Timelike killing vektors

  • Thread starter befj0001
  • Start date
43
0
If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.

How can one prove that?
 

WannabeNewton

Science Advisor
5,774
530
Consider the vector field ##\xi^{\mu}## which in the coordinates for which ##\partial_0 g_{\mu\nu} = 0## takes the form ##\xi^{\mu} = \delta^{\mu}_0##.

Then ##\mathcal{L}_{\xi}g_{\mu\nu} = \partial_0 g_{\mu\nu} = 0## so ##\xi^{\mu}## is a time-like Killing field for this metric.

But this isn't the correct way to think about the time-like Killing field. Rather one first looks for the existence of a time-like vector field ##\xi^{\mu}## satisfying ##\nabla_{(\mu}\xi_{\nu)} = 0##. If such a vector field exists then there necessarily exists a coordinate system ##\{x^{\mu}\}## in which ##\partial_0 g_{\mu\nu} = 0## and ##\xi^{\mu} = \delta^{\mu}_0##.
 

Bill_K

Science Advisor
Insights Author
4,155
194
But this isn't the correct way to think about the time-like Killing field. Rather one first looks for the existence of a time-like vector field ##\xi^{\mu}## satisfying ##\nabla_{(\mu}\xi_{\nu)} = 0##. If such a vector field exists then there necessarily exists a coordinate system ##\{x^{\mu}\}## in which ##\partial_0 g_{\mu\nu} = 0## and ##\xi^{\mu} = \delta^{\mu}_0##.
Why is this the wrong way to think of it?? It seems to me a perfectly reasonable way to think of it.

Someone hands you an unfamiliar metric. What's the first thing you do -- write out Killing's Equation and try to solve it?

Or do you say, aha, I see it doesn't depend on one of the coordinates ζ, so there's an obvious Killing vector.
 
26,915
7,236
If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.
You're assuming that the "0" coordinate is timelike. That's not always the case. Consider, for example, Schwarzschild coordinates on Schwarzschild spacetime inside the horizon; there the "0" (i.e., ##t##) coordinate is spacelike, not timelike, but the metric is still independent of ##t##, so ##\partial / \partial t## is still a KVF--just not a timelike one.
 

Related Threads for: Timelike killing vektors

Replies
21
Views
4K
Replies
3
Views
562
  • Last Post
Replies
16
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
17
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
5K
Top