Calculus of variations applied to geodesics

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Homework Statement



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.

Homework Equations



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

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?
 
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Answers and Replies

  • #2
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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?
 
  • #3
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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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