Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 0 like - 0 dislike
1291 views

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, 2022 in Theoretical Physics by anonymous [ no revision ]
recategorized Apr 15, 2022 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, 2022 by Flamma (110 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. 

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...