Calculus of variations applied to geodesics

1. Oct 22, 2014

JakobM

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 $$\frac{D}{Ds} T^\mu = 0$$, where $$T^\mu \left(s,a\right) = \frac{\partial z^\mu}{\partial s}$$ 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

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

3. The attempt at a solution

Variation:
$$\delta \left( \frac{D}{Ds} T^\mu = 0 \right)$$
I tried the following:
$$\frac{D}{Ds} T^\mu = T^\nu \nabla_\nu T^\mu =0$$
and
$$\delta \rightarrow \delta a \frac{\partial z^\lambda}{\partial a} \nabla_\lambda$$
I then get
$$\delta a \frac{\partial z^\lambda}{\partial a} \left( \left( \nabla_\lambda T^\nu \right) \nabla_\nu T^\mu + T^\nu \nabla_\lambda \nabla_\nu T^\mu \right) =0$$
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. Oct 27, 2014