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Schwarzenchild vs Minkowski

  1. Jul 17, 2013 #1
    The Schwarzenchild Metric can be the Minkowski Tensor with the correct terms in 4-Space. If not Schwarzenchild Metric must have EigenValues are all real and the Matrix is symmetrical.
  2. jcsd
  3. Jul 17, 2013 #2


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    I'm not really sure what it is you're saying but if you're asserting that there exists a coordinate system in which the Schwarzschild metric becomes the Minkowski metric everywhere on the open subset the chart is defined on then that's obviously false; Schwarzschild space-time is not flat. What you can do is put the Schwarzschild metric in Minkowski form at a given point; this is a simple consequence of the spectral theorem.
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