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Timelike Killing vector field and stationary spacetime

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  1. May 21, 2013 #1
    I am trying to understand why in the definition of a stationary spacetime the Killing vector field has to be timelike.

    It is required that the metric is time independent, i.e. the time translations [itex]x^0 \to x^0 + \epsilon[/itex] leave the metric unchanged. So the Killing vector is [itex]\xi^{\mu}=\delta_{0}^{\mu}[/itex]. In other words it is required that there exists a Killing vector that has only [itex]x^{0}[/itex] component and no space components (they are 0). But in general, a timlike vector can also have non zero space components. For example, the vector [itex](1,1/2)[/itex] in two dimensional Minkowski space, is timelike because [itex](1,1/\sqrt{2})^2=-1/2<0[/itex]. This vector is a linear combination of the two independent killing vectors (1,0) (time translations) and (0,1) (space translations) of the 2-d Minkowski spacetime, so it's also a killing vector and it generates translations in time and also in space.

    So, if there exists a timelike killing vector with non zero space components, then necessarily there exists the killing vector [itex]\delta_{0}^{\mu}[/itex]. So saying that there should exist timelike Killing vector is because of this? It is simply a more general statement than saying that "spacetime is stationary if it admits the Killing vector field [itex]\delta_{\mu}^{0}[/itex]"?
     
    Last edited: May 21, 2013
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  3. May 21, 2013 #2

    PeterDonis

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    Um, because that's how the term "stationary spacetime" is defined?

    No, that's not true in general. It's only true if you adopt an appropriate coordinate chart. The correct definition of a KVF is that it satisfies Killing's equation:

    [tex]\nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu} = 0[/tex]

    This will be true, if it's true, regardless of the coordinate chart you adopt.
     
  4. May 21, 2013 #3

    PeterDonis

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    This is also not correct in general; it happens to be true in Minkowski spacetime for the particular KVFs you picked, but it's not true in a lot of other spacetimes, or even for other KVFs in Minkowski spacetime.
     
  5. May 21, 2013 #4
    Thnx Peter! Indeed, the given definition is coordinate independent.
    By a coordinate transformation a general timelike vector (with non zero space components) can be transformed to the vector (1,0,0,0), and in this coordinate system, if the metric is independent of the time coordinate, then the vector (1,0,0,0) is a killing vector of this spacetime. But in an arbitrary coordinate system, the metric of stationary a spacetime may depend on time.
     
  6. May 21, 2013 #5

    WannabeNewton

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    You should realize that in an arbitrary coordinate system, "time" has no meaning. You should also realize that by a coordinate transformation, you are taking ##\xi^{a}## to ##\xi^{a} = (\frac{\partial }{\partial t})^{a}##. This is the vector, ##\delta^{\mu}_{t}## is just the coordinate representation - there's a difference.
     
  7. May 21, 2013 #6

    PeterDonis

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    As WannabeNewton points out, "time" in an arbitrary coordinate system doesn't have any direct physical meaning. (In fact, it's possible to have a coordinate chart without any of the coordinates even being timelike.) That's one good reason why coordinate-independent definitions are preferred.
     
  8. May 21, 2013 #7
    What do you mean that time has no meaning? When I say time coordinate and space coordinates, I mean ##x^0## and ##x^i## respectively. Should I write timelike coordinate instead of time coordinate? Is there any difference?

    On the second I agree, the (1,0,0,0) is the representation of the vector ##\partial/\partial t## in that specific coordinate system, so I should have wrote what you say.
     
  9. May 21, 2013 #8
    When does it have a physical meaning?
     
  10. May 21, 2013 #9

    WannabeNewton

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    If your coordinate system is constructed in such a way that it represents space-time being foliated by a one-parameter family of space-like hypersurfaces ##\Sigma_{t}##, with ##t## being the time-like coordinate of the coordinate system and the vector field ##\nabla^{a}t## being hypersurface orthogonal to ##\Sigma_{t}##, you can interpret the space-like hypersurfaces as "snapshots" of the space-time at different "times" ##t = \text{const.}## and interpret ##\nabla^{a}t ## as the "flow of time".
     
  11. May 21, 2013 #10
    "A spacetime is stationary when all the metric components are time independent in some special coordinate system, ##\partial g_{ab}/\partial x^0=0## "

    In this special coordinate system the ## x^0## coordinate is the ordinary time t which has meaning :) ? It says time independent...
     
  12. May 21, 2013 #11

    WannabeNewton

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    A space-time is stationary if there exists a one-parameter group of isometries whose orbits are time-like curves. This one parameter family of isometries generates a vector field whose flows lie transport the metric tensor. We interpret this as "time-translation invariance" on account of the orbits being time-like curves. This is as much physical meaning as you can give to the word "time" in a general context with regards to the above.
     
  13. May 21, 2013 #12

    PeterDonis

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    Are you quoting from something? If so, please give a link to the source. I strongly suspect it's not a very rigorous source.
     
  14. May 21, 2013 #13

    PeterDonis

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    It depends on the spacetime and on what particular objects in that spacetime are of interest to you.

    For example, in some cases, you can find a coordinate chart such that objects of interest to you have constant values for 3 out of 4 coordinates in that chart. If the 4th coordinate is timelike, then that coordinate will often have a direct physical interpretation as the proper time of the objects of interest. But there's no guarantee that you can always find such a chart.
     
  15. May 21, 2013 #14
    It's from Ray D'Inverno's "Introducing Einstein's Relativity", chapter 14.1 Stationary solutions.
    Then he says that ##x^0## is timelike. Is this different from time t? What does it mean that a coordinate is timelike?
     
    Last edited: May 21, 2013
  16. May 21, 2013 #15

    PeterDonis

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    In what chart? As I said, there are charts for which there is not even a timelike coordinate at all, including the one labeled ##x^0##. Just labeling a coordinate as ##x^0## doesn't make it a time coordinate, or a timelike coordinate.

    (As far as "time" vs. "timelike" is concerned, I don't think there's much difference between them, but that's more a matter of terminology than physics. I tend to prefer "timelike" because it has an obvious mathematical definition: a coordinate ##x^{\mu}## is timelike if its corresponding partial derivative ##\partial_{\mu}## defines a timelike vector field. People use "time" to mean all sorts of things, so that term seems less clear to me.)

    (Note, btw, that the definition of "timelike" I just gave makes it coordinate-dependent; the same coordinate can be timelike in one chart and not in another, because the partial derivatives are taken holding the other coordinates constant, and how that is done depends on the chart.)
     
  17. May 21, 2013 #16

    PeterDonis

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    Ah, ok. I can't exactly say that's a non-rigorous source :wink:, but it is (unfortunately, IMO) true that a lot of textbooks are not as rigorous as they could be about definitions. Here's how I would rephrase the definition:

    "A spacetime is stationary if it possesses a timelike Killing vector field, which is a vector field ##\xi^{\mu}## that satisfies Killing's equation. It can be shown that, if there exists such a vector field, one can always find a coordinate chart such that the following is true: (i) there is one coordinate, which we can without loss of generality label ##x^0##, such that ##\partial / \partial x^0## is the coordinate representation of ##\xi^{\mu}##--i.e., the components of ##\xi^{\mu}## in this chart are ##(1, 0, 0, 0)##; and (ii) all of the metric coefficients are independent of ##x^0##."

    This should make the logical progression clearer.
     
  18. May 21, 2013 #17

    WannabeNewton

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    Coordinate systems do not necessarily have to be related in any way to one setup by an observer in his/her frame. What you are thinking of is more along the lines of a special tetrad ##(e_{\alpha})^{a}## of basis vectors transported along the worldline of an observer in space-time. In this frame, the observer uses the proper time along his/her worldline to define the "time" coordinate so that ##(e_0)^{a} = u^{a}## where ##u^{a}## is his/her 4-velocity. This lends itself to a more operational interpretation of the "time" coordinate. This tertrad is used, for example, in Fermi-Walker transport.
     
  19. May 21, 2013 #18
    Actually later he defines a stationary spacetime more rigorously as you did, so the book is ok :)

    I think I understand it now (at least much better than before). Thank you very much for your help, both of you!
     
  20. May 21, 2013 #19

    WannabeNewton

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    Yeah the word time-like has precise meaning. I feel like the word "time" is thrown around so much in such a non-chalant way even though it doesn't always have the meaning one would think in the everyday connotation of the word.
     
  21. May 21, 2013 #20

    WannabeNewton

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    No problem, and keep asking the questions like you have thus far. It is really nice! I would strongly recommend the book "General Relativity" - Wald if you want to see more mathematically rigorous definitions of everything in GR as you seem to like seeing. Most of the definitions Peter and I gave in this and other threads of yours (like the spherically symmetric space-times one) come from Wald.

    By the way, I have D'Inverno as well and I would say while it is a good book, it is not quite as modern as Wald (although Wald is itself outdated in a sense).
     
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