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I Equivalent Klein-Gordon Lagrangians and equations of motion

  1. May 8, 2017 #1
    Suppose one starts with the standard Klein-Gordon (KG) Lagrangian for a free scalar field: $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}$$ Integrating by parts one can obtain an equivalent (i.e. gives the same equations of motion) Lagrangian $$\mathcal{L}=-\frac{1}{2}\phi\left(\Box +m^{2}\right)\phi$$ where ##\Box:=\partial_{\mu}\partial^{\mu}##.

    My question is, what is the correct form for the Euler-Lagrange equations in this second case? Naively, I find that $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=-m^{2}\phi-\frac{1}{2}\Box\phi +0=-m^{2}\phi-\frac{1}{2}\Box\phi=0$$ which clearly isn't correct. Should the Euler-Lagrange equation be modified to $$\frac{\partial\mathcal{L}}{\partial\phi}-\Box\left(\frac{\partial\mathcal{L}}{\partial(\Box\phi)}\right)=0\;?$$ such that $$\frac{\partial\mathcal{L}}{\partial\phi}-\Box\left(\frac{\partial\mathcal{L}}{\partial(\Box\phi)}\right)=-m^{2}\phi-\frac{1}{2}\Box\phi
    -\frac{1}{2}\Box\phi=-(\Box+m^{2})\phi =0$$
    Apologies for a potentially silly question, but I'm a bit stuck on this issue.
     
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  3. May 8, 2017 #2

    Orodruin

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    The EL equation you quote is only valid when the Lagrangian only contains first order derivatives. For the case when the Lagrangian contains higher order derivatives, you need to do additional partial integrations when deriving the corresponding equations of motion. I suggest going through the entire derivation of the EL equations in this case to see the difference.
     
  4. May 8, 2017 #3
    Ah ok. So even if they are equivalent Lagrangians, their Euler-Lagrange equations will be different because the former contains up to first-order derivatives whereas the latter contains up to second-order derivatives.

    I've had a go at deriving the Euler-Lagrange equations for a Lagrangian containing up to second-order derivatives and the form I get is: $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)+\partial_{\mu}\partial_{\nu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\right)=0$$ assuming that ##\delta\phi=0=\delta(\partial_{\mu}\phi)## on the boundary of the spacetime volume.
    Would this be correct?
     
  5. May 8, 2017 #4

    Orodruin

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    The resulting EL equations will be exactly the same if you include the additional terms that you need to include when there are second order derivatives in the Lagrangian. What terms in the EL equations that are non-zero will depend on which form of the Lagrangian you use - but the end result will be the same.

    Yes. Note that this will give you exactly the same equation of motion in the end (the KG equation).
     
  6. May 8, 2017 #5
    Ok cool, thanks.

    How does one compute the second-order variation? Naively, if I expand the action around the stationary solution, ##\varphi## to second order, I have: $$S[\phi]=S[\varphi]+\int\,d^{4}x\frac{\delta S[\phi]}{\delta\phi(x)}\delta\phi(x)
    +\int\,d^{4}xd^{4}y\frac{\delta^{2} S[\phi]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)+\cdots$$ The first-order term is just ##\frac{\delta S[\phi]}{\delta\phi(x)}=\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=(\Box +m^{2})\varphi=0##, however, what is the form of the second-order term? I assume that the second-order variation of the Lagrangian is $$\delta^{2}\mathcal{L}=\frac{\delta^{2}\mathcal{L}}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)+\frac{\delta^{2}\mathcal{L}}{\delta\phi(x)
    \delta(\partial_{\mu}\phi(y))
    }\delta\phi(x)\delta(\partial_{\mu}\phi(y))
    +\frac{\delta^{2}\mathcal{L}}{
    \delta(\partial_{\mu}\phi(x))
    \delta(\partial_{\mu}\phi(y))
    }
    \delta(\partial_{\mu}\phi(x))
    \delta(\partial_{\mu}\phi(y))$$ but I'm not sure.
     
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