4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
3
+150
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.