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A GR question about null surfaces, vectors and coordinates

  1. Oct 8, 2013 #1
    I wondered anyone can explain the significance of the above as applied to metrics in the context of general relativity. This came up when the video lecturer in GR mentioned that r for example, was null or this or that vector or surface was null, say in the context of the eddington finkelstein coordinate system. I am unable to grasp the meaning or significance of the word null. I think null geodesics are clear however but even here not sure if there is any connection with null vectors etc.
     
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  3. Oct 8, 2013 #2

    PeterDonis

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    A null vector is a vector whose length is zero; that is, a vector ##k^a## such that ##g_{ab} k^a k^b = 0##, where ##g_{ab}## is the metric. One of the things that makes spacetime different from an ordinary Riemannian space is that you can have vectors with zero length that are not the zero vector.

    A null surface is a surface that has null tangent vectors; in other words, there are null vectors that lie completely within the surface. For example, the event horizon of a black hole is a null surface.

    A null geodesic is just a geodesic whose tangent vector is null; for example, the worldline of a light ray traveling through a vacuum is a null geodesic. So there is indeed a close connection between null vectors and null geodesics.
     
  4. Oct 8, 2013 #3
    Ok thanks I especially found helpful your comment about 4 vectors that have zero length but that are not the zero vector. A follow on question if ok : what then is a null coordinate.
     
  5. Oct 8, 2013 #4

    WannabeNewton

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    It means the vector field associated with the coordinate is null.
     
  6. Oct 8, 2013 #5
    Ok many thanks to you both for clarifying.
     
  7. Oct 8, 2013 #6

    WannabeNewton

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    By the way, in case you wanted an example, consider the surface ##\Sigma## given by ##r = 2M## in Schwarzschild space-time. The vector field ##\nabla r## is a normal vector field to ##\Sigma##. Now ##\nabla r = g^{\mu r}\partial_{\mu} = (1 - \frac{2M}{r})\partial_r## so ##g(\nabla r, \nabla r) = (1 - \frac{2M}{r})##. On ##\Sigma## then, ##g(\nabla r, \nabla r) = 0## so ##\nabla r## is null on ##\Sigma## meaning ##\Sigma## is a null hypersurface ("hyper" because it is of codimension 1); this is of course just the event horizon of the Schwarzschild black hole.

    Note that for any tangent vector ##v \in T_p \Sigma##, ##v \perp \nabla r## implies that ##v## is itself null so this is equivalent to Peter's definition above.
     
  8. Oct 9, 2013 #7
    Hi again, I am a bit unfamiliar with the notation here, how is g(del-r, del-r) defined?
     
  9. Jan 13, 2015 #8
    That's the scalar product (the dot product) of the vector field del-r with itself.
     
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