Timelike geodesic

barnflakes

My lecturer has written:

$\ddot x^{\mu} + \Gamma^{\mu}{}_{\alpha \beta} \dot x^{\alpha} \dot x^{\beta} = 0$ where differentiation is with respect to some path parameter $\lambda$.

If we choose $\lambda$ equal to proper time $\tau$ then it can be readily proved that

$c^2 = g_{\mu \nu}(x) \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau}$

Only problem is I can't quite see how to go from the first to the second, can someone explain for me please?

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hamster143

The second does not follow from the first. The second is just a statement that proper time is normalized in such a way that the magnitude of the 4-velocity $dx^{\mu} / d\tau$ is c.

barnflakes

Ahar, thank you hamster, makes sense now.

My lecturer has written something like this:

$R_{\mu \nu} - \frac{1}{2}R g_{\mu \nu} + \Lambda g_{\mu \nu} = 0$

"Now contract indices on both sides:

$R^{\mu}{}_{\mu} - \frac{1}{2} g^{\mu}{}_{\mu}R + \Lambda g^{\mu}{}_{\mu} = 0$

Can someone explain exactly what "contraction" he has done he? I assume he means multiplying by the metric tensor but I'm not sure exactly what metric tensor multiplication has gone on here?

nicksauce

Homework Helper
It is just multiplying both sides by $g_{\mu\nu}$.

barnflakes

It is just multiplying both sides by $g_{\mu\nu}$.
I was just coming online to say don't bother replying I figured it out but you beat me to it haha, thank you.

I figured out it was just multiplying by $g_{\mu\nu}$ and then the fact you have nu's instead of mu's makes no difference since it's just a dummy index. Thanks anyway :)

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