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Einsten notation, 4-gradients

  1. Jan 6, 2012 #1
    I'm trying to get into quantum field theory, I've seen Einstein notation before and I think I'm grasping it fairly well, but I've ran into two things that I just could not figure out and they both involved derivatives, or gradients.
    My problem is just mathematical, doesn't have to do with the physics at all.

    Can anyone please help me understand the following from the book An introduction to quantum field theory by Peskin and Schroeder. I have another issue, but I don't want this post to get too long and I'm hoping that if I understand this I will understand my other problem as well:

    ==================================
    The book gives the following:
    [tex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}[/tex]
    Now, the thing I don't understand is that the book says if we use this lagrangian with the Euler-Lagrange equation:
    [tex]\partial_{\mu}\left( \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)-\frac{\partial\mathcal{L}}{\partial\phi}=0[/tex]
    and they say that plugging that lagrangian into the above will give:
    [tex](\partial^{\mu}\partial_{\mu}+m^{2})\phi=0[/tex]

    Now, when I try to work this out I get this instead:
    [tex](\partial_{\mu}\partial_{\mu}+m^{2})\phi=0[/tex]
    which is obviously wrong I think, but I don't understand where the raised subscript comes from in their answer.
     
  2. jcsd
  3. Mar 12, 2013 #2
    It is better if you write the lagrangian in the form

    [itex] \mathcal{L} = \frac{1}{2} \eta^{\mu\mu} (\partial_\mu \phi) (\partial_\mu \phi) - \frac{1}{2} m^2 \phi^2 [/itex]

    where [itex]\eta^{\mu\mu}[/itex] is the metric for flat space

    The metric raises the index of the first [itex] \partial_\mu [/itex] of the Euler-Lagrange equation as [itex] \partial^\mu = \eta^{\mu\mu} \partial_\mu [/itex] and you get the final equation correctly
     
  4. Mar 12, 2013 #3

    HallsofIvy

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    Very roughly, the derivative has an index, [itex]\mu[/itex], as a subscript in the denominator. Without the derivative notation, that becomes a superscript just as a negative power in the denominator of a fraction becomes a positive power in the numerator.
     
  5. Mar 12, 2013 #4

    WannabeNewton

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    vik's notation is incorrect because it does not obey the Einstein summation convention and I suspect he just made some typos but his idea is the correct one. We can rewrite the klein gordon lagrangian as [itex]\mathcal{L} = -\frac{1}{2}\eta ^{\mu \nu }(\partial _{\mu }\phi)( \partial _{\nu }\phi )-\frac{1}{2}m^{2}\phi ^{2}[/itex]. Then, [itex]\frac{\partial \mathcal{L} }{\partial \phi } = -m^{2}\phi , \frac{\partial \mathcal{L} }{ \partial (\partial _{\mu }\phi)} = -\eta ^{\mu \nu }\partial _{\nu }\phi [/itex] so the equations of motion are just [itex]\partial ^{\mu }\partial _{\mu }\phi - m^{2}\phi = 0[/itex] as usual.
     
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