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Confusion over Einstien summation convention and metric tensors.

  1. Oct 22, 2011 #1
    My understanding of the Einstein Summation convention is that you sum over the repeated indices. But when I look at the metric tensor for a flat space I know that

    g[itex]^{λ}_{λ}[/itex] = 1

    But the summation convention makes me think that it should equal the trace of the matrix g_{μσ}. So it should be the number of dimensions of the space?
     
  2. jcsd
  3. Oct 22, 2011 #2

    Pengwuino

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    [itex]g_{\alpha}^{\; \alpha} = N[/itex] where N is the dimension of the space. Why do you think it's 1?
     
  4. Oct 22, 2011 #3
    A[itex]^{\nu}[/itex] = g[itex]^{μ \nu}[/itex]A[itex]_{μ}[/itex]

    A[itex]_{μ}[/itex] = g[itex]_{μ\nu}[/itex]A[itex]^{\nu}[/itex]

    A[itex]_{μ}[/itex] = g[itex]_{μ\nu}[/itex]g[itex]^{\nu\rho}[/itex]A[itex]_{\rho}[/itex]

    g[itex]_{μ\nu}[/itex]g[itex]^{\nu\rho}[/itex] = g[itex]^{\rho}_{\mu}[/itex]

    g[itex]^{\rho}_{\mu}[/itex] = 1 for μ = ρ

    g[itex]^{\rho}_{\mu}[/itex] = 0 for μ [itex]\neq[/itex] ρ

    taken from Dirac's book
     
  5. Oct 22, 2011 #4

    Pengwuino

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    Yes but [itex]g^{\rho}_{\mu}[/itex] is not a summation. It simply indicates what the components are. So it's saying when [itex]\rho = \mu[/itex], the component is 1, else it is 0. However, [itex]g^{\rho}_{\rho} = \sum^{N}_{\alpha = 1} g_{\alpha}^{\alpha}[/itex].

    It's like how sometimes people get confused with the delta function [itex]\delta_{ij}[/itex]. In a 3-dimensional space, you can have [itex]\delta_{ij} = 1[/itex] if [itex]i=j[/itex], else it is 0. However, this is different from [itex]\delta_{ii}[/itex] which is the actual summation which is equal to 3.
     
  6. Oct 22, 2011 #5
    ok thanks
     
  7. Oct 22, 2011 #6
    I asked the question because I'm getting g[itex]^{λ}_{λ}[/itex] after some contractions of a certain curvature tensor. For Example I have a term g[itex]^{μ}_{σ}[/itex] and I'm contracting the tensor by setting μ and σ to λ. The term will now equal the dimension D?
     
  8. Oct 22, 2011 #7

    Pengwuino

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    Yes it will equal D if you're contracting the metric in the way you asked. This is true only for the metric, however.
     
    Last edited: Oct 22, 2011
  9. Oct 22, 2011 #8

    WannabeNewton

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    Not in general. If [itex]g_{\mu \nu } = \delta _{\mu \nu }[/itex] on the n - manifold in question then [itex]g^{\lambda }_{\lambda } = \delta ^{\lambda }_{\lambda } = n[/itex]. But this doesn't have to be true for arbitrary metrics on their respective manifolds.
     
  10. Oct 22, 2011 #9

    Matterwave

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    No, in general (at least in GR, I don't know about some really weird cases or w/e),

    [tex]g^\mu_\nu=\delta^\mu_\nu[/tex]

    This is because the contravariant metric must be the inverse of the covariant metric. The two must be inverses of each other or else you won't get the same vector back if you lower it's indices and then raise them.

    The trace of the Kronecker delta is obviously the dimension, and therefore so is the trace of the (1,1) form of the metric.
     
  11. Oct 22, 2011 #10

    Pengwuino

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    No, in general it is true if [itex]g_{ab}[/itex] and [itex]g^{ab}[/itex] are inverses of each other. Remember, [itex]g_{a}^{a} = \sum_{i,j=1}^N g_{ij}g^{ij} =N[/itex]
     
  12. Oct 22, 2011 #11

    WannabeNewton

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    Well it still isn't true in general because what if g isn't diagonal. But yeah I get your point if g is diagonal.
     
  13. Oct 22, 2011 #12
    ORLY? Where did you see that?
     
  14. Oct 22, 2011 #13

    Pengwuino

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    This is in general, the metric does not need to be diagonal since [itex]g_{ab}[/itex] and [itex]g^{ab}[/itex] by definition are inverses of each other. If [itex]g^{ab}[/itex] is the inverse of [itex]g_{ab}[/itex], then by definition, [itex]g_a^b = \sum_{i=1}^n g_{ai}g^{ib} = \delta_{ab}[/itex]. Remember, this [itex]g_a^b[/itex] component is a n-term summation and the inverse metric must be defined so this is true.
     
  15. Oct 22, 2011 #14
    But what if it's the fundamental metric tensor? Where g[itex]_{00}[/itex] is -1 and the rest are 1? Would I get D-2 ?
     
  16. Oct 22, 2011 #15

    Matterwave

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    Here:

    [tex]A_\mu=g_{\mu\nu}A^\nu[/tex]

    [tex]A^\mu=g^{\mu\nu}A_\nu[/tex]

    Which implies:

    [tex]A^\mu=g^{\mu\nu}g_{\nu\tau}A^\tau[/tex]

    If it is true that [tex]A^\mu=\delta^\mu_\tau A^\tau[/tex]

    Then it must be that:

    [tex]g^{\mu\nu}g_{\nu\tau}=g^\mu_\tau=\delta^\mu_\tau[/tex]
     
  17. Oct 22, 2011 #16

    Matterwave

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    No, you can only trace the (1,1) form of the metric. You can't trace the purely covariant metric. the (1,1) form of the metric, almost by definition, MUST be diag(1,1,1,1...). See my proof above.
     
  18. Oct 22, 2011 #17

    WannabeNewton

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    So you're telling me that for the metric [itex]g_{\mu \nu } = \begin{pmatrix}
    1 + 4c^{2}p^{2} & 2cp\\
    2cp & 1
    \end{pmatrix}[/itex] that [itex]g_{\mu \nu }g^{\mu \nu } = 2[/itex] (c = const.).
     
  19. Oct 22, 2011 #18

    Pengwuino

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    Yes. And this is because the inverse metric must be defined such that it is true. The inverse for that metric won't just look like 1 over the components of the metric because that does not satisfy [itex]g_{am}g^{mb} = \delta_{ab}[/itex]
     
  20. Oct 22, 2011 #19

    Matterwave

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    Yes, WannabeNewton, see my previous proof...why does this bother you?
     
  21. Oct 22, 2011 #20

    WannabeNewton

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    yeah my bad forgot to include [itex]2g_{12}g^{12}[/itex] lol
     
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