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Lets consider the line element:

$$ds^2=dr^2+r^2[d\theta^2+\sin^2\theta d\phi^2]$$

There are three variables r,theta and phi. If we use a surface constraint like r=constant the number of independent variables is reduced by one--now we have two independent variables.These surfaces[corresponding to r=const] may be embedded in a three dimensional space.

Now lets consider Schwarzschild's metric:

$$ds^2=(1-2m/r)dt^2 - (1-2m/r)^{-1} dr^2 - r^2[d \theta^2+\sin^2 \theta d\phi^2]$$

If we use a surface constraint[for example: t=constant] we have three independent variables.The resulting time slices are three dimensional surfaces which are naturally embedded in a 4D space.

The General Relativity metric has four variables: one relating to time and three relating to the spatial coordinates.Any surface constraint would reduce the number of variables to three.

In fact any arbitrary spacetime curve[world-line] may be made to lie on a 3D- Surface obtained by applying some suitable constraint on 4D space.The constraint may not be a simple one like t= constant or r= constant. It may be of a complicated nature. For the purpose of embedding in GR a 4D space seems to be sufficient.


  1. Is it essential the we should consider a 4D surface embedded in 5D space to understand or interpret GR?
  2. It appears that the curve[4D path] is more important than the surface on which it is lying since the same curve may be made to lie on several distinct surfaces at the same time[you may extend this to the case of 4D surf embedded in 5D space].Is this interpretation correct?
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IMO, Christophs answer is quite complete and should close the question – Ilja Mar 21 at 21:50
A tangent vector of finite length protrudes out of the manifold.The null vector is parallel to and perpendicular to itself .It is perpendicular to the manifold . It protrudes out of the surface--into what? – Anamitra Palit Mar 24 at 13:58

Any enclosing space is outside of the problem domain of GR: All results can be obtained from within the space-time. Physically, it makes no sense to talk about an enclosing space which has no impact whatsoever on measurements. In particular, even though we say space-time is curved, the question Where does it curve to? makes no sense in the framework of GR.

Also, 5 dimensions are not enough to contain arbitrary 4-dimensional pseudo-Riemannian manifolds of index 1 if you want to preserve the metric. Quoting C. J. S. Clarke, On the Global Isometric Embedding of Pseudo-Riemannian Manifolds:

The space-time of general relativity can be embedded isometrically in $E^{2,q+2}$ (pseudo-Euclidean space of signature $q-2$) where $q=46$ or $q=87$ for compact or non-compact space-time, respectively.

However, the result is only valid for finite $k$ and not $C^\infty$.

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I suppose you need a second time in order to deal with closed timelike loops. If you use time-oriented spacetimes the dimension should go down a lot. Also, these types of results tend to not be at all close to optimal. – Ron Maimon Dec 6 '11 at 17:54

No, it is not essential to consider a surface to be embedded in a higher-dimensional space. In fact the embedding is pretty much irrelevant. GR is based on differential geometry, and part of the reason differential geometry was developed the way it was is that it allows you to analyze surfaces without having to consider their embeddings in higher-dimensional spaces. You can calculate the geodesics or whatever else you want using only properties that can be measured from "within" the space.

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But we always think in terms of a space-time surface.The same 4D path relevant to the motion of a particle, may be included in several[distinct] surfaces. The path seems to be more important than the surface including it.How do you explain that? – Anamitra Palit Dec 6 '11 at 3:30
@Anamitra Palit: Nobody thinks in terms of a "space-time surface". The space time is not sitting in a bigger space time, it's all there is. – Ron Maimon Dec 6 '11 at 4:00
In your language (where "path" is the thing that is embedded and "surface" is the thing that it is embedded in), no, we do not always think in terms of a spacetime surface. As I was saying, the surface is completely irrelevant. There doesn't even have to be a surface. Not all 4D paths are embeddable in 5D surfaces. – David Z Dec 6 '11 at 4:01
In fact, you have to work very hard if you want to represent your spacetime as an embedded surface in a flat space. "Hard" in the sense that the flat space has, in general, to have very large number of dimensions. I can't remember the dimensionality for embedding 3+1 Lorentzian spacetimes, but the calculations are related to the Whitney Embedding Theorem. Does anyone know the number ? - something not much less than 100 rings a bell..... – twistor59 Dec 6 '11 at 7:56
@twistor59: for 4-dimensional Riemannian manifolds, the upper bounds of the dimension of the enclosing Euclidean space are 46 for compact manifolds and 230 for non-compact manifolds by Nash's embedding theorem; similarly, any pseudo-Riemannian manifold can be embedded into a pseudo-Euclidean space, but I don't know the specific bounds; I'd be surprised if they were smaller than the one's for the Riemannian case, though... – Christoph Dec 6 '11 at 9:26

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