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Metric tensor and spacetime

  1. Oct 7, 2008 #1
    Hi guys. I'm taking a GR course right now, my first one. I was reading the textbook and I was wondering if you guys could help me out just to make sure I'm getting things straight here. I'm reading about the metric tensor, and I'm pretty sure I am expected to know what the metric tensor for a given spacetime is, if the line element if given to me. So, say we have this line element:

    ds2=ev(r)dt2-e[tex]\lambda[/tex](r)dr2-r2(d[tex]\theta[/tex]2+sin2[tex]\theta[/tex]d[tex]\phi[/tex]2)
    Is the metric tensor simply the coefficients of each differential arranged in the diagonal of a matrix, with the rest of the elements in it equal to zero? Or am I way off here?

    NOTE: The Thetas and the phi in the expression are not supposed to be exponentials, I don't know why they came out that way.
     
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2

    cristo

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    The line element is related to the metric tensor as follows:

    [tex]ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}[/tex].

    So, yes, for this line element, the metric tensor is diagonal. Note that, in general, the metric tensor can have off diagonal terms.

    Inline latex is obtained by using [itex] brackets.
     
  4. Oct 7, 2008 #3

    tiny-tim

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    Hi student85! :smile:

    (type "itex" instead of "tex", and it gives you in-line tex, which fits better. :wink:)

    Yes … the coefficients of dsomething2 go into the diagoanl positions in the matrix.

    And if the line element contained a mixed term, for example, dxdt, then half of the coefficent would go into each of the two corresponding off-diagonal positions.
     
  5. Oct 7, 2008 #4
    Thanks guys!
    I love this forum!
     
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