# Show that angular momentum is independent of the angle

I was thinking about the Bohr atomic model which states, that the angular momentum (L) must be an integral multiple of the reduced Planck constant, this implies that $$L=mvr$$ must be constant for a given orbit:

$$mvr=n\frac{h}{2\pi}.$$

Bohr only considered circular orbits so $$r$$ is constant as well as $$v$$ and of course $$m$$. Later elliptical orbits were considered by Sommerfeld and it was proposed, that the integral of $$L$$ over one period must be an integral multiple of the Planck constant:

$$\oint L d\phi = nh.$$

Now apparently $$L$$ is independent of $$\phi$$ and integration yields:

$$L 2\pi = nh \Rightarrow L=n\frac{h}{2\pi}$$

Which is the same as before. Now I was wondering how I can show that L is independent of $$\phi$$ because $$r$$ and $$v$$ are arguably functions of $$\phi$$ in the case of elliptical orbits. The velocity $$v$$ is given by:

$$v=\omega r(\phi)= \frac{d\phi}{dt}r(\phi).$$

And the radius is given by:

$$r(\phi)=\frac{ab}{\sqrt{a^2sin^2(\phi)+b^2cos^2(\phi)}}.$$

I imagine I would have to solve the following equation by integration and all terms containing $$\phi$$ would cancel out?

$$Ldt=mr(\phi)^2d\phi$$

I'm not very proficient in physics or math, is my line of thinking right? If so it would be great if someone has a suggestion on where I could find this derivation so I can comprehend how $$\phi$$ cancels out.

The $$1/r$$ potential interaction is, fundamentally, the same as what's in gravitationally orbiting bodies. How do you prove that $$L$$ is constant for those? (a couple of videos on the topic that might be helpful: a 3Blue1Brown video presenting an argument by Feynman showing why orbits are elliptical, an AP Physics level lecture on angular momentum and elliptical orbits.
If nothing else, you can work out \begin{align} \vec{L} &= \vec{r}\times (m\vec{v}) \\ &= r v \sin\theta \hat{n}, \end{align} where $$\theta$$ is the angle between $$\vec{r}$$ and $$\vec{v}$$, and $$\hat{n}$$ is unimportant (it's perpendicular to the ellipse). The crucial part is that the origin is at one focus of the ellipse, so the relationship between $$\phi$$ and $$\theta$$ will be very specific.
Also, you're wrong about the formula for the velocity. That is just the component of the velocity around the origin (we call that the $$\hat{\phi}$$ direction). The velocity will also have a component inward and outward (the $$\hat{r}$$ direction) that is given by $$\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}r}{\mathrm{d}\phi}\times \frac{\mathrm{d}\phi}{\mathrm{d}t}$$. The total velocity vector is given by $$\vec{v} = \frac{\mathrm{d}r}{\mathrm{d}t} \hat{r} + r \frac{\mathrm{d}\phi}{\mathrm{d}t} \hat{\phi},$$ and its magnitude will be \begin{align} v & = \sqrt{\left(\frac{\mathrm{d}r}{\mathrm{d}t} \right)^2 + \left(r \frac{\mathrm{d}\phi}{\mathrm{d}t}\right)^2} \\ & = \sqrt{\left(\frac{\mathrm{d}r}{\mathrm{d}\phi} \right)^2 + r ^2} \times \frac{\mathrm{d}\phi}{\mathrm{d}t}. \end{align} Although, all of that is only useful if you have an expression for $$\frac{\mathrm{d}\phi}{\mathrm{d}t}$$. Better is to use energy conservation to say that $$E = \frac{1}{2} \mu v^2 - k_C \frac{q_e q_p}{r}, \tag1$$ where $$\mu$$ is the reduced mass of the atom.
If you solve (1) for $$v$$ you'll get an expression for the total $$v$$ in terms of the total energy and $$r$$ constants. Then, if you work out $$\theta$$ in terms of $$r$$ and $$\phi$$, everything works out much more nicely than if you worry about time.