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Minkowski space time is defined in terms of a flat pseudo-Riemannian manifold. I have wondered if it can be redefined as Riamannian manifold and in the case what type of curvature would there appear.

Formally:

Let M be a semi-Riemannian manifold of dimension 4, corresponding to the Minkowski space, and let g be the metric tensor (non positive definite), T be the Riemann curvature tensor and P a generic point of M.

Question 1

Which (if any) of the following is true?

a. at every P there exists one system of coordinates for which the metric g becomes Riemannian (positive definite) in a ball of radius R non infinitesimal centred in P

b. there exists one system of coordinates for which g is Riemannian (positive definite) at all P of M

Comment: in words, can we, with a change of coordinates, get rid of semi-Riemannianity – either in finite region or globally?

If this is the case, how do we pay it in terms of curvature? This the next question:

Question 2

c. if previous a) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

d. if previous b) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

Thanks a lot

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    $\begingroup$ Um...*homeomorphic* does not mesh well with Euclidean, since homeomorphy does not care for metrics or inner product, so all your questions about homeomorphy make no sense, since the very definition of a manifodl is that it is locally homeomorphic to $\mathbb{R}^n$. For the others, Minkowski space is flat but it is not Euclidean. You should start at the basics of differential geometry and SR/GR before asking such huge questions which are borderline unanswerable. $\endgroup$ – ACuriousMind Aug 2 '14 at 15:33
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    $\begingroup$ Comment to the question (v2): Echoing @ACuriousMind's comment, it seems that OP might be using non-conventional terminology. For example, Minkowski spacetime usually refers to $\mathbb{R}^4$ endowed with the Minkowski metric, not general Lorentzian manifolds. See also e.g. this Phys.SE answer. $\endgroup$ – Qmechanic Aug 2 '14 at 15:35
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    $\begingroup$ Dear ACuriousMind, homeomorphic was the only transformation I could think of for: isometric would not make sense, because it is "the change of metric" which would enable you to pass from a flat to a curved representation. When you represent the geometry of a curved emisphere (inheriting tridimensional euclidean metric) onto a flat euclidean plan, by orthogonal projection of the first on the second, your metric is no longer Euclidean. $\endgroup$ – massimo Aug 2 '14 at 15:48
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    $\begingroup$ Dear Qmechanic, you said right: my use of non-euclidean may not correspond to common use. Not only: it is exactly the analysis of how it changed the non euclidean concept from Gauss/ Lobacevski/ Bolyai to Riemann that made me rethink of Minkowski's model interpretation. $\endgroup$ – massimo Aug 2 '14 at 16:09
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    $\begingroup$ This is not too broad anymore, so I'll vote to reopen, but the answer is simply: No $\endgroup$ – ACuriousMind Aug 8 '14 at 21:16
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We interpret OP's question (v3) as essentially asking

Can a Lorentzian manifold (with Minkowski signature) by coordinate transformations be redefined as a Riemannian manifold (with Euclidean signature)?

The answer is No since the metric signature of a pseudo-Riemannian manifold is invariant under general coordinate transformations. This follows e.g. from Sylvester's law of inertia. Recall that the metric tensor in any coordinate system is a real symmetric matrix, and therefore diagonalizable.

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  • $\begingroup$ Dear Qmechanic, thks for your answer BUT my question was significantly different in one point: I did not ask that the Riemannian manifold be "with Euclidean signature"! otherwise I would not have even asked about curvature tensor. Thks anyhow, M $\endgroup$ – massimo Aug 9 '14 at 13:25
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    $\begingroup$ Dear @massimo: A Riemannian manifold has by definition Euclidean signature. A pseudo-Riemannian manifold can have other signatures. $\endgroup$ – Qmechanic Aug 9 '14 at 13:27
  • $\begingroup$ sorry Qmechanic, may be we use different terminologies here: Sylvester applies to diagonalisable matrices; for me only euclidean riemannian are diagonalisable; "curved" riemannian are not and are those I am after. .... $\endgroup$ – massimo Aug 9 '14 at 13:31
  • $\begingroup$ Dear Qmechanic, may be I have understood you now  you would say: every Riemannian is in an infinitesimal ball of its points Euclidean; if Minkowski manifold were changeable into a Riemannian manifold in the infinitesimal ball of one of its points it would therefore be also Euclidean - that it its metric signature would be Euclidean; but this cannot be because of Sylvester ... Have I got you right? thanks $\endgroup$ – massimo Aug 9 '14 at 13:40
  • $\begingroup$ Dear @massimo: You are right that small enough neighbourhood of an $n$-dimensional manifold $M$ are diffeomorphic to a neighbourhood of $\mathbb{R}^n$. But focusing on neighbourhoods seems like a distraction here. Consider for simplicity just a single point $p\in M$. Study the metric tensor $g_p: T_p M \times T_p M \to \mathbb{R}$. In any coordinate system the metric $g_{\mu\nu}(p)=g_p(\partial_{\mu},\partial_{\nu})$ is a real symmetric matrix. $\endgroup$ – Qmechanic Aug 9 '14 at 15:00
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As mentioned by Qmechanic, the answer to your questions is no.

However, assuming space-time is oriented, we have the following:

For any pseudo-Riemannian metric $g$, there exists a normalized time-like one-form $h^0$ and a Riemannian metric $g^R$ so that $$ g = 2h^0\otimes h^0 - g^R $$

This yields a locally Euclidean topology compatible with the manifold topology.

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