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How can we model intrinsic curvature?

+ 0 like - 0 dislike
40 views

Can it only be done in Euclidean space? Doesn't Euclidean space only model extrinsic curvature?


This post imported from StackExchange Physics at 2014-03-22 16:54 (UCT), posted by SE-user Ocsis2

asked Apr 11, 2012 in Mathematics by Ocsis2 (20 points) [ revision history ]
retagged Mar 25, 2014 by dimension10

2 Answers

+ 4 like - 0 dislike

I'm not quite sure what you mean by the term "model" in this context, but:

If a space is a Euclidean space, in the sense that it has a Euclidean metric, then its Levi Civita connection (the connection compatible with its metric) has no intrinsic curvature (for example a flat plane is like this). However, it may be given some extrinsic curvature by means of an embedding into a higher dimensional space (the flat plane may be rolled up into a cylinder in $\mathbb{R}^3$).

But if you were a two dimensional organism living on the cylinder, you couldn't detect this extrinsic curvature by locally measuring angles and distances.

This post imported from StackExchange Physics at 2014-03-22 16:54 (UCT), posted by SE-user twistor59
answered Apr 11, 2012 by twistor59 (2,490 points) [ no revision ]
Very clear. How can one relate both metrics?

This post imported from StackExchange Physics at 2014-03-22 16:54 (UCT), posted by SE-user phoenix
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No, Euclidian space is not necessary. You can "model" intrinsic curvature using the beautiful language of Riemannian geometry, whose great triumph was formulating a vocabulary that lets you talk about the curvature of a space without making reference to an extrinsic space in which the curved space is embedded: hence the term intrinsic.

This is crucial to general relativity, since embedding a curved 4-space in flat Euclidean space requires that the Euclidean space be ten-dimensional, and dealing with that embedding would suck.

This post imported from StackExchange Physics at 2014-03-22 16:54 (UCT), posted by SE-user Rory
answered Apr 12, 2012 by Rory (0 points) [ no revision ]

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