1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotation around x-axis of a sphere

  1. Jul 21, 2016 #1
    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.
    ________________________________________________________________________________________________________
    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
     
  2. jcsd
  3. Jul 21, 2016 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Jul 22, 2016 #3
    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.
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Rotation around x-axis of a sphere
  1. Fixed Axis Rotating (Replies: 2)

  2. Axis of rotation (Replies: 5)

Loading...