Derivation of Euler-Lagrange equation from principle of least action

When deriving the Euler-Lagrange equation for a field $\phi$ the term $$\int\textrm{d}x^{\mu}~\partial_{\mu}\left( \dfrac{\partial \mathcal{L} }{\partial(\partial_{\mu}\phi)}\right)\delta\phi$$ is assumed to vanish as it can be turned into a surface integral and $\delta\phi$ is assumed to vanish on the surface. What are the exact steps to turn this into a surface integral?

123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means that, in the formula

$$\delta S=\int_{\Omega}d^{d}x\ \bigg[\bigg(\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\bigg)\,\delta\phi\bigg]+\int_{\Omega}d^{d}x\ \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)$$ ($d=\dim \Omega$), $\Omega$ must be compact. Now, as for the last term, we get $$\int_{\Omega}d^{d}x\ \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)=\int_{\partial\Omega}d^{d-1}x_{\mu}\ \bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)$$ where I denoted by $d^{d-1}x_{\mu}$ the (oriented) volume element of the boundary $\partial\Omega$. The identity follows from Stoke's theorem, which (in one of its many forms) states that if you have a function $f$ defined on a compact set $\Omega$, then $$\int_{\Omega} d^{d}x\ \big(\partial_{\mu}\,f\big)=\int_{\partial\Omega} d^{d-1}x\ \big(f\ n_{\mu}\big)$$ where $\partial\Omega$ is the boundary of $\Omega$ and $n_{\mu}$ are the components of the vector field normal to $\partial \Omega$ (notice that $n_{\mu}\,d^{d-1}x$ and my definition of $d^{d-1}x_{\mu}$ are the same thing). The proof of the theorem can be easily found on standard textbooks or on the internet. Going back to our integral, as $\delta\phi$ is by definition (i.e. as part of the hypotheses of the theorem) zero on the boundary of $\Omega$, $$\int_{\Omega}d^{d}x\ \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)=\int_{\partial\Omega}d^{d-1}x_{\mu}\ \bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)=\int_{\partial\Omega}d^{d-1}x_{\mu}\ 0=0$$ hence, as $\delta S=0$, $$\delta S=\int_{\Omega}d^{d}x\ \bigg[\bigg(\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\bigg)\,\delta\phi\bigg]=0$$ and since this must hold for any compact $\Omega$ and any compactly supported $\delta\phi$, $$\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}=0$$ The compactness of $\Omega$ (and in turn the compactly supportedness of the variation of the fields) can indeed be removed from the hypotheses of the theorem, as long as the action integral is well-defined on $\Omega$. On the other hand, the fact that the variation must vanish on the boundary of the domain of integration cannot be removed from the hypotheses. Hence the former is still valid for non-compact $\Omega$'s.

When proving Noether's theorem (which is different from proving the equivalence between minimization and the Euler-Lagrange equations, and ultimately depends on this very proof), one allows for variations that do not vanish on the boundary of the domain of integration; moreover, in its usual formulation, Noether's theorem allows for the coordinates of the domain of integration to be varied. It is in this context that the Noether current arises as a divergenceless field, and the Noether current is defined as $$j^{\mu}=-T^{\mu}_{\nu}\ \delta x^{\nu}+\frac{\partial\mathcal{L}}{\partial\,\partial_{\mu}\phi}\ \delta\phi$$ where $$T^{\mu}_{\nu}=\frac{\partial\mathcal{L}}{\partial\,\partial_{\mu}\phi}\ \partial_{\nu}\phi-\mathcal{L}\ \delta^{\mu}_{\nu}$$ is the canonical energy-momentum tensor.

• While I really enjoy your answers, I think you slid right over the step that OP is asking about - they ask "What are the exact steps to turn this into a surface integral?", i.e. it appears they are not familiar with Stokes' theorem, which you just use without even mentioning its name. – ACuriousMind Apr 29 '16 at 12:45
• I had assumed it was a known result, given he's playing around with field theory. I'll slightly modify my answer, but it is not as one can give a proof of Stoke's theorem, given the answer and the setting. – Giorgio Comitini Apr 29 '16 at 12:52
• @ACuriousMind Done. More than this, only a proof of Stoke's theorem would do :-) – Giorgio Comitini Apr 29 '16 at 13:02
• @GiorgioComitini I always support compactly supported arguments, so for me....+1 :-) – magma Apr 29 '16 at 18:16
• @GiorgioComitini I also ask you: what kind of Tex editor (if any) did you use? – magma Apr 29 '16 at 18:18

The real answer is: it doesn't really. Or rather: we can still extract some physics out of it! Lets derive $$S[\phi] = \int d^d x \mathcal{L}[\phi,\partial\phi]\\ \delta S[\phi] = \int d^dx \delta \mathcal{L} = \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta (\partial_\mu\phi) \right)$$ we first use that $\delta$ and $\partial$ commute and then Leibniz rule to get $$= \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \partial_\mu(\delta \phi) \right)\\ = \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi \right) - \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi}\right) \delta \phi \right)\\ = \int d^dx \left( \underbrace{\left( \frac{\partial\mathcal{L}}{\partial \phi} - \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi}\right) \right)}_\text{EoM = 0} \delta \phi + \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi \right) \right) = 0\\$$ So we are left with $$\delta S [\phi] = \int_{\mathbb{R}^n} d^d x \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi \right) = 0 \\$$ There are 2 ways this can be $0$.

• First we see that by stokes we get the GLOBAL statement $$\delta S [\phi] = \int d \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi \right) = \left.\frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi \right|_{\partial \mathbb{R}^n} =0 \\$$ where we usually assume all fields vanish at $\partial \mathbb{R}^n$, i.e. at infinity.

• Another way to get $\delta S = 0$ (and in a sense a more physical and mathematically rigorous way) is to demand the LOCAL statement $\partial_\mu \left( \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi \right) = 0$. This is exactly the conserved current demand: $\partial_\mu j^\mu = 0$. So we find that $$j^\mu = \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta \phi$$ is the conserved current associated with variation $\delta\phi$.

• but... but you assumed the EoM, which is what we were supposed to prove! – AccidentalFourierTransform Apr 29 '16 at 11:33
• Ah, yes! The fact that we can put that part I labelled as "EoM" $= 0$ is because $\delta S =0$ must hold for ANY $\delta \phi$. So by arbitrary change of $\delta \phi$ we must still have $\delta S =0$. This can only be true if the "EoM" part $= 0$. – 123hoedjevan Apr 29 '16 at 11:38
• I suppose if you have to derive the EoM, using the GLOBAL statement to let the $\partial_\mu (...)$ part vanish before using my argument about $\delta S =0$ for any $\delta \phi$ is more rigorous. – 123hoedjevan Apr 29 '16 at 11:39
• The Action must be minimized wrt. compactly supported variations of the fields which vanish on the contour of the support. There is no need to go to infinity, hence your answer is misleading, and ultimately wrong. – Giorgio Comitini Apr 29 '16 at 12:09
• @123hoedjevan, Giorgio is correct. His answer is the rigorous statement and proof of the least action principle and EOMs. Nevertheless I do not think is useful to down vote your answer, but rather encourage you to change your arguments, so that your answer is correct too. I like your TEXing very much, what tex editor (if any) did you use? For me...+1 – magma Apr 29 '16 at 18:15