Efficiently deriving Maxwell from Euler-Lagrange equations

Question

From the Lagrangian density for electromagnetism

$$\mathcal{L} = -\frac14 F_{\mu\nu} F^{\mu\nu}$$

and the Euler-Lagrange equations

$$\frac{\partial \mathcal{L}}{\partial A_\alpha} = \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\alpha)}$$

one can derive a Maxwell's equations in a few lines of algebra

$$-\partial_\mu F^{\mu\nu} = (g^{\mu\nu}\partial^2 - \partial^\mu \partial^\nu)A_\mu = 0$$

where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. However it looks like there should be an even simpler way to derive the above. Using integration by parts (inside the action $S = \int d^4x \mathcal{L}$), one can show that the following Lagrangian is equivalent

$$\mathcal{L} = A_\mu (g^{\mu\nu}\partial^2 - \partial^\mu \partial^\nu)A_\nu.$$

This is done in Peskin & Schroeder Eq. 9.51. Notice the similarity to Maxwell's equations above. Can one infer Maxwell's equations directly from the above somehow? Naively one would need to use the Euler-Lagrange equations involving second derivatives, and it again seems like the derivation would be at least a few lines rather than following immediately.

A similar pattern is seen in the Lagrangian for the Dirac equation, which makes it seem like this is not a coincidence.

Answer 1

Using the antisymmetry of $F^{\mu \nu}$ and partial integration one finds \begin{align} \delta S &= -\frac{1}{4}\, \delta \!\int \! d^4x \, F^{\mu \nu} F_{\mu \nu} =-\frac{1}{2}\int \! d^4x \, F^{\mu \nu} \delta F_{\mu \nu}=-\frac{1}{2}\int \! d^4x \, F^{\mu \nu} (\partial_\mu \delta A_\nu-\partial_\nu\delta A_\mu) \\[5pt] &= \int \! d^4 x \, (\partial_\mu F^{\mu \nu}) \, \delta A_\nu=0 \quad \Rightarrow \quad \partial_\mu F^{\mu \nu}=0. \end{align}

Edit: If you wish to start directly from $S= \int d^4x \, A_\mu(x) (g^{\mu \nu} \partial^2 - \partial^\mu \partial^\nu) A_\nu(x)$, consider the variation \begin{align} \delta S&= \int \! d^4x \, [\delta A_\mu(x)(g^{\mu \nu}\partial^2-\partial^\mu \partial^\nu) A_\nu(x)+A_\mu(x)(g^{\mu \nu}\partial^2 -\partial^\mu \partial^\nu)\delta A_\nu(x)]\\[5pt] &=2 \int \!d^4x \, \delta A_\mu(x) (g^{\mu \nu} \partial^2 -\partial^\mu \partial^\nu) A_\nu(x)=0 \\[5pt] &\Rightarrow \quad (g^{\mu \nu}\partial^2-\partial^\mu \partial^\nu)A_\nu(x)=0, \end{align} where partial integration was used in the step from the first to the second line.

Answer 2

Let $\xi:E\rightarrow M$ be a vector bundle, $\xi^+:E^+\rightarrow M$ its adjoint bundle, defined as $E^+=E^\ast\otimes\Lambda^m(M)$ (where $m=\dim M$), whose sections are interpreted as the set of all vector bundle morphisms $E\rightarrow \Lambda^m(M)$.

Let $D:\Gamma(E)\rightarrow\Gamma(E^+)$ be a linear differential operator. Due to its signature, its formal adjoint $D^+:\Gamma(E)\rightarrow\Gamma(E^+)$ has the same signature, so it makes sense to talk about $D$ being formally selfadjoint.

It is well-known that the linear differential equation $D(\phi)=0$ admits a Lagrangian description if and only if $D$ is formally selfadjoint. A corresponding Lagrangian is given by $$ L[\phi]=\langle D(\phi),\phi\rangle, $$ where $\langle -,-\rangle$ is the pairing of $E^+$ and $E$ (it is $\Lambda^m(M)$-valued, so this is a Lagrangian $m$-form).

It is fairly easy to prove that this is the case. Take the variation $$ \delta L[\phi]=\langle D(\phi),\delta\phi\rangle + \langle D(\delta\phi),\phi\rangle, $$ and in the second term we use $$ \langle D(\delta\phi),\phi\rangle=\langle\delta\phi, D^+(\phi)\rangle + dB(\phi,\delta\phi) = \langle \delta\phi, D(\phi)\rangle + dB(\phi,\delta\phi), $$ where at the first equality we used the definition of the formal adjoint ($B$ is an $m-1$-form valued bidifferential operator whose exact form is not relevant now), and at the second equality we used the fact that $D$ is self-adjoint. Since $E^{++}=E$, the pair bracket $\langle-,-\rangle$ can be regarded as symmetric, so we get $$ \delta L = 2\langle D(\phi),\delta\phi\rangle + d(\cdots), $$ so the Euler--Lagrange derivative is $2D(\phi)$, i.e. the EL equation is equivalent to $D(\phi)=0$.

Now in the case in the OP it is sufficient to recognize that the linear differential operator $g^{ij}\partial^2-\partial^i\partial^j$ is formally selfadjoint.

This also explains why essentially all "free field" equations have Lagrangians of this form.

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Edit: Let $D:\Gamma(E)\rightarrow\Gamma(F)$ be a linear differential operator between vector bundles $E,F\rightarrow M$. The formal adjoint of $D$ is the unique linear differential operator $D^+:\Gamma(E^+)\rightarrow\Gamma(F^+)$ such that the equation $$ \langle \psi,D(\phi)\rangle-\langle D^+(\psi),\phi\rangle = dB(\phi,\psi) $$holds, where $\phi\in\Gamma(E)$, $\psi\in\Gamma(F^+)$, and $B:\Gamma(E)\times\Gamma(F^+)\rightarrow\Omega^{m-1}(M)$ is any bivariate linear differential operator with values in $m-1$-forms. So the formal adjoint is defined such that $D$ and $D^+$ are adjoint with respect to the bracket $\langle -,-\rangle$ "up to an exact term".

The thing is that $B$ is not uniquely defined as $B^\prime = B + dC$ - where $C:\Gamma(E)\times\Gamma(F^+)\rightarrow\Omega^{m-2}$ is any bivariate differential operator - is also an equivalently good $B$.

So the answer can be understood simply as "$dB$ is some surface term".