the 4th dimension?

[scope]

hi,

what do you think that in general relativity, since space is curved, the curvature could be interpreted as a 4th space dimension "hyperspace"?
in other words, the 4th space dimension would take the place of time, for a coordinate observer?

[Nabeshin]

No. When thinking of curved surfaces, we as humans often picture them embedded in a higher dimensional space. That is, we draw a curvy line (1D) in a plane (2D). We see a wavy blanket (2D) in real life (3D). So we assume that our 3D universe must be in a 4D (spatial!) similar situation. This is only our limitations as humans, however. These objects, the line, the curvy blanket, and our own universe, have intrinsic curvature which exists completely independent of their embedding in a higher dimensional space. Mathematically, that is, it is perfectly fine to talk about a 3D curvy object on its own with no reference to a fourth dimension -- only if we tried to visualize such a situation would we need one! Therefore, it is not necessary to posit the existence of such extra dimensions in order to have a curved universe.

[TylerH]

Curvature is the effect of 2+ dimensions, not a dimension in itself. Think of a dimension as a measurement, and the count of dimensions as the minimum measurements you need to describe the position of an object, length, width, height, and time. Curvature can be thought of as the effect of the independent variance of 2+ measurements, like how there can be many heights for the same width of an object, thus making the object curved on that slice.

Also, given the above definite of a dimension, you could eliminate curvature as a possible dimension because it doesn't help determine position.

[scope]

at the horizon, space is infinitely contracted in the reference frame and then a fourth space dimension(and no time dimension, since the horizon exists only for t=infinite), would be welcome

[DaleSpam]

Nothing pathological happens to the manifold at the horizon. Only the Schwarzschild coordinates become pathological there. This is easily remedied by introducint a different set of coordinates. A new dimension is certainly not needed and I don't see how it would be helpful in any way for that.

[scope]

its not pathological but the curvature.

[DaleSpam]

The curvature at the horizon is always finite and can be made arbitrarily small by using a sufficiently large mass.

[scope]

i did not mean the scalar curvature but the curvature that is observer-dependent

[DaleSpam]

What is that? I have never heard of such a thing.

[Mueiz]

What is that? I have never heard of such a thing.

It is the full Riemannian tensor (of order 4) which is a physical characteristic of space-time in Einstein Theory ..it is an absolute property and has nothing to do with any observer's world-line.

[Mueiz]

hi,

what do you think that in general relativity, since space is curved, the curvature could be interpreted as a 4th space dimension "hyperspace"?
in other words, the 4th space dimension would take the place of time, for a coordinate observer?

It is a sign to deep understaning of GR to try to think of another formulation to the theory but this new formulation would be a true progress in two cases;
first:It is more simple than the classical formulation.
second:It makes possiple to solve the problem of unifying GR and QP or other problems of GR
Otherwise the classical formulation of GR in four-dimension space characterized by the curvature is a good final theory for gravitation and need not be interpreted.

[DaleSpam]

It is the full Riemannian tensor (of order 4) which is a physical characteristic of space-time in Einstein Theory ..it is an absolute property and has nothing to do with any observer's world-line.Exactly, so it is not observer-dependent. I have never heard of an observer-dependent curvature like scope was talking about.

[Mentz114]

It is the full Riemannian tensor (of order 4) which is a physical characteristic of space-time in Einstein Theory ..it is an absolute property and has nothing to do with any observer's world-line.
It is true that scalar measures of curvature Raa and RabcdRabcd are absoluely invariant ( being scalars) but the components of the Riemann tensor will change if calculated wrt a frame field representing some observer.

See for instance equations (1.4.16a),(1.4.16b) and (1.4.16c) in arXiv:0904.4184v3 [gr-qc] 4 Nov 2010 in the section "1.4.2 Tetrad transformations".

[DaleSpam]

Right, but tensors are geometric objects that do not depend on the coordinates. Of course, their representation in a given coordinate system or basis will depend on the coordinates, but the geometrical object itself does not.