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Why the metric is covariantly constant?

  1. Nov 20, 2007 #1
    [SOLVED] Why the metric is covariantly constant?

    you can show that the metric is covariantly constant by writing:
    V_a;b=g_acV^c;b

    for linearity V_a;b=(g_acV^c);b=g_ac;bV^c+g_acV^c;b

    than must be g_ac;b=0

    is there an alternative argument that show that is true?

    if I calculate g_ac;b considering that g_ac is a (0,2) tensor than I will write the 2 connections in form of the metric, but I have obtained this form using the fact that g_ac;b=0 so it seems to me like I'm just turning around

    any help?
     
  2. jcsd
  3. Nov 20, 2007 #2

    Ben Niehoff

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    As far as I can tell, the explanation given in Misner, Thorne, and Wheeler is circular, yes. There are other, external ways to arrive at the conclusion that the metric is covariantly constant. It is also possible to obtain the formula for the Christoffel symbols without assuming [itex]\nabla_{\gamma}g_{\alpha \beta} = 0[/itex], but it involves some tricky algebra and index-juggling. I'm not sure why MTW didn't bother to include it.

    You can try working it out yourself...I was able to after a little thought, but it did stump me for a while.
     
    Last edited: Nov 20, 2007
  4. Nov 21, 2007 #3
    I just need an argument (even shorter than what I wrote) that shows that the metric is truly covariantly constant.
     
  5. Nov 21, 2007 #4

    Ben Niehoff

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    Hmm...I know that the reason we want the metric to be covariantly constant is so that the operation of raising/lowering indices commutes with the covariant derivative...otherwise, it would make all of our calculations at least twice as difficult. However, I fail to see why this must be so, except in the roundabout method of proving the formula for [itex]\Gamma^{\alpha}_{\mu \nu}[/itex] first.

    Maybe someone more knowledgeable can help. :|
     
  6. Nov 21, 2007 #5

    robphy

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  7. Nov 21, 2007 #6
    can't be something related to the properties that the metric must have to represent univocally a particular frame of reference?
     
  8. Nov 22, 2007 #7
    my teacher gave me a little hint: it's something related to tensor equations :uhh::rolleyes:
     
  9. Nov 22, 2007 #8

    nrqed

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    That does not tell much!
    have you looked at the thread referenced by robphy? In general, one could have a torsion field which would increase the fields describing the theory. Assuming a torsion free theory is, as far as I know, an assumption based on simplicity and economy, not a requirement from a mathematical point of view. The fact that there is no torsion in "real life" is something determined by experiment.

    EDIT: You may want to post in the General relativity subforum to get more replies.
     
    Last edited: Nov 22, 2007
  10. Nov 22, 2007 #9
    yes, I have but I haven't studied yet what a torsion field is
     
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