Deriving the FLRW Metric: 4D Euclidean Space Needed?

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Discussion Overview

The discussion centers around the necessity of considering a 4D Euclidean space in the derivation of the Friedmann-Lemaître-Robertson-Walker (FLRW) metric. Participants explore the implications of using a 4D Euclidean framework versus a Lorentzian manifold, particularly in the context of cosmological models and symmetry in general relativity.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question the need for a 4D Euclidean space, suggesting that the FLRW metric is locally Lorentzian and does not require such a framework.
  • Others argue that certain derivations introduce a 4D Euclidean space to define a 3D sphere, which represents the constant cosmological time surface.
  • A participant references two derivations that utilize a 4D Euclidean space, indicating a belief that it is necessary for the discussion.
  • Another participant mentions that symmetry plays a crucial role in deriving maximally symmetric spaces, suggesting that studying symmetry is beneficial in the context of general relativity.
  • There is a discussion about the embedding of maximally symmetric spaces in higher-dimensional Euclidean spaces, with some participants noting that negative-curvature cases cannot be fully embedded in Euclidean space.
  • Participants also discuss the methods used by Weinberg in deriving these spaces, including the use of the Killing equation and the uniqueness of spaces with constant curvature.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of a 4D Euclidean space for the FLRW metric. There is no consensus on whether such a framework is essential, with multiple competing perspectives presented throughout the discussion.

Contextual Notes

Some arguments depend on the definitions of curvature and the nature of the spaces being discussed. The discussion also highlights the complexity of embedding spaces and the implications of symmetry in the context of general relativity.

kent davidge
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Why is it needed to consider a 4D Euclidean space to introduce the FLRW metric? Is it because with a fourth parameter, we can set the radius of the 4D sphere formed with the four parametres as constant?
 
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Why do you need to consider a 4D Euclidean space to introduce the FLRW metric? It's a Lorentzian manifold, and you just ask for the spacetimes with maximally symmetric spaces.
 
vanhees71 said:
Why do you need to consider a 4D Euclidean space to introduce the FLRW metric? It's a Lorentzian manifold, and you just ask for the spacetimes with maximally symmetric spaces.
Thanks
 
kent davidge said:
Why is it needed to consider a 4D Euclidean space to introduce the FLRW metric?

We don't. The FLRW metric is locally Lorentzian, not Euclidean. Why do you think a 4D Euclidean space is needed?

(Btw, vanhees71 was asking you the same question. He was not answering your question in the OP; he was telling you the question was based on a false assumption.)
 
PeterDonis said:
(Btw, vanhees71 was asking you the same question. He was not answering your question in the OP; he was telling you the question was based on a false assumption.)
Ah :biggrin:

PeterDonis said:
Why do you think a 4D Euclidean space is needed?
Because I found two derivations where the author introduces such a space.
One of them is here:
http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px436/notes/lecture20.pdf

The another derivation I found in a book I've read.
 
kent davidge said:
Because I found two derivations where the author introduces such a space.
One of them is here:
http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px436/notes/lecture20.pdf

But you need to read it more carefully. The 4D Euclidean space is introduced in order to define a 3D sphere, it is not the space-time in the FRWL model. That 3D sphere is the constant cosmological time surface. Think of the sphere through all time, that forms the space-time manifold, which is not Euclidean.
 
As usual, Weinberg is much more straight forward. He just derives the maximally symmetric 3D spaces of constant curvature using symmetry. Symmetry is anyway the key to all of physics. So it's well invested time to study it also in the context of GR (keyword: Killing vectors).
 
In the section "Maximally symmetric spaces. Constructions", Weinberg constructs them the same way, via embedings in a ##n+1## dimensional Euclidean space.
 
I guess you mean n+1-dimensional affine (flat) space. The negative-curvature case cannot be completely embedded in an Euclidean space. If I remember right, Weinberg discusses this within the mentioned chapter too.
 
  • #10
It is the hyperboloid ##-x_0^2+x_1^2+\cdots+x_n^2=0##.

My point was that he also does that. I assumed that when you said that his method was more straight forward you meant that he doesn't construct them the same way i.e. as hypersurfaces in an Euclidean space of one more dimension.
 
  • #11
Maybe I'm wrongly attributing the derivation to Weinberg (I've to check as soon as I'm back home), but hasn't he also given the derivation of finding systematically the spaces with maximal symmetry via the Killing equation?
 
  • #12
vanhees71 said:
Maybe I'm wrongly attributing the derivation to Weinberg (I've to check as soon as I'm back home), but hasn't he also given the derivation of finding systematically the spaces with maximal symmetry via the Killing equation?

Well, he shows that they have constant curvature and that the curvature and the signature of the metric uniquely determine the space (at least locally and up to an isometry). But to find them he says that since they are unique it suffices to construct them any way we want, and he does the same thing as in the lectures in the link above.
 
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