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A Definition of hypersurface orthogonal

  1. Oct 9, 2013 #1
    Definition of "hypersurface orthogonal"

    Hi all!
    I'm not sure if the thread belongs more to General Relativity or Differential Geometry, but I guess the border is labile.

    I've come across the term "hypersurface orthogonal" many times, but I still haven't found a clear definition. Apparently, if a flow is geodesic and irrotational, it is also hypersurface orthogonal, whatever this means.

    Could you please give me a definition and explain how the hypersurface orthogonality is linked to the geodesy and irrotationality of a flow?

    Thanks in advance!
     
  2. jcsd
  3. Oct 9, 2013 #2

    WannabeNewton

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    https://www.physicsforums.com/showpost.php?p=4483918&postcount=8

    Note that a hypersurface orthogonal vector field need not have integral curves that are geodesics; it simply needs to be irrotational. An example of a hypersurface orthogonal vector field that doesn't have geodesic integral curves is the 4-velocity field of the family of Rindler observers in flat space-time.
     
  4. Oct 10, 2013 #3
    Thank you for your reply!
    If I got it correct, I would say the definition is: "A (timelike?) vector [itex]\frac{\partial}{\partial\lambda}[/itex] is said to be hypersurface orthogonal if there exist a foliating of space-time into hypersurfaces of constant [itex]\lambda[/itex] such that [itex]\frac{\partial}{\partial\lambda}[/itex] is everywhere orthogonal to such hypersurfaces". Am I right?

    Is it possible to apply the term to non-timelike congruences?
    For example, I found this sentence in a paper: "It is proved that the vectors of this eigenframe are hypersurface orthogonal and consequently that a coordinate system exist in which the metric [...] is diagonal"
    For the metric to be diagonal, this must mean that you can foliate space-time so that, for instance, the y vector is orthogonal to the surfaces of constant y.
     
  5. Oct 10, 2013 #4

    WannabeNewton

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    Yes it can be applied to any vector field although and what you said is fine except that the term only applies to vector fields, not to vectors, At any given event in space-time there always exists a hypersurface that is orthogonal to a given vector so there isn't really anything interesting there.
     
  6. Oct 10, 2013 #5
    Thanks!
     
  7. Oct 10, 2013 #6

    WannabeNewton

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    No problem! By the way, I should probably mention that the simplest definition of hypersurface orthogonality of any vector field ##\xi^a## is simply that ##\xi^a = \alpha \nabla^a \beta## for smooth scalar fields ##\alpha,\beta##. It turns out that locally this is equivalent to ##\xi_{[a}\nabla_{b}\xi_{c]} = 0## (this is a result of Frobenius' theorem for integral subbundles) and this in turn is equivalent to ##\xi^a## being irrotational i.e. ##\epsilon^{abcd}\xi_b \nabla_c \xi_d = 0## which can be shown through a simple calculation.
     
  8. Oct 10, 2013 #7
    It makes sense. Thanks again!
     
  9. Oct 10, 2013 #8

    WannabeNewton

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    Anytime :) cheers!
     
  10. May 23, 2014 #9
    WBN: do you know if there are any existence theorem on hypersurface orthogonal vector fields?

    I.e. if you have a foliation by codimension one hypersurfaces - what must be satisfied for there to exists an everywhere hypersurface-orthogonal vector field?

    I recognize that this would be equivalent the existence of foliations by codimension one surfaces, and one-dimensional curves. The existence of the first does depend on the topology of the manifold. Ex: http://www.map.mpim-bonn.mpg.de/Foliations states a theorem that says that the existence of codimension one foliations depend on the relevant topological space having Euler characteristic 0!

    Thus, I wonder what the topological conditions on the first and the latter coexisting. And what if they are orthogonal to each other?
     
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