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Calculus of variations applied to geodesics

  1. Oct 22, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm working on a problem from my gravitation book. The question is the following:

    Given \begin{equation} \frac{D}{Ds} T^\mu = 0 \end{equation}, where \begin{equation} T^\mu \left(s,a\right) = \frac{\partial z^\mu}{\partial s} \end{equation} is the tangent vector to a certain geodesic from a family of geodesics, indicated by parameter "a", at a distance s (coordinates are $$z^\mu$$ ). I now have to vary this equation with respect to a and point out the term in the resulting equation that is not covariant. I worked out a solution, but cannot find this noncovariant term.

    2. Relevant equations

    \begin{equation} \delta \left( \left( \frac{D}{Ds} T^\mu \right) \right)= 0 \end{equation}

    3. The attempt at a solution

    Variation:
    \begin{equation} \delta \left( \frac{D}{Ds} T^\mu = 0 \right) \end{equation}
    I tried the following:
    \begin{equation} \frac{D}{Ds} T^\mu = T^\nu \nabla_\nu T^\mu =0 \end{equation}
    and
    \begin{equation} \delta \rightarrow \delta a \frac{\partial z^\lambda}{\partial a} \nabla_\lambda \end{equation}
    I then get
    \begin{equation} \delta a \frac{\partial z^\lambda}{\partial a} \left( \left( \nabla_\lambda T^\nu \right) \nabla_\nu T^\mu + T^\nu \nabla_\lambda \nabla_\nu T^\mu \right) =0 \end{equation}
    All the terms between brackets seem however covariant. What is wrong in the derivation above? Should there be a partial derivative instead of a covariant derivative in the chain rule I used for the variation?
     
    Last edited: Oct 22, 2014
  2. jcsd
  3. Oct 27, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 30, 2014 #3
    I asked a fellow student yesteray and he was able to shed some light on the matter. So my question is pretty much answered :)
     
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