general relativity - Induced metric on a null hypersurface

Consider a metric $g_{\mu\nu}(x)$ and a hyper surface ${\cal H}$ defined by $~f(x) = c$. One (or at least I) usually finds the "induced metric" on ${\cal H}$ by solving$~f(x) = c$ for one of the coordinates and plugging it back into the metric to give us $\gamma_{ij}$. For null hypersurfaces, I seem to be finding that the resultant induced metric is singular.

1. Is it true in general that the induced metric as defined above is singular on a null hypersurface?


2. How does one define the induced metric on a null hypersurface?

Answer

You are exactly right. The intrinsic metric of a null submanifold is going to have determinant zero (which makes sense when you consider null as a transition from spacelike to timelike). This is going to have a bunch of consequences--the metric is no longer invertible, so you will no longer have a natural mapping from vectors to one-forms (in fact, the correct null tangent vector is a different vector than the null cotangent one-form that is mapped to it by the enveloping 4-metric).

In order to proceed here, you really need to apply the pull-back, push-forward stuff that they teach in a good differential topology class. The lazy route is that you find a basis of two spacelike vectors (I'll call them $\theta^{a}$ and $\phi^{a}$ that span the spacelike subset of the null manifold. There will be (up to a rescaling) two distinct null vectors $\ell^{a}$ and $k^{a}$ normal to both of these vectors satisfying $\ell_{a}k^{a} = -1$. Then, the push-forward of the metric of your 3-space into the enveloping 4-space will be given by $q^{ab} = g^{ab} + \ell^{a}k^{b} + \ell^{b}k^{a}$, while you find the lowered version of $q_{ab}dx^{a}dx^{b}$ by the usual technique that you would if you were just solving (for example) $r=2M$ in the Schwarzschild metric in Kerr coordinates, and taking the pull-back by eliminating all of the $dr$ components in the case of the Schwazschild horizon in Kerr coordinates.

You can then think of one of $\ell^{a}$ as your null tangent to the horizon (this would be the one proportional to $\partial_{r}$ in the Schwarzschild example above, and the other null vector as the null normal to the horizon.

With care, you can do a lot of this stuff, and even go as far as generating curvature equations and the like, which is necessary when trying to do things like double-null decompositions of the enveloping 4-geometry.

Does that make sense?