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Raising/lowering using the metric tensor

  1. Apr 21, 2009 #1

    trv

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    1. The problem statement, all variables and given/known data
    Given a N-dimensional manifold, let gab, be a metric tensor.
    Compute
    (i) gabgbc
    (ii)gabgab

    Also, just need a clarification on something similar.
    gcdTcd=gcdTdc=tr T?
    I'm pretty sure its yes. Probably even a stupid question but a clarification would be useful.

    2. Relevant equations


    3. The attempt at a solution

    (i)tr g ?
    (ii)Here I'm caught between tr g, since I'm thinking its still the tensor and its inverse, and the indices don't matter.

    The other option is gac.

    Which is correct?
     
  2. jcsd
  3. Apr 21, 2009 #2

    Hurkyl

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    Because g is symmetric,

    By symmetry of g: gcdTcd = gdcTcd
    By renaming summed indices twice: gdcTcd = gzpTpz = gcdTdc

    (I only used p and z to make things overly clear. You could do the renaming one step, and would normally do so)
     
  4. Apr 21, 2009 #3

    trv

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    thanks. Any thoughts on the other bits?

    i.e. gabgbc
    gabgab?
     
  5. Apr 21, 2009 #4

    Dick

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    g_ab*g^bc=delta_a^c, right? The g with upper indices is defined as the inverse of the one with lower indices.
     
  6. Apr 21, 2009 #5

    trv

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    Checked notes...yes it is.
     
  7. Apr 21, 2009 #6
    g_{ab}g^{ab} = g_{ab}g^{ba} = delta_a^a = tr delta = N.

    also, I don't see why g^{cd}T_{cd} would be tr T-- what does the metric have to do with the trace of a tensor?
     
  8. Apr 22, 2009 #7

    trv

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    Good point.

    It should simply be

    gcdTdc=Tdc

    right.

    I actually needed these for another problem. And, it was much easier if gcdTdc=tr T :p. Guess I'll have to have a look at it again.
     
  9. Apr 22, 2009 #8

    Dick

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    The metric has a lot to do with the trace of a tensor. You are right the second time. g^cd*T_dc is tr(T). You are forgetting the index balance again. In the first form the left side has no free indices and the right side has two. How can that be??
     
  10. Apr 22, 2009 #9
    Ah...ok, sorry for the wrong advice-- I think I see why I was wrong. You can't really get the trace of a (2,0) tensor directly so tr T is the trace of the corresponding (1,1) tensor, so the computation is like g^{cd} T_{dc} = g^{cd} g_{ac} T_d^a = delta_a^d T_d^a = T_d^d = tr T, right?
     
  11. Apr 22, 2009 #10

    Dick

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    Well, I would say g_cd*T^cd=tr(T) is pretty direct, but yes, what you say is also correct.
     
  12. Apr 23, 2009 #11

    trv

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    thanks for the clarification Dick. And interesting way of looking at what's happening eok.
     
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