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Understanding the metric tensor

  1. Nov 12, 2012 #1
    I've read that the metric tensor is defined as
    [tex]{{g}^{ab}}={{e}^{a}}\cdot {{e}^{b}} [/tex]

    so does that imply that?
    [tex]{{g}^{ab}}{{g}_{cd}}={{e}^{a}}{{e}^{b}}{{e}_{c}}{{e}_{d}}={{e}^{a}}{{e}_{c}}{{e}^{b}}{{e}_{d}}=g_{c}^{a}g_{d}^{b}[/tex]
     
  2. jcsd
  3. Nov 12, 2012 #2

    TSny

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    No, that can't be correct. For example, let ##a = 1, b = 2, c = 1, d = 2## and assume we're dealing with the standard Minkowski metric.

    Then ##g^{ab}g_{cd} = g^{12}g_{12} = 0\cdot 0 = 0##

    But, ##g^a_cg^b_d = g^1_1g^2_2 = 1\cdot 1 = 1##
     
  4. Nov 12, 2012 #3

    Dick

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    Nope. You lost the dot product completely when you went to the next expression. If your metric is diagonal is [itex]g^{01}g_{01}=g^0_0 g^1_1[/itex]?
     
  5. Nov 13, 2012 #4
    The easiest way to understand the metric tensor is to use dyadic notation:

    I = (ei[itex]\cdot[/itex]ej)eiej = gij eiej = (ei[itex]\cdot[/itex]ej)eiej = gij eiej

    Any vector or tensor dotted with the metric tensor returns that vector or tensor unchanged. Thus, the metric tensor can be regarded as the identity tensor.
     
  6. Nov 14, 2012 #5
    Makes sense. thanks for all the help
     
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