# Phase space of spherical coordinates and momenta

1. Oct 23, 2016

### ShayanJ

1. The problem statement, all variables and given/known data

(a) Verify explicitly the invariance of the volume element $d\omega$ of the phase space of a single particle under transformation from the Cartesian coordinates $(x, y, z, p_x , p_y , p_z)$ to the spherical polar coordinates $(r, θ, φ, p_r , p_θ , p_φ )$.

(b) The foregoing result seems to contradict the intuitive notion of “equal weights for equal solid angles,” because the factor $\sin θ$ is invisible in the expression for $d\omega$. Show that if we average out any physical quantity, whose dependence on $p_θ$ and $p_φ$ comes only through the kinetic energy of the particle, then as a result of integration over these variables we do indeed recover the factor $\sin θ$ to appear with the subelement $(dθ \ dφ)$.

(This is in the context of the invariance of the phase space volume element under a canonical transformation.)

2. Relevant equations

$L=\frac 1 2 m(\dot r^2+r^2 \dot \theta^2+r^2\sin^2 \theta \dot \varphi^2)-V(r,\theta,\varphi)$
$p_a=\frac{\partial L}{\partial \dot q_a}$

3. The attempt at a solution

For the first part, I used the definition of momenta as the derivative of the Lagrangian in spherical coordinates w.r.t. the time derivatives of coordinates and then using the formulas for spherical coordinates, I wrote the momenta in spherical coordinates, in terms of coordinates and momenta in Cartesian coordinates:

$p_r=\frac{xp_x+yp_y+zp_z}{\sqrt{x^2+y^2+z^2}}$
$p_\theta=z\frac{xp_x+yp_y}{\sqrt{x^2+y^2}}-p_z\sqrt{x^2+y^2}$
$p_\varphi=xp_y-yp_x$

Using Maxima, I verified that the Jacobian determinant of this transformation is equal to 1 as it should be. But its really tedious to actually write the Jacobian matrix of this transformation and the calculate the determinant. Is there an easier way of doing it that I'm missing?

My next question is about the part b of the problem. I have no idea what he's talking about. When calculating the average of a quantity using the phase space integral, we use the dependence of that quantity on the coordinates and momenta and, in general, the kinetic energy of the particle doesn't appear in the integrand at all. So I really don't understand the question. Can anyone clarify?

Thanks

2. Oct 23, 2016

### TSny

For (a) I don't see a way to avoid a messy determinant. If you consider the reverse transformation from spherical coordinates to rectangular coordinates, you can avoid dealing with square roots. But the Jacobian still looks very messy. Maybe someone else can help here.

For (b), the question assumes that the physical quantity depends on $p_{\theta}$ and $p_{\phi}$ only through the kinetic energy $T$. If you express $T$ in spherical coordinates in terms of $p_r$, $p_{\theta}$, $p_{\phi}$, $r$ and $\theta$, you can see that the physical quantity must depend on $p_{\theta}$ and $p_{\phi}$ only in a certain way. Then, when you integrate the quantity over phase space you can deduce that the integration over the spatial variables $r$, $\theta$, and $\phi$ will include a factor of $r^2 \sin \theta$.

3. Oct 23, 2016

### ShayanJ

But I still have problem with it. The kinetic energy in the spherical coordinates and in terms of the momenta has the form $T=\frac 1 2 m\left[ (\frac{p_r}m)^2+(\frac{p_\theta}{mr})^2+(\frac{p_\varphi}{mr\sin\theta})^2 \right]$ so I should consider an integral of the form:

$\int F(r,\theta,\varphi,p_r,(\frac{p_\theta}{mr})^2+(\frac{p_\varphi}{mr\sin\theta})^2) dr \ d\theta \ d\varphi \ dp_r \ dp_\theta \ dp_\varphi$

Surely $\sin\theta$ appears in the integrand but, in general, it can be anywhere and there is no guarantee that it gives the form $r^2 \sin\theta dr \ d\theta \ d\varphi \ dp_r \ dp_\theta \ dp_\varphi$.

4. Oct 23, 2016

### TSny

OK.

You can try a substitution $u = \frac{p_{\theta}}{mr}$ and a similar substitution for $p_{\varphi}$ .