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Index manipulations

  1. May 28, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove the following: ##(\partial_{\mu} \phi)^{2} = \dot{\phi}^{2} - (\nabla \phi)^{2}##.

    2. Relevant equations

    3. The attempt at a solution

    ##(\partial_{\mu} \phi)^{2}##
    ## = (\partial_{\mu} \phi)(\partial_{\mu} \phi)##
    ## = (\partial_{0} \phi)(\partial_{0} \phi) + (\partial_{1} \phi)(\partial_{1} \phi) + (\partial_{2} \phi)(\partial_{2} \phi) + (\partial_{3} \phi)(\partial_{3} \phi)##

    Am I on the right track?
     
  2. jcsd
  3. May 28, 2015 #2

    DEvens

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    Nearly there. But you seem to be missing a minus sign. And the definitions of ()^2 has a small issue.

    Look in your course material or text or whatever you are using. There should be something called a metric. For the purposes of this question it looks like the metric should be a diagonal square matrix, zeroes off axis, and (1, -1, -1, -1) on the diagonal.
     
  4. May 28, 2015 #3
    Is this correct?

    ##(\partial_{\mu} \phi)^{2}##
    ##= (\partial_{\mu}) (\phi \partial^{\mu} \phi)##
    ## = (\partial_{0} \phi) (\partial^{0} \phi) + (\partial_{1} \phi) (\partial^{1} \phi) + (\partial_{2} \phi) (\partial^{2} \phi) + (\partial_{3} \phi) (\partial^{3} \phi) ##
     
  5. May 28, 2015 #4

    DEvens

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    Not quite. ##\partial_{\mu} \phi ## is a vector. The index is "down." So to get the "square" of such a vector you need the inner product.

    ##g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi ##
     
  6. May 28, 2015 #5

    nrqed

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    Yes, this is absolutely correct (I disagree with Devens on that point) . But it is true that it may also be written as [itex] g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi [/itex] although here we are not doing general relativity so we would usually write it as [itex] \eta^{\mu \nu} \partial_\mu \phi \partial_\nu \phi [/itex]. But you don't need that notation if you know what [itex] \partial^0, \partial_0, \partial^i [/itex] and [itex] \partial_i [/itex] mean.
     
  7. May 29, 2015 #6

    DEvens

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    I see that I forgot to say what was actually wrong. My bad. It's this line. This is taking the derivative ## (\partial_{\mu}) ## of ## (\phi \partial^{\mu} \phi) ##. That is, this is not the same as the previous line. It should be ## (\partial_{\mu} \phi) ( \partial^{\mu} \phi)##.
     
  8. May 29, 2015 #7

    nrqed

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    Sorry, I had misunderstood your point. I thought you were referring to the last line, I had actually ignored the second line, assuming that it was a typo. I agree completely with you.


    Regards,


    Patrick
     
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