How to Prove Rotation Around x-Axis of a Sphere?

[IanBerkman]

I have posted this question earlier but I think it was ill-stated. I try to give it in a simpler fashion in this thread.

In the problem it is stated that there a rotation around the x-axis of a (stereographic) sphere is given by
$$\delta \phi = \cot \phi \cot \theta \delta \theta$$
where $\theta$ is the angle from the z-axis and $\phi$ the angle from the x-axis.
The rotation maps the sphere on itself.

I need to prove that this equation indeed represents a rotation around the x-axis.
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In spherical coordinates we have
$$ x = \rho\sin\theta\cos\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho\cos\theta$$
I changed the $\delta$ in the equation for the infinitesimally small d. (It is nowhere stated what $\delta$ is, it could also be that I have to substitute this in $\delta S = \delta\int L dt = 0$ but this will give a huge integral. I can give this integral in a comment).

This gives
$$
\tan\phi d\phi = \cot\theta d \theta$$
Integration gives
$$sin\phi = const\cdot\csc\theta\tan\phi$$
Where the constant follows from the integration.

Substituting this equation in the spherical coordinates will give a constant $x$, but no constant for $y^2+z^2$ which must follow from a rotation around the x-axis.

I am stuck with this problem for over a week, I am really thankful if someone can provide some insight.

Thanks in advance,
Ian

 

[andrewkirk]

Unfortunately the question is still unclear. It does not state what the equation represents. Getting a clear statement of a problem is often half the battle in solving it.

I'll have a guess as to what it might be intended to mean. If you can confirm or otherwise that that's correct then at least there will be a clear problem to solve.

The reference to a sphere is a red herring. The problem does not concern a sphere. It is about rotation of a point around the x-axis. The question is:

Let P be any point in Euclidean 3-space. Let C be the circle mapped out by rotating P around the x axis. Parametrise C in a constant-speed fashion by function $\gamma:[0,2\pi]\to C$ such that the maximum $z$ coordinate occurs at $\gamma(0)$.

Under the constraint of remaining on the circle C, find $\frac{\partial \phi}{\partial\theta}$. More formally, that means we want to find:
$$\lim_{\delta t\to 0}\frac{\phi(\gamma(t+\delta t))-\phi(\gamma(t))}{\theta(\gamma(t+\delta t))-\theta(\gamma(t))}$$
where $\theta$ and $\phi$ are the zenithal and azimuthal coordinate functions of the spherical coordinate system.

We would expect to find (if the interpretation is correct) that the limit is equal to
$$\cot\Big(\phi(\gamma(t))\Big)\cot\Big(\theta(\gamma(t))\Big)$$

It is also possible that the problem is instead asking about a rotation of the spherical coordinate system, rather than of the point P. If so then the formulas in the problem will be very similar, with just a sign changed here and there.

 

[IanBerkman]

Thank you kindly, it looks like you are more familiar with this problem than me. I think that is the whole problem with this assignment, not knowing what the equation represents. I have uploaded the "tale" which corresponds to this problem and I will give a brief summary:

The tale is about Kepler's problem and later on this is used to explain degeneracy in a hydrogen atom.The Hamiltonian of Kepler's problem is given by $H = \textbf{p}^2/2m - \alpha/r$ where $\alpha$ is the constant corresponding to the gravitational pull. Since this is equal to the energy, we could associate this with a constant momentum $p_0$ by $$
E = - \frac{p_0}{2m} = \frac{\textbf{p}^2}{2m}-\frac{\alpha}{r}$$ (The energy is negative).
Which consequently gives the radius by
$$ r = \frac{2m\alpha}{p_0^2+\textbf{p}^2}$$
The infinitesimal action is given by $dS = -\textbf{r}d\textbf{p}$, but since $\textbf{r}$ and $\dot{\boldsymbol p}$ are parallel, this becomes
$$ dS = (2m\alpha)^2 \frac{dp_x^2+dp_y^2}{(p_0^2+\textbf{p}^2)^2}$$
And from here on they perform a stereographic project by letting
$$ p_x = p_0\cos \phi \cot \theta/2\\
p_y = p_0 \sin\phi \cot \theta/2$$
Which results in the metric on a 2D sphere
$$ dS^2 = \left(\frac{m\alpha}{p_o^2}\right)^2(d\theta^2+\sin^2\theta d\phi^2)$$
Setting up the anti-Lagrangian gives the equation $\sin\theta_0 = \frac{v_\phi}{v}\sin\theta$ (eq 16) which tells explicitly that the trajectory on a stereographic sphere belongs to the plane which has angle $\theta_0$ with the vertical axis.
Then it tells, out of the blue, that there are two more rotations of the stereographic sphere, around the x-axis and y-axis given by equation (17) and (18), introducing the $\delta$. The one around the x-axis is given by the first formula in the original post. Since the problem is also talking about action, I would not know if they mean a variation of this action with the $\delta$.

I understand if this is all too much since it needs quite some concepts to arrive at this point. However, I would also be satisfied enough if you could explain how the limit in your post reduces to the two cotangent terms.