Geodesic vs null geodesic

  • Thread starter kkz23691
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  • #26
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I'm just wondering about the specific example in the textbook...
Sorry to be a bit scattered, I have a number of relativity texts and have been looking at them, just don't have the time today to go through them to find a close example. I thought the problem and the question made sense in general, as martinbn said:
Yes, but the question was in principle. If the point and the vector at the point are given, there is a geodesic through that point with the given vector as a tangent vector, for any metric.
This is what I thought martinbn meant in his initial post#2. My issue is with the null vector. I understand the meaning of "initial point"; what is the meaning of initial "null vector"? I would assume this is an initial "zero tangent". Is this a correct assumption? The expression for the zero tangent is
##g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}=0## (the vector has zero length)
which is confusing, because in it I see derivatives of the coordinates w/respect to the parameter ##\lambda##, while the initial conditions given in that example were time derivatives ##\dot{r}=\dot{z}=0##.
I guess then we can still write
##g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}=g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}(\frac{dt}{d\lambda})^2=0##

But the vector is null, which is one more condition.
martinbn, could you explain the condition for the "null vector"...
 
  • #27
martinbn
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I think the derivatives, denoted by the dots, are derivatives with respect to the affine paramter of the curve. That parameter need not have any interpretation as time. As PeterDonis said, perhaps there is more in the example, may be something is understood from the context it was given in.
 

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