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I would appreciate if someone tell me how I should go about finding eom. for the following Lagrangian:$$L=-\frac{1}{2}\phi(\Box + m^2)\phi$$

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  • $\begingroup$ That usually goes by the name of the Klein-Gordon Lagrangian --- it's the simplest relativistic field theory that exists. $\endgroup$ – gj255 Sep 17 '16 at 22:07
  • $\begingroup$ this lagrangian has a second order derivative $\Box$. as far as I know klein-Gordon lagrangian is $L=\frac{1}{2}(\partial_{\mu}\phi \partial^{\mu}\phi -m^2 \phi)$ $\endgroup$ – MSB Sep 17 '16 at 22:11
  • $\begingroup$ If we consider the integral of this Lagrangian over all space, we can integrate by parts to turn the KG Lagrangian that you're familiar with into the Lagrangian in your question (up to an overall sign). In other words, they describe the same theory (up to surface terms). $\endgroup$ – gj255 Sep 17 '16 at 22:13
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    $\begingroup$ By integrating by parts, you're merely showing that whichever Lagrangian you use, the action is precisely the same. This means they define the same theory. If I had one theory with action $S$, and another theory with action $(2S+2)/2 - 1$, they might look different at first glance, but by 'performing the division and subtraction', you would see that they are in fact the same. That's all we're doing when we perform the integration by parts. $\endgroup$ – gj255 Sep 17 '16 at 22:24
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    $\begingroup$ As I show below this gives the Klein-Gordon equation for the Euler-Lagrange equation extended to second order terms in the Lagrangian. This produces nothing interesting in particular. If you had instead a lagrangian of the form ${\cal L}~=~\frac{1}{2}\sqrt{\partial_\mu\phi\partial^\mu\phi}\square\phi~+~\frac{1}{2}\phi m^2\phi$, or some other deviation from standard Lagrangians, you would get something different $\endgroup$ – Lawrence B. Crowell Sep 17 '16 at 22:33
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The Euler-Lagrangian (E-L) equation is $$ \frac{\partial}{\partial x_\mu}\frac{\partial{\cal L}}{\partial(\partial^\mu\phi)}~-~\frac{\partial{\cal L}}{\partial\phi}~=~0. $$ For the standard Klein-Gordon field the Lagrangian, ignoring complex valued nature of the field, is $$ {\cal L}~=~\frac{1}{2}\partial_\mu\phi\partial^\mu\phi~+~\frac{1}{2}m^2\phi^2 $$ the equation of motion is the Klein-Gordon equation $\square\phi~-~m^2\phi~=~0$.

The Lagrangian proposed here is a bit different and this requires a modified E-L equation. The Lagrangian is dependent on the second derivatives of the field or ${\cal L}~=~{\cal L}(\phi,~\partial_\mu\phi,~\partial_\mu\partial_\nu\phi)$. I am not going to reproduce the derivation, but it is a second order generalization of the derivation of the E-L equation from the action. This E-L equation is then $$ \frac{\partial^2}{\partial x_\mu\partial x_\nu}\frac{\partial^2{\cal L}}{\partial(\partial^\mu\phi)\partial(\partial^\nu\phi)}~+~\frac{\partial}{\partial x_\mu}\frac{\partial{\cal L}}{\partial(\partial^\mu\phi)}~-~\frac{\partial{\cal L}}{\partial\phi}~=~0. $$. The calculation is then comparatively simple, where the $\frac{\partial^2{\cal L}}{\partial(\partial^\mu\phi)\partial(\partial^\nu\phi)}$ will eliminate the $\square\phi$ and the second order derivative will then reproduce $\square\phi$. This then returns the Klein-Gordon equation. There is no first order differential of the field in the Lagrangian, so the $\frac{\partial}{\partial x_\mu}\frac{\partial{\cal L}}{\partial(\partial^\mu\phi)}$ returns nothing.

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    $\begingroup$ Remark: it is often useful to just start from the stone age and derive EOMs directly from variational principle. Sometimes this is much faster than writing down the Euler-Lagrange equation. $\endgroup$ – Blazej Sep 17 '16 at 23:41

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