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 $\psi$ ?


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 $ \ \ $ $J(E, L) = (\frac {∂x}{dE} \frac {∂y}{∂L} - \frac {∂x}{dL} \frac {∂y}{∂E})$.


(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 $1/J(x, y)$.

  • $\begingroup$ Thank you for your answer. In fact, I believe that your first relation $f(x,y) dx dy = h(E,L) dE dL$ is correct only when the mapping $(x,y) \mapsto (E,L)$ is bijective. However, here this is not the case. The starting point is instead that $h(E',L') = \int d^{3}x \, d^{3} v \, \delta_{\rm D} (E'\!-\!E(x,v)) \, \delta_{\rm D} (L'\!-\!L(x,v)) F(x,v)$. Do you agree ? Using this formula, I believe I am now able to compute $h(E,L)$ and will take the time tonight to post my solution so that it will be easier to discuss ! $\endgroup$ – jibe Jan 13 '15 at 16:03
  • $\begingroup$ @jibe what means bijective? Does it mean that we can move from $x, v$ to $E, L$ and vice-versa? And what is $F(x, v)$ ? In your calculi appears $f(x, v)$. $\endgroup$ – Sofia Jan 13 '15 at 16:17
  • $\begingroup$ Yes, bijective means that the mapping is unambiguous and in the both ways. For example, the polar mapping $(x,v) \mapsto (r,\theta)$, with $\theta \in [0,2 \pi]$ and $r > 0$ is bijective. Concerning $F(x,v)$, this is a mistake, and it should always read $f(x,v)$. In my current case, one of the difficulty comes from the fact that ${ (E(x_1,v_1) , L(x_1,v_1)) = (E(x_{2},v_{1}) , L(x_{2},v_{2})) }$ does not at all imply that $(x_{1},v_{1}) = (x_{2},v_{2})$. $\endgroup$ – jibe Jan 13 '15 at 16:26
  • $\begingroup$ @jibe : also, I didn't pay attention until now, but what is $v_t$ ? Anyway, for passing from a distribution $x, y$ to a distribution $E, L$ you have to apply the constraint that the number of configurations in the unit surface has to be the same, but also to pass from one unit surface to the other: $dx \ dv$ is not equal to $dE /dL$.The relation between the unit surfaces is the Jacobian. You can't evade it. Thus, what you need to calculate for finding the new distribution is not $h(E, L)$ but $h(E, L) \ dE \ dL$. $\endgroup$ – Sofia Jan 13 '15 at 16:28
  • $\begingroup$ I totally agree that the Jacobian is essential, and controls how the unit volumes are modified ! $v_{t}$ stands for the tangential component of velocity, so that the angular momentum $L$ is given by $L = r |v_{t}|$. Let me maybe write down my solution and we could discuss about it ? $\endgroup$ – jibe Jan 13 '15 at 16:32

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