## FANDOM

153 Pages

Thermal eccentricity distribution refers to the distribution of eccentricities of a population of binary stars, which is "thermalized" in some sense. This basic result in dynamics was derived by J. H. Jeans in his paper from 1919.

## Problem set up Edit

Consider a population of binary systems, given sufficient time to achieve Boltzmann distribution in energy. This scenario is somewhat artificial, since interactions between binaries will likely to dissociate some pairs, while others will form triples. In this case, there could not exist a thermal distribution of binaries. The last paragraph will describe the consequences of this simplification.

## Derivation Edit

The energy distribution:

$f\sim exp(-E/T)$

Where $E$ is the energy of a binary system, and $T$ is some temperature measure.

$E=\frac{1}{2} \mu v^2 - GM\mu/r$

Where $M$ and $\mu$ are the total and reduced mass respectively. $v$ is the relative velocity, and $r$ is the distance between the components. We write $v^2 = v_\perp^2 + v_\parallel^2$, where $v_\parallel$ is the velocity component parallel to the rotation axis, and $v_\perp$ is the perpendicular velocity component. In a phase space volume $d\tilde{V}$, the number of systems is:

$f d \tilde{V} \sim exp(-\frac{1}{T} (\frac{1}{2} \mu v^2 - GM\mu/r) )v_\perp dv_\perp d\phi dv_\parallel r^2 dr d\Omega$

Where $d\Omega$ is an infinitesimal solid angle. We wrote the perpendicular velocity component in polar coordinates. The volume element related to perpendicular velocity is written as $v_\perp dv_\perp d\phi$.

We integrate over $\Omega$ and $\phi$, which leaves us with a factor of $8\pi^2$.

We switch to new coordinates, related to the specific energy and angular momentum. It turns out that working with the inverse angular momentum simplifies the derivation. The original coordinates describing the system are $r, v_\parallel$ and $v_\perp$.

In therms of the physical coordinates:

$E = \frac{1}{2}(v_\parallel^2+v_\perp^2) - GM/r$

$K = \frac{1}{rv_\perp}$

And:

$1/r = GMK^2(1+(A/GM\mu^2)cos\theta)$

Where $A=GM\mu^2e$ is the magnitude of the Runge-Lenz vector, and $cos\theta$ is the true anomaly, specifying the phase of the stars in their orbit.

$(Ak\mu^2)cos\theta = v_\perp-\frac{GM}{rv_\perp}$

To get the distribution in terms of the new coordinates ($E, k$ and $cos\theta$), we need to calculate the determinant of the Jacobi of the transformation.

Luckily, $k$ and $\theta$ are independent of $v_\parallel$.

$\begin{vmatrix} \partial E/\partial r & \partial E/\partial v_\parallel & \partial E/\partial v_\perp \\ \partial k/\partial r & \partial k/\partial v_\parallel & \partial k/\partial v_\perp \\ \partial (cos\theta)/\partial r & \partial (cos\theta)/\partial v_\parallel & \partial (cos\theta)/\partial v_\perp \\ \end{vmatrix}$

$= v_\parallel \frac{\mu^2}{Ar} = \frac{\mu^2 v_\parallel}{Akr^2v_\perp}$

And so the volume element transforms to:

$dv_\parallel dv_\perp dr = \frac{Akr^2v_\perp}{\mu^2 v_\parallel} dE dk d cos\theta$

Transforming the coordinates:

$v_\perp = \frac{Ak}{\mu^2} cos\theta + GMk$

$r = \frac{mu^2}{k^2(Acos\theta+GM\mu^2)}$

$v_\parallel = \sqrt{ 2E + (GMk)^2 - (Ak/\mu^2)^2 cos^2\theta}$

Now, the relationship between $A$ and the energy and specific angular momentum ($l$) is:

$A^2 = (GM\mu^2)^2 + 2E(l\mu^2)^2$

And we can write:

$v_\parallel = (Ak/\mu^2) \sqrt{1-cos^2\theta} = (Ak/\mu^2)sin\theta$

Plugging into the distribution expression, we get:

$f d \tilde{V} \sim exp(-\frac{E}{T}) \frac{\mu^4}{(Acos\theta+GM\mu^2)^2sin\theta}d cos\theta \frac{1}{k^6}dkdE$

Integrating over $\theta$, between $0$ and $2\pi$ gives:

$f d \tilde{V} \sim exp(-\frac{E}{T}) \frac{GM\mu^6}{((GM\mu^2)^2-A^2)^{3/2}} \frac{1}{k^6}dkdE$

We rewrite $A$ using the eccentricity, $e$.

$A = GM\mu^2e$

And we get:

$f d \tilde{V} \sim exp(-\frac{E}{T}) \frac{1}{(1-e^2)^{3/2}} \frac{1}{k^6}dkdE \frac{1}{(GM)^2}$

The next step is to switch from the inverse specific angular momentum, $k$ to the eccentricity.

$k = \sqrt{2E}{GM} \frac{1}{(1-e^2)^{1/2}}$

$\partial k/\partial e = \frac{\sqrt{2E}}{GM}\frac{e}{(1-e^2)^{3/2}}$

This simplifies the distribution to this nice expression:

$f d \tilde{V} \sim exp(-\frac{E}{T}) \frac{(GM)^3}{(2E)^{5/2}} dE e de$

Here we see that the dependence of the distribution function on eccentricity is just through the term $e de$, independent of the energy. Therefore, the distribution as a function of eccentricity, $f(e)$, after normalizing is:

$f(e) de = 2e de$