# Calculus of variations applied to geodesics

## 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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