Is Poincare wrong about no preferred geometry?


by jmarshall
Tags: geometry, poincare, preferred
jmarshall
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#1
Feb18-13, 08:56 PM
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I heard that some physicists are trying to determine the spacial/geometric curvature of the universe by measuring the angles of distant stars (a very large triangle).

Is this possible? Or is Poincare correct when he said that there is no preferred geometry and that there is no experiment that will show one(Euclidean vs. non-Euclidean) the truest?
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jedishrfu
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#2
Feb18-13, 09:27 PM
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Buried in this document is something about the geometry of the universe and the information was just too good to not post:

http://pages.towson.edu/zverev/universe/Universe.htm

There's a bunch of other stuff as well.

I think the geometry of the universe is still an open question.

http://en.wikipedia.org/wiki/Shape_of_the_Universe
jmarshall
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#3
Feb19-13, 10:53 AM
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Thank you for your response. Sadly the geometry of space section does not cover the mathematical problem of there being no preferred geometry.

I believe that a curved space geometry has been solved that has all the logical consistencies that Euclidean geometry has. If this is true then can't this geometry be used in place of the standard Euclidean geometry? And if this is true then can't we make the argument that we live in a curved space just as easily as a flat (Euclidean) one?

lavinia
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Feb19-13, 08:49 PM
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Is Poincare wrong about no preferred geometry?


Quote Quote by jmarshall View Post
Thank you for your response. Sadly the geometry of space section does not cover the mathematical problem of there being no preferred geometry.

I believe that a curved space geometry has been solved that has all the logical consistencies that Euclidean geometry has. If this is true then can't this geometry be used in place of the standard Euclidean geometry? And if this is true then can't we make the argument that we live in a curved space just as easily as a flat (Euclidean) one?
in Physics today there is no fixed geometry of space but rather a time evolving geometry that is determined by the distribution of matter.

In mathematics there are many geometries and Euclidean geometry has not preferred.

Historically it was beleived that Euclidean geometry was the natural geometry of space that in fact it was an intrinsic feature of the idea of space itself. In the 18'th century another plane geometry was discovered so Euclidea geometry was seen not to be intrinsic and scientists treid to make measurements to determine which of the two was true in space. today it is realized that neither of these two describe the actual geometry of space.
atyy
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Feb19-13, 09:42 PM
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Quote Quote by jmarshall View Post
Thank you for your response. Sadly the geometry of space section does not cover the mathematical problem of there being no preferred geometry.

I believe that a curved space geometry has been solved that has all the logical consistencies that Euclidean geometry has. If this is true then can't this geometry be used in place of the standard Euclidean geometry? And if this is true then can't we make the argument that we live in a curved space just as easily as a flat (Euclidean) one?
Under some circumstances a flat spacetime and a curved spacetime are physically equivalent. However the flat spacetime geometry is not Euclidean, but Minkowskian. This is discussed in Chapter 11 of http://books.google.com/books?id=Gzl...gbs_navlinks_s .
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Feb19-13, 10:19 PM
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Quote Quote by atyy View Post
Under some circumstances a flat spacetime and a curved spacetime are physically equivalent.
What circumstances are you thinking of? This doesn't sound correct in general, so I assume you are thinking of some exceptional circumstances.
jmarshall
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#7
Feb19-13, 10:46 PM
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Hi,

Is the geometry really Minkowskian? Or is applying a Minkowskian geometry the most simple mathematical model to describe the relation between light, masses, and moving bodies in those special cases?

This is the core of my original question. Is there really a fundamental geometry?
atyy
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Feb19-13, 11:42 PM
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Quote Quote by DaleSpam View Post
What circumstances are you thinking of? This doesn't sound correct in general, so I assume you are thinking of some exceptional circumstances.
I am not sure about the exact conditions for this to hold. I have variously heard that spacetime should be coverable by harmonic coordinates, or that its topology is R3 X R. I know the former is sufficient, but I am not sure it is necessary. I am not sure whether the latter is true. Anyway, the basic idea is that a curved spacetime where the metric is the degree of freedom, can also be physically equivalent to a field (the metric perturbation, not necessary small or linear) on a flat spacetime.
WannabeNewton
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#9
Feb19-13, 11:57 PM
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When you say harmonic coordinates do you mean coordinate functions [itex]\alpha ^{(i)}[/itex] that satisfy [itex]\triangledown ^{a}\triangledown _{a}\alpha ^{(i)} = 0[/itex]? If so, I know that given a 2 dimensional manifold [itex]M[/itex] with a lorentzian metric and a harmonic function [itex]\alpha :M\rightarrow \mathbb{R}[/itex] together with the harmonic function [itex]\beta :M\rightarrow \mathbb{R}[/itex] conjugate to [itex]\alpha[/itex], [itex]\forall p\in M[/itex] there exists a neighborhood [itex]U[/itex] of [itex]p[/itex] on which we can transform the lorentzian metric to harmonic coordinates but I'm not sure what this has to do with flatness. This is actually a problem in Wald (chapter 3 problem 7) and assuming I didn't make calculation errors (which I very well may have!) I certainly didn't get an identically vanishing Riemann curvature tensor. Hopefully I didn't misunderstand what you were saying atyy. Cheers.
atyy
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Feb20-13, 12:09 AM
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Quote Quote by WannabeNewton View Post
When you say harmonic coordinates do you mean coordinate functions [itex]\alpha ^{(i)}[/itex] that satisfy [itex]\triangledown ^{a}\triangledown _{a}\alpha ^{(i)} = 0[/itex]? If so, I know that given a 2 dimensional manifold [itex]M[/itex] with a lorentzian metric and a harmonic function [itex]\alpha :M\rightarrow \mathbb{R}[/itex] together with the harmonic function [itex]\beta :M\rightarrow \mathbb{R}[/itex] conjugate to [itex]\alpha[/itex], [itex]\forall p\in M[/itex] there exists a neighborhood [itex]U[/itex] of [itex]p[/itex] on which we can transform the lorentzian metric to harmonic coordinates but I'm not sure what this has to do with flatness. This is actually a problem in Wald (chapter 3 problem 7) and considering I didn't make calculation errors I certainly didn't get an identically vanishing Riemann curvature tensor. Hopefully I didn't misunderstand what you were saying atyy!
The basic idea is that instead of considering the basic degree of freedom to be the metric g, we consider the basic degree of freedom h, where g=h+η.

I understand poorly the exact conditions for this equivalence to hold, so let me give a reference: http://arxiv.org/abs/gr-qc/0411023 .

A similar idea is in http://relativity.livingreviews.org/.../fulltext.html, Eq 62.
WannabeNewton
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Feb20-13, 12:18 AM
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Oh are you talking about the usual procedure of having a background flat space - time and having a perturbation field propagate on the background flat space - time?
atyy
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Feb20-13, 12:19 AM
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Quote Quote by WannabeNewton View Post
Oh are you talking about the usual procedure of having a background flat space - time and having a perturbation field propagate on the background flat space - time?
Yes, but with the perturbation not necessarily small or linear.
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Feb20-13, 12:31 AM
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Quote Quote by atyy View Post
Yes, but with the perturbation not necessarily small or linear.
Ah ok. The second link looks rather extensive and interesting, thanks for that! Till next time.
atyy
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Feb20-13, 12:59 AM
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Quote Quote by WannabeNewton View Post
Ah ok. The second link looks rather extensive and interesting, thanks for that! Till next time.
The first link is also a classic. I hope to understand it properly some day:)
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Feb20-13, 01:10 AM
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Quote Quote by atyy View Post
The first link is also a classic. I hope to understand it properly some day:)
I was reading until spin and Yang - Mills came up and I said yeah...this is where it ends for me sigh
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Feb20-13, 06:36 AM
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Quote Quote by jmarshall View Post
I heard that some physicists are trying to determine the spacial/geometric curvature of the universe by measuring the angles of distant stars (a very large triangle).

Is this possible? Or is Poincare correct when he said that there is no preferred geometry and that there is no experiment that will show one(Euclidean vs. non-Euclidean) the truest?
My understanding of the contemporary meaning of "no perferred geometry" is that it means that the distribution of matter determines the geometry, not that the geometry can't ever be measured.

So I don't see any issue with measuring the geometry of the universe, I don't think this contradicts there being "no preferred geometry".

It's possible that Poincare's meaning is different than the contemporary one, I suppose. But it would be odd to say that we couldn't measure a geometry, unless one insists that distances are arbitrary. There might be a philosophy that claims this, I suppose, but it gets into metaphysics rather than physics.

For the most part, physics insists that the Lorentz interval is well defined, the Lorentz interval plus a notion of simultaneity (which is mostly regarded as conventional) defies distance, and that the geometry of space-time is the geometry of the Lorentz interval. Thus the geometry of space-like slices (the usual notion of distance) can be deterined from the geometry of space-time plus the details of the method used to slice it into space + time, i.e from the geometry of space-time plus some notion of simultaneity.
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#17
Feb20-13, 07:43 AM
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Quote Quote by jmarshall View Post
I heard that some physicists are trying to determine the spacial/geometric curvature of the universe by measuring the angles of distant stars (a very large triangle).

Is this possible? Or is Poincare correct when he said that there is no preferred geometry and that there is no experiment that will show one(Euclidean vs. non-Euclidean) the truest?
If I'm not mistaken post #1 is referring only to the spatial geometry of the universe ("spacial,measure of angles...") not to the spacetime geometry as it seems to be interpreted by many here but it is confusing since in a later post he refers to Minkowskian spacetime.

I'll address here the easier spatial case only:the spatial geometry according to mainstream cosmology is indeed not exactly determined but it is highly constrained to three models, following the FRW model the three only possible spatial geometries are the Euclidean, the hyperbolic and the elliptic geometry with respectively 0, negative and positive constant curvature.
Empirically the curvature cannot ever be exactly measured because there is a limit of precision in detection and an inherent error that means we could be missing very small curvatures, so far the observations indicate a near flat geometry but with very wide error bars that don't allow us to discard either positive or negative small curvatures.

I would like to see that comment of Poincare in context, can you give a reference?
madness
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Feb20-13, 07:49 AM
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I think people are getting confused in this discussion. Poincare's philosophical views did not agree with the standard modern interpretation of GR. From wikipedia:

Poincaré believed that Newton's first law was not empirical but is a conventional framework assumption for mechanics. He also believed that the geometry of physical space is conventional. He considered examples in which either the geometry of the physical fields or gradients of temperature can be changed, either describing a space as non-Euclidean measured by rigid rulers, or as a Euclidean space where the rulers are expanded or shrunk by a variable heat distribution. However, Poincaré thought that we were so accustomed to Euclidean geometry that we would prefer to change the physical laws to save Euclidean geometry rather than shift to a non-Euclidean physical geometry.

This is what Poincare meant by no preferred geometry.


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