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Using Euler Lagrange in General Relativity

  1. Mar 13, 2008 #1
    I'm trying to follow a professor's notes for finding Christoffel symbols for a two-sphere. He gives the following two equations:

    The Lagrangian for a two sphere:[tex]L = \left( \frac{d\theta}{ds} \right)^2 + sin^2\theta \left( \frac{d\phi}{ds} \right)^2[/tex]

    The Euler Lagrange equation:[tex] \frac{d}{ds} \left( \frac{\partial L}{\partial (dx^\mu/ds)} \right ) - \frac{\partial L}{\partial x^\mu} = 0[/tex]

    Using these, the professor magically gets:

    for [tex]x^\mu = \theta[/tex]:[tex]2\frac{d^2\theta}{ds^2} - 2 sin\theta cos\theta \left(\frac{d\phi}{ds^2}\right)[/tex]

    for [tex]x^\mu = \phi[/tex]:[tex]2sin^2 \theta \frac{d^2\phi}{ds^2} + 4 sin\theta cos\theta \left(\frac{d\theta}{ds}\frac{d\phi}{ds}\right)[/tex]

    And the Christoffel symbols can be found with minimal effort. The problem is that I can't follow the derivation to get the equations. I feel like this is a really simple thing, yet I'm having trouble getting the same answer as he showed.


    For example, in the [tex]\theta[/tex] case:

    [tex] \frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left( \left( \frac{d\theta}{ds} \right)^2\right)\right )
    +
    \frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right )
    -
    \frac{\partial}{\partial \theta}\left(\left( \frac{d\theta}{ds} \right)^2\right)
    -
    \frac{\partial}{\partial \theta}\left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right) = 0[/tex]

    [tex] 2 \frac{d^2\theta}{ds^2}
    +
    \frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right )
    -
    \frac{\partial}{\partial \theta}\left(\left( \frac{d\theta}{ds} \right)^2\right)
    -



    2 sin\theta cos\theta \left(
    \frac{d\phi}{ds}
    \right)^2


    = 0[/tex]

    But I don't see how to drive the other terms to zero.

    The [tex]\phi[/tex] case is even worse:

    [tex]
    \frac{d}{ds}
    \left(
    \frac{\partial}{\partial (d\phi/ds)}
    \left(
    \left( \frac{d\theta}{ds} \right)^2
    \right)
    \right )
    +
    \frac{d}{ds}
    \left(
    \frac{\partial}{\partial (d\phi/ds)}
    \left(
    sin^2\theta \left( \frac{d\phi}{ds} \right)^2
    \right)
    \right )

    -
    \frac{\partial}{\partial \phi}
    \left(
    \left( \frac{d\theta}{ds} \right)^2
    \right)

    -
    \frac{\partial}{\partial \phi}
    \left(
    sin^2\theta \left( \frac{d\phi}{ds} \right)^2
    \right)

    = 0[/tex]

    [tex]
    \frac{d}{ds}
    \left(
    \frac{\partial}{\partial (d\phi/ds)}
    \left(
    \left( \frac{d\theta}{ds} \right)^2
    \right)
    \right )
    +
    \frac{d}{ds}
    \left(
    2 sin^2\theta \left( \frac{d\phi}{ds} \right)
    \right )

    -
    \frac{\partial}{\partial \phi}
    \left(
    \left( \frac{d\theta}{ds} \right)^2
    \right)

    -
    \frac{\partial}{\partial \phi}
    \left(
    sin^2\theta \left( \frac{d\phi}{ds} \right)^2
    \right)

    = 0[/tex]

    I'm not sure how to simplify it from there, since each time I try a method I get the wrong answer. I especially don't see where his [tex]+4sin\theta cos\theta[/tex] came from.

    All my attempts have failed. It seems like such a trivial thing, since the professor left it out. And obviously the results are correct since they give the correct Christoffel terms. I would be eternally grateful if someone with a working knowledge of how to combine ordinary and partial derivatives could give me a step by step.
     
  2. jcsd
  3. Mar 15, 2008 #2

    Mentz114

    User Avatar
    Gold Member

    [tex]\frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right ) \equiv \frac{d}{ds} \left( \frac{\partial}{\partial y} \left(sin^2\theta \left( z \right)^2\right)\right )[/tex]

    This term is zero, because [tex]\left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right ) [/tex]
    is not explicitly dependent on [tex]\frac{d\theta}{ds}[/tex].

    All your troublesome terms disappear for similar reasons.
     
  4. Mar 15, 2008 #3

    kdv

    User Avatar

    Mentz114 already gave you the explanation. Are you familiar with the Lagrangian approach in ordinary mechanics? If so, recall that is the time derivative of a variable does not appear in the lagrangian the corresponding conjugate momentum is conserved because [tex] \frac{\partial L}{\partial \dot q} = 0 [/tex] . It's the same type of thing here.

    Maybe it would be useful to write the Lagrangian as

    [tex] L = \dot{\theta}^2 + \sin^2 \theta (\dot{\phi})^2 [/tex]


    where a dot indicates a derivative with respect to s, obviously.

    And now you must think of [itex] \theta [/itex] and [itex] \dot{\theta} [/itex] as independent variable (same for [itex] \phi[/itex] and [itex] \dot{\phi} [/itex]).

    For example,

    [tex] \frac{\partial L}{\partial \dot{\theta}} = 2 \dot{\theta} [/tex]
    simply.

    Then it's a snap to get the correct result from the E-L equations
     
  5. Mar 17, 2008 #4
    That was my problem. Thanks guys.
     
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