Gravity in R5?

  • #1
718
9

Main Question or Discussion Point

Can a 4-dimensional manifold with the Schwarzschild metric be embedded into a flat manifold of 5 (or more if necessary) dimensions? In other words, are there functions of [tex]t,r,\theta , \phi [/tex] and [tex] M[/tex] such that if

[tex]
x_1 = f_1 (t,r,\theta ,\phi ,M)[/tex]
[tex]
x_2 = f_2 (t,r,\theta ,\phi ,M)
[/tex]
.
.
etc.

then
[tex] ds^2=dx_1 ^2 +dx_2 ^2 +dx_3 ^2 +dx_4 ^2 +dx_5 ^2[/tex]

[tex]=(1-\frac{2GM}{c^2 r})c^2 dt^2-\frac{dr^2}{1-(2GM/c^2 r)} - r^2 sin^2 \theta d\phi ^2 - r^2 d\theta ^2?[/tex]

I like the idea of a Euclidean (or Minkowskian) hyperspace that contains gravitational fields, even if it turns out to have no practical application.
 

Answers and Replies

  • #2
718
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I guess the gravitational constant [tex]G[/tex] should be an independent variable of the functions, too.
 
  • #3
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Well what you have there is the idea that a curved space can always be embedded in a flat space of some higher dimension. This is known as Nash's embedding theorem.

The other possibility is that the higher dimensional space is still curved, but small. This is known as the Kaluza-Klein reduction, and leads to interesting consequences.
 
  • #4
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Can a 4-dimensional manifold with the Schwarzschild metric be embedded into a flat manifold of 5 (or more if necessary) dimensions?
Yes, a Schwarzschild solution can be expressed in exactly 5 flat dimensions.
 
  • #5
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Yes, a Schwarzschild solution can be expressed in exactly 5 flat dimensions.
The problem being, of course, that that's not what the question was. He/she asked whether or not the four-dim Schwarzschild manifold could be isometrically embedded in [itex]\mathbb{R}^5[/itex], not whether there exists a Schwarzschild solution in [itex]\mathbb{R}^5[/itex].
 
  • #6
1,997
5
The problem being, of course, that that's not what the question was. He/she asked whether or not the four-dim Schwarzschild manifold could be isometrically embedded in [itex]\mathbb{R}^5[/itex], not whether there exists a Schwarzschild solution in [itex]\mathbb{R}^5[/itex].
I do not know what your problem is with my statement. Again a Schwarzschild solution can be described in terms of 5 flat dimensions.
 
  • #7
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I do not know what your problem is with my statement. Again a Schwarzschild solution can be described in terms of 5 flat dimensions.
The problem is obvious. A Schwarzschild solution in five dimensions is not equivalent to an isometric embedding of the four-dimensional Schwarzschild solution in five dimensions. The OP's question was related to the latter, not the former.

For what it's worth, there are technicalities regarding the embedding of non-compact solutions to Einstein's equations in four dimensions into regions of compact support in [itex]\mathbb{R}^{d\ge5}[/itex] which make the OP's question more subtle than it first appears.
 
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  • #8
1,997
5
The problem is obvious. A Schwarzschild solution in five dimensions is not equivalent to an isometric embedding of the four-dimensional Schwarzschild solution in five dimensions. The OP's question was related to the latter, not the former.
Being able to describe the Schwarzschild solution in 5 flat dimensions implies that the 4D curved manifold can be embedded.

For what it's worth, there are technicalities regarding the embedding of non-compact solutions to Einstein's equations in four dimensions into regions of compact support in [itex]\mathbb{R}^{d\ge5}[/itex] which make the OP's question more subtle than it first appears.
Irrelevant for the Schwarzschild solution, since this solution describes a static spacetime.
 
  • #9
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Perhaps I can clarify my question by restating it as an analogy.

This

[tex]
ds^2 = r^2 d\theta ^2 + r^2 sin^2 \theta d\phi ^2
[/tex]

is to this

[tex]

x=r sin \theta cos \phi[/tex]
[tex]y=r sin \theta sin \phi[/tex]
[tex]z=r cos \theta
[/tex]

as this

[tex]ds^2 =(1-\frac{2GM}{c^2 r})c^2 dt^2 - \frac{1}{1-(2GM/c^2 r)}dr^2 - r^2 sin^2 \theta d\phi ^2 - r^2 d\theta ^2[/tex]

is to

[tex]?[/tex]
 
  • #10
A.T.
Science Advisor
9,910
1,685
Can a 4-dimensional manifold with the Schwarzschild metric be embedded into a flat manifold of 5 (or more if necessary) dimensions?
Yes it can, but I cannot provide you the exact embedding for the 4d-Schwarzshild metric. Since the metric is symmetric and the angular coordinates don't affect the t & r components, they can be omitted if considering one spatial dimension is sufficient.

For such a 2d-Schwarzshild metric(interior & exterior) the embedding in 3d is described here:
http://fy.chalmers.se/~rico/Webarticles/2001GRG-Jonsson33p1207.pdf

An interactive 3d-visualization of a similar embedding can be found here:
http://www.adamtoons.de/physics/gravitation.swf
 
Last edited by a moderator:
  • #11
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Thanks, A.T.

Unfortunately, when I click on the first link I only get what looks like alphabet soup (along with some drawings). Perhaps I'll have better luck using a different computer.
 
  • #12
A.T.
Science Advisor
9,910
1,685
  • #13
718
9
I tried the adobe reader link, but it looks like my operating system is too old. Maybe a computer at a local library can do the job. I will let you know.
 
  • #14
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I'm delurking because one or two claims have been made by MeJennifer in this thread that are hugely misleading. To wit:

MeJennifer said:
Being able to describe the Schwarzschild solution in 5 flat dimensions implies that the 4D curved manifold can be embedded.
Untrue. Shoehorn is perfectly correct in his claim that the four-dimensional Schwarzschild solution cannot be embedded in a flat five-dimensional manifold. This is an extremely well-known result, and was discovered decades ago by Tangherlini (I don't have the reference to hand, but if memory serves, the paper in which this result was originally announced was published in Il Nuovo Cim. during the first half of the sixties). The general statement is that the four-dimensional Schwarzschild solution can be embedded in a flat [itex]N[/itex]-dimensional manifold if and only if [itex]N\ge 6[/itex].

MeJennifer said:
Irrelevant for the Schwarzschild solution, since this solution describes a static spacetime.
Nope. The requirement that a given spacetime is static is of little or no help when one is discussing embeddings, which is what the original question was about. This is because Birkhoff's theorem breaks down for [itex]N>4[/itex]. This is deeply related to the more fundamental result that any general solution of the four-dimensional Einstein equations can be embedded in a higher-dimensional flat space if and only if that space is [itex](N\ge10)[/itex]-dimensional.
 
  • #15
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Untrue. Shoehorn is perfectly correct in his claim that the four-dimensional Schwarzschild solution cannot be embedded in a flat five-dimensional manifold. This is an extremely well-known result, and was discovered decades ago by Tangherlini (I don't have the reference to hand, but if memory serves, the paper in which this result was originally announced was published in Il Nuovo Cim. during the first half of the sixties). The general statement is that the four-dimensional Schwarzschild solution can be embedded in a flat [itex]N[/itex]-dimensional manifold if and only if [itex]N\ge 6[/itex].
Precisely. This is a well known result and is something I'd expect to be covered in any introductory grad-level course in GR.

I'd hazard a guess that somebody here will now claim you know nothing about the subject and attempt to browbeat you into admitting you're wrong. :-)

coalquay404 said:
Nope. The requirement that a given spacetime is static is of little or no help when one is discussing embeddings, which is what the original question was about. This is because Birkhoff's theorem breaks down for [itex]N>4[/itex]. This is deeply related to the more fundamental result that any general solution of the four-dimensional Einstein equations can be embedded in a higher-dimensional flat space if and only if that space is [itex](N\ge10)[/itex]-dimensional.
Again, this is precisely the point. Now that I think about it a little bit, you wouldn't have to know just GR to be aware of this result. It's also very important in higher-dimensional theories due to its relationship with protective theorems like Campbell-Magaard.
 
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