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The unit vector normal to the curvature of spacetime

  1. Nov 11, 2011 #1
    Let us say there is a curved region of spacetime whose curvature is [tex]\kappa(s)[/tex]. How does one find the coordinates of the unit vector normal to a certain point on the region of spacetime? I tried searching Hamilton's principle and the general theory of relativity but I could not find any equation which describes the coordinates of the normal vector to the curved region of spacetime.

    Thanks.
     
  2. jcsd
  3. Nov 11, 2011 #2

    George Jones

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    In classical general relativity, there no normal to spacetime curavture. Since spacetime is all there is, how could something be normal to it.

    Perhaps you mean the normal to a "curved" hypersurface. For example, if a 3-dimensional hypersurface is picked out by the level set [itex]\Phi \left( x^\mu \right) = 0[/itex], then [itex]\partial \Phi / \partial x^\mu[/itex] is normal to the hypersurface.

    Perhaps you're thinking of spacetime as a brane in a higher-dimensional bulk, in which case there is a normal to spacetime sticking out into the bulk.
     
  4. Nov 11, 2011 #3
    Yes. That is what I am asking.
     
  5. Nov 12, 2011 #4

    Bill_K

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    There's a considerable body of work relating to the differential geometry of a surface embedded in a manifold. I can only summarize.

    You need to use two sets of coordinates: uα in the surface and xi in the manifold. You need the relationship between the two: xiα ≡ ∂xi/∂uα. You need metric tensors aαβ in the surface and gij in the manifold. You need a covariant derivative that includes both sets of Christoffel symbols: one for the x's and one for the u's.

    Then you show that xiα;β = bαβ ni, where ni is the unit normal vector and bαβ is a symmetric surface tensor, the "second fundamental form". One can then derive the Weingarten formulas, ni = - aβγbβαxiγ and solve for n.
     
  6. Nov 12, 2011 #5
    Thanks.
     
  7. Nov 12, 2011 #6
    How can there be a unique vector, unless the (locally isometric) product space is R^5?
     
  8. Nov 13, 2011 #7

    Bill_K

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    Phrak, You're right, this assumes we're dealing with an n-dimensional surface embedded in an n+1-dimensional space. In the general case one has a set of equations generalizing the Serret-Frenet formulas for a curve, namely there will be a principal normal, second normal, etc.
     
  9. Nov 14, 2011 #8
    Ok, assuming we have a embedded 4 surface in 5 dimenions-

    I don't follow your math, but I would not start with contravariant vectors but densities.

    [tex]A_a = {\epsilon_a}^{bcde} \sqrt{-g} B_{bcde}[/tex]

    It seems that the dual of A is a normal vector to the surface at any point where B=B(w,x,y,z,t) is any 4-form. I couldn't say if this is equivalent to your formulation.

    Edit: I should have specificed that B is a 4-form at a point lying in the 4-surface.
     
    Last edited: Nov 14, 2011
  10. Nov 14, 2011 #9

    Bill_K

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    The problem is we don't have the embedding. We start given only the intrinsic geometry of the surface and try to construct an embedding. It may not exist, and if it does exist it may not be unique. For instance a flat 4-space immersed in 5-space could be rolled up like a piece of paper.

    In the internal coordinate system we have only εabcd to deal with, and a 4-form is equivalent to a scalar. Constructing the normal vector requires 5-dimensional coordinates, and we aren't given the relationship between them, xiα ≡ ∂xi/∂uα.
     
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