Metrics on a manifold, gravity waves, gauge freedom

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Discussion Overview

The discussion revolves around the properties of metric tensors on manifolds, specifically addressing whether any twice-covariant tensor can serve as a metric on a finite patch of a manifold. The inquiry touches on concepts of gauge freedom in the context of gravitational perturbations and the implications of coordinate transformations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Michael questions whether for any twice-covariant tensor \(\gamma\), there exists a chart on a finite patch of a manifold such that \(\gamma\) acts as a metric.
  • Michael suggests that if a coordinate chart can always be constructed, then gravitational perturbations could be interpreted solely as gauge effects, leading him to suspect that the answer to his question is negative.
  • Michael requests a mathematical argument to support or refute his suspicion regarding the existence of such coordinate charts.
  • A participant challenges the interpretation of what it means for a tensor to "act as a metric," stating that the conditions for a tensor to be a metric are independent of coordinates and must include symmetry, nondegeneracy, and a Lorentz signature.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which a tensor can be considered a metric. Michael's suspicion about the existence of coordinate charts remains unresolved, as does the broader question of the relationship between gauge effects and physical variations.

Contextual Notes

The discussion does not resolve the mathematical conditions required for a tensor to act as a metric, nor does it clarify the implications of gauge freedom in the context of gravitational perturbations.

Michael_1812
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Suppose I have a manifold. I say that it can support a certain configuration of gravity field described by metric tensor \gamma. I do not write \gamma_{\mu\nu}, because that would immediately imply a reference to a particular chart. A tensor field, however, exists on a manifold unrelated to this manifold's parameterisation. It is a box, in which you deposit two vectors, and which then spits out a scalar.

Can I always assume that, for any twice-covariant tensor \gamma, there exists (not necessarily global) a chart, for which this tensor acts as a metric?

Let me now express my question a bit more carefully. We always introduce a manifold via local charts. Suppose some patch of the manifold is parametrisable by some grid z^\mu with metric w_{\mu\nu}. Can I always introduce on this patch a new grid x^\alpha, whose metric is exactly the desired \gamma? (whom I shall now call \gamma_{\alpha\beta} ) Please be mindful that I am talking not about an infinitesimal neighbourhood, but about a finite patch.

I suspect that the answer to my question is negative, though I am not 100% sure. Here is my argument. Suppose that such a coordinate chart can always be assembled. I take a "background metric" \gamma and build the appropriate grid x^\alpha. I take a "perturbed metric" g and construct an appropriate chart y^\alpha. Then, if g and \gamma are "close", their difference being h, I shall say that the difference between g and \gamma is solely due to a coordinate transformation. Hence the gravitational perturbation h is always a gauge effect.

If h can be an actual physical variation, then the answer to my above question must be negative.

This logic, if correct, presents a physical answer to the question.
Still, I would love to see a mathematical argument.

Great many thanks,

Michael
 
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http://www.personal.soton.ac.uk/jav/karlhede.html
 
Dear atyy,
Thanks for the interesting link. While it is indeed relevant, it still does not provide an immediate answer to the question: given an arbitrary twice-covariant tensor field, is it possible to build, on a *finite* patch, a coordinate grid, for which this tensor field act as a metric?
Thanks again,
Michael
 
What do you mean "act as a metric"? Whether a given tensor can be a metric or not doesn't depend on any coordinates. The only condition is that it must be symmetric, nondegenerate, and have a Lorentz signature.
 

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