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


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).


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?

  • $\begingroup$ 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. $\endgroup$ – honeste_vivere Sep 14 '16 at 23:03
  • $\begingroup$ Hi @honeste_vivere ; J is the determinant jacobian $\endgroup$ – Otto Sep 19 '16 at 3:45
  • $\begingroup$ Then I think that solves the unit issue. $\endgroup$ – 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.

  • $\begingroup$ I see, so you are saying that the two jacobian determinants cancel out? $\endgroup$ – Otto Sep 20 '16 at 8:47
  • $\begingroup$ But wouldnt the determinant from $d^3v$ contribute as $\frac{\partial E}{\partial v}$? $\endgroup$ – Otto Sep 20 '16 at 11:48
  • $\begingroup$ I should add that the above form is only valid for non-relativistic cases. Cancel out? No, I was showing that the units are okay between the two versions. I went through this exercise back in grad school and do recall (though vaguely) those two versions. $\endgroup$ – honeste_vivere Sep 20 '16 at 12:56
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    $\begingroup$ I've not calculated this, but whether J has the right units or not, and assuming f in both equations really means f of X and v equal to g of E and L (i.e., different functions of their arguments but the same numerical result of equivalent arguments) and that g is really what you mean on the right side of either equation, the real question is whether J = 1? Otherwise I don't see how the two could be equal. What am I missing if the answer posted is right? $\endgroup$ – Bob Bee Sep 24 '16 at 20:01
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    $\begingroup$ @BobBee - I see what you mean and yes, you are correct. The change in the arguments of the function derive from which parameters are a constant of motion. A similar type of approach is used to model radiation belt particles that are in relatively stable bouncing/drifting/gyrating orbits. $\endgroup$ – honeste_vivere Sep 25 '16 at 0:08

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