What is the Unique Metric Tensor in this Line Element?

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Discussion Overview

The discussion revolves around the derivation and properties of the metric tensor associated with a specific line element in a cosmological context. Participants explore the structure of the metric tensor, its components, and the implications of symmetry in the matrix representation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a line element and asks for the corresponding metric tensor matrix.
  • Another participant questions how to derive the metric tensor from the provided space-time interval.
  • A proposed metric tensor matrix is presented, but it is challenged for missing off-diagonal terms and symmetry.
  • Participants discuss the implications of squaring terms in the line element and the resulting contributions to the metric tensor.
  • There is a debate about whether off-diagonal terms need to be explicitly included in the metric tensor representation.
  • Some participants suggest that the metric tensor must be symmetric and discuss how to handle terms that arise from the squaring of the first term in the line element.
  • Further contributions refine the metric tensor matrix, with discussions on the necessity of including symmetric terms for off-diagonal components.
  • Participants express uncertainty about the correct formulation of the metric tensor and the implications of the terms derived from the line element.
  • There are suggestions to diagonalize the metric tensor and to explore the relationship between the metric and vierbein fields.
  • Some participants propose methods for transforming the metric into a more standard form, while others question the need for such transformations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct form of the metric tensor, with multiple competing views on the inclusion of off-diagonal terms and the symmetry of the matrix. The discussion remains unresolved regarding the final structure of the metric tensor.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the terms in the metric tensor and the dependence on the definitions of the components involved. The mathematical steps for deriving the metric tensor from the line element are not fully resolved.

Who May Find This Useful

This discussion may be of interest to those studying general relativity, cosmology, or differential geometry, particularly in the context of metric tensors and their applications in theoretical physics.

  • #31
Breo said:
The same metric with different coords?

$$ ds^2 = dp^2 - \delta_{ij} a^2dx^idx^j $$

Which is diagonal. This seems too much easy :-/

I wrote this on mobile. Hope it worked xD

It seems too easy because Pervect basically gave it to you haha.

You have diagonalized the metric, but you still need to put it in terms of diag(1,-1,-1,-1) so you have to do one more transformation.
 
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  • #32
Mmm as it seems to work with the signs I bet you are talking about this:

$$ ds^2 = dp^2 - a^2 [(dx^1)^2 + (dx^2)^2 + (dx^3)^2] $$

removing the delta expression?
 
  • #33
You still have the ##a^2## there. What you wrote is just explicitly what you had in post #30, you just wrote out every term. It is identical with the one in #30 though. You need to go 1 more step.
 
  • #34
I am still wondering :-/
 
  • #35
Breo said:
I am still wondering :-/

I'm not sure what other hints I can give other than just telling you the answer haha. Recall what you did for the spherical coordinates? What did the metric look like there, and how did you find a triad?
 
  • #36
Matterwave said:
I'm not sure what other hints I can give other than just telling you the answer haha. Recall what you did for the spherical coordinates? What did the metric look like there, and how did you find a triad?

But using ##dp^2## ?
 
  • #37
Breo said:
But using ##dp^2## ?

Hmm, let's try this, if you had a metric ##ds^2=dr^2+r^2d\theta^2## (polar coordinates! This space is flat by the way), what are the two ortho-normal forms one could make? (do not transform back to x and y, just turn this into an orthonormal set).
 
  • #38
$$e^1 = dr \\ e^2 = rd\theta $$

But I still do not see the point which you are aiming to. The only relation with the minkowskian metric I could find was ## g_{\mu\nu} = \eta_{ab} e^a_{\mu}e^b_{\nu} ## which is satisfied by the metric I already found and it is diagonal aswell. My mind will blow-up xD

I think the problem should be something that I made which is wrong...

EDIT: Are, maybe, you talking about this: ##ds^2 = \eta_{ab} e^a \otimes e^b ## ?
 
Last edited:
  • #39
So you were able to find really quickly ##e^1=dr,~e^2=rd\theta##, why not apply this same idea to ##ds^2=dp^2-a^2[(dx^1)^2+(dx^2)^2+(dx^3)^2]##? Do you see the similarity? What happens if I switch ##r\leftrightarrow a##?
 
  • #40
$$e^1= dp \\
e^2= adx^1 \\
e^2 = adx^2 \\
e^3 = adx^3 \\

$$

Oh wait. Are you trying to tell me that if I make a transformation that involves some coordinates like dp does with ##dx^i## should I change also the rest of ##dx^i## in the metric? So I can not write ##a^2(dx^i)^2## ?
 
  • #41
No, you're good. You found all 4 (but you numbered them wrong).
 
  • #42
$$e^0= dp \\
e^i= adx^i

$$

I noticed this error. Still do not find what is that next step you are talking about. I have just found almost the same metric in Carroll's book: pg.490 and still do not know what to do.
 
  • #43
? I think you are done...you asked to find the 4 orthonormal forms, and you have.
 
  • #44
Maybe my last post sounds a bit aggresive. Was not my intention, I am not good at english. Sorry if so.

Is just I can not find the next step which you referred to.
 
  • #45
Breo said:
Maybe my last post sounds a bit aggresive. Was not my intention, I am not good at english. Sorry if so.

Is just I can not find the next step which you referred to.

I don't think I referred to any "next step". You asked to find an ortho-normal set of one-forms basis, and I think you have in post #42.
 
  • #46
Matterwave said:
You have diagonalized the metric, but you still need to put it in terms of diag(1,-1,-1,-1) so you have to do one more transformation.

Matterwave said:
You still have the ##a^2## there. What you wrote is just explicitly what you had in post #30, you just wrote out every term. It is identical with the one in #30 though. You need to go 1 more step.

Must be all a misscomunication haha I thought I must to do something to the line element
 
  • #47
Breo said:
Must be all a misscomunication haha I thought I must to do something to the line element

You already did. Now the line element is ##ds^2=(e^0)^2-(e^1)^2-(e^2)^2-(e^3)^2## there is no longer that ##a^2## because you subsumed them into ##e^i##.
 

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