The Jeans mass

I trust you have already encountered the concept of Jeans mass when studying stars. Let me just phrase the same thing slightly differently, starting with the difference between stable and unstable equilibrium.

When a system is in equilibrium, forces balance. In a stable equilibrium, small changes remain small, whereas in an unstable equilibrium, small changes are amplified and grow.

For example, consider small perturbations in an infinite, self-gravitating fluid, with density $\rho$ and temperature $T$. If the scale $\lambda$ of the perturbation is small, then you know what happens: the perturbation will be a sound wave. This is why you can hear what I am saying. But what happens for perturbations on larger scales? I'll explain why eventually gravity will become dominant.

Consider the material inside one perturbation length $\lambda$. The thermal energy $K$ enclosed within it, is of order $K\sim M\hbox{${\rm k}_{\sc\rm B}$} T$, where $M=\rho\lambda^3$ is the mass within the perturbation (I'm neglecting constants of order unity, like the $4\pi/3$ and such - we'll do better later). The gravitational energy $U$ within the perturbation, is of order $U\sim G M^2/\lambda\sim G M\rho\lambda^2$. For sufficiently small $\lambda$, the gravity energy is small with respect to the thermal energy, and can be neglected: these are the usual sound waves. But for larger $\lambda$, the ratio $U/K\sim \lambda^2$ gets bigger, and so eventually, gravity will dominate! The Jeans mass is the critical mass above which gravity dominates. For perturbations below the Jeans mass, pressure forces dominate, and so the perturbation will re-expand when being compresses. But for perturbation more massive than the Jeans mass, gravity dominates, and so a perturbation will collapse even further when compressed, leading to run-away collapse.

Now the proper way to derive the criterion where gravity dominates requires some more mathematics that I don't want to go into. What you should do, is compute the dispersion relation for sounds waves, i.e. the relation between the wave-length $\lambda$ and the sound speed $c_s$. For small wavelengths, gravity can be neglected, and $c_s$ is independent of $\lambda$. However, as I explained, for larger and larger $\lambda$, eventually gravity becomes important and $c_s$ starts to depend on $\lambda$, and for the critical wave-length $\lambda_J$, the sound speed becomes zero. Now, gravity dominates, and as the wave stalls its amplitude will grow, and the perturbation will collapse. I hope you will have the chance to investigate this nice problem in more detail somewhere else.

Here we will do a different derivation that follows CO.^4.8 Putting in all the constants, we find that $K$ and $U$ are given by

$$K = M{\hbox{${\rm k}_{\sc\rm B}$}T\over (\gamma-1)\mu{\hbox{m$_{\rm H}$}}}$$

$$U = -{3\over 5} {GM^2\over\lambda} .$$ (4.5)

Here, $\hbox{${\rm k}_{\sc\rm B}$}$ is Boltzmann's constant, $\gamma$ the ratio of specific heats of the gas (e.g. $\gamma=5/3$ for a mono-atomic gas), and $\mu$ the mean molecular weight ($\mu=1$ for a pure neutral hydrogen gas, $\mu=0.5$ is the hydrogen is fully ionised). $K$ is the product of the mass of the cloud, $M$, with the thermal energy per unit mass, $u=(\hbox{${\rm k}_{\sc\rm B}$}T)/((\gamma-1)\mu{\hbox{m$_{\rm H}$}})$. The potential energy $U\sim GM^2/\lambda$. The factor in front, $3/5$ is appropriate for a spherical, homogeneous density perturbation. Now, in virial equilibrium, $2K=\vert U\vert$, So whenever $\vert U\vert$ is larger, gravity dominates, and so the critical length $\lambda_J$ is whenever

$$2K=\vert U\vert .$$ (4.6)

Now we can use $M=(4\pi/3)\rho\lambda^3$ in Eqs.(4.5+7.1) to obtain the critical Jeans length $\lambda_J$, and the Jeans mass $M_J$ as

$$\lambda_J = \left({5\hbox{${\rm k}_{\sc\rm B}$}T\over 2\pi(\gamma-1)\mu{\hbox{m$_{\rm H}$}}\rho G}\right)^{1/2}$$ (4.7)

$$M_J \equiv {4\pi\over 4}\rho\lambda_J^3=\left({10\hbox{${\rm k}_{\sc\rm B}$}... ...mu{\hbox{m$_{\rm H}$}}G}\right)^{3/2} \left({3\over 4\pi\rho}\right)^{1/2} .$$ (4.8)

So note that in a homogeneous fluid, the Jeans mass is the mass contained in a volume with radius the Jeans length. The latter is such that perturbations larger than the Jeans length will collapse under their own gravity.

Fragmentation A nice application of this concept is to investigate what happens during collapse of a cloud. Let's say that the Jeans mass is given by $M_J=A T^{3/2}/\rho^{1/2}$, where I've assembled all constants into a new constant $A$. Now assume you have a cloud with mass $M=M_J$ that starts to collapse, and hence $\rho$ increases. In general $T$ will rise as well, and so $M_J$ will change. Now suppose $M_J$ decreases to $M_J'$. Following our earlier discussion, this would mean that, if there were smaller perturbations within the cloud already (substructure), then those with masses larger than $M_J'$ will start collapsing on their own - the cloud may fragment. Given our expression for the Jeans mass, $M_J'< M_J$ will happen whenever $T^{3/2}$ increases slower than $\rho^{1/2}$. Let's parametrise the dependence of $T$ on $\rho$ as $T\propto\rho^{\gamma-1}$. Then $M_J'< M_J$ requires $\gamma<4/3$. For adiabatic, mono-atomic gas, $\gamma=5/3$ and you don't expect fragmentation. But if radiative cooling can keep the gas isothermal, $\gamma=1$ and you expect fragmentation.

So you've learnt how important cooling is for the formation of stars. After all, the mass of the GMCs in which stars form, is far bigger than stellar masses, and so we need to understand why stars are so much smaller than GMCs.