Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metrics on a manifold, gravity waves, gauge freedom

  1. Mar 28, 2010 #1
    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
     
  2. jcsd
  3. Mar 28, 2010 #2

    atyy

    User Avatar
    Science Advisor

  4. Mar 28, 2010 #3
    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
     
  5. Mar 28, 2010 #4

    dx

    User Avatar
    Homework Helper
    Gold Member

    What do you mean "act as a metric"? Whether a given tensor can be a metric or not doesnt depend on any coordinates. The only condition is that it must be symmetric, nondegenerate, and have a Lorentz signature.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook