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Static + Hypersurface Orthogonality

  1. Jun 28, 2014 #1
    Static spacetimes can be defined as having no [itex]g_{tx}[/itex] component of the metric.

    Alternatively we can say that they are foliated by a bunch of spacelike hypersurfaces to which the Killing vector field [itex]\frac{\partial}{\partial t}[/itex] is orthogonal.

    How are these two statements consistent?

    [itex]g_{tx}=0 \Rightarrow g(\frac{\partial}{\partial t}, \frac{\partial}{\partial x})=0[/itex] but I always thought this meant there was no distance between timelike and spacelike vectors rather than a statement about them being orthogonal?

    Can someone please clear this up for me.

    Thanks.
     
  2. jcsd
  3. Jun 28, 2014 #2

    PAllen

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    √g(v1,v2) is the definition of dot product. Dot product=0 is the definition of orthogonality, in this case, 4-orthonality.
     
  4. Jun 28, 2014 #3
    Why do you have a square root? This would correspond to "ds" rather than ds^2 and I though ds was an infinitesimal line element i.e. a measure of distance?

    If the metric is just a statement about orthogonality, how is it connected to length? Or is it only a statement about orthogonality when it vanishes? I seem to be getting myself very confused about this haha
     
  5. Jun 28, 2014 #4

    Bill_K

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    No, for example the dear old Friedmann cosmology has gtx = 0 but is not static. And contrariwise you can take a static spacetime like Schwarzschild and write it in a coordinate system for which gtx ≠ 0
     
    Last edited: Jun 28, 2014
  6. Jun 28, 2014 #5
    So the correct defn is the hypersurface stuff because it is coordinate invariant.

    Is it correct to say that if it's static then it's always possible to find a coordinate system with [itex]g_{tx}=0[/itex]?
     
  7. Jun 28, 2014 #6

    WannabeNewton

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    Static means stationary and hypersurface orthogonal. In geometric terms this means there exists a time-like Killing field ##\xi## for the space-time and this Killing field is hypersurface orthogonal i.e. (locally) there exists scalar fields ##f,g## such that ##\xi^{\flat} = g df##. It is very easy to show that this implies there exists a coordinate system ##\{x^{\mu}\}## on the space-time such that ##\partial_t g_{\mu\nu} = 0## and ##g_{ti} = 0##. You will find this shown in e.g. Wald, see chapters 6 and 7.
     
  8. Dec 8, 2014 #7
    there might be a notational problem in the above, g(V,V)=g^{ab}V_aV_b
     
  9. Dec 9, 2014 #8

    PeterDonis

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    You have the indexes backwards, at least if you are referring to the implicit statement in the OP that ##g_{tx} = g \left( \partial / \partial t, \partial / \partial x \right)##. That translates to ##g \left( V, V \right) = g_{ab} V^a V^b##, with the metric indexes lower and the vector indexes upper. Also, the two vectors aren't the same, so you should really be writing ##g \left( V, W \right) = g_{ab} V^a W^b##.

    Then, if ##V## and ##W## are coordinate basis vectors, as ##\partial / \partial t## and ##\partial / \partial x## are, then the components ##V^a## and ##W^b## are either 1 or 0, depending on which basis vector we are looking at. Thus, ##\partial / \partial t = (1, 0, 0, 0)## and ##\partial / \partial x = (0, 1, 0, 0)##, so the formula reduces to ##g \left( \partial / \partial t, \partial / \partial x \right) = g_{ab} \left( \partial / \partial t \right)^a \left( \partial / \partial x \right)^b = g_{tx}##, as the OP wrote.
     
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