10
$\begingroup$

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?
$\endgroup$
  • 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. $\endgroup$ – Luboš Motl 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
7
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jerry Schirmer 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 $\endgroup$ – Jerry Schirmer Aug 16 '13 at 18:19
  • 2
    $\begingroup$ @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 $\endgroup$ – Jerry Schirmer Aug 16 '13 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.