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  Is it incorrect to say that the four dimensional curved space-time is globally curved and locally flat?

+ 0 like - 0 dislike
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In General Relativity, defining the fourth coordinate as time is necessary to establish similarity with the special relativity.  In General Relativity, there are no "bilocal vectors" (See the book by Misner C. W, Thorne K.  S., Wheeler J. A.: Gravitation)  and therefore it is not possible to define the velocity or force vectors. But, the special relativity deals with velocity and force vectors. Therefore, the space-time is supposed to be globally curved and locally flat so that principles of special relativity can be locally applicable.

See the preprint on ResearchGate:

Vector analysis in the curved space-time: The problems, the remedies and some concerns

(https://www.researchgate.net/publication/357875376_Vector_analysis_in_the_curved_space-time_The_problems_the_remedies_and_some_concerns)

The classical vector analysis ("bilocal vectors" method) method gives all the desired results the same as tensor analysis dealing with only the relations between scalar components of a tensor. But, this  vector analysis also raises some serious geometrical concerns about the curved space-time. We see that locally the curved space can be only called as approximately flat which is not same as flat. Hence, it will be difficult to define vectors, such as force, even locally. 

asked Apr 10 in Theoretical Physics by anonymous [ no revision ]
recategorized Apr 15 by Dilaton

1 Answer

+ 1 like - 0 dislike

Yes, it is incorrect. If you have a spacetime with the curvature scalar \(R=g^{ab}R_{ab}\neq0\), with \(g^{ab}\) the metric tensor and $R_{ab}$ the Ricci-tensor, then this spacetime is not flat, in particular it cannot be called flat in any point where $R\neq0$. As $R$ is a scalar, no change of coordinates gets rid of $R\neq0$. 

In general relativity velocity and force can be defined. These quantities, at a given point in spacetime, are elements of the tangent space to the spacetime at the given point. No "bilocal vectors" (vector as an arrow connecting two points in spacetime) are needed.

answered Apr 10 by Flamma (80 points) [ no revision ]

Thanks for your reply. 

The classical "bilocal vectors" method gives exactly same mathematical answers as obtained by tensor analysis in GR. But it becomes difficult to give any geometrical framework as it raises many serious questions. The change in definition of vectors in GR can bypass some of these concerns.

The advantage of using classical vectors is that: we can attempt to draw a classical picture of curved space time. Then we realise that even when the Ricci tensor is zero, the space is not locally flat! The curved space is not flat for any condition!  This conclusion is based on the classical picture attempted using the bilocal vectors. This method also gives all the desired mathematical results and identities. The difference arises in geometrical interpretations. 

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