Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Null Hypersurface

  1. Oct 15, 2014 #1
    Hi

    Why is it possible to be able to pick a spacelike 2 surface S that lies in a null hypersurface N? We know that all the tangents vectors to N are either spacelike or parrelel to the normal vector. I imagine we want to build up S as the surface that is tangent to all the spacelike vectors in N, but I'm not quite sure how to build this up.. A hint rather than the exact answer would be much appreciated (although I will be grateful for either!)

    Thanks
     
    Last edited: Oct 15, 2014
  2. jcsd
  3. Oct 15, 2014 #2

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The light cone is an example of a null hypersurface.
     
  4. Oct 15, 2014 #3

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let ##\left\{ e_0, e_1, e_2, e_3 \right\}## be an orthonormal basis. WLOG, let ##n = e_0 + e_1## be a null vector. What spacetime vectors are orthogonal to ##n##?
     
  5. Oct 15, 2014 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This is true. Now ask yourself: how many linearly independent tangent vectors to N are there? (In other words, if I want to construct a basis of tangent vectors to N at a particular point, how many vectors will I need?) Of this total number of linearly independent tangent vectors, how many are spacelike?
     
    Last edited: Oct 15, 2014
  6. Oct 16, 2014 #5
    Using notation as George above, the basis for [itex] T_pN [/itex] would be [itex] (n, e_2, e_3 )[/itex] .The two spacelike vectors would be [itex] e_2 , e_3 [/itex], not [itex] e_1 [/itex] since this is not orthogonal to [itex] n [/itex]. Hence, at this particular point [itex] p [/itex] where we have this ortho basis, I would like to have [itex] T_pS = sp(e_2, e_3) [/itex] for the desired surface [itex] S [/itex]. This would induce a Riemannian metric at [itex] p [/itex] for [itex] S [/itex]. I would like to do this for all [itex] p\in N [/itex] in a smooth way so as to build up a spacelike surface [itex] S [/itex]. But how would this be done? I'm aware that the orthormal basis we picked only holds at the point [itex] p [/itex] so the question is how to combine this method together over all of [itex] N [/itex] .I'm still not entirely sure either how I would get a form for [itex] S [/itex] from this anyway. I'm guessing somehow using this operation can help me to define [itex] S [/itex] as the level set of some function [itex] f:N\rightarrow \mathbb{R} [/itex] ?

    Thanks for your comments!
     
    Last edited: Oct 16, 2014
  7. Oct 16, 2014 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    You've got the right idea, but note that if you do it (i.e., construct an induced Riemannian metric at each point ##p## using the two spacelike basis vectors ##e_2## and ##e_3##) for all ##p \in N##, you won't get one spacelike surface; you will get an infinite family of them. You might want to think about how this infinite family of spacelike surfaces can be parametrized.
     
  8. Oct 16, 2014 #7

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Maybe you should work this out for the example that robphy suggested, a future lightcone in Minkowski spacetime. It seems to me (without thinking much, or putting pen to paper, so I could be wrong) that, in this case, each member of (as PeterDonis pointed out) the family of 2-dimensional hypersurfaces is (diffeomorphic to) ##S^2##.

    I base this on analogy. Consider a 3-dimesional version of Minkowski spacetime that has one timelike dimension and two spacelike dimensions. A lightcone is a 2-dimensional surface in this spacetime. There is a family of 1-dimensional spacelike hypersurfaces, i.e., the closed circles that result when ##t## is held constant.
     
  9. Oct 16, 2014 #8
    Yes I did that example just now, I must apologize to robphy I thought he was just giving an example of a null hypersurface. Indeed, doing it for the Minkowski light cone things follow through as I had hoped (in this case it was clear that the spacelike surface would be 2 spheres, but it was good to see that the tangent space was indeed spanned by the two "angle" spacelike vectors). These 2 spheres are also the level set of [itex] f:N\rightarrow\mathbb{R}, f(X) = r [/itex] on the light cone, parametrised by [itex] r (=t)[/itex]. Maybe I need to generalise this so that my (I'll be really sketchy here) "null" coordinate is the one that defines by function [itex] f [/itex] (so that the spacelike surface is given by [itex] f = constant [/itex] . But I also notice that [itex] r [/itex] appears when we redefine the angle coordinates to have unit length ( i.e choosing our ortho basis) so maybe this is another thing I need to try and generalize? This is what I'll try to do now (unless I've got it horribly wrong!) just thought I would give an update as to my progress!
     
    Last edited: Oct 16, 2014
  10. Oct 16, 2014 #9
    okay a sort of different approach but how about this. I'll start by the motivation with the light cone for 3D Minkowski (in polar coordinates). The light cone has normal [itex] n^a = (\frac{\partial}{\partial t} + \frac{\partial}{\partial r} )^a [/itex] . We know that the integral curves of [itex] n^a [/itex] are null geodesics that lie in [itex] N [/itex] . It's clear from say drawing a picture that these generate all of the light cone. Let's label each null geodesic as [itex] X^a(\lambda) [/itex] where [itex] \lambda [/itex] is an affine parameter. It is clear that for fixed [itex] \lambda = \lambda _0, X^a(\lambda _0) [/itex] defines a circle, which is spacelike. Hence, for the light cone the spacelike surfaces (which are circles) are parametrised by the affine parameter.

    Let's do this for our general null hypersurface [itex] N [/itex] . With the same notation as above, set [itex] S = X^a(\lambda _0) [/itex]. Pick [itex] p\in S [/itex]. Then knowing what we have above from our orthonormal representation, we have [itex] T_pS = span(e_2. e_3) [/itex] since [itex] n^a = (\frac{d}{d\lambda})^a [/itex] vanishes as [itex] \lambda = \lambda _0 [/itex] is constant. Thus we have an induced Riemannian metric for [itex] S [/itex] at [itex] p [/itex] and since [itex] p [/itex] was arbitrary we are done. Then the spacelike hypersurfaces would be parametrised by the affine parameter [itex] \lambda [/itex] along the generators of [itex] N [/itex].

    This seems right to me. The only issue is whether [itex] S [/itex] as defined above actually defines a surface. What do people think?
     
  11. Oct 16, 2014 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    If we pick a point ##p## in a given ##S##, do the integral curves of ##e_2## and ##e_3##, starting from ##p##, lie in ##S##?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Null Hypersurface
  1. Null coordinates (Replies: 1)

Loading...