Does such a set of equations exist or does our universe have an intrinsic curvature that can't be explained by an embedding in a flat Minkowski space of 1 higher dimension? Even if general relativity can be explained by such equations, it still doesn't prove our universe is embedded in a flat Minkowski space of higher dimension.
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3$\begingroup$ From Whitney's embedding theorem, you might need as many as $2n$ dimensions to embed a manifold of dimension $n$, so you will need at least 8 dimensions if you want to do this $\endgroup$– SlereahCommented Jul 13, 2016 at 20:37
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9$\begingroup$ @slereah: Whitney's embedding theorem is doubly irrelevant here. First, there is no reason to think that the embeddings guaranteed by Whitney preserve the metric, so you might need a whole lot more than 8 dimensions. Second, if you ignore the metric, Whitney's bound is a worst-case scenario, and it's perfectly possible that the requirements of general relativity allow you do even better --- so you might need a whole lot less than 8 dimensions. $\endgroup$– WillOCommented Jul 13, 2016 at 20:53
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1$\begingroup$ Oh yes, I meant "at most", not "at least" $\endgroup$– SlereahCommented Jul 13, 2016 at 20:54
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3$\begingroup$ @sleareah: "At most" and "at least" are equally wrong. $\endgroup$– WillOCommented Jul 13, 2016 at 20:55
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1$\begingroup$ @Slereah They certainly don't guarantee the metric. Why? Don't ask me, I've tried to understand the Nash embedding theorem but its proof is well beyond me. I find that absolutely maddenning that I can't given that I can understand the proof of the Whitney theorem, and that, on the face of it, would seem a harder (more general thing) to prove, yet the reality for me is very much the other way around. $\endgroup$– Selene RoutleyCommented Jul 14, 2016 at 0:19
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5 Answers
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For Riemannian manifolds, I believe the best result currently known is that a manifold of dimension $n$ can be isometrically embedded in a euclidean space of dimension $2(2n+1)(3n+7)$. So, for example, a 3-dimensional spacelike slice of spacetime can be embedded in a flat euclidean space of at most 224 dimensions. Maybe in low-dimensional cases like this one can do better, but if so I'm not aware of it.
So much for space. If you want to embed all of spacetime, I think the best known result is that every Lorentzian manifold can indeed be embedded in a flat Lorentzian manifold, but I don't think any bound is known on the necessary number of dimensions.
Edited to add: I see by the reference I posted in the comments that there is in fact a known bound for Lorentzian manifolds: $2(2n+1)(2n+6)$. So you can imbed all of (four dimensional) spacetime in a copy of $R^{252}$, with signature $(126,126)$
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1$\begingroup$ ??? Really? Is it that bad? I knew it was more than ten, but that's an insane bound. Could you link to the theorem, please? I was always looking for something like that. $\endgroup$ Commented Jul 13, 2016 at 23:10
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$\begingroup$ @CuriousOne: Reference here: ams.org/journals/bull/1969-75-06/S0002-9904-1969-12407-9/… . I briefly posted a comment saying one can in fact do better, but that comment was mistaken and I've deleted it. $\endgroup$– WillOCommented Jul 13, 2016 at 23:27
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1$\begingroup$ @CuriousOne: As far as it being "that bad", I don't think that's known. This is an upper bound, but there might be a much better upper bound that hasn't yet been discovered. $\endgroup$– WillOCommented Jul 13, 2016 at 23:31
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3$\begingroup$ @CuriousOne Incidentally, these theorems were highly important in the history of mathematics, since mathematicians, including Cartan, often liked to think of manifolds as embedded in higher dimensional flat spaces but the modern definition is much easier to work with. Everyone worked under the conjecture that the two notions were ultimately the same, however it wasn't until the the 1940s Whitney embedding theorem that people knew they were the same. $\endgroup$ Commented Jul 14, 2016 at 0:04
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1$\begingroup$ @WetSavannaAnimalakaRodVance: I am fully aware of the importance of these theorems, but I have yet to see a nice enumeration of them in a textbook. They seemed to be scattered in the literature some 20+ years ago, when I looked at the matter the last time (could be 30+, by now), depending on the individual author's needs. I'll take a look at the citation by WillO. The general case is insanely bad, for sure. $\endgroup$ Commented Jul 14, 2016 at 0:08
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This is an afternote to WillO's answer which cites:
Robert E. Greene, "Isometric Embeddings," Bull. AMS 1969
which addressed known bounds on the dimension required of flat Euclidean / Minkowsian space if it is to be an embedding for a solution of the Einstein field equations, which of course is a four-dimensional signatured manifold.
It's worth noting that important special cases one can be embedded in much lower dimensions than the insanely loose bounds defined by the Nash embedding theorem and its Lorentzian equivalents. Such simplifications happen in cases of high symmetry. For example, the large scale homogeneous/ isotropic universe defined by the Friedmann–Lemaître–Robertson–Walker (FLRW) metric can indeed be thought of as being embedded in five dimensional space, with signature $(1,\,4)$, because it can be partitioned into foliations of space, with the foliations indexed by a universal time co-ordinate, and the spatial foliations are isometric to three dimensional spheres / hyperboloids in $\mathbb{R}^4$ with only the scale factor and energy / pressure evolving over time.
Indeed, the lecture notes:
Balša Terzić, Lecture notes for PHYS 652, Old Dominion University
take the unusual approach of lifting well known 19th century geometry results on the hypersphere / hyperboloid and linking them to a homogeneous, isotropic stress energy tensor. The Ricci tensor is of course diagonal in this case, so the student is relieved of the full complexity of GR and gets to see the direct link between stress energy (in the diagonal case) and curved spaces.
Lawrence B. Crowell's answer cites two other examples very like FRLW which can be split into spatial foliations such that the whole spacetime is embedded in 5 dimensions with a $(1,\,4)$ signature - the de Sitter and anti-de Sitter spaces, which are like FLRW but with no matter and pressure (vacuum solution) and a special value for the cosmological constant.
answered Jul 14, 2016 at 0:44
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1$\begingroup$ Unfortunately such an important case as the Schwarzschild black hole requires at least 6 dimensions. $\endgroup$– RuslanCommented Feb 11, 2020 at 20:40
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$\begingroup$ Note Lawrence B. Crowell's answer has one embedding in 1+4 and one in 2+3 dimensions. Also this paper claims that FLRW can require up to 2+4 dimensions, though I think 1+4 is enough when $ρ\ge0$ and $Λ\ge0$. $\endgroup$– benrgCommented Oct 2, 2020 at 22:10
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$\begingroup$ @Ruslan Source, please! thanks And what about pp-wave and wormhole solutions? $\endgroup$ Commented Nov 16, 2021 at 23:01
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1$\begingroup$ @BastamTajik see e.g. the discussion here, in particular references [9] (E. Kasner, Am. J. Math., v. 43, no. 2, pp. 126, 1921), [14] (E. Kasner, Am. J. Math., v. 43, no. 2, pp. 130, 1921), [6] (C. Fronsdal, Phys. Rev., v. 116, no. 3, pp. 778, 1959) $\endgroup$– RuslanCommented Nov 17, 2021 at 8:29
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General relativity in four dimensions does not need to be embedded in a larger space of any sort. Curvature in general relativity is completely defined according to curvatures that are intrinsic induced by parallel translation of vectors. One does not need to have the spacetime in four dimensions embedded in some higher dimension spacetime.
There is general relativity in five dimensions with the flat metric $$ ds^2~=~dt^2~\pm~du^2~-~dx^2~-~dy^2~-~dz^2. $$ The constraint $$ t^2~\pm~u^2~-~x^2~-~y^2~-~z^2~=~\alpha^2 $$ defines hyperboloids embedded in this flat spacetime. The condition $\pm u^2$ defines the anti-deSitter $(+du^2)$ and de Sitter spacetimes $(-du^2)$. The constant $\alpha$ defines the cosmological constant. So this is a special case of what you are talking about.
answered Jul 13, 2016 at 20:58
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To start with, a manifold is not always able to be embed in higher dimension, especially when singularity (black hole) involves.
I would more agree if it is described by a 3-d gravity-free field theory. This is similar to the idea named AdS/CFT duality. Of course here is not AdS space, but the spirit is similar, I think.
But I'm not an expert in this, so...
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$\begingroup$ I don't see why the singluarity couldn't travel faster than light in the higher dimensional space without the fabric of space itself travelling faster than light. The only problem I see with the embedding is that the equations also for a black hole of any mass no matter how small. $\endgroup$– TimothyCommented Jul 13, 2016 at 20:58
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$\begingroup$ @user46757 First, I never say anything related to light speed. Second, flat spacetime has a simple zero curvature while singularities have divergent. That's the reason it cannot be embed in higher flat spacetime. $\endgroup$ Commented Jul 13, 2016 at 21:05
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$\begingroup$ Think of the region near the singularity as a cone where its sperical cross sections are shrimking slower than light. That could still cause the tip of the cone to travel faster than light. Maybe the fabric of space is rapidly accelerating and Einstein's field equations are an approximation of the correct equations that predict that a sufficienly small mass if compressed enough will form a black hole and disappear at the singularity which will cease to stay a singularity destroying the black hole. $\endgroup$– TimothyCommented Jul 13, 2016 at 21:18
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1$\begingroup$ You can still embed the open manifold with the singularities excised; indeed that is the only meaningful way to treat them in the manifold framework. $\endgroup$ Commented Jul 14, 2016 at 1:20
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1$\begingroup$ @WetSavannaAnimalakaRodVance I see, after read some reference other answers recommend. $\endgroup$ Commented Jul 14, 2016 at 1:22
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Maybe our universe is a uniformly accelerating fabric embedded in a 5-dimensional Minkowski space and doesn't actually obey the laws of general relativity. If gravity is entirely the result of objects making a dent in it with their weight, then for objects with an escape velocity much slower than the speed of light like Earth, the gravitational time dilation is exactly the amount general relativity predicts. The emission of a gravitional wave from merging black holes which we detected is also consistent with this theory of gravity. If that really is how our universe works, an object in free fall will still follow a geodesic of the fabric of space but there will be no gravitomagnetism and an object of sufficiently low mass if it's compressed dense enough might form a singularity then disappear at it and then the singularity will rush towards the rest of the fabric of space faster than light and cease to exist once its speed goes down to the speed of light, destroying matter and its gravitational field. Maybe at the quantum level, the information of what went into the singularity is preserved as disordered vacuum fluctuations outside the fabric of space. How do we know our universe follows the laws of general relativity if we haven't made detailed enough measurements of the behaviour of objects in a gravitational field or observed gravitomagnetism? Why should general relativity be true? Just because angular momentum has been proven to be conserved according to simplified laws of physics doesn't mean it's also conserved in the formation of a black hole. The gravitational constant is determined by the acceleration of the fabric and its resistance to bending but we already know what the gravitational constant is. Given what the gravitation constant is, the faster the fabric is accelerating, the lower the minimum mass that can be compressed to form a permanent black hole. Although Einstein's field equations are probably the simplest possible equations that are consistent with observations and are preserved at any point at any velocity below c at any orientation and are consistent with observations, there's no reason to be sure the universe isn't an embedded fabric whose equations do not simplify to such simple equations that described only the space itself According to https://en.wikipedia.org/wiki/Binary_star#Cataclysmic_variables_and_X-ray_binaries, one member of a binary system is believed to be a black hole. It's probably because of detection of movement of the other stat in the system that it's believed that there's a black hole in the system. If our universe does work the way I described, the existence of a black hole of that mass rules out the possibility that the fabric is accelerating below a certain acceleration.