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I Lagrangian - surface of sphere

  1. Mar 1, 2016 #1
    I have a free particle moving on the surface of a sphere of fixed radius R. Gravity is ignored and m/2 is left out since its constant.
    The lagrangian is [itex] L = R^2 \dot{\theta^2} + R^2 sin^2{\theta} \dot{\phi^2}[/itex]

    Using the Euler Lagrange equations I obtain
    [itex] sin^2{\theta} \dot{\phi} = A = const \ (1) \\ \ddot{\theta} - sin{\theta}cos{\theta} \dot{\phi^2} = 0 \ (2) [/itex]
    and by substituting 1 into 2
    [itex] \ddot{\theta} = A^2 cos{\theta}/sin^3{\theta} [/itex]

    By integrating w.r.t time and using the fact [itex] dt = d{\theta}/\dot{\theta}[/itex] and that theta and its time derivative are treated as independent coordinates i get
    [itex] \dot{\theta} + A^2/ 2 \dot{\theta} sin^2{\theta} \ = c_1 [/itex]
    integrating w.r.t time again i get. [Using omega instead of theta dot now.]

    [itex] \theta - A^2 cot{\theta}/\omega^2 \ = t c_1 + c_2[/itex]

    I cant see anything wrong but I am supposed to get this into the form

    [itex] \theta(t) = arccos ( \sqrt{1 - A^2/\omega^2} cos(\omega t + \theta_0)[/itex]

    and I cant. If anyone can help I would appreciate it a lot. Thanks
     
  2. jcsd
  3. Mar 1, 2016 #2

    Orodruin

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    Your integration of the ##\ddot\theta## equation looks suspicious.
     
  4. Mar 1, 2016 #3
    You're right it was. What I get now which Im pretty sure is correct is this horrible integral

    [itex] \theta = \frac{1}{\omega} \int \sqrt[]{ c_1 - c_2/sin^2{\theta} } \ d \theta [/itex]

    Which I also have no idea how to solve :S
     
  5. Mar 1, 2016 #4

    Orodruin

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    You cannot have ##\theta## as the integration variable and the result!
     
  6. Mar 1, 2016 #5
    wow I'm being stupid today.

    If i separate the variables I get [itex] \int \frac{1}{ \sqrt[]{(c_1 - c_2/sin^2{\theta} )} }d \theta \ = \ \int dt [/itex]
     
  7. Mar 1, 2016 #6

    Orodruin

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    Indeed, so I suggest a change of variables in the left integral. (Doing this change of variables from the beginning significantly simplifies the differential equation. You might try it instead.)
     
  8. Mar 1, 2016 #7
    I start again and get [itex] \dot \theta = c_1 \int \frac{u}{(1-u^2)^{3/2}} \frac{dt}{du}du[/itex] But still abit stuck. If I sub in du/dt I keep getting a theta dot term. But I cant just assume u dot is independent of u and take it out the integral either. Brain is really hitting brick wall after brick wall with this
     
  9. Mar 1, 2016 #8

    Orodruin

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    So start by solving the integral then.
     
  10. Mar 1, 2016 #9
    Solving the integral IS the question. If i could solve the integral I wouldn't be asking for help on solving the integral.
     
  11. Mar 1, 2016 #10

    Orodruin

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    Well, so use the substitution in it. It is a standard integral.
     
  12. Mar 1, 2016 #11
    The integral is now [itex] \int d\theta (c_2 + \frac{c_1^2}{u^2 -1})^{-1/2}= \int dt[/itex] As I find it. But i get this simplifying to [itex] \int du \frac{1}{u^2 - a^2} [/itex] for the LHS where a^2 is a combination of the other constants. But this is a logarithm when integrating which doesnt make sense for the answer.
     
  13. Mar 1, 2016 #12
    [itex] \int 1/(c - 1/2sin^2x)^{1/2} dx[/itex] is what ive got now
     
  14. Mar 1, 2016 #13

    Orodruin

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    You are not doing the arithmetics properly then. Please show your work.
     
  15. Mar 2, 2016 #14
    That was the correct integral bar a constant, which has now been solved. Thanks anyway
     
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