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Contraction of metrices

  1. Aug 15, 2013 #1
    I'm starting to doubt about something:

    In 4 dimension, what value has δ[itex]^{λ}_{μ}[/itex] in following equation:
    ημ[itex]\nu[/itex]η[itex]\nu[/itex]λ = δ[itex]^{λ}_{μ}[/itex]

    is it 4 or 1?

    and IF it's 1, what is the difference that this equation:


    = 4?
  2. jcsd
  3. Aug 15, 2013 #2


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    ##\delta_\mu{}^\lambda = 1## if ##\mu = \lambda## and 0 otherwise.

    Only repeated indices are contracted, perhaps writing out the summations explicitly makes it clearer:

    $$\eta_{\mu\nu} \eta^{\nu\lambda} \equiv \sum_{\nu = 1}^4 \eta_{\mu\nu} \eta^{\nu\lambda}$$
    which is equal to 1 if ##\mu = \lambda## and 0 otherwise. In matrix notation, it basically says ##\eta^{-1} \eta = \mathbb{I}##.

    If you also contract the two free indices, you get
    $$\eta_{\mu\nu} \eta^{\nu\mu} \equiv \sum_{\mu = 1}^4 \sum_{\nu = 1}^4 \eta_{\mu\nu} \eta^{\nu\mu}$$
    Replacing the inner sum by the delta, this is
    $$\cdots = \sum_{\mu = 1}^4 \delta_\mu{}^\mu$$
    and since the upper- and lower index are the same, the delta evaluates to 1 for every value of mu:
    $$\cdots = \sum_{\mu = 1}^4 1 = 4$$

    As $$\eta_{\mu\nu} \eta^{\nu\lambda}$$ is the ##(\mu, \lambda)## component of the matrix ##\eta^{-1} \eta##, in matrix notation the double sum this basically says
    $$\sum_{i = 1}^4 (\eta^{-1} \eta)_{ii} = 4$$
    which is indeed the trace of the 4x4 identity matrix.
  4. Aug 15, 2013 #3
    Wow thanks, very clear explenation.
    Thank you very much! :)
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