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Stationary/ Static Conditions Metric?

  1. Mar 14, 2015 #1
    What are the sufficient / necessary conditions for a metric to be stationary / static?

    - If the metric components are independent of time in some coordinate system , is this sufficient for stationary?

    - I've read for static if a time-like killing vector is orthogonal to a family of hyper-surfaces than it's static and that this condition is equivalent to there being no cross terms containing ##dt##. How are these 2 conditions equivalent? I have no idea how to go about showing this?

    Many thanks in advance.
     
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  3. Mar 14, 2015 #2

    Matterwave

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    1. Yes, that condition is sufficient as the basis vectors associated with the time coordinate will be a killing field.

    2. Well, if there are no cross terms involving ##dt## then certainly the killing field will be orthogonal to the other 3 coordinate basis vectors right? So that part is easy to see. Showing that the other 3 coordinate basis vectors integrate to form a family of hypersurfaces which foliate the manifold requires the Frobenius theorem. Frobenius's theorem basically says that if a set of vector fields X,Y,Z, etc., have lie brackets which are all contained in this set, then the vector fields will define a foliation. Coordinate basis vectors definitely fulfill this requirement since they all have lie bracket 0 with each other.

    Actually, some authors (e.g. Schutz) prefers to use the definition of hypersurface as being a subset of the manifold for which one can find a coordinate system on the manifold where the hypersurface is described by setting one of the coordinates to 0. If that is your definition, then property 2 above should be self-evident.
     
    Last edited: Mar 14, 2015
  4. Apr 14, 2015 #3
    I haven't heard/looked into lie derivatives, but, from what I've read I think we need to check that ##[Lx,Ly]\partial x,\partial y, \partial z=0##, ##[Lx,Lz]\partial x,\partial y, \partial z=0## and ##[Ly,Lz]\partial x,\partial y , \partial z=0##, so nine cases. This is made easier for the case of vector fields, as the basis vectors are, since we have the property ##Lx (Y) =[X,Y]##.

    All commutators for the basis vectors, ##[\partial_{i},\partial_{j}] =0## and hence the lie bracket ##[Lx,Ly]## is.

    Are my thoughts on the right track?

    Thanks.
     
  5. Apr 14, 2015 #4

    Matterwave

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    I'm not sure about your notations, but it is definitely true that coordinate basis vectors have 0 Lie bracket with each other. That is one of the defining qualities of a coordinate basis. A non-coordinate basis will have non-0 Lie brackets between (at least some of) the basis vectors.
     
  6. Apr 19, 2015 #5
    Frobenius's theorem basically says that if a set of vector fields X,Y,Z, etc., have lie brackets ...[/QUOTE]

    Just a quick question..this is equivalent to 'have commutators' right,, for the case of vector fields?
    Why do you state lie brackets- does Frobenius's theorem generalise to any tensor field?

    Thanks.
     
  7. Apr 19, 2015 #6

    Matterwave

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    Just a quick question..this is equivalent to 'have commutators' right,, for the case of vector fields?
    Why do you state lie brackets- does Frobenius's theorem generalise to any tensor field?

    Thanks.[/QUOTE]

    Lie brackets work on vector fields...commutators (terminology originating from QM) work for operators which vector fields do not necessarily have to be. This is a terminology thing, but this is the terminology that I found in most of my sources.

    I don't know what the Lie bracket of tensor fields is, only the Lie derivative of tensor fields.

    There's also a Frobenius's theorem that applies to a set of one forms (fields) rather than vector fields.
     
  8. Apr 19, 2015 #7

    PeterDonis

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    Yes, as Matterwave says, provided that the coordinate you're calling "time" is actually timelike. If it isn't, then while the basis vector for that coordinate still defines a Killing vector field, it is not a timelike one. This happens, for example, at and inside the horizon in Schwarzschild spacetime.

    Remember that if two vectors ##A^a## and ##B^b## are orthogonal, that means their inner product is zero, i.e., ##g_{ab} A^a B^b = 0##. If ##g_{ab}## is of the form ##g_{00}, g_{ii}##, i.e., no cross terms between the "0" coordinate (time, assumed timelike per the above) and the "i" coordinates (the space coordinates), then the vector ##\partial / \partial t##, given by ##A^a = (1, 0, 0, 0)##, must be orthogonal to any vector of the form ##B^b = (0, B^1, B^2, B^3)##. And any vector that lies in a surface of constant ##t## must be of the latter form.
     
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