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Degenerate Metric at semi-regularizable Schwarzschild singularity

  1. Jun 7, 2014 #1

    ChrisVer

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    I am reading this paper
    http://arxiv.org/pdf/1111.4837.pdf
    and I came across under eq12 that the new metric is degenerate...
    How can someone see that from the metric's form?
    Degeneracy for a metric means that it has at least 2 same eigenvalues (but isn't that the same for the Minkowski metric since it's diagonal with 3 times -1 eigenvalues)? Or that you can define two different metric tensors [itex]g[/itex] and [itex]\bar{g}[/itex] which keep [itex]ds^{2}[/itex] invariant?

    Thanks
     
  2. jcsd
  3. Jun 7, 2014 #2

    Bill_K

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    Degeneracy for a metric means that det(gab) = 0. In this case, gττgξξ - gτξ2 = 0.
     
  4. Jun 7, 2014 #3

    ChrisVer

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    Oh well then that means that my first choice was the correct one about Eigenvalues...
    and why is the degenerate metric singular? or put in another manner, what does singular metric mean?
     
    Last edited: Jun 7, 2014
  5. Jun 7, 2014 #4

    Bill_K

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    For the Minkowski metric, the eigenvalues are -1, +1, +1, +1 and det(g) = -1. Det(g) is a scalar density, transforming under a coordinate transformation as Det(g') = |∂x/∂x'|2 Det(g) where |∂x/∂x'| is the Jacobian of the transformation. Consequently there exists no nonsingular transformation that can take a singular metric to Minkowski. Wherever Det(g) = 0, the spacetime is not locally Minkowskian.
     
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