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?
  • 3
    $\begingroup$ Well, a null hypersurface is called null because, for example, its proper volume is zero i.e. "null". If a vanishing proper volume is enough to call it singular, then it's singular by definition. $\endgroup$ Aug 16 '13 at 16:03
  • $\begingroup$ @LubošMotl - I was using the term singular to mean $\det \gamma = 0$. I guess a vanishing proper volume would imply that. Thanks a lot! In these cases, how does one define the induced metric (if there is such a definition) $\endgroup$
    – Prahar
    Aug 16 '13 at 16:05

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?

  • $\begingroup$ Assuming you are in $d=4$, does $a=1,2,3$ here? $\endgroup$
    – Prahar
    Aug 16 '13 at 16:43
  • $\begingroup$ @Prahar: yes, I'm implicitly assuming $d=4$. $a$ represents 4-indices. If you actually go and do the calculation for $q^{ab}$ in a reasonable coordinate system, you'll find that it will be nothing but zeroes except for the bit that acts on $\theta_{a}$ and $\phi_{a}$, at which point you can just merrily drop one of your rows of zeroes, and go to 3-indices. $\endgroup$ Aug 16 '13 at 16:47
  • $\begingroup$ Just to be sure I'm understanding everything properly. You are taking $g_{ab}$ to be the spacetime metric and $q_{ab}$ to be the induced metric? $\endgroup$
    – Prahar
    Aug 16 '13 at 17:28
  • $\begingroup$ @Prahar: yes, that's right $\endgroup$ Aug 16 '13 at 18:19
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    $\begingroup$ @Prahar: other than my dissertation, no. I had to work it all out for myself, because I couldn't find anything decent. Have at it: arxiv.org/abs/1009.0934 $\endgroup$ Aug 16 '13 at 19:22

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