General relativity, geodesic, KVF, chain rule covariant derivatives

[binbagsss]

Homework Statement

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##

Homework Equations

see above

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

 

[Orodruin]

In normal calculus when you use the chain rule, the order doesn't matter does it?

Yes it does.

 

[binbagsss]

Yes it does.

I never knew this, what is the order?