# General Relativity- geodesics, killing vector, Conserved quantity Schwarzschild metric

## Homework Statement

Conserved quantity Schwarzschild metric.

## The Attempt at a Solution

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##\partial_u=\delta^u_i=k^u## is the KVF ##i=1,2,3##

We have that along a geodesic ##K=k^uV_u## is constant , where ##V^u ## is the tangent vector to some affinely parameterised geodesic.

For example if we take the Schwarzschild metric, ##K^u=(1,0,0,0)## , ##V^u=(\dot{t},\dot{r},\dot{\theta},\dot{\phi})## so we get ##K= (1-\frac{2GM}{r})\dot{t}## is conserved, for example.

where dot denotes a derivative with respect to some affine parameter ##s##

QUESTION:

What here is to say that ##s## is an affine parameter?
I.e- why is ##V^u=(\dot{t},\dot{r},\dot{\theta},\dot{\phi})## the tangent vector to an affinely parameterised geodesic?

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## Answers and Replies

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You assumed it to be an affine parameter and then wonder why it is affine?

If it is not an affine parameter ##k^u V_u## will not be a conserved quantity.

PeterDonis
Mentor
2020 Award
Moderator's note: moved to homework.

You assumed it to be an affine parameter and then wonder why it is affine?

If it is not an affine parameter ##k^u V_u## will not be a conserved quantity.

ahh apologies got it yes, it is in the proof showing that ##k^u V_u## will not be a conserved quantity is a conserved quantity.