# Is it necessary to embed a 4D surface in 5D space?

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.

Queries:

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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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