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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?
Qmechanic
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Prahar
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    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. – Luboš Motl Aug 16 '13 at 16:03
  • @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) – Prahar Aug 16 '13 at 16:05

1 Answers1

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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?

Zo the Relativist
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  • Assuming you are in $d=4$, does $a=1,2,3$ here? – Prahar Aug 16 '13 at 16:43
  • @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. – Zo the Relativist Aug 16 '13 at 16:47
  • 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? – Prahar Aug 16 '13 at 17:28
  • @Prahar: yes, that's right – Zo the Relativist Aug 16 '13 at 18:19
  • Great! I'll try and work it out. Do you have any good references for this? Thanks for your help. – Prahar Aug 16 '13 at 18:20
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    @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: http://arxiv.org/abs/1009.0934 – Zo the Relativist Aug 16 '13 at 19:22
  • Thanks a lot for your comments. I have a couple of more questions. Suppose I have a null hypersurface ${\cal H}$ on which $\lambda$ is setup to be the affine parameter and $x^a$ are the directions normal to the null geodesics on ${\cal H}$. The intrinsic metric on ${\cal H}$ is of course of the form $ds^2 \big|{\cal H} = q{ab} dx^a dx^b$. In general $q_{ab}$ is a function of $(v,x^a)$. My question is that is it possible to perform a coordinate transformation of some sort to set $\partial_v q_{ab} = 0$. Is this true in general? – Prahar Feb 03 '16 at 20:26
  • Sorry, I take that back. I don't think it's possible. For instance when ${\cal H}$ corresponds to the horizon of a black hole, the area of the black hole is defined as $A(v) = \int dx\det q$. Of course, in any physical process where particles enter the black hole, the area generically increases so we must have $\partial_v A > 0$ (assuming $v$ is future-directed). This negates my requirement -- is that right? – Prahar Feb 03 '16 at 20:31
  • It's not true in general. Generally, the condition that the horizon is isolated, rather than dynamic, is equivalent to the null tangent vector being a killing field of the induced metric, which gives you the condition you're looking for. – Zo the Relativist Feb 04 '16 at 00:37
  • however, I'm now trying to think through ane example where the statement is false.

    Note that when particles enter the black hole, the black hole horizon becomes a spacelike, not a null, surface, so this null geometry is no longer valid.

     – Zo the Relativist Feb 04 '16 at 00:39
  • It's been a long time since I've looked at this stuff, and I"m thinking of it to give you a better answer, but you should look at Sean Hayward's articles on this stuff. – Zo the Relativist Feb 04 '16 at 00:40
  • I had a question. Consider null geodesics in d = 4, then it's transverse metric is 2 dimensional, as opposed to the naively expected answer to be 3d. Can you intuitively tell me why this is so. I know the math, but the physical intuition is not clear. – Tushar Gopalka May 07 '18 at 05:18
  • @TusharGopalka I don't think I understand your question. What null geodesics are you taking the transverse of? The standard version of this usually requires making a choice of either two special null vectors or choosing a null vector and a time coordinate. – Zo the Relativist May 07 '18 at 15:22
  • @JerrySchirmer I just found your answer related to my question in https://physics.stackexchange.com/questions/408857/integrating-along-a-null-geodesic Do you think you could help me here? Thanks! – blackhole1511 May 29 '18 at 19:12
  • @JerrySchirmer taking from your answer, the det(q) for a spacelike submanifold would be $>0$, right? Sorry, if the question is a bit silly. – quirkyquark Mar 21 '20 at 22:49
  • @NamanAgarwal, if you're using the (-,+,+,+) signature, yes. – Zo the Relativist Mar 21 '20 at 23:41
  • @JerrySchirmer Extending the question by TusharGopalka choosing 2 special null vectors $K^\alpha$ and $N^\alpha$ such that $K^\alpha N_\alpha = -1$ we define 2 dimensional transverse metric (which seems quite arbitrary to me atleast). Can you please explain is this the unique way of defining transverse metric and what is the physical motivation behind defining like this? I know that the transverse metric will not be unique as choosing $N^\alpha$ is not unique but what I am specifically asking is if there is totally different way of defining transverse metric without taking 2 null vectors? – Ashley Chraya Sep 17 '20 at 19:01
  • @RandomXYZ: You only really need one null vector field to do this -- because that null vector field will determine a unique span of two spacelike normals, which then gets you a null 3-surface. Once youré there, the null vector is unique up to a choice for scale for the null vector, because mixing the two spacelike directions into it will give a spacelike direction. Then, going back to the 4-space, there is only one null vector that satisfies $n^{a}k_{a} = -1$ – Zo the Relativist Sep 18 '20 at 21:00