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Euler-Lagrange eqns confusion

  1. May 23, 2010 #1
    I have a Lagrangian [tex]L = \frac{R^2}{z^2} ( -\dot{t}^2 +\dot{x}^2 +\dot{y}^2 +\dot{z}^2)[/tex] and I want to find the Euler-Lagrange equations [tex]\frac{\partial L}{\partial q} = \frac{d}{ds} \frac{\partial L}{\partial \dot{q}}[/tex]
    I'm fine with the LHS and the partial differentiation on the RHS, but when it comes to the [tex]\frac{d}{ds} [/tex] I'm not sure which coordinates I'm supposed to consider as a function of s.

    Is it all of them (ie t, x, y, z, and their derivatives)
    Or is it only the one I'm doing the equation for (so for the z-equation that's just z and its derivative)?
     
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  3. May 23, 2010 #2

    George Jones

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    Consider the RHS of the z equation in two steps. First step: What is

    [tex]\frac{\partial L}{\partial \dot{z}}?[/tex]
     
  4. May 23, 2010 #3
    [tex]\frac{2R^2}{z^2}\dot{z}[/tex]
     
  5. May 23, 2010 #4

    George Jones

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    So

    [tex]\frac{d}{ds} \frac{\partial L}{\partial \dot{z}} = \frac{d}{ds} \left( \frac{2R^2}{z^2}\dot{z} \right) = ? [/tex]
     
  6. May 23, 2010 #5
    Well this is it - I don't know what should be considered a function of s.

    If it's just z-dot then [tex]\frac{2R^2}{z^2}\ddot{z}[/tex]

    If it's z-dot & z then [tex]\frac{2R^2}{z^2}\ddot{z} - \frac{4R^2}{z^3}\dot{z}^2[/tex]
     
  7. May 23, 2010 #6

    George Jones

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    If [itex]z[/itex] weren't a function of [itex]s[/itex], then [itex]\dot{z}[/itex] would always be zero.
     
  8. May 23, 2010 #7
    oops, haha good point. Thanks.
     
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