# Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates

I am following these notes: Dynamics and Astrophysics of Galaxies.

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?

Same question on Math.SE & Astronomy.SE.

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 & Astronomy.SE.