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Relativistic index notation del-operator

  1. Jan 22, 2013 #1
    Hi,

    I've been wondering about this forever and I finally decided to ask on the forums. In relativistic index notation (with c= [itex]\hbar[/itex] =1) with the minkowski metric g[itex]\mu[/itex][itex]\nu[/itex]=diag(1,-1,-1,-1), the 4-vector [itex]x^{\mu}=(t,x,y,z)=(x^0,\vec{x})[/itex], and with the del operator defined as [itex]\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}=(\partial_{t},\nabla)[/itex]. I should have that:
    [itex]\partial^{\mu} x_{\mu}=\partial_{\mu} x^{\mu}=1[/itex]
    but this is inconsistent with the way I usually think of vector calc because I should have
    [itex]\partial_{\mu} x^{\mu}=\partial_{t}t+\nabla\bullet\vec{x}[/itex]
    and
    [itex]\partial_{t}t=1[/itex]
    [itex]\nabla\bullet\vec{x}=3[/itex]
    so, with the way I normally think, I should have:
    [itex]\partial_{\mu} x^{\mu}=4[/itex]
    Where am I going wrong here?

    -Adam

    P.S. all of this notation is straight out of Peskin and Schroeder's QFT text
     
  2. jcsd
  3. Jan 22, 2013 #2
    You are not going wrong at all. In 4D flat spacetime, it should be 4.
     
  4. Jan 22, 2013 #3

    WannabeNewton

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    Hi arwright3! It's because there is an implied summation involved so [itex]\partial _{\mu }x^{\mu } = \partial _{0}x^{0} + \partial _{1}x^{1} + \partial _{2}x^{2} + \partial _{3}x^{3} = 4[/itex]. Don't forget when you have an index on the top and the same index on the bottom there is an implied summation! Cheers!
     
  5. Jan 23, 2013 #4
    Thanks for your help! One more quick question. Can someone derive this:
    [itex]\partial^{\mu} x^{\nu}=g^{\mu \nu}[/itex]?

    P.S. I'm trying to pick up this 4-vector stuff on my own because I need it for QFT and I never had an undergrad class in special relativity. If anyone has a suggestion for a good resource I'd appreciate it. I've read a few pdfs that I found with various google searches, and I understand how the Lorentz transformations work and everything, but what I'd really like is something that explains what is and is not allowed when working with four-vectors. There are many derivations in my QFT book that are ~2 lines long and I don't know how to work out the intermediate steps. Most of these things have nothing to do with QFT, even the classical E&M field equations/lagrangian are hard for me to understand and work with in their index notation form.
     
  6. Jan 23, 2013 #5

    WannabeNewton

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    [itex]\partial _{\mu }x^{\nu } = \delta _{\mu }^{\nu }[/itex] so [itex]g^{\sigma \mu }\partial _{\mu }x^{\nu } = \partial ^{\sigma }x^{\nu } = g^{\sigma \mu }\delta _{\mu }^{\nu } = g^{\sigma \nu }[/itex] and you can just relabel sigma to nu from there if you would like. I have a good book for you that gives you a ton of practice with 4 - vectors and Einstein summation and tensor calculus and all the other fun stuff which I will pm to you however I have never done QFT so I do not know the depth at which you need to know these things - I am basing it off of what I know from GR.
     
  7. Jan 23, 2013 #6
    Thanks, that's very nice of you. and BTW I like your profile picture!

    -Adam
     
  8. Jan 23, 2013 #7

    WannabeNewton

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    All hail the mighty zep =D
     
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