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Quick question: 4 gradient

  1. Sep 20, 2008 #1
    ok, quick and dirty and stupid question about calculation rules with 4 gradients:


    consider the Klein Gordon Lagrangian [tex]L_{KG} = \frac{1}{2} \partial_{\mu}\Phi\partial^{\mu} \Phi - \frac{1}{2} m^2 \Phi^2 [/tex].

    Why is

    [tex] \partial_{\mu} \left( \frac{\partial L_{KG} }{\partial(\partial_{\mu} \Phi)} \right) = \partial_{\mu}\partial^{\mu} \Phi[/tex]

    Where does the factor 2 come from that cancels out the 1/2 ?
     
  2. jcsd
  3. Sep 20, 2008 #2

    malawi_glenn

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    have you taken the lagrangian and 4gradient from same source?

    I have always written KG lagrangian (density) as: [tex]L_{KG} = (\partial_{\mu}\Phi) ^{\dagger}\partial^{\mu} \Phi - m^2 |\Phi |^2 [/tex]

    Then the 4gradient is the one you have written.
     
  4. Sep 20, 2008 #3
    same source.

    the factors 1/2 are there throughout, and it certainly makes sense for the mass term where a factor 2 comes from differentiating.

    But where does the factor 2 come from when differentiating by [tex] \partial_{\mu} \Phi[/tex] ?? Probably I miss out a very simple thing...
     
  5. Sep 20, 2008 #4

    cristo

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    I can't see where it comes from either, but then I often miss basic things.

    Is there some reason you feel the 2 should be there?
     
  6. Sep 20, 2008 #5
    well yes, since applying the Euler Lagrange equation on the KG Lagrangian should produce the KG equation:

    EL: [tex] \frac{\partial L}{\partial \Phi} - \partial_{\mu} \left( \frac{\partial L}{\partial(\partial_{\mu} \Phi} \right) = 0 [/tex]

    KG equation: [tex] (\square + m^2) \Phi(x, t) = 0 [/tex]
     
  7. Sep 20, 2008 #6

    haushofer

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    Maybe I'm overlooking something, but as far as I can see the factor 2 comes from the product rule. It gives you 2 delta functions.
     
  8. Sep 20, 2008 #7

    cristo

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    Ok, I see. Well, as I said above, I always miss obvious things: note that [itex]\partial^{\mu}\varphi[/itex] and [itex]\partial_{\mu}\varphi[/itex] are not independent, thus your derivative will include two terms. We can rewrite the Lagrangian as [tex]\mathcal{L}=\frac{1}{2}g^{\mu\nu}\partial_{\mu}\varphi\partial_{\nu}\varphi-\frac{1}{2}m^2\varphi^2[/tex]. Differentiating wrt [itex]\partial_{\mu}\varphi[/itex] then yields [tex]\frac{1}{2}\left[\partial_{\nu}\varphi g^{\mu\nu}+\delta_{\mu\nu}\partial_{\mu}\varphi g^{\mu\nu}\left]=\frac{1}{2}\left[2\partial^{\mu}\varphi\left][/tex], which yields the result.

    Does that make sense?

    Edit: Looks like I was beaten to it!
     
  9. Sep 21, 2008 #8
    Thank you so much!

    I never really liked the covariant picture, although it looks very elegant. It always leads to me missing out basic things.
    I really have to dig into it now...
     
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