# Question about Jean's Equation and Liouville's Theorem

TL;DR

Is this true: $$\int f(\mathbf{x},\mathbf{v})d^3 x d^3 v = \int f(E,L) dE dL \, ?$$

Intro

One of the common things that is done in e.g. Binney and Tremaine's book is transforming between $f(\mathbf{x},\mathbf{v})$ and $f(E,L)$, where $E$ and $L$ are orbital energy and angular momentum (constants of motion along particle paths).

Application

In e.g. here, they use the expression of phase space density $f(E,L)$ to retrieve the density

$$\rho = \int f(\mathbf{x},\mathbf{v}) d^3 v = \int J f(E,L) dE dL$$

where $J$ is some Jacobian matrix to relate $d^3v$ and $dE dL$.

However, now it is very unclear to me whether or not both $f(E,L)$ and $f(\mathbf{x},\mathbf{v})$ are normalized the same way. Should integrating over all "states" give the same result?

In other words is the following true:

$$\int f(\mathbf{x},\mathbf{v})d^3 x d^3 v = \int f(E,L) dE dL$$ or should there be a jacobian as well?

• What are the units of J? Is it the matrix or its determinant? If the latter, I am guessing it has units of inverse volume, thus making it appropriate. – honeste_vivere Sep 14 '16 at 23:03
• Hi @honeste_vivere ; J is the determinant jacobian – Otto Sep 19 '16 at 3:45
• Then I think that solves the unit issue. – honeste_vivere Sep 19 '16 at 12:39

The following expression: $$\rho = \int \ f\left(\mathbf{x},\mathbf{v}\right) d^{3}v = \int \ J \ g\left(E,L\right) \ dE \ dL$$ is okay and consistent with: $$\int \ f\left(\mathbf{x},\mathbf{v}\right) \ d^{3}x \ d^{3}v = \int \ g\left(E,L\right) \ dE \ dL$$ because $J$ is the determinant of the Jacobian, which means there will be factors of the form $\left(\partial/\partial x_{i} \ \partial/\partial x_{j} \ \partial/\partial x_{k}\right)$ in each term of the determinant.
The units of the $\left(\partial/\partial x_{i} \ \partial/\partial x_{j} \ \partial/\partial x_{k}\right)$ terms are proportional to inverse volume. The units of $\ f\left(\mathbf{x},\mathbf{v}\right)$ are $\left[ \# \ s^{3} \ m^{-6} \right]$ and the units of $\ g\left(E,L\right)$ are $\left[ \# \ s^{2} \ m^{-3} \ kg^{-1}\right]$.
Therefore, there is nothing wrong with either integral version. However, as @BobBee pointed out, it should be $\ g\left(E,L\right)$ not $\ f\left(\mathbf{x},\mathbf{v}\right)$ in the $dE \ dL$ integral.
• But wouldnt the determinant from $d^3v$ contribute as $\frac{\partial E}{\partial v}$? – Otto Sep 20 '16 at 11:48