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Signature of a metric

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a metric and I need to find the signature.

    2. Relevant equations

    [itex]ds^{2} = -(1-e^{-x^{2}})\ dt^{2} + 6x\ dy^{2} + 9\ dx\ dy + y^{2}\ dx^{2}[/itex]

    3. The attempt at a solution

    In matrix form, the metric is
    -(1-e^{-x^{2}}) & 0 & 0 \\
    0 & y^{2} & 9 \\
    0 & 9 & 6x

    Now, I'm pretty sure my lecturer hasn't told us what a signature is, but after a bit of wikipedia-ing I see that I need to diagonalise this matrix.
    What I want to know is, is that really what I need to do? 'cos it's a really horrible equation I'd have to solve to find the eigenvalues...
    [itex]\lambda ^{3} + (-6x -y^{2} + 1 - e^{-x^{2}})\lambda ^{2} + (6xy^{2} + 9^{2} - 6x(1-e^{-x^2}}) - y^{2} (1-e^{-x^{2}}))\lambda + (1 - e^{-x^{2}})(6xy^{2} + 9^{2}) = 0[/itex]
  2. jcsd
  3. Dec 5, 2009 #2


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    The signature is a constant....

    That said, that doesn't look like a metric. Are there restrictions on x,y that you haven't told us about?
  4. Dec 5, 2009 #3
    No, but a later question is "Is the metric well-defined for all [itex]0<x<\infty[/itex]?"
    Haven't though about it, but I assume the answer is no, and that suitable restrictions will make it OK.

    yeeees, but how do I find it?

    The wikipedia way is to diagonalise the matrix (by finding eigenvalues, then eigenvectors, and then doing a bit of mtx multiplication) and count the number of +ve and -ve entries. Is that really what I need to do? I see no easy way of solving the (cubic!) characteristic eigenvalue equation.
  5. Dec 5, 2009 #4


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    You aren't being precise enough! I think you have made two implicit assumptions:
    * You want the general solution
    * You want an exact solution
    But you don't need either of those to answer the question you're really interested in, do you?
  6. Dec 5, 2009 #5
    uh, sorry, I can't see what you're getting at... are you saying I don't need to find the eigenvalues at all?
  7. Dec 5, 2009 #6


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    No, I'm saying that the solution you need doesn't have to be valid for all t,x,y, nor does it need to be an exact one.
  8. Dec 5, 2009 #7
    Ah, I think I see! (I hope)
    You mean that, since the signature is constant, I can take any x,y,t that I like and calculate it for them?
  9. Dec 6, 2009 #8
    Thanks for your help - I think I've got it.

    As [itex]x \rightarrow \infty[/itex] the signature is 0
    As [itex]x \rightarrow 0[/itex] the signature is 1

    So this implies that the metric is not well defined for all [itex]0<\infty[/itex], since the signature should be constant.
    Last edited: Dec 6, 2009
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