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If you draw a big triangle in earth 2D surface you will have an aproximated spherical triangle, this will be a non euclidean geometry.

but from a 3D perspective, for example the same triangle from space, it could be depicted as euclidean in 3D.

then why we talk about non euclidean instead of adding 1 dimension?

It is easier the non-euclidean approach? I don't see how!

Sorry if the question is perhaps naive, but this is a long time doubt for me, so I will be grateful for an answer, thanks

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This is a maths question with no physics content that should be answered at maths.stackexchange.com. Please vote to close. –  John McVirgo Apr 20 '11 at 20:14
    
The question is about why GR and others theories (of physics) uses non euclidean space, instead of euclidean. That is a physics question. –  HDE Apr 20 '11 at 21:05
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Questions about the mathematical structure of physical theories -- even to the level of aesthetics, IMO -- are definitely physics questions, and we'd be doing a disservice to Physics.SE to close them. This is a sensible question, and Ted Bunn provides a fantastic answer. +1 on both counts. –  wsc Apr 20 '11 at 23:43
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2 Answers

up vote 14 down vote accepted

You have to add more than one dimension, in general. Mathematicians have studied in great detail the question of how many extra dimensions you need in order to embed a curved manifold in a flat one. One key result is the Nash Embedding Theorem, which says that you can isometrically embed an $m$-dimensional Riemannian manifold in $n$-dimensional flat space, for some $n\le m(m+1)(3m+11)/2$. (Isometric embedding means embedding in a way that preserves lengths, etc., which is what seems to be relevant here.)

That's 120 dimensions for 3-dimensional space!

This theorem only applies to Riemannian manifolds, not Lorentzian ones -- that is, it applies to space, not spacetime with its annoying minus sign in the metric. If it did apply to spacetime, so that we could apply the result with $m=4$, you'd need 230 dimensions.

As far as I know, there's not a comparably clean result for Lorentzian spacetime. There are bunches of references here. But it certainly can't be any easier to embed spacetime than to embed space!

Anyway, I think this illustrates why people don't like to think of general relativity in terms of embeddings in higher-dimensional space. It's going to be way, way harder than the standard approach.

In addition, many people have a philosophical preference not to populate a theory with unobservable entities. If you don't need those extra dimensions, why postulate them?

One addendum, after thinking about Marek's comment that one might expect to get by generically with only one (or maybe a few) extra dimensions, rather than the large number in the Nash theorem. I mentioned in the comments that my intuition was different, although I wasn't sure. I just want to expand on that a bit.

You can run into trouble when using two-dimensional intuition to make guesses about higher-dimensional manifolds. In two dimensions, the Riemann curvature tensor only has one component -- that is, the curvature is described by a single number at each point. I conjecture that that's the reason why it seems intuitive that you can embed 2 dimensions in 3: you just need to "bend" the surface one way at each point to account for the curvature. (Even so, it turns out that you can't embed 2 dimensions in 3, even in relatively simple cases.) But the number of components of the curvature tensor grows rapidly with dimension. In 3 dimensions, there are 6 components, and in 4 there are 20. It's wildly implausible, to me, that you could "usually" account for all of those extra degrees of freedom with one or two extra dimensions.

(A bit of a digression, just because I think it's cool: Another example of how 2-D intuition can be a bad guide to higher dimensions. The problem of topologically classifying 2-D manifolds was solved ages ago. One might have guessed that the problem of classifying 3-D manifolds would be similar, but it turns out to be vastly harder. Last time I checked, it was thought that this problem had been solved, but there was some doubt about whether the solution was correct. And in 4-D or more, the problem is apparently known to be undecidable!)

One more point: even if it's true that you can "usually" get by with fewer dimensions, I'm not sure how relevant that is. Any manifold is a possible solution to Einstein's equation (for some stress tensor), so if you try to recast the theory in terms of extra dimensions, you'll need enough extra dimensions to account for all possibilities, not just the simple ones.

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I'd imagine that the bound occurs only for veeery weird spaces. For the usual ones (like sphere) one usually gets away with $m+1$. But I guess this obvious fact (that actually goes a long way in helping intuition) is not as impressive as showing numbers like 120 and 230 :) –  Marek Apr 20 '11 at 20:30
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@Marek: Even 2-D hyperbolic space doesn't embed into $\mathbb{R}^{3}$. You probably don't need the extreme upper bound, but you probably need more than $n+1$. I spent six months looking for an isometric embedding of a 2-d section of the Kerr Horizon into $\mathbb{R}^4$. It's an extremely non-trivial problem. Even if you don't care about geoemtry and are trying to embed things topologically, the Whitney embedding theorem tells you that you need a 2d euclidean space to contain a general d dimensional space. –  Jerry Schirmer Apr 20 '11 at 20:47
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@Marek: 2D hyperbolic space is the third simplest manifold there is. There is a very large class of d dimensional manifolds that don't embed in d+1 space. Yes, you can scaffold a lot by talking about embeddings of simple spaces, in 3d Euclidean space, but that doesn't even begin to touch on the big picture. The intrinsic picture is much cleaner. –  Jerry Schirmer Apr 20 '11 at 21:06
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Marek's conjecture, that you can "usually" get by with $m+1$ is, to say the least, not obvious to me. If I had to guess, I'd guess that "generic" manifolds require dimensionality not that different from the upper bound, but I don't actually know. (Indeed, I'm not sure I even know how to make such a statement precise enough to be either true or false.) –  Ted Bunn Apr 21 '11 at 2:08
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@Jerry, @Ted: I am not going to argue with you as we are all obviously talking about different things. Once again, I never said that every manifold can be embedded in this way. I only said that for usual ones you can do much better (and even if it's not $m+1$ it would still be much closer to $m$ than $n$). If any of you don't agree, please let me know about your favorite day-to-day example where you use embedding of codimension much higher than 1 or 2 ;) –  Marek Apr 22 '11 at 8:09
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I think there is some pedagogical value in the embedding approach you described. First courses on differential geometry often go through a detailed study of plane and space curves, and surfaces in $\mathbb{R}^3$ to build up intuition and then by observing that some geometric quantities are intrinsic to the surface, start the study of intrinsic geometry of manifolds. Dirac wrote a book on general relativity using this embedding approach. But in the end, as far as general relativity is concerned, this is only a pedagogical device since one has to abandon the extra dimensions as nonphysical. It is also true for intrinsic geometry of manifolds (as mathematical subject) since there can be countless different ways to choose the embedding.

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