# General relativity, geodesic, KVF, chain rule covariant derivatives

1. Jun 25, 2017

### binbagsss

1. The problem statement, all variables and given/known data

To show that $K=V^uK_u$ is conserved along an affinely parameterised geodesic with $V^u$ the tangent vector to some affinely parameterised geodesic and $K_u$ a killing vector field satisfying $\nabla_a K_b+\nabla_b K_a=0$

2. Relevant equations
see above
3. The attempt at a solution

Proof:

So the proof is to use the chain rule that $\frac{d}{ds}=$ , where $s$ is some affine parameter:

$\implies \frac{dK}{ds}=V^{\alpha}\nabla_{\alpha}(K_uV^u)= V^u V^{\alpha}\nabla_{\alpha}K_u + K_u V^{\alpha}\nabla_{\alpha}V^u$ ; first term is zero from KVF equation - antisymetric tensor multiplied by a symmetric tensor, and the second term is zero from the geodesic equation

MY QUESTION - this may be a stupid question, but concerning the order of the chain rule application since the covariant derivative operates on everything to the right...

How do you know to write $\frac{d}{ds}=V^{u}\nabla_u$ as a pose to $\frac{d}{ds}=\nabla_uV^u$

In normal calculus when you use the chain rule, the order doesn't matter does it? For e.g $\frac{d}{ds}=\frac{dx}{ds}\frac{d}{dx} = \frac{d}{dx}\frac{dx}{ds}$ ?

But if i try to apply the above proof writing $\frac{d}{ds}=\nabla_uV^u$ I get an extra non-zero term : $K_uV^u\nabla_{\alpha}V^{\alpha}$

so the proof fails.

Many thanks

2. Jun 26, 2017

### Orodruin

Staff Emeritus
Yes it does.

3. Jun 26, 2017

### binbagsss

I never knew this, what is the order?