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Naive tensor question?

  1. Aug 20, 2008 #1
    I'm doing a reading of Carroll's General Relativity text, and I am a bit confused about one bit of tensors. Particularly the trace of a tensor.

    For a (1, 1) tensor we just sum up the diagonal elements, but for other tensors this does not work. For the flat Euclidean metric for Minkowski space the trace is 4, not 2.

    So how exactly do you calculate the trace of an arbitrary tensor? I.e., let's say we wanted to take the trace of a (2, 0) tensor [tex]X^{\mu \nu}[/tex]?

    Carroll mentions that if we lower or raise an index, we are taking the trace of a different tensor. In the case of a (2, 0) we could lower an index and get a (1, 1) tensor, but that trace is not the trace of a (2, 0) tensor.

    What exactly is the proper procedure?
  2. jcsd
  3. Aug 20, 2008 #2
    This makes no sense. Specifically, the "Euclidean metric for Minkowski space" is meaningless. That said, the trace of the Euclidean metric in four dimensions is, unsurprisingly, four.

    The trace of [itex]X^{\mu\nu}[/itex] is simply [itex]X = g_{\mu\nu}X^{\mu\nu}[/itex]. No mysteries.

    If [itex]T_{ab}[/itex] is a (0,2) tensor, we can raise an index with the metric to obtain, say, [itex]T_a^{\phantom{a}b} = g^{bc}T_{ac}[/itex], a (1,1) tensor. However, [itex]T_a^{\phantom{a}b}[/itex] is not a trace. By definition, the trace is a scalar quantity, i.e., a (0, 0) tensor. Hence to construct a (0, 0) tensor from [itex]T_{ab}[/itex] we need to contract indices: [itex]T = g^{ab}T_{ab} = T_a^{\phantom{a}a}[/itex]. Your difficulty seems to be that you don't appreciate the difference between raising/lowering indices and contracting indices.

    I haven't got the patience to read Carroll's book since it is, to be frank, appalling, but I do seem to recall that it discusses index contraction. I suggest you go and read it a bit more closely.
    Last edited: Aug 20, 2008
  4. Aug 21, 2008 #3

    George Jones

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    I wanted to wait until I could look at what Carroll had written before replying.

    Carroll defines trace as shoehorn did and then gives a warning (equivalent to): "Be careful! Trace so defined gives a result different than that obtained by introducing a basis and taking the trace of the matrix of values [itex]T_{\mu \nu}[/itex]. In particular, the trace of the metric is 4, not 2."
    Wow! This is by far the most negative comment I have seen/heard about Carroll's book.
  5. Aug 21, 2008 #4
    The trace should be 'defined' as [itex]X = X^{\mu}_{\mu}=g_{\mu\nu}X^{\mu\nu}[/itex]
    Something like [itex]\sum_{\mu}X^{\mu\mu}[/itex] is not invariant under 'rotation'(real rotation in R^3 or Lorentz transformation in the Minkowski space) and can't represent a physical quantity, although you can define it in a purely mathematical sense.

    All these hassles occur because the space-time isn't Euclidean. i.e. the 'length' is not [itex]\sum_{\mu}V^{\mu}V^{\mu}[/itex] but [itex]g_{\mu\nu}V^{\mu}V^{\nu}[/itex] .(V is a vector)
    Last edited: Aug 21, 2008
  6. Aug 21, 2008 #5
    I don't see why you can't take some arbitrary rank 2 tensors


    where the superscript k labels the tensors and is not an index, and then define the traces:

    [tex]\operatorname{Tr}_{k}X = \eta^{k}_{\alpha,\beta}X^{\alpha,\beta}[/tex]
  7. Aug 21, 2008 #6
    To Count Iblis

    It doesn't really matter mathematically, but can't be a physical quantity. It isn't invariant under 'rotation'.

    For example, in the special relativity, in which the Lorentz transformation is the 'coordinate rotation', we can define a (2,0) tensor such that
    [itex] A^{\mu\nu} \equiv P^{\mu}P^{\nu} [/itex]. (P: momentum)

    If you define the trace as [itex]\operatorname{Tr}A = g_{\mu\nu} A^{\mu\nu}[/itex] it corresponds to an physical quantity(mass). If you replace the [itex]g_{\mu\nu}[/itex] with an arbitrary tensor, it no more is invarant.
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