (where \(R^{\prime} > R_1\)) due to the extra magnetic pressure. The exact amount by which \(R^{\prime}\) is larger than \(R_1\) will depend on how much energy is in magnetic fields compared to gravitational potential energy.

Since the energy density in magnetic fields in the ISM is comparable to the thermal energy density (and pretty much everything else, as it happens), we can a priori guess that in a protocluster with strong magnetic fields, the magnetic fields will have an order unity effect on the evolution of the cluster. So, for the cluster above where we estimated that only 10% of mass loss was necessary to unbind the cluster, it might be more like 15% or 20%. The more careful analysis of Hills (1980) reveals that this guess appears to be correct.