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FLRW metric tensor

  1. May 22, 2009 #1

    trv

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    1. The problem statement, all variables and given/known data

    Not really homework, but thought this might be the best place to get a quick answer.

    Question
    Calculate g^{ca}g{ab} for the FLRW metric.

    I would have thought this would be

    [itex]
    g^{ca}g{ab}=\delta^c_b=4
    [/itex]

    I thought 4 because I assumed there should be "1" for each non-zero element in the metric tensor g_{ab}.

    Apparently however it should be g^{ca}g_{ab}=1

    Can anyone please explain why?


    2. Relevant equations

    3. The attempt at a solution
     
  2. jcsd
  3. May 22, 2009 #2

    Cyosis

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    Homework Helper

    This is just the Kronecker delta which is 1 if b=c and zero if not, there is no summation going on. If it had been [itex]\delta_b^b[/itex] Einstein summation would have been implied, which would yield 4.

    Mathematically:

    [tex]
    A^\gamma=g^{\gamma \alpha}A_\alpha= g^{\gamma \alpha}g_{\alpha \nu}A^\nu=\delta_\nu^\gamma A^\nu=A^\gamma
    [/tex]

    If it were four, you would get [itex]A^\gamma=4A^\gamma[/itex] which would make little sense.
     
    Last edited: May 22, 2009
  4. May 22, 2009 #3

    trv

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    Thanks again Cyosis that clears things up.
     
  5. May 23, 2009 #4

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    Actually, another quick question.

    So the tensor, g^a_b for the FLRW metric, simply has components 1 along the diagonal?

    Or for that matter, to generalise it further. Given any metric tensor r_{ab} with all elements being non-zero, r^a_b has components 1 everywhere?

    To add to that, given,

    [itex]
    L=-0.5(\frac{d\phi}{dt})^2+V(\phi)
    [/itex]

    does g^a_bL = L?
     
    Last edited: May 23, 2009
  6. May 23, 2009 #5

    Cyosis

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    Correct.

    No, proof:

    [tex]
    r^\alpha_{\;\;\beta}=r^{\alpha \gamma}r_{\gamma \beta}=\delta^\alpha_\beta
    [/tex]

    In words, any mixed metric is equal to the Kronecker delta. You basically have a matrix with 1's on the diagonal and zeros elsewhere.
     
  7. May 23, 2009 #6

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    Oh yeh, true the kronecker delta would imply zeros everywhere apart from the diagonal. I think where I'm going wrong is assuming we multiply each element by its inverse. Giving 1 at all places where the metric tensor had non-zero components. Obviously that's not how it works.
     
  8. May 23, 2009 #7

    cristo

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    Think of these tensors as matrices, where g^{ab} is the inverse of g_{ab}. Just as multiplying a matrix with its inverse yields the identity matrix (AA^{-1}=I), multiplying the metric with its inverse yields the kronecker delta.
     
  9. May 23, 2009 #8

    trv

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    Thanks the two of you.
     
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