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Confusion with Einstein tensor notation

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm confused about writing down the equation: [itex] \Lambda \eta \Lambda^{-1} = \eta [/itex] in the Einstein convention.

    2. Relevant equations

    The answer is: [itex] \eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho\sigma}[/itex]

    However it's strange because there seems to be no distinction between [itex]\Lambda[/itex] and [itex]\Lambda^{-1}[/itex] if we write it this way.
    However we know that:

    [itex](\Lambda^{-1})^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu} [/itex]

    3. The attempt at a solution

    If the equation was instead [itex] \Lambda B \Lambda^{-1} = B [/itex]

    Where [itex] B [/itex] is a tensor given in the form [itex] B^{\mu}{}_{\nu}[/itex] then it's clear to me how to write it:

    [itex] \Lambda^{\rho}{}_{\mu} B^{\mu}{}_{\nu} \Lambda_{\sigma}{}^{\nu} = B^{\rho}{}_{\sigma}[/itex]

    But [itex] \eta [/itex] is given in the form [itex] \eta^{\mu\nu} [/itex] and I don't understand how I can contract it with both [itex] \Lambda^{\mu}{}_{\nu} [/itex] and [itex] \Lambda_{\nu}{}^{\mu} [/itex] in order to arrive eventually at the result quoted in (2).
     
  2. jcsd
  3. May 28, 2013 #2
    Is there an actual question? :tongue:

    So, your confusion is how (2) works?
     
  4. May 28, 2013 #3
    Haha sorry :tongue:

    I would like to know why (2) works, and possibly how I could arrive at it, starting from an expression that has both [itex]\Lambda^{\mu}{}_{\nu}[/itex] and [itex]\Lambda_{\nu}{}^{\mu}[/itex].
     
  5. May 28, 2013 #4

    Dick

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    Well, just raise the ##\mu## index and lower the ##\rho## index on the first ##\Lambda## in your form with the B tensor using the metric tensor.
     
    Last edited: May 28, 2013
  6. May 29, 2013 #5
    Thanks,

    Like that: ?

    [itex] \Lambda_{\rho}{}^{\mu} \eta_{\mu}{}_{\nu} \Lambda_{\sigma}{}^{\nu} = \eta_{\rho}{}_{\sigma} [/itex]

    But then again both [itex]\Lambda[/itex]'s are of the same form - this time they both seem to be inverses.
     
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