Phase space of spherical coordinates and momenta

[ShayanJ]

Homework Statement

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(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.)

Homework Equations

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

The Attempt at a Solution

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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

 

[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 $r^2 \sin \theta$.

 

[ShayanJ]

Oh...careless reading! Thanks.
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 $r^2 \sin\theta dr \ d\theta \ d\varphi \ dp_r \ dp_\theta \ dp_\varphi$.

 

[TSny]

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

OK.

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

 

[aliens123]

For part A:
We need to look at the Jacobian, which is the matrix of partial's:
$$J=\frac{\partial(x,y,z,p_x,p_y,p_z)}{\partial(r,\theta,\phi,p_r,p_\theta,p_\phi)}.$$
\begin{align*}
x&=r\sin \theta \cos \phi \\
y&=r\sin \theta \sin \phi \\
z&=r\cos \theta \\
p_x = m\dot{x} &= m\dot{r} \sin \theta \cos \phi + mr \dot{\theta}\cos \theta \cos \phi - mr \dot{\phi} \sin \theta \sin \phi \\
p_y = m\dot{y} &= m\dot{r} \sin \theta \sin \phi + mr \dot{\theta} \cos \theta \sin \phi + mr \dot{\phi} \sin \theta \cos \phi \\
p_z =m\dot{z} &=m\dot{r} \cos \theta - mr \dot{\theta} \sin \theta \\
\end{align*}
$$J= \left| \begin{matrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial p_r} & \frac{\partial x}{\partial p_\theta} & \frac{\partial x}{\partial p_\phi} \\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial p_r} & \frac{\partial y}{\partial p_\theta} & \frac{\partial y}{\partial p_\phi} \\

\frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial p_r} & \frac{\partial z}{\partial p_\theta} & \frac{\partial z}{\partial p_\phi} \\
\frac{\partial p_x}{\partial r} & \frac{\partial p_x}{\partial \theta} & \frac{\partial p_x}{\partial \phi} & \frac{\partial p_x}{\partial p_r} & \frac{\partial p_x}{\partial p_\theta} & \frac{\partial p_x}{\partial p_\phi} \\
\frac{\partial p_y}{\partial r} & \frac{\partial p_y}{\partial \theta} & \frac{\partial p_y}{\partial \phi} & \frac{\partial p_y}{\partial p_r} & \frac{\partial p_y}{\partial p_\theta} & \frac{\partial p_y}{\partial p_\phi} \\
\frac{\partial p_z}{\partial r} & \frac{\partial p_z}{\partial \theta} & \frac{\partial p_z}{\partial \phi} & \frac{\partial p_z}{\partial p_r} & \frac{\partial p_z}{\partial p_\theta} & \frac{\partial p_z}{\partial p_\phi}
\end{matrix}\right|$$Now, we use a lemma: for a block matrix, $$\det \begin{bmatrix} A & 0 \\ B & C \end{bmatrix} = \det A \cdot \det C.$$ It is easy to see that our matrix is of this form. Note that
$$A= \left| \begin{matrix}
x/r & x\cos \theta/\sin \theta & -y \\

y/r & y\cos \theta/\sin \theta & x \\

\cos \theta & -r\sin \theta & 0 \\\end{matrix}\right|$$
And so
$$\det A =0 +\frac{x^2 \cos^2 \theta}{\sin \theta} + \frac{y^2 \sin^2 \theta}{\sin \theta} + \frac{x^2 \sin^2 \theta}{\sin \theta}-0+\frac{y^2 \cos^2 \theta}{\sin \theta}$$
$$\det A=\frac{x^2 \cos^2 \theta + x^2\sin^2 \theta + y^2 \sin^2 \theta + y^2 \cos^2 \theta}{\sin \theta}$$
$$\det A=\frac{x^2+y^2}{\sin \theta}=r^2 \sin \theta$$
Now we need to find $C.$ We use
\begin{align*}
p_r &= m\dot{r} \\
p_\theta &=mr^2 \dot{\theta} \\
p_\phi&=mr^2 \sin^2 \theta \dot{\phi}
\end{align*}
So that
\begin{align*}
p_x &= p_r \sin \theta \cos \phi + \frac{p_\phi}{r}\cos \theta \cos \phi - \dfrac{p_\phi}{r \sin \theta}\sin \phi \\
p_y &= p_r \sin \theta \sin \phi + \frac{p_\phi}{r} \cos \theta \sin \phi + \dfrac{p_\phi}{r \sin \theta} \cos \phi \\
p_z &=p_r \cos \theta - \frac{p_\phi}{r} \sin \theta \\
\end{align*}
Hence
$$C= \begin{bmatrix}
\frac{\partial p_x}{\partial p_r} & \frac{\partial p_x}{\partial p_\theta} & \frac{\partial p_x}{\partial p_\phi} \\
\frac{\partial p_y}{\partial p_r} & \frac{\partial p_y}{\partial p_\theta} & \frac{\partial p_y}{\partial p_\phi} \\
\frac{\partial p_z}{\partial p_r} & \frac{\partial p_z}{\partial p_\theta} & \frac{\partial p_z}{\partial p_\phi}
\end{bmatrix} = \begin{bmatrix}
x/r & \frac{x \cos \theta }{\sin \theta r^2} & -\frac{y}{r^2 \sin^2 \theta} \\
y/r & \frac{y \cos \theta }{\sin \theta r^2} & \frac{x}{r^2 \sin^2 \theta} \\
\cos \theta & -\frac{\sin \theta}{r} & 0\end{bmatrix}$$
$$\det C = 0 + \frac{x^2 \cos^2 \theta}{\sin^3 \theta r^4} + \frac{y^2 }{r^4 \sin \theta} - \frac{-x^2}{r^4 \sin \theta} -0 -\frac{-y^2 \cos^2 \theta}{r^4 \sin^3 \theta}$$
$$\det C = \frac{(x^2 + y^2)\cos^2 \theta}{\sin^3 \theta r^4}+\frac{x^2 + y^2}{r^4 \sin \theta}= \frac{x^2 + y^2}{r^4 \sin^3 \theta}=\frac{1}{r^2 \sin \theta}$$
So $$\det A \cdot \det C =0$$ as we wished to show.

 

[aliens123]

OK.

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

Sure, but this type of substitution could be done for any F We would still get the factor of $$r^2 \sin \theta$$ if we do this substitution, regardless of what F looks like.

 

[TSny]

Sure, but this type of substitution could be done for any F We would still get the factor of $$r^2 \sin \theta$$ if we do this substitution, regardless of what F looks like.

Yes, the substitutions $u = \frac{p_\theta}{r}$ and $v = \frac{p_\phi}{r\sin \theta}$ will give a Jacobian factor of $r^2 \sin \theta$ when switching from the variables $(p_\theta, p_\phi)$ to the variables $(u, v)$.

However, for an arbitrary function $F(r, \theta, \phi, p_r, p_\theta, p_\phi)$, there can be additional factors involving $r$ and $\theta$ that arise when integrating over the new variables $u$ and $v$. This can mess up the $r^2 \sin \theta$ that we want to have for the spatial integrations over $r$, $\theta$, and $\phi$. But if $F$ is special in that it involves $p_\theta$ and $p_\phi$ only in the combination $\frac{p_\theta^2}{r^2} + \frac{p_\phi^2}{r^2 \sin^2\theta}$, then switching to the new variables $u$ and $v$ and integrating over $u$ and $v$ does not introduce any additional factors involving $\theta$ and $\phi$. So, we end up getting just the $r^2 \sin \theta$ that we want.

If this does not make sense, we can look at a specific example.

 

[TSny]

For part A:
We need to look at the Jacobian, which is the matrix of partial's:
$$J=\frac{\partial(x,y,z,p_x,p_y,p_z)}{\partial(r,\theta,\phi,p_r,p_\theta,p_\phi)}.$$
\begin{align*}
x&=r\sin \theta \cos \phi \\
y&=r\sin \theta \sin \phi \\
z&=r\cos \theta \\
p_x = m\dot{x} &= m\dot{r} \sin \theta \cos \phi + mr \dot{\theta}\cos \theta \cos \phi - mr \dot{\phi} \sin \theta \sin \phi \\
p_y = m\dot{y} &= m\dot{r} \sin \theta \sin \phi + mr \dot{\theta} \cos \theta \sin \phi + mr \dot{\phi} \sin \theta \cos \phi \\
p_z =m\dot{z} &=m\dot{r} \cos \theta - mr \dot{\theta} \sin \theta \\
\end{align*}
$$J= \left| \begin{matrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial p_r} & \frac{\partial x}{\partial p_\theta} & \frac{\partial x}{\partial p_\phi} \\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial p_r} & \frac{\partial y}{\partial p_\theta} & \frac{\partial y}{\partial p_\phi} \\

\frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial p_r} & \frac{\partial z}{\partial p_\theta} & \frac{\partial z}{\partial p_\phi} \\
\frac{\partial p_x}{\partial r} & \frac{\partial p_x}{\partial \theta} & \frac{\partial p_x}{\partial \phi} & \frac{\partial p_x}{\partial p_r} & \frac{\partial p_x}{\partial p_\theta} & \frac{\partial p_x}{\partial p_\phi} \\
\frac{\partial p_y}{\partial r} & \frac{\partial p_y}{\partial \theta} & \frac{\partial p_y}{\partial \phi} & \frac{\partial p_y}{\partial p_r} & \frac{\partial p_y}{\partial p_\theta} & \frac{\partial p_y}{\partial p_\phi} \\
\frac{\partial p_z}{\partial r} & \frac{\partial p_z}{\partial \theta} & \frac{\partial p_z}{\partial \phi} & \frac{\partial p_z}{\partial p_r} & \frac{\partial p_z}{\partial p_\theta} & \frac{\partial p_z}{\partial p_\phi}
\end{matrix}\right|$$Now, we use a lemma: for a block matrix, $$\det \begin{bmatrix} A & 0 \\ B & C \end{bmatrix} = \det A \cdot \det C.$$ It is easy to see that our matrix is of this form. Note that
$$A= \left| \begin{matrix}
x/r & x\cos \theta/\sin \theta & -y \\

y/r & y\cos \theta/\sin \theta & x \\

\cos \theta & -r\sin \theta & 0 \\\end{matrix}\right|$$
And so
$$\det A =0 +\frac{x^2 \cos^2 \theta}{\sin \theta} + \frac{y^2 \sin^2 \theta}{\sin \theta} + \frac{x^2 \sin^2 \theta}{\sin \theta}-0+\frac{y^2 \cos^2 \theta}{\sin \theta}$$
$$\det A=\frac{x^2 \cos^2 \theta + x^2\sin^2 \theta + y^2 \sin^2 \theta + y^2 \cos^2 \theta}{\sin \theta}$$
$$\det A=\frac{x^2+y^2}{\sin \theta}=r^2 \sin \theta$$
Now we need to find $C.$ We use
\begin{align*}
p_r &= m\dot{r} \\
p_\theta &=mr^2 \dot{\theta} \\
p_\phi&=mr^2 \sin^2 \theta \dot{\phi}
\end{align*}
So that
\begin{align*}
p_x &= p_r \sin \theta \cos \phi + \frac{p_\phi}{r}\cos \theta \cos \phi - \dfrac{p_\phi}{r \sin \theta}\sin \phi \\
p_y &= p_r \sin \theta \sin \phi + \frac{p_\phi}{r} \cos \theta \sin \phi + \dfrac{p_\phi}{r \sin \theta} \cos \phi \\
p_z &=p_r \cos \theta - \frac{p_\phi}{r} \sin \theta \\
\end{align*}
Hence
$$C= \begin{bmatrix}
\frac{\partial p_x}{\partial p_r} & \frac{\partial p_x}{\partial p_\theta} & \frac{\partial p_x}{\partial p_\phi} \\
\frac{\partial p_y}{\partial p_r} & \frac{\partial p_y}{\partial p_\theta} & \frac{\partial p_y}{\partial p_\phi} \\
\frac{\partial p_z}{\partial p_r} & \frac{\partial p_z}{\partial p_\theta} & \frac{\partial p_z}{\partial p_\phi}
\end{bmatrix} = \begin{bmatrix}
x/r & \frac{x \cos \theta }{\sin \theta r^2} & -\frac{y}{r^2 \sin^2 \theta} \\
y/r & \frac{y \cos \theta }{\sin \theta r^2} & \frac{x}{r^2 \sin^2 \theta} \\
\cos \theta & -\frac{\sin \theta}{r} & 0\end{bmatrix}$$
$$\det C = 0 + \frac{x^2 \cos^2 \theta}{\sin^3 \theta r^4} + \frac{y^2 }{r^4 \sin \theta} - \frac{-x^2}{r^4 \sin \theta} -0 -\frac{-y^2 \cos^2 \theta}{r^4 \sin^3 \theta}$$
$$\det C = \frac{(x^2 + y^2)\cos^2 \theta}{\sin^3 \theta r^4}+\frac{x^2 + y^2}{r^4 \sin \theta}= \frac{x^2 + y^2}{r^4 \sin^3 \theta}=\frac{1}{r^2 \sin \theta}$$
So $$\det A \cdot \det C =0$$ as we wished to show.

OK. So, cranking through the Jacobian wasn't so bad. Nice.

I noticed a couple of typos:

$$p_z =p_r \cos \theta - \frac{p_\phi}{r} \sin \theta$$

$p_\phi$ should be $p_\theta$ here. But, this did not affect your work.

So $$\det A \cdot \det C =0$$

Of course you meant 1 instead of 0.

 

[aliens123]

OK. So, cranking through the Jacobian wasn't so bad. Nice.

I noticed a couple of typos:

$p_\phi$ should be $p_\theta$ here. But, this did not affect your work.Of course you meant 1 instead of 0.

Yes.

Yes, the substitutions $u = \frac{p_\theta}{r}$ and $v = \frac{p_\phi}{r\sin \theta}$ will give a Jacobian factor of $r^2 \sin \theta$ when switching from the variables $(p_\theta, p_\phi)$ to the variables $(u, v)$.

However, for an arbitrary function $F(r, \theta, \phi, p_r, p_\theta, p_\phi)$, there can be additional factors involving $r$ and $\theta$ that arise when integrating over the new variables $u$ and $v$. This can mess up the $r^2 \sin \theta$ that we want to have for the spatial integrations over $r$, $\theta$, and $\phi$. But if $F$ is special in that it involves $p_\theta$ and $p_\phi$ only in the combination $\frac{p_\theta^2}{r^2} + \frac{p_\phi^2}{r^2 \sin^2\theta}$, then switching to the new variables $u$ and $v$ and integrating over $u$ and $v$ does not introduce any additional factors involving $\theta$ and $\phi$. So, we end up getting just the $r^2 \sin \theta$ that we want.

If this does not make sense, we can look at a specific example.

Well, mostly, I think that the problem just needs to be stated more precisely. "Show that if we average out any physical quantity, whose dependence on $p_\theta$ and $p_\phi$ comes only through the kinetic energy of the particle." However, a function $R(r, \theta, p_\phi)=r\sin\theta+p_\phi$ could be rewritten $$R\left(r\sin\theta, \frac{p_\phi^2}{r^2 \sin^2 \theta}
\right) = r\sin\theta \left(1 + \sqrt{\frac{p_\phi^2}{r^2\sin^2 \theta}}\right).$$

What is meant, I assume, is that the function $$R(r, \theta, \phi, p_r, p_\theta, p_\phi) = F(r, \theta, \phi, p_r) + C\cdot \left(\frac{p_\theta^2}{r^2} + \frac{p_\phi^2}{r^2 \sin^2\theta} \right) $$

Where $C$ is a constant. But you also need to be a little more precise, because you want to show that when you perform the ($p_\phi, p_\theta$) integral, you will just be leftover with $F+D$ where $D$ doesn't depend on $r, \theta, \phi, p_r$. This means that your limits of integration for $\frac{p_\theta^2}{r^2}, \frac{p_\phi^2}{r^2 \sin^2\theta}$ should have no $r, \theta, \phi, p_r$ dependence. Let me know if you agree...

 

[TSny]

But you also need to be a little more precise, because you want to show that when you perform the ($p_\phi, p_\theta$) integral, you will just be leftover with $F+D$ where $D$ doesn't depend on $r, \theta, \phi, p_r$. This means that your limits of integration for $\frac{p_\theta^2}{r^2}, \frac{p_\phi^2}{r^2 \sin^2\theta}$ should have no $r, \theta, \phi, p_r$ dependence. Let me know if you agree...

Yes, I agree. Consider the substitution $v = \frac{p_\phi}{r\sin\theta}$ that I mentioned. If we're integrating over all of phase space, then $p_\phi$ would range over 0 to ∞. Then $v$ would also range over 0 to ∞, which is good. But if, for example, the upper limit of integration for $p_\phi$ were some finite number $a$, then the upper limit for $v$ would be $ar\sin\theta$. The result of the $v$ integration would then produce some function of $\sin \theta$, which would not be good. I believe this is similar to what you are saying.