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Klein-Gordon Lagrangian

  1. Apr 8, 2013 #1
    Hi, I hope I put this in the right place!

    I'm having trouble with some of the calculus in moving from the Klein-Gordin Lagrangian density to the equations of motion. The density is:

    L = \frac{1}{2}\left[ (\partial_μ\phi)(\partial^\mu \phi) - m^2\phi ^2 \right]

    Now, to apply the Euler-Lagrange equations one needs to find:

    \frac{\partial L}{\partial(\partial_\mu\phi)}

    Which to me, looked like it could be:
    \frac{\partial L}{\partial(\partial_\mu\phi)} = \frac{1}{2}\partial^\mu\phi

    But that gives the wrong equation of motion - the half shouldn't be there. I guess my mistake is that this can sort of be thought of as being like:

    L = \frac{1}{2}\left[ (\partial_μ\phi)^2 - m^2\phi ^2 \right]

    Which works out fine, but I'm just not quite sure what's happening here. What is the best way to think of this? In particular, how does the positioning of μ change how to think of it. This seems really basic but I'm pretty lost! Can anyone help me understand how it works?

  2. jcsd
  3. Apr 8, 2013 #2

    Ben Niehoff

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    You should write the Lagrangian as

    [tex]\mathcal{L} = \frac12 \Big( \eta^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - m^2 \phi^2 \Big)[/tex]
  4. Apr 8, 2013 #3
    Wow that makes so much sense! So the upstairs index is just a short-hand that allows you to avoid writing the metric explicitly. Thank you!
  5. Apr 8, 2013 #4
    Actually, I'm still not completely on board. When I take the derivative, I use the product rule? And that's where the factor of 2 comes from?

    I think maybe the dummy indices are what are confusing me. If I think of it as:

    \mathcal{L} = \frac{1}{2}\left(\eta^{ab}\partial_a\phi\partial_b\phi - m^2\phi^2\right)

    Then I get it better.
  6. May 18, 2013 #5
    Exact same doubt here.
  7. May 18, 2013 #6
    That's exactly right. You can't use the same index for the summation as for the derivative; one is a dummy, the other is free. With the form you've written it, the product rule sorts it all out.
  8. May 18, 2013 #7


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    [tex]\frac{\partial \mathcal{L}}{\partial (\partial_{a}\varphi)} = \frac{1}{2}(\eta^{bc}\delta^{a}_{b}\partial_{c}\varphi + \eta^{bc}\delta^{a}_{c}\partial_{b}\varphi) = \partial^{a}\varphi[/tex] so [tex]\partial_{a}(\frac{\partial \mathcal{L}}{\partial (\partial_{a}\varphi)}) - \frac{\partial \mathcal{L}}{\partial \varphi} = \partial_{a}\partial^{a}\varphi + m^{2}\varphi = 0[/tex].
  9. May 22, 2013 #8
    Should I understand [itex]\partial_{\mu}\partial_{\nu}[/itex] like [itex](\partial_{\mu})^{2}[/itex]?

    Otherwise I'm kind of lost, feeling too stupid asking basic questions here... :cry:
  10. May 22, 2013 #9


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    As you seem to have problems with the index convention, why don't you write out the sum explicitly to see what is going on?
  11. May 22, 2013 #10


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    Don't feel stupid, the notation can be quite confusing at first (trust me it was insanely confusing for me when I first saw it). ##\partial_{\mu}\partial_{\nu}## isn't the same as ##\partial^{\mu}\partial_{\mu}##. Note that in the second expression, there is an implied summation over the ##\mu## index whereas in the first case there is no such implication. If we are given a Klein-Gordon field ##\varphi## propagating across a background Minkowski space-time, the effect of applying ##\partial^{\mu}\partial_{\mu}## can be seen by choosing a global inertial coordinate system ##(t,x,y,z)## for the background metric. We then have [tex]\partial^{\mu}\partial_{\mu}\varphi = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}\varphi = \eta^{tt}\partial^{2}_{t}\varphi + \eta^{xx}\partial^{2}_{x}\varphi + \eta^{yy}\partial^{2}_{y}\varphi + \eta^{zz}\partial^{2}_{z}\varphi = -\partial_{t}^{2}\varphi + \nabla^{2}\varphi[/tex]

    On the other hand, ##\partial_{\mu}\partial_{\nu}\varphi## just represents any possible second partial derivative of the Klein-Gordin field e.g. it could be ##\partial_{t}\partial_{x}\varphi## or ##\partial_{y}\partial_{z}\varphi## depending on what you choose for the indies ##\mu,\nu##.
  12. May 22, 2013 #11
    I've already done, just like Griffiths did in Introduction to Elementary Particles, unnumbered equation between 11.12 and 11.13, and in this fashion I felt convinced. Problem is that I really wanted to understand this way with the indexes , which I think is more practical and usual.. (sorry my lame english).

  13. May 22, 2013 #12
    Perhaps varying the action directly instead of using the E-L equations will make more sense to you:

    [tex]S = \frac{1}{2} \int \left [ \eta^{ab} \partial_a \phi \partial_b \phi-m^2 \phi^2 \right ] d^4 x[/tex]

    When varying the action, the metric is constant so:

    [tex]\delta S = \frac{1}{2} \int \left [ \eta^{ab} \left [\partial_a \phi ~ \delta (\partial_b \phi)+ \partial_b \phi~ \delta (\partial_a \phi) \right ]-2m^2 \phi \delta \phi \right ] d^4 x[/tex]

    The two terms involving variations of the gradient of the field are symmetric, i.e. [itex]\partial_a \phi ~ \delta (\partial_b \phi) = \partial_b \phi~ \delta (\partial_a \phi)[/itex]. Thus, they can be combined into a single term:

    [tex]\delta S = \int \left [ \eta^{ab} \partial_a \phi ~ \delta (\partial_b \phi)-m^2 \phi \delta \phi \right ] d^4 x[/tex]

    Next you can integrate the first term by parts (using the fact that partial derivatives and variations commute):

    [tex]\int \eta^{ab} \partial_a \phi ~ \delta (\partial_b \phi) d^4 x= \int \partial_b (\eta^{ab} \partial_a \phi~ \delta \phi ) d^4 x - \int \partial_b (\eta^{ab} \partial_a \phi) \delta \phi d^4x[/tex]

    The first integral is of a total derivative, which can be converted into a surface integral via Stokes' Theorem. Since the variation at the boundary is zero (by definition), the whole first integral is zero. The variation of the action becomes:

    [tex]\delta S = -\int \left [ \eta^{ab} \partial_b \partial_a \phi +m^2 \phi \right ] \delta \phi d^4 x=0[/tex]

    Since this holds for any [itex]\delta \phi[/itex], what you're left with is:

    [tex]\eta^{ab} \partial_b \partial_a \phi + m^2 \phi = 0[/tex]
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