The Jeans Instability

Linear perturbative analyis on the self-gravitating fluid equations under the assumption of infinite, constant (v_0), static background density (\rho_0) results in the following 2nd order PDE in the density perturbation (\rho_1):

\frac{\partial^2\rho_1}{\partial t^2} - C_s^2 \triangledown^2 \rho_1 - 4\pi G \rho_0 \rho_1 = 0

where C_s is the sound speed.

This is a wave equation, so we propose a solution of the form:

\rho_1(\vec{x},t) = C e^{i(\vec{k} \cdot \vec{x} - \omega t)}

where the wave number k is given by k = 2 \pi /\lambda, with \lambda specifying the perturbation oscillation wave length, and \omega is the angular frequency of oscillation given by \omega = 2 \pi f. Plugging this into the wave equation above gives the following dispersion relation:

\omega ^2 = C_s^2 k^2 - 4 \pi G \rho_0

When

\omega^2 > 0

or

4 \pi G \rho_0 < C_s^2 k^2

\omega is real, and the density perturbation has the form:

\rho_1 = e^{i(\vec{k} \cdot \vec{x})} e^{-i(^+_- \omega t)}

which is oscillatory. Thus the density perturbation propagates through the medium as a sound wave, i.e. pressure gradients are strong enough to support the medium against self-gravitational collapse. On the other hand, when

\omega^2 < 0

or

4 \pi G \rho_0 > C_s^2 k^2

we see that \omega is imaginary. This implies that there is an exponential growth or decay factor multiplying the oscillatory e^{i(\vec{k}\cdot \vec{x})} in the solution for \rho_1. That is,

\rho_1 = e^{i(\vec{k} \cdot \vec{x})} e^{-i(^+_- i \omega t)} =

e^{i(\vec{k} \cdot \vec{x})} e^{^+_- \omega t}

Taking the real part, we have:

\rho_1 = \cos(kx)e^{^+ _-\omega t}

Thus, the perturbation takes on a form that is oscillatory in x and exponential in time.

The Jeans Length

To get the critical length scale (aka Jeans length) for the oscillations, we write the oscillatory condition

C_s^2 k^2 > 4 \pi G \rho_0

Substituting 2 \pi / \lambda for k and solving for \lambda we have

\lambda < \frac{2 \pi C_s}{(4 \pi G \rho_0)^{1/2}}

That is, for wavelengths smaller than 2 \pi C_s / (4 \pi G \rho_0), the perturbation is stable to collapse. You can think of this wavelength as length scales over which the average \rho = \rho_0. The critical length is known as the Jeans length and is given by:

\lambda_J = Cs(\frac{\pi}{G \rho_0})^{1/2}

Unstable Modes

What we'd like to do now is determine, for which frequencies is the perturbation unstable (exponentially growing). In order to determine this we can rewrite \omega into a more intuitive form.

\omega ^2 = -(4 \pi G \rho_0 - C_s^2 k^2) \Rightarrow

\omega = ^+_- [-4 \pi G \rho_0(1-\frac{C_s^2 k^2}{4 \pi G \rho_0})]^{1/2} \Rightarrow

\omega = ^+ _- i[4 \pi G \rho_0 (1 - \frac{C_s^2 k^2}{4 \pi G \rho_0})]^{1/2} \Rightarrow

\omega = ^+ _- i[4 \pi G \rho_0 (1 - \frac{\lambda_J ^2}{\lambda^2})]^{1/2}

Taking the positive root and plugging into \rho_1, we see

\rho_1 \propto e^{-i i [4 \pi G \rho_0 (1-\frac{\lambda_J^2}{\lambda^2})]^{1/2}t}

which gives

\rho_1 = C \cos(kx) e^{[4 \pi G \rho_0 (1-\frac{\lambda_J^2}{\lambda^2})]^{1/2}t}

Thus we have found the condition on \omega which gives unstable perturbation modes. Now, the growth rate of these modes is given by

\Gamma = [4 \pi G \rho_0 (1-\frac{\lambda_J^2}{\lambda^2})]^{1/2}

and so the characteristic timescale for exponential growth, that is the time in which the perturbation increases by a factor of e, is given by

\tau = \frac{1}{\Gamma}

High lambda limit, i.e. pressure-less collapse, or 'uniform collapse'

From

\rho_1 = C \cos(kx) e^{[4 \pi G \rho_0 (1-\frac{\lambda_J^2}{\lambda^2})]^{1/2}t}

it is easy to see that in the limit

\lambda \rightarrow \infty

for finite \lambda_J (i.e. pressure gradients are null/gravity dominates), the characteristic collapse time becomes

\tau = [\frac{1}{4 \pi G \rho0}]^{1/2} ~ \frac{1}{2} t_{ff}

where t_ff is the freefall time for a spherical region of uniform gas.

Writing a module to test self-gravity using the Jeans Instability

I go through the procedure of setting up the problem on this page here.