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

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Main Question or Discussion Point

Homework Statement



Conserved quantity Schwarzschild metric.

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The Attempt at a Solution


[/B]
##\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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  • #2
Orodruin
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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.
 
  • #4
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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.
 

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