Density of states and anisotropic distribution functions We consider a $3D$ dynamical system. Its distribution function is given by the function ${  (\mathbf{x},\mathbf{v}) \mapsto f (\mathbf{x},\mathbf{v})}$, so that
$$
\mathrm{d}^{3} \mathbf{x} \, \mathrm{d}^{3} \mathbf{v} \, f(\mathbf{x}, \mathbf{v})
$$
corresponds to the number of particles in the elementary volume ${\mathrm{d}^{3} \mathbf{x} \, \mathrm{d}^{3} \mathbf{v}}$ around the position $(\mathbf{x}, \mathbf{v} )$.
We now suppose that the system is spherical and can be represented by an anistropic distribution function, so that one has
$$
f(\mathbf{x} , \mathbf{v}) = f (E(\mathbf{x} , \mathbf{v}),L(\mathbf{x} , \mathbf{v})) \, ,
$$
where $E = \mathbf{v}^{2}/2 + \psi(\mathbf{x})$ and $L = r \, |\mathbf{v}_{t}|$ are respectively the energy and the angular momentum. I would now like to sample my particles in the ${(E,L)-}$domain. Therefore, I would like to determine the function ${(E,L) \mapsto h (E,L)}$ such that
$$
\mathrm{d} E \, \mathrm{d} L \, h (E , L) 
$$
corresponds to the number of particles in the elementary volume $\mathrm{d} E \, \mathrm{d} L$ around the location ${(E,L)}$. How can I determine ${h (E,L)}$ ?
It seems that one often introduces the density of states ${g (E,L)}$ such that
$$
h(E,L) = g(E,L) \, f(E,L)
$$
How can I determine effectively ${g (E,L)}$ as a function of the exterior potential $\psi$ ?
 A: The distribution in the $x, y$ representation has to lead to the same number of states per unit surface as the distribution in the $E, L$ representation, i.e.
$f(x, y) \ dx \ dy = h(E, L) \ dE \ dL$.
From now on the calculus is almost simple. You have to express the element of surface $dx \ dy$ in terms of $dE \ dL$, and that is done by the determinant of the Jacobian
$dx \ dy = J(E, L) dE \ dL$, $\ \ $ where $ \ \ $ $J(E, L) = (\frac {∂x}{dE} \frac {∂y}{∂L} - \frac {∂x}{dL} \frac {∂y}{∂E})$.
Therefore,
(1) $f(E(x,v),L(x,v)) \ dx \ dy = f(E,L) J(E, L) dE \ dL $.
Now, to calculate the derivatives $\frac {∂x}{dE}$ and all the other derivatives in the Jacobian $J(E, L)$, will be probably difficult for you because you know the functions $E(x, y), L(x, y)$ but not $x(E, L), y(E, L)$. Then, try the following trick :
$dE \ dL = J(x, y) dx \ dy$, $\ \ $ where $ \ \ $ $J(x, y) = (\frac {∂E}{dx} \frac {∂L}{∂y} - \frac {∂E}{dy} \frac {∂L}{∂x})$ .
From this you get
$dx \ dy = \frac {dE \ dL}{J(x, y)}$ .
Therefore, in the formula (1), replace $J(E,L)$ by $1/J(x, y)$. Your desired function $g$ is $1/J(x, y)$.
