<p>After correspondence with the author of those notes, I arrived at the following conclusion.</p>
<p>When we do <span class="math-container">$\frac{\partial f}{\partial r}$</span>, we require that all other (other than <span class="math-container">$r$</span>) variables which f is dependent on, are held constant. That is: <span class="math-container">$\theta, \phi, p_r, p_{\theta}, p_{\phi}$</span> are constants.</p>
<p>Note that:</p>
<p><span class="math-container">$(p_r, p_{\theta}, p_{\phi})=(\dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta})$</span></p>
<p>Keep in mind that:</p>
<ul>
<li><p><span class="math-container">$p_r$</span>, ie <span class="math-container">$\dot{r}$</span> is held constant: it does not mean <span class="math-container">$r$</span> is held constant (in fact, we are differentiating with respect to <span class="math-container">$r$</span>).</p>
</li>
<li><p><span class="math-container">$p_{\theta}$</span>, ie <span class="math-container">$r^2\sin^2\theta\dot{\phi}$</span> is held constant. This does not mean that either <span class="math-container">$r$</span>, <span class="math-container">$\theta$</span> or <span class="math-container">$\dot{\phi}$</span> is held constant.</p>
</li>
<li><p><span class="math-container">$p_{\phi}$</span>, ie <span class="math-container">$r^2\dot{\theta}$</span> is held constant. We are not claiming this about <span class="math-container">$r$</span> or <span class="math-container">$\dot{\theta}$</span>.</p>
</li>
</ul>
<p>Writing the steps emphasizing this statement above:</p>
<p><span class="math-container">$$\int d\vec{p} p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$</span></p>
<p>Writing out explicitly what we mean by <span class="math-container">$d\vec{p}$</span>:</p>
<p><span class="math-container">$$=\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$</span></p>
<p><strong>We can move out the partial differentiation because all other term than <span class="math-container">$\frac{\partial f}{\partial r}$</span> is held constant: [THIS IS THE MAIN STEP]</strong></p>
<p><span class="math-container">$$=\frac{\partial}{\partial r}\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r^2 f \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$</span></p>
<p>Changing from the <span class="math-container">$p$</span>-nomenclature to the variables they represent:</p>
<p><span class="math-container">$$\frac{\partial}{\partial r} \int \mathrm{d}(\dot{r}, r^2\dot{\theta}, r^2\sin^2\theta\dot{\phi})  v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$</span></p>
<p>ie</p>
<p><span class="math-container">$$\frac{\partial}{\partial r} \int \mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$</span></p>
<p>This step I am unsure about (ie why I don't have a <span class="math-container">$dr$</span> and <span class="math-container">$d\theta$</span> term, for example), but we say:</p>
<p><span class="math-container">$$\mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) = r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}$$</span></p>
<p>Rewrite integral</p>
<p><span class="math-container">$$\frac{\partial}{\partial r} \int r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$</span>
<span class="math-container">$$=\frac{\partial}{\partial r} r^2 \sin\theta \int \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$</span>
<span class="math-container">$$=\frac{\partial}{\partial r} \left( r^2 \sin\theta \nu \bar{v_r^2} \right)$$</span></p>
<p>As claimed by the notes.</p>
<hr />
<p>Same answer on <a href="https://math.stackexchange.com/a/3970297/590567">Math.SE</a> &amp; <a href="https://physics.stackexchange.com/a/604848/212053">Physics.SE</a>.</p>

CC BY-SA 4.0
2021-01-02T22:25:18.293Z
2021-01-02T22:25:18.293Z
zabop
<p>I am following these notes: <a href="http://astro.utoronto.ca/%7Ebovy/AST1420/notes/notebooks/05.-Equilibria-Spherical-Collisionless-Systems.html" rel="nofollow noreferrer">Dynamics and Astrophysics of Galaxies</a>.</p>
<p>After equation 6.37, we have:
<span class="math-container">\begin{equation*}
    p_r\,\frac{\partial f}{\partial r} + \frac{p_\theta}{r^2}\,\frac{\partial f}{\partial \theta} +
    \frac{p_\phi}{r^2\,\sin^2\theta}\,\frac{\partial f}{\partial \phi}
    -\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{p_\theta^2}{r^3}-\frac{p_\phi^2}{r^3\,\sin^2\theta}\right)\,\frac{\partial f}{\partial p_r}
    +\frac{p_\phi^2\,\cos\theta}{r^2\,\sin^3\theta}\,\frac{\partial f}{\partial p_\theta} = 0\,.\\
\end{equation*}</span></p>
<p>This is the Collisionless Boltzmann Equation in Spherical Polar Coordinates.</p>
<p>Then</p>
<blockquote>
<p>We now multiply this by <span class="math-container">$p_r$</span> and integrate over all <span class="math-container">$(p_r,p_{\phi},p_{\theta})$</span> using
that <span class="math-container">$\mathrm{d}p_r\,\mathrm{d}p_\phi\,\mathrm{d}p_\theta = r^2\,\sin\theta\,\mathrm{d}v_r\, \mathrm{d}v_\phi\,\mathrm{d}v_\theta$</span> and using partial integration to deal
with the derivatives of f with respect to the momenta</p>
</blockquote>
<p><span class="math-container">\begin{align}\label{eq-spher-jeans-penult}
    \frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2_r})}{\partial r} + \frac{\partial (\sin\theta\,\nu\,\overline{v_r\,v_\theta})}{\partial \theta} &amp; +
    \frac{\partial (\nu\,\overline{v_r\,v_\phi}/\sin\theta)}{\partial \phi}\\
    &amp;
    +r^2\,\sin\theta\,\nu\,\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{\overline{v_\theta^2}}{r}-\frac{\overline{v_\phi^2}}{r}\right) = 0\nonumber\,.
\end{align}</span></p>
<p>I would like to arrive to this result myself. Multiplying the SPC CBE by <span class="math-container">$p_r$</span> &amp; going ahead with the suggested volume element, considering the <em>first</em> term in the above equation, not being worried about the integration limits gives us:</p>
<p><span class="math-container">\begin{equation}
\int p_r \frac{\partial f}{\partial r} p_r r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
=\int v_r^2 \frac{\partial f}{\partial r} r^2 \sin \theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
=r^2 \sin \theta \int v_r^2 \frac{\partial f}{\partial r} \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
\end{equation}</span></p>
<p>Using 6.32 from the mentioned notes,</p>
<p><span class="math-container">\begin{equation}
\frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2_r})}{\partial r} = \frac{\partial}{\partial r} \left( r^2 \sin\theta \int v^2_r f \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} \right)
\end{equation}</span></p>
<p>Which is not equal to what I have found just a line above.</p>
<p>What am I doing wrong?</p>
<hr />
<p>Some question on <a href="https://math.stackexchange.com/q/3968607/590567">Math.SE</a> &amp; <a href="https://physics.stackexchange.com/q/604530/212053">Physics.SE</a>.</p>

<galactic-dynamics><stellar-dynamics>
Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates
zabop
# Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates

(Asked by: username_0 on 2021-01-01)

I am following these notes: [Dynamics and Astrophysics of Galaxies](http://astro.utoronto.ca/%7Ebovy/AST1420/notes/notebooks/05.-Equilibria-Spherical-Collisionless-Systems.html).

After equation 6.37, we have:
\begin{equation\*}
p\_r\,\frac{\partial f}{\partial r} + \frac{p\_\theta}{r^2}\,\frac{\partial f}{\partial \theta} +
\frac{p\_\phi}{r^2\,\sin^2\theta}\,\frac{\partial f}{\partial \phi}
-\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{p\_\theta^2}{r^3}-\frac{p\_\phi^2}{r^3\,\sin^2\theta}\right)\,\frac{\partial f}{\partial p\_r}
+\frac{p\_\phi^2\,\cos\theta}{r^2\,\sin^3\theta}\,\frac{\partial f}{\partial p\_\theta} = 0\,.\\
\end{equation\*}

This is the Collisionless Boltzmann Equation in Spherical Polar Coordinates.

Then

> We now multiply this by $p\_r$ and integrate over all $(p\_r,p\_{\phi},p\_{\theta})$ using
> that $\mathrm{d}p\_r\,\mathrm{d}p\_\phi\,\mathrm{d}p\_\theta = r^2\,\sin\theta\,\mathrm{d}v\_r\, \mathrm{d}v\_\phi\,\mathrm{d}v\_\theta$ and using partial integration to deal
> with the derivatives of f with respect to the momenta

\begin{align}\label{eq-spher-jeans-penult}
\frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2\_r})}{\partial r} + \frac{\partial (\sin\theta\,\nu\,\overline{v\_r\,v\_\theta})}{\partial \theta} & +
\frac{\partial (\nu\,\overline{v\_r\,v\_\phi}/\sin\theta)}{\partial \phi}\\
&
+r^2\,\sin\theta\,\nu\,\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{\overline{v\_\theta^2}}{r}-\frac{\overline{v\_\phi^2}}{r}\right) = 0\nonumber\,.
\end{align}

I would like to arrive to this result myself. Multiplying the SPC CBE by $p\_r$ & going ahead with the suggested volume element, considering the *first* term in the above equation, not being worried about the integration limits gives us:

\begin{equation}
\int p\_r \frac{\partial f}{\partial r} p\_r r^2 \sin\theta \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi}
=\int v\_r^2 \frac{\partial f}{\partial r} r^2 \sin \theta \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi}
=r^2 \sin \theta \int v\_r^2 \frac{\partial f}{\partial r} \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi}
\end{equation}

Using 6.32 from the mentioned notes,

\begin{equation}
\frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2\_r})}{\partial r} = \frac{\partial}{\partial r} \left( r^2 \sin\theta \int v^2\_r f \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi} \right)
\end{equation}

Which is not equal to what I have found just a line above.

What am I doing wrong?

---

Some question on [Math.SE](https://math.stackexchange.com/q/3968607/590567) & [Physics.SE](https://physics.stackexchange.com/q/604530/212053).

**Tags:** galactic-dynamics, stellar-dynamics


---

## 1 answers

### Answer by username_0 on 2021-01-02. Score: 1 | Accepted answer

After correspondence with the author of those notes, I arrived at the following conclusion.

When we do $\frac{\partial f}{\partial r}$, we require that all other (other than $r$) variables which f is dependent on, are held constant. That is: $\theta, \phi, p\_r, p\_{\theta}, p\_{\phi}$ are constants.

Note that:

$(p\_r, p\_{\theta}, p\_{\phi})=(\dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta})$

Keep in mind that:

* $p\_r$, ie $\dot{r}$ is held constant: it does not mean $r$ is held constant (in fact, we are differentiating with respect to $r$).
* $p\_{\theta}$, ie $r^2\sin^2\theta\dot{\phi}$ is held constant. This does not mean that either $r$, $\theta$ or $\dot{\phi}$ is held constant.
* $p\_{\phi}$, ie $r^2\dot{\theta}$ is held constant. We are not claiming this about $r$ or $\dot{\theta}$.

Writing the steps emphasizing this statement above:

$$\int d\vec{p} p\_r \frac{\partial f}{\partial r} p\_r \Bigr|\_{\theta, \phi, p\_r, p\_{\theta}, p\_{\phi}}$$

Writing out explicitly what we mean by $d\vec{p}$:

$$=\int \mathrm{d}(p\_r,p\_{\theta},p\_{\phi}) p\_r \frac{\partial f}{\partial r} p\_r \Bigr|\_{\theta, \phi, p\_r, p\_{\theta}, p\_{\phi}}$$

**We can move out the partial differentiation because all other term than $\frac{\partial f}{\partial r}$ is held constant: [THIS IS THE MAIN STEP]**

$$=\frac{\partial}{\partial r}\int \mathrm{d}(p\_r,p\_{\theta},p\_{\phi}) p\_r^2 f \Bigr|\_{\theta, \phi, p\_r, p\_{\theta}, p\_{\phi}}$$

Changing from the $p$-nomenclature to the variables they represent:

$$\frac{\partial}{\partial r} \int \mathrm{d}(\dot{r}, r^2\dot{\theta}, r^2\sin^2\theta\dot{\phi}) v\_r^2 f \Bigr|\_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$

ie

$$\frac{\partial}{\partial r} \int \mathrm{d}(v\_r, r \sin\theta v\_{\phi}, r v\_{\theta}) v\_r^2 f \Bigr|\_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$

This step I am unsure about (ie why I don't have a $dr$ and $d\theta$ term, for example), but we say:

$$\mathrm{d}(v\_r, r \sin\theta v\_{\phi}, r v\_{\theta}) = r^2 \sin\theta \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi}$$

Rewrite integral

$$\frac{\partial}{\partial r} \int r^2 \sin\theta \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi} v\_r^2 f$$
$$=\frac{\partial}{\partial r} r^2 \sin\theta \int \mathrm{d}v\_r \mathrm{d}v\_{\theta} \mathrm{d}v\_{\phi} v\_r^2 f$$
$$=\frac{\partial}{\partial r} \left( r^2 \sin\theta \nu \bar{v\_r^2} \right)$$

As claimed by the notes.

---

Same answer on [Math.SE](https://math.stackexchange.com/a/3970297/590567) & [Physics.SE](https://physics.stackexchange.com/a/604848/212053).


---

