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Derivation of Lorentz algebra commutation relation

  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data

    1. Show that the Lorentz algebra generator ##J^{\mu \nu} = i(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})## lead to the commutation relation ##[J^{\mu \nu}, J^{\rho \sigma}] = i(g^{\nu \rho}J^{\mu \sigma} - g^{\mu \rho}J^{\nu \sigma}-g^{\nu \sigma}J^{\mu \rho}+g^{\mu \sigma}J^{\nu \rho})##.

    2. Show that one particular representation of the the Lorentz group given by ##(J^{\mu \nu})_{\alpha \beta} = i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})##.

    2. Relevant equations

    3. The attempt at a solution

    ##[J^{\mu \nu}, J^{\rho \sigma}]##

    ##= J^{\mu \nu}J^{\rho \sigma}-J^{\rho \sigma}J^{\mu \nu}##

    ##= -(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})(x^{\rho}\partial^{\sigma}-x^{\sigma}\partial^{\rho})+(x^{\rho}\partial^{\sigma}-x^{\sigma}\partial^{\rho})(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})##.

    How do I get the metric tensor from here?
     
  2. jcsd
  3. Oct 30, 2015 #2

    fzero

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    When you distribute terms you will find various factors like ##\partial^\nu x^\rho##.
     
  4. Oct 30, 2015 #3
    How are those factors related to the metric tensor?

    As far as I know, a term like ##\partial_{\nu}x^{\rho}## gives the Kronecker delta ##\delta_{\nu}^{\rho}##, not the metric tensor.
     
  5. Oct 30, 2015 #4

    fzero

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    Now raise the index.
     
  6. Oct 30, 2015 #5
    So, you say ##\delta_{\nu}^{\rho}## equals ##g^{\nu \rho}##?
     
  7. Oct 30, 2015 #6
    Or, do I use ##\delta_{\nu}^{\rho} = g^{\nu \rho}g_{\nu \rho}##?
     
  8. Oct 30, 2015 #7

    fzero

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    Not exactly, ##\partial^\nu x^\rho = g^{\nu\mu} \partial_\mu x^\rho##. So use the result for ##\partial_\mu x^\rho## here.
     
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