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Question about tensor derivative

  1. Jan 15, 2010 #1

    I'm beginning to learn GR and ran across the following in the beginning of Stephani's book.

    In deriving the equations of motion for a force-free particle, we have the Lagrangian:
    [tex]L = \frac{m}{2} g_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta[/tex]

    To construct the equations of motion we need
    [tex]\frac{\partial L}{\partial \dot{x}^\nu} = m g_{\alpha \nu} \dot{x}^\alpha[/tex]


    [tex]\frac{\partial L}{\partial x^\nu} = L_{,\nu} = \frac{m}{2} g_{\alpha \beta, \nu} \dot{x}^\alpha \dot{x}^\beta[/tex]

    Then from the Lagrange equations we get

    [tex]g_{\alpha \nu }\ddot{x}^\alpha + g_{\alpha \nu, \beta} \dot{x}^\alpha \dot{x}^\beta - \frac{1}{2}g_{\alpha \beta, \nu}\dot{x}^\alpha \dot{x}^\beta =0[/tex]

    I don't understand where the second term in the last equation comes from.
  2. jcsd
  3. Jan 15, 2010 #2

    George Jones

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    It comes from the product and chain rules:

    [tex]\frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{x}^\nu} \right) = m \frac{d}{d \tau} \left( g_{\alpha \nu} \dot{x}^\alpha \right) = m \frac{d}{d \tau} \left( g_{\alpha \nu} \right) \dot{x}^\alpha + m g_{\alpha \nu} \frac{d}{d \tau} \left( \dot{x}^\alpha \right). [/tex]

    In the second-last term,

    [tex]\frac{d}{d \tau} = \frac{d x^\beta}{d \tau}\frac{\partial}{\partial x^\beta}[/tex]

    is used.
  4. Jan 15, 2010 #3
    Oh ok. Thank you.

    I wasn't realizing the metric tensor was a function of time. Now that I think about it, it makes sense that it is.
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