Timelike geodesic

  • Thread starter barnflakes
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  • #1
barnflakes
156
4
My lecturer has written:

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

If we choose [itex]\lambda[/itex] equal to proper time [itex]\tau[/itex] then it can be readily proved that

[itex]c^2 = g_{\mu \nu}(x) \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau}[/itex]

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

Answers and Replies

  • #2
hamster143
908
2
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 [itex]dx^{\mu} / d\tau[/itex] is c.
 
  • #3
barnflakes
156
4
Ahar, thank you hamster, makes sense now.

My lecturer has written something like this:

[itex]R_{\mu \nu} - \frac{1}{2}R g_{\mu \nu} + \Lambda g_{\mu \nu} = 0[/itex]

"Now contract indices on both sides:

[itex]R^{\mu}{}_{\mu} - \frac{1}{2} g^{\mu}{}_{\mu}R + \Lambda g^{\mu}{}_{\mu} = 0[/itex]

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?
 
  • #4
nicksauce
Science Advisor
Homework Helper
1,272
5
It is just multiplying both sides by [itex]g_{\mu\nu}[/itex].
 
  • #5
barnflakes
156
4
It is just multiplying both sides by [itex]g_{\mu\nu}[/itex].

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 [itex]g_{\mu\nu}[/itex] 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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