Timelike Geodesic: Proving c^2 from $\ddot x^{\mu}$

[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?

 

[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]

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 :)