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Lagrangian in General Relativity

  1. Mar 7, 2009 #1
    It's not really a homework question, but it is a question I have in my mind while studying for my midterm and I can't get it out of my head. Somehow, it would change the speed at which I'm solving problems with lagrangian, studying for my exam.

    The book we are using in the course is Hartle's Gravity.

    Hartle defines the Lagragian and Euler-Lagrange's equations as :

    [itex]
    \begin{equation}
    L (\frac{dx^{\alpha}}{d\sigma},x^{\alpha})=(-g_{\alpha \beta}\frac{dx^{\alpha}}{d\sigma} \frac{dx^{\beta}}{d\sigma})^{\frac{1}{2}}
    \end{equation}
    \begin{equation}
    -\frac{d}{d\sigma}(\frac{\partial L}{\partial(\frac{dx^{\alpha}}{d\sigma})})+\frac{\partial L}{\partial x^{\alpha}}=0
    \end{equation}
    [/itex]
    Where [itex]$\sigma$[/itex] is a parameter which varies between 0 and 1, respectively at point A and B.

    Now, one example further, for the wormhole geometry,

    [itex]
    \begin{equation}
    ds^2=-dt^2+dr^2+(b^2+r^2)(d\theta^2+sin^2\theta d \phi^2)
    \end{equation}
    [/itex]
    Now, he finds the lagrangian for this metric as

    [itex]
    \begin{equation}
    L=[(\frac{dt}{d\sigma})^2-(\frac{dr}{d\sigma})^2-(b^2+r^2)( (\frac{d\theta}{d\sigma})^2+sin^2\theta(\frac{d\phi}{d\sigma})^2)]^{\frac{1}{2}}
    \end{equation}
    [/itex]
    Now, he mentions "In writing out Lagrange's equations, differentiating the square root in (8.12) produces a factor of 1/L. However, from 8.8, the value of L is [itex]$ d\tau / d\sigma$ [/itex]. The inverse factors of L can, therefore, be used to trade derivatives with respect to [itex] $\sigma$ [/itex] for derivatives with respect to [itex] $\tau$[/itex]."

    My question is : is that similar to write the new lagrangian as


    [itex]
    \begin{equation}
    L (\frac{dx^{\alpha}}{d\tau},x^{\alpha})=(-g_{\alpha \beta}\frac{dx^{\alpha}}{d\tau} \frac{dx^{\beta}}{d\tau})^{\frac{1}{2}}
    \end{equation}
    \begin{equation}
    -\frac{d}{d\tau}(\frac{\partial L}{\partial(\frac{dx^{\alpha}}{d\tau})})+\frac{\partial L}{\partial x^{\alpha}}=0
    \end{equation}
    [/itex]

    Is that formulation general enough? Is it the same? Because, well... how come tau varies only between 0 and 1 between two points A and B (that was a property of sigma)? The two equations somehow gives the same results if I want my geodesic equations in terms of tau. In an example or two, I have the same answers as my teacher.

    How about it?

    I have this question in mind because my teacher never explicitly wrote the equations above. Neither does Hartle. What puzzle me is that Euler-Lagrange's equations in terms of tau are, I think, more general and more conventional to use in place of sigma. But as they are not written explicitly, I assume I'm wrong somewhere.

    Hartle also mentionned that there are many parameters having the properties of sigma. One problem with this formulation, replacing sigma as tau in my Lagrangian, gives me the equation [itex] $ d\tau=d\tau L$[/itex], which gives L=1, which is somehow weird.

    Thank you!
     
  2. jcsd
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