<p>Poission equaiton:
<span class="math-container">$$\nabla^{2} \Phi=4 \pi G S_{0}$$</span></p>
<p>In cylindrical polar coordinates:
<span class="math-container">$$\frac{1}{r} \frac{d}{dr}\left(r \frac{d \Phi}{d r}\right)+\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}$$</span></p>
<p>We also know from <span class="math-container">$\ddot{\vec{r}}=-\nabla\Phi$</span> that <span class="math-container">$\ddot{r}-r \dot{\phi}^{2}=-\frac{\partial\Phi}{\partial r}$</span>, so for circular orbit, we have: <span class="math-container">$r\dot{\phi}^2=\frac{\partial \Phi}{\partial r}$</span>. This means that we can write <span class="math-container">$v_c=r^2\dot{\phi}^2=r\frac{\partial \Phi}{\partial r}$</span></p>
<p>Rewrite Poisson above:</p>
<p><span class="math-container">$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}-\frac{1}{r} \frac{d}{d r}\left(v_{c}^{2}\right) = 4 \pi G \rho_{0}- \frac{2v_c}{r}\frac{dv_c}{dr}$$</span></p>
<p>Keeping in mind that  <span class="math-container">$A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$</span>, rewrite RHS:</p>
<p><span class="math-container">$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}+2(A^2-B^2)$$</span></p>
<p>which is what we wanted.</p>

CC BY-SA 4.0
2021-05-10T10:13:38.613Z
2021-05-10T10:13:38.613Z
zabop
<p>I am tackling this problem, from a <a href="https://www.ast.cam.ac.uk/sites/default/files/Papers_1-4_2016.pdf" rel="nofollow noreferrer">2016 Cambridge Astrophysics Tripos past paper</a>:</p>
<p>Let <span class="math-container">$\Phi(R, z)$</span> be the axi-symmetric Galactic potential. At the Solar location, <span class="math-container">$(R, z) = (R_0, 0)$</span>, prove that
<span class="math-container">$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$</span>
where <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$\rho_0$</span> is the density in the Solar neighborhood and <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are Oort’s constants.</p>
<p>What I do is:</p>
<p><strong>Rewrite equation we want to prove</strong></p>
<p>Note:</p>
<p><span class="math-container">$A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$</span></p>
<p>Using definiton of <span class="math-container">$A$</span> and <span class="math-container">$B$</span>. (Which are: <span class="math-container">$\left.A \equiv \frac{1}{2}\left(\frac{v_{c}}{R}-\frac{d v_{c}}{d R}\right)\right|_{R_{0}}$</span> and <span class="math-container">$B \equiv-\left.\frac{1}{2}\left(\frac{v_{c}}{R}+\frac{d v_{c}}{d R}\right)\right|_{R_{0}}$</span>).</p>
<p>Rewrite what we want to prove:</p>
<p><span class="math-container">$$
\bbox[5px,border:3px solid green]{
\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}
}
$$</span></p>
<p><strong>Use Poisson's equation:</strong></p>
<p><span class="math-container">$$\nabla^2\Phi=4\pi G\rho_0$$</span></p>
<p>Expand LHS of Poisson:</p>
<p><span class="math-container">$$\frac{\partial^2}{\partial R^2}\Phi+\frac{\partial^2}{\partial z^2}\Phi=4\pi G \rho_0$$</span></p>
<p><strong>Use equation of motion for circular orbit:</strong></p>
<p><span class="math-container">$$\frac{v_c^2}{R}=\frac{\partial}{\partial R}\Phi$$</span></p>
<p>Differentiate once more wrt <span class="math-container">$R$</span>:</p>
<p><span class="math-container">$$\frac{\partial^2}{\partial^2 R}\Phi=\frac{\partial}{\partial R}\frac{v_c^2}{R}=\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2$$</span></p>
<p><strong>Collect results</strong></p>
<p>Substitute result into Poisson's equation expanded form:</p>
<p><span class="math-container">$$\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2+\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0$$</span></p>
<p>Rearrange:</p>
<p><span class="math-container">$$
\bbox[5px,border:3px solid green]{
\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}+\bbox[5px,border:3px solid red]{\frac{1}{R^2}v_c^2}
}$$</span></p>
<p><strong>Conclude</strong></p>
<p>Compare with what we wanted to prove (ie the two green-boxed equations). We have an extra term (red boxed) which shouldn't be there. <strong>How do I fix the proof?</strong></p>

CC BY-SA 4.0
2021-05-06T11:34:41.807Z
Are you sure about your expansion of $\nabla^2\Phi$ in cylindrical polars? Ought there not be a term $\frac{1}{R}\frac{\partial}{\partial R}\Phi$? [But this from 60 years ago..]
ancient mathematician
CC BY-SA 4.0
2021-05-06T13:39:06.470Z
What is the value of $v_c$ at $R_0$? Is it zero? Notice that the quantities A and B are evaluated at $R = R_0$.
Daddy Kropotkin
CC BY-SA 4.0
2021-05-06T14:18:49.533Z
Oh yepp, thanks @ancient mathematician, that is wrong. Rewriting it...
zabop
CC BY-SA 4.0
2021-05-06T14:19:52.517Z
@Daddy Kropotkin, thanks, will fix the expansion first, if it doesn't solve the problem will dive into this. (I don't think it's zero.)
zabop
CC BY-SA 4.0
2021-05-06T19:04:19.853Z
The problem is that when I tried it it doubled the error! Clearly I've got a sign wrong or the sign of $A^2-B^2$ isn't correct.
ancient mathematician
CC BY-SA 4.0
2021-05-09T10:29:17.707Z
@ancientmathematician did it now I think! The correct term for Laplacian turns out to be $\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \Phi}{\partial r}\right)+\frac{d}{dz^2}\Phi$. Good way of remembering the form of it in different coordinate systems: https://math.stackexchange.com/questions/4129584/connection-between-laplacian-and-volume-element-in-spherical-and-cylindrical-pol
zabop
CC BY-SA 4.0
2021-05-09T12:22:31.230Z
I hope you'll post your solution. I think before I did finals my brain was hardwired with div grad curl and the laplacian in "the three coordinate systems" - no ned for mnemonics.
ancient mathematician
CC BY-SA 4.0
2021-05-10T10:13:54.893Z
Posted it is! :)
zabop
<mathematics>
Poisson's equation-like formula in axis-symmetric Galactic potential in terms of Oort constants
zabop
# Poisson's equation-like formula in axis-symmetric Galactic potential in terms of Oort constants

(Asked by: username_0 on 2021-05-05)

I am tackling this problem, from a [2016 Cambridge Astrophysics Tripos past paper](https://www.ast.cam.ac.uk/sites/default/files/Papers_1-4_2016.pdf):

Let $\Phi(R, z)$ be the axi-symmetric Galactic potential. At the Solar location, $(R, z) = (R\_0, 0)$, prove that
$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho\_{0}+2\left(A^{2}-B^{2}\right)$$
where $G$ is the gravitational constant, $\rho\_0$ is the density in the Solar neighborhood and $A$ and $B$ are Oort’s constants.

What I do is:

**Rewrite equation we want to prove**

Note:

$A^2-B^2=-\frac{v\_c}{R}\frac{\mathrm{d}v\_c}{\mathrm{d}R}$

Using definiton of $A$ and $B$. (Which are: $\left.A \equiv \frac{1}{2}\left(\frac{v\_{c}}{R}-\frac{d v\_{c}}{d R}\right)\right|\_{R\_{0}}$ and $B \equiv-\left.\frac{1}{2}\left(\frac{v\_{c}}{R}+\frac{d v\_{c}}{d R}\right)\right|\_{R\_{0}}$).

Rewrite what we want to prove:

$$
\bbox[5px,border:3px solid green]{
\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho\_{0}-2\frac{v\_c}{R}\frac{\mathrm{d}v\_c}{\mathrm{d}R}
}
$$

**Use Poisson's equation:**

$$\nabla^2\Phi=4\pi G\rho\_0$$

Expand LHS of Poisson:

$$\frac{\partial^2}{\partial R^2}\Phi+\frac{\partial^2}{\partial z^2}\Phi=4\pi G \rho\_0$$

**Use equation of motion for circular orbit:**

$$\frac{v\_c^2}{R}=\frac{\partial}{\partial R}\Phi$$

Differentiate once more wrt $R$:

$$\frac{\partial^2}{\partial^2 R}\Phi=\frac{\partial}{\partial R}\frac{v\_c^2}{R}=\frac{2v\_c}{R}\frac{\mathrm{d}v\_c}{\mathrm{d}R}-\frac{1}{R^2}v\_c^2$$

**Collect results**

Substitute result into Poisson's equation expanded form:

$$\frac{2v\_c}{R}\frac{\mathrm{d}v\_c}{\mathrm{d}R}-\frac{1}{R^2}v\_c^2+\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho\_0$$

Rearrange:

$$
\bbox[5px,border:3px solid green]{
\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho\_0-2\frac{v\_c}{R}\frac{\mathrm{d}v\_c}{\mathrm{d}R}+\bbox[5px,border:3px solid red]{\frac{1}{R^2}v\_c^2}
}$$

**Conclude**

Compare with what we wanted to prove (ie the two green-boxed equations). We have an extra term (red boxed) which shouldn't be there. **How do I fix the proof?**

**Tags:** mathematics
> **Comments:**
> **username_1**: Are you sure about your expansion of $\nabla^2\Phi$ in cylindrical polars? Ought there not be a term $\frac{1}{R}\frac{\partial}{\partial R}\Phi$? [But this from 60 years ago..]  
> **username_2**: What is the value of $v_c$ at $R_0$? Is it zero? Notice that the quantities A and B are evaluated at $R = R_0$.  
> **username_0**: Oh yepp, thanks @ancient mathematician, that is wrong. Rewriting it...  
> **username_0**: @Daddy Kropotkin, thanks, will fix the expansion first, if it doesn't solve the problem will dive into this. (I don't think it's zero.)  
> **username_1**: The problem is that when I tried it it doubled the error! Clearly I've got a sign wrong or the sign of $A^2-B^2$ isn't correct.  
> **username_0**: @ancientmathematician did it now I think! The correct term for Laplacian turns out to be $\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \Phi}{\partial r}\right)+\frac{d}{dz^2}\Phi$. Good way of remembering the form of it in different coordinate systems: https://math.stackexchange.com/questions/4129584/connection-between-laplacian-and-volume-element-in-spherical-and-cylindrical-pol  
> **username_1**: I hope you'll post your solution. I think before I did finals my brain was hardwired with div grad curl and the laplacian in "the three coordinate systems" - no ned for mnemonics.  
> **username_0**: Posted it is! :)  

---

## 1 answers

### Answer by username_0 on 2021-05-10. Score: 2 | Accepted answer

Poission equaiton:
$$\nabla^{2} \Phi=4 \pi G S\_{0}$$

In cylindrical polar coordinates:
$$\frac{1}{r} \frac{d}{dr}\left(r \frac{d \Phi}{d r}\right)+\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho\_{0}$$

We also know from $\ddot{\vec{r}}=-\nabla\Phi$ that $\ddot{r}-r \dot{\phi}^{2}=-\frac{\partial\Phi}{\partial r}$, so for circular orbit, we have: $r\dot{\phi}^2=\frac{\partial \Phi}{\partial r}$. This means that we can write $v\_c=r^2\dot{\phi}^2=r\frac{\partial \Phi}{\partial r}$

Rewrite Poisson above:

$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho\_{0}-\frac{1}{r} \frac{d}{d r}\left(v\_{c}^{2}\right) = 4 \pi G \rho\_{0}- \frac{2v\_c}{r}\frac{dv\_c}{dr}$$

Keeping in mind that $A^2-B^2=-\frac{v\_c}{R}\frac{\mathrm{d}v\_c}{\mathrm{d}R}$, rewrite RHS:

$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho\_{0}+2(A^2-B^2)$$

which is what we wanted.


---

