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Induced metric on the brane

by atrahasis
Tags: brane, induced, metric
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atrahasis
#1
Mar8-12, 07:41 AM
P: 11
Hello,

I have a problem to understand what people say by "induced metric". In many papers, it is written that for brane models, if we consider the metric on the bulk as [itex]g_{\mu\nu}[/itex] hence the one in the brane is [itex]h_{\mu\nu}=g_{\mu\nu}-n_\mu n_{\nu}[/itex] where [itex]n_{\mu}[/itex] is the normalized spacelike normal vector to the brane. I agree that it defines a projection tensor since [itex]h_{\mu\nu}n^{\mu}=0[/itex] but I don't understand how this can be the induced metric on the brane.

For example, if we consider a flat spacetime in spherical coordinates:

[itex]ds^2=-dt^2+dr^2+r^2\Bigl(d\theta^2+sin^2\theta d\phi^2\Bigr)[/itex]

and we consider the surface defined by the equation [itex]r=a(t)[/itex], hence we have

[itex]ds^2=-\Bigl(1-\dot a^2\Bigr)dt^2+a^2\Bigl(d\theta^2+sin^2\theta d\phi^2\Bigr)[/itex]

which is for me the induced metric on the surface. But it doesn't match with the metric [itex]h_{\mu\nu}[/itex] where [itex]n_\mu=(0,1,0,0)[/itex]

which would give [itex]h_{00}=-1\neq -\Bigl(1-\dot a^2\Bigr)[/itex] ????????
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atrahasis
#2
Mar9-12, 03:43 AM
P: 11
Ok I have half of the answer, the normal vector is wrong, because [itex]r=a(t)[/itex], we have [itex]dr-\dot a dt=0[/itex], which gives for the normal vector [itex]n^\mu=n(-\dot a,1,0,0)[/itex] with [itex]n[/itex] a normalization factor in the goal to have [itex]g_{\mu\nu}n^\mu n^\nu=+1[/itex].
But I still don't have the right induced metric
Sam Gralla
#3
Mar9-12, 07:59 AM
P: 95
There are two ways of thinking about the induced metric. One is the way you've given, as a 4D metric that is "tangent" to the surface. The other is as a genuine 3D object. What I think you are asking is why, when you use the 4D definition, you don't suddenly see a 3D metric pop out in front of you. You will see this only in "adapted" coordinates in the sense that the hypersurface is described by holding one of the coordinates fixed. If you want to translate between the 4D induced metric and a genuine 3D metric when you're not working in adapted coordinates, you have to effectively work out the coordinate transformation. For the 3D viewpoint you should read the excellent treatment given in Eric Poisson's book, "A Relativist's Toolkit".

This was a bit vague I think it will still help you answer your questions. Also note that I worked with the example of a 3D surface in a 4D spacetime, but the same logic works generally. (Note that there are new subtleties with null surfaces, however.)

atrahasis
#4
Mar10-12, 12:31 AM
P: 11
Induced metric on the brane

Thanks for the reply,
I checked on Poisson's book and also Gourgoulhon's review but I couldn't found the reason.

I finally understood my mistake, [itex]h_{\mu\nu}[/itex] is not the induced metric but only the projection tensor. For to have the induced metric we have to look to the tangential components of the tensor and not to [itex]h_{00}[/itex].
In fact the 3 vectors orthogonal to the normal vector and which define a basis on the hypersurface are
[itex]V1^\mu=(1,\dot a,0,0)[/itex]
[itex]V2^\mu=(0,0,1,0)[/itex]
[itex]V3^\mu=(0,0,0,1)[/itex]

so it is perfectly fine to look for [itex]h_{22}[/itex] and [itex]h_{33}[/itex]. But the last component is not [itex]h_{00}=h_{tt}[/itex] but [itex]h_{V1 V1}[/itex]

So now we have [itex]\partial_{V1}=\partial_t+\dot a \partial_\rho[/itex] which implies that

[itex]h_{V1V1}=h_{00}+2\dot a h_{01}+\dot a^2 h_{11}[/itex] which gives the correct result [itex]h_{V1V1}=-1+\dot a^2[/itex].

So it is a modification of the coordinates ...

Thanks


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