# Null Hypersurface

1. Oct 15, 2014

### tommyj

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. Oct 15, 2014

### robphy

The light cone is an example of a null hypersurface.

3. Oct 15, 2014

### George Jones

Staff Emeritus
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 $n$?

4. Oct 15, 2014

### 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
5. Oct 16, 2014

### tommyj

Using notation as George above, the basis for $T_pN$ would be $(n, e_2, e_3 )$ .The two spacelike vectors would be $e_2 , e_3$, not $e_1$ since this is not orthogonal to $n$. Hence, at this particular point $p$ where we have this ortho basis, I would like to have $T_pS = sp(e_2, e_3)$ for the desired surface $S$. This would induce a Riemannian metric at $p$ for $S$. I would like to do this for all $p\in N$ in a smooth way so as to build up a spacelike surface $S$. But how would this be done? I'm aware that the orthormal basis we picked only holds at the point $p$ so the question is how to combine this method together over all of $N$ .I'm still not entirely sure either how I would get a form for $S$ from this anyway. I'm guessing somehow using this operation can help me to define $S$ as the level set of some function $f:N\rightarrow \mathbb{R}$ ?

Last edited: Oct 16, 2014
6. Oct 16, 2014

### 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.

7. Oct 16, 2014

### George Jones

Staff Emeritus
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.

8. Oct 16, 2014

### tommyj

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 $f:N\rightarrow\mathbb{R}, f(X) = r$ on the light cone, parametrised by $r (=t)$. Maybe I need to generalise this so that my (I'll be really sketchy here) "null" coordinate is the one that defines by function $f$ (so that the spacelike surface is given by $f = constant$ . But I also notice that $r$ 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
9. Oct 16, 2014

### tommyj

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 $n^a = (\frac{\partial}{\partial t} + \frac{\partial}{\partial r} )^a$ . We know that the integral curves of $n^a$ are null geodesics that lie in $N$ . It's clear from say drawing a picture that these generate all of the light cone. Let's label each null geodesic as $X^a(\lambda)$ where $\lambda$ is an affine parameter. It is clear that for fixed $\lambda = \lambda _0, X^a(\lambda _0)$ 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 $N$ . With the same notation as above, set $S = X^a(\lambda _0)$. Pick $p\in S$. Then knowing what we have above from our orthonormal representation, we have $T_pS = span(e_2. e_3)$ since $n^a = (\frac{d}{d\lambda})^a$ vanishes as $\lambda = \lambda _0$ is constant. Thus we have an induced Riemannian metric for $S$ at $p$ and since $p$ was arbitrary we are done. Then the spacelike hypersurfaces would be parametrised by the affine parameter $\lambda$ along the generators of $N$.

This seems right to me. The only issue is whether $S$ as defined above actually defines a surface. What do people think?

10. Oct 16, 2014

### 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 $S$?