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Metric for a surface of a cone

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data
    The metric for this surface is [itex]ds^2 = dr^2 + r^2\omega^2d\phi^2[/itex], where [itex]\omega = sin(\theta_0)[/itex].
    Solve the Euler-Lagrange equation for phi to show that [itex]\dot{\phi} = \frac{k}{\omega^2r^2}[/itex]. Then sub back in to the metric to get [itex]\dot{r}[/itex]


    2. Relevant equations
    [itex]L = 1/2 g_{ab} \dot{x}^a \dot{x}^b[/itex]


    3. The attempt at a solution
    I've solved it to get [itex]\ddot{\phi} + 2\frac{\dot{r}}{r}\dot{\phi} = 0 [/itex]
    and
    [itex]\ddot{r} - r\omega^2\dot{\phi}^2 = 0[/itex]

    So how on earth do you get that answer?
     
  2. jcsd
  3. Feb 16, 2014 #2

    TSny

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    Note how you got this equation. Back up a step where you must have had ##\frac{d}{dt} (\rm { expression}) = 0##

    What can you conclude about the expression?
     
  4. Feb 16, 2014 #3
    Ok I get it! [itex] d/dt (\dot{\phi} r^2 \omega^2)=0[/itex]. So [itex]\dot{\phi} = k/(r^2\omega^2)[/itex]
    Thanks!!
     
  5. Feb 16, 2014 #4

    TSny

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    That's it. Good!
     
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