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Normal to a null surface

  1. Jul 21, 2015 #1
    I am trying to use the Israel junction conditions for a null surface, but I am running into complications with defining a normal vector for a null surface.

    As I understand it the normal vector is defined to be perpendicular to the surfaces tangent vectors [itex]n\cdot e_i=0[/itex], as well as satisfying [itex]n\cdot n=0[/itex].

    However, this does not fix [itex]n[/itex] completely, it can still be rescaled by an overall factor (as opposed to the case for a time like surface where this is fixed by the normalization [itex]n\cdot n=1[/itex]). Is this correct? Or is there another convenient constraint one should impose as well to fix it completely?

    The issue is then that when I try to use it in the junction formalism my result seems to depend on this overall arbitrary normalization. (see eg http://iopscience.iop.org/0264-9381/14/5/029/pdf/q70520.pdf , equation (5) )
     
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  3. Jul 21, 2015 #2

    bcrowell

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    I've deleted an earlier wrong post. What seemed a little odd to me about this was the following:


    As an example of a null surface S, let's take a light cone in Minkowski space. If you pick a point P on S, and a linearly independent set of tangent vectors at P, then you can make one of these tangent vectors, say ##e_1##, null. Let's say you also make the ##e_i## all perpendicular to each other. Then by this definition, ##e_1## also qualifies as a normal vector. (Since it's null, it's perpendicular to itself.) This seemed weird to me, since it meant that ##e_1## was *both* a normal vector and a tangent vector. However, I think it's correct.

    If you have any 3-surface S, then at any given point P on S, without appealing to the metric, you automatically get a covector that is normal to the surface. This should be pretty obvious if you use the visualization of a covector as a stack of parallel planes. If you then take this covector and raise an index, you get a vector that we could say is normal to the surface -- but in the example above, that vector is also a tangent. (I think this holds in general, not just in that example.) What this probably tells us is that it's more natural to talk about normal covectors, not normal vectors.

    The paper is paywalled. Could you transcribe the relevant material?
     
  4. Jul 21, 2015 #3

    PeterDonis

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    I agree; AFAIK this is a well-known property of null surfaces, that any null vector tangent to them is also normal to them.

    I want to be sure I understand the concept of a "normal covector" in the absence of a metric. The usual definition of a covector (without involving a metric) is that it's a linear map from vectors to numbers. What additional property would this map have to make it a "normal" covector to a surface? The obvious one is that it maps every vector tangent to the surface to the number zero, and every vector not tangent to the surface to some nonzero number. Is that the definition you're using?
     
  5. Jul 21, 2015 #4

    bcrowell

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    I think this works, although I guess it would take a little work to show that such a map exists, can be nonzero, and is unique up to a scaling factor.

    Another way of defining it would be as in section 6.6 of my SR book: http://www.lightandmatter.com/sr/ . For a given surface S and point P, pick any three linearly independent vectors ##\{a,b,c\}## at P that are tangent to S. For any vector ##d##, define ##V## as the four-volume of the parallelepiped spanned by ##\{a,b,c,d\}##. Since four-volume is Lorentz-invariant, this definition is unambiguous. Then the map that takes ##d## to ##V## is a covector that is normal to S.
     
  6. Jul 21, 2015 #5

    PeterDonis

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    Ah, I see. And this method will obviously map any vector tangent to ##S## to zero, and any vector not tangent to ##S## to some nonzero number, so it's consistent with the definition I gave.
     
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