[Math] Maxwell equations as Euler-Lagrange equation without electromagnetic potential

calculus-of-variationsmp.mathematical-physicsphysics

In (mathematical) physics many equations of motion can be interpreted as Euler-Lagrange (EL) equations. The Maxwell equation for electromagnetic (EM) field (say in vacuum and in absence of charges) seems to me quite unusual in comparison to examples known in classical mechanics. This is a system of first order PDE on 6 components of EM field. To get the Lagrangian density, one takes the first pair of the Maxwell equations and deduces from it existence of electromagnetic potential. Substituting the potential into the second pair of Maxwell equations, one gets second order equations for the potential. They can be presented as EL-equations for the potential.

I am wondering if there is a way to present the Maxwell equations as an EL-equation in terms of electromagnetic field only rather than potential.

I think I can prove that this is impossible if one requires in addition that the Lagrangian density is quadratic in fields and their first derivatives and invariant under the Poincare group.

ADDED 1: By the Maxwell equations I mean
$$\operatorname{div}\vec B=\operatorname{div} \vec E=0,\, \operatorname{rot}\vec E=-\frac{1}{c}\dot{\vec B},\, \operatorname{rot}\vec B=\frac{1}{c}\dot{\vec E}.$$
Thus I am looking for a Lagrangian depending on $\vec E,\vec B$ as independent fields and their derivatives such that the EL-equation is equivalent to all these equations.

ADDED 2: Let me reformulate my question on the language of some of the answers and comments below. The standard approach to interpret Maxwell equations as EL-equation in electrodynamics is as follows. One selects out of 8 Maxwell equations four equations and declares them to be constrains, and the other four – equations of motions. One considers the variations of the action functional $\int L$ only in the class of electromagnetic fields $(\vec B,\vec E)$ satisfying the constrains. Its extrema recover the four equations of motion. This separation of the Maxwell equations into two halves seems to me to be artifical and different from all other examples I know. I am wondering if there is a way to consider all 8 Maxwell equations on equal footing for this purpose.

Best Answer

Yes indeed, the Maxwell's equations are Euler-Lagrange equations. And this is quite interesting. Let me give here a presentation within Special Relativity, in which the light speed is set to $c=1$. The ambiant space is therefore a Minkowski space $\mathbb R^{1+3}$ with metric $dt^2-d{x_1}^2-d{x_2}^2-d{x_3}^2$. I restrict myself to the case of a vacuum.

The electromagnetic field is by definition a closed two-form $\Omega$. The components of the field can be retrieved, in a given coordinates frame, by $$\Omega=\vec E\cdot dt\times dx+\vec B\cdot(dx\times dx).$$ The constraint $d\Omega=0$ writes $$\partial_t\vec B+{\rm curl}_x\vec E=0,\qquad{\rm div}_x\vec B=0.$$

The rest of the Maxwell's equations are obtain by writing $$\delta\int\int L(\Omega)dxdt=0,$$ still under the constraint that the variations of $\Omega$ are compatible with the closedness. Writing $L$ as a function of $(\vec B,\vec E)$, we obtain $$\partial_t\vec D-{\rm curl}_x\vec H=0,\qquad{\rm div}_x\vec D=0,\qquad(\dagger)$$ where $$\vec D=\frac{\partial L}{\partial\vec E},\qquad\vec H=\frac{\partial L}{\partial\vec B}.$$ An important point is that $L$ must be invariant under Lorentz transformation. This translates the following way: there exists a function $\ell$ of two scalar variables only, such that $$L(\Omega)=\ell\left(\frac12(\vec E^2-\vec B^2),\vec E\cdot\vec B\right).$$ For instance, the choice $L=\frac12(\vec E^2-\vec B^2)$ yields the standard, linear Maxwell's equations (in which $D=E$ and $H=-B$).

The energy density is a partial Legendre transform, $$W=\vec D\cdot\vec E-L.$$ The Poynting vector is $\vec E\times\vec H$. It also equals $\vec D\times\vec B$. The fact that both formulas give the same quantity is equivalent to the Lorentz invariance.

Edit. Here are some of the details. The variational principle $\delta{\cal L}=0$, where ${\cal L}(\Omega)=\int\int L(\Omega)dxdt$, means that $$\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}\int\int L(\Omega+\epsilon\alpha)dxdt=0$$ for every closed $2$-form $\alpha$. Equivalently, we have $$\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}\int\int L(\Omega+\epsilon d\beta)dxdt=0$$ for every $1$-form (say smooth, compactly supported) $\beta$. Let us write $\beta=\phi dt-\vec A\cdot dx$, then $d\beta=(\partial_t\vec A+\nabla\phi)\cdot dt\times dx+{\rm curl}\vec A \cdot dx\times dx$. We therefore have $$\int\int(\vec H\cdot{\rm curl}\vec A+\vec D\cdot(\partial_t\vec A+\nabla\phi))dxdt=0$$ for every test function $\phi$ and field $\vec A$. This gives $\partial_t\vec D-{\rm curl}\vec H=0$ and ${\rm div}\vec H=0$.

Edit. In a recent paper, I explore a variant of the variational principle, in which the admissible variations run over the same class as $\Omega$, modulo pullback composition by a diffeomorphism: $\delta\Omega=\Omega-\phi^*\Omega$. This is narrower than the additive perturbation considered above. The resulting equations ar interesting. We don't obtain the full ($\dagger$), but we do obtain the conservation of energy $$\partial_tW+{\rm div}(\vec E\times\vec H)=0,$$ and that of momentum, which are perhaps more natural.