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Can one diagonalize the Kerr metric?

  1. Jul 31, 2008 #1
    Is it possible to diagonalize the Kerr metric in the Boyer-Lindquist coordinates? If so then I think calculations with the metric will become easier. I forget under what condition a matrix can be diagonalized. Can anybody remind me?
  2. jcsd
  3. Jul 31, 2008 #2


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    I think if you transform to a co-rotating frame, the off-diagonal term will be zero. I might try it when I have more time.

    [edit] I have since discovered that this transformation does not diagonalise the metric.

    Last edited: Jul 31, 2008
  4. Sep 4, 2010 #3
    Unfortunately, it is not possible. The reason that one can get as close as one off-diagonal term is a theorem by Achille Papapetrou. This requires that the metric be well-behaved on the axis of rotation. (Roy Kerr)
  5. Oct 15, 2012 #4
    Sorry to bump an old thread, but is it possible to diagonalize the metric every where except on the axis of rotation? In other words, is the problem that no general diagonalization exists, or that it is simply a non-diagonalizable matrix? The second seems improbable to me, since the metric is symmetric and thus always diagonalizable, even unitarily, but I don't know much about the Kerr solution so maybe something weird is going on.
  6. Oct 15, 2012 #5

    George Jones

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    Notice who made post #3.

    Every spacetime metric is diagonalizable, but not necessarily by a coordinate basis, i.e., there does not necessarily exist a coordinate system that diagonalizes the metric.
  7. Oct 15, 2012 #6


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    Isn't diagonalizability in a certain choice of coordinates equivalent to staticity? Since the Kerr metric isn't static, it seems to me that we don't need no fancy theorems as claimed in #3. Am I oversimplifying or getting something wrong?
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