I am tackling this problem, from a 2016 Cambridge Astrophysics Tripos past paper:

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



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:

$$\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2+\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0$$


$$ \bbox[5px,border:3px solid green]{ \frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}+\bbox[5px,border:3px solid red]{\frac{1}{R^2}v_c^2} }$$


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?

  • $\begingroup$ 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..] $\endgroup$ Commented May 6, 2021 at 11:34
  • $\begingroup$ What is the value of $v_c$ at $R_0$? Is it zero? Notice that the quantities A and B are evaluated at $R = R_0$. $\endgroup$ Commented May 6, 2021 at 13:39
  • $\begingroup$ Oh yepp, thanks @ancient mathematician, that is wrong. Rewriting it... $\endgroup$
    – zabop
    Commented May 6, 2021 at 14:18
  • $\begingroup$ @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.) $\endgroup$
    – zabop
    Commented May 6, 2021 at 14:19
  • $\begingroup$ 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. $\endgroup$ Commented May 6, 2021 at 19:04

1 Answer 1


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.


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