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Free Particle on spherical surface

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data
    By finding the Lagrangian and using the metric:
    [tex]\left(\begin{array}{cc}R^2&0\\0&R^2sin^2(\theta)\end{array}\right)[/tex]
    show that:
    [tex]\theta (t)=arccos(\sqrt{1-\frac{A^2}{\omega^2}}cos(\omega t +\theta_o))[/tex]
    2. Relevant equations


    3. The attempt at a solution
    So I got the lagrangian to be [itex] L=R^2 \dot{\theta^2} +R^2sin^2(\theta)\dot{\phi^2}[/itex] and then used the E-L equation to find the equations of motion and the fact that [itex] 2R^2sin^2(\theta) \dot{\phi}=const=p [/itex].
    Using this and substituting into the equation i get for [itex]\theta[/itex] I get:
    [tex]\frac{d}{dt}(2R^2\dot{\theta})=\frac{p^2}{2R^2}cot(\theta)csc^2(\theta)[/tex]
    which I then integrate using the substitution [itex]dt=d\theta / \dot{\theta}[/itex] to get:
    [tex]\dot{\theta}=\frac{p}{2R^2}\sqrt{c-\frac{1}{2}sin^{-2}(\theta)}[/tex]
    Where c is the integration constant. Now if I seperate variables to attempt to get a solution for [itex]\theta[/itex] i get:
    [tex]\int _{\theta_o}^{\theta} \frac{d\theta}{\sqrt{c-\frac{1}{2}sin^{-2}}}=\frac{tp}{2R^2}[/tex]
    But i have absolutely no idea how to solve that integral. Please any pointers would be appreciated.
     
    Last edited: Mar 1, 2016
  2. jcsd
  3. Mar 1, 2016 #2

    Orodruin

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  4. Mar 1, 2016 #3
    Ah okay, I did just try [itex]u=cos(\theta)[/itex] but it gives:
    [tex] \int \frac{du}{\sqrt{c(1-u^2)-1/2}}[/tex]
    It didn't prove to be any easier to solve.
    Also tired doing [itex]u=cos(\theta)[/itex] from the beginning just now as you suggested in the other post (although this could be the wrong substitution) and I must be doing something wrong because I get a complex square root on the LHS of the differential equation for [itex] \dot{u} [/itex].
     
  5. Mar 1, 2016 #4

    Orodruin

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    This is a quite standard integral. It is of the form
    $$
    \int \frac{dx}{\sqrt{1 - x^2}}.
    $$
     
  6. Mar 1, 2016 #5
    Thankyou so much! I managed to get the answer now, i think it was just the fact i hadn't noticed that it was a standard integral.
     
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