# Maxwell equations as Euler-Lagrange equation without electromagnetic potential

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.

• I suppose you are aware of Jackson's book on classical electrodynamics and the Lagrangian density relation (12.85), p.599, 3rd edition ? And you are not satisfied by this, due to the the second summand $\frac{1}{c}j_\alpha A^\alpha$ ? Dec 26, 2020 at 20:09
• x-posted on physics.SE: physics.stackexchange.com/q/603014/84967 Dec 26, 2020 at 21:53
• If you are going to think of the electromagnetic field $F_{\alpha \beta}$ as an arbitrary two-form (as opposed to the curvature of a connection one-form $A_\alpha$), without any other additional fields present, then any Euler-Lagrange equation can only achieve six out of the eight Maxwell equations at best. If however you constrain $F_{\alpha \beta}$ to be a curvature, then the Yang-Mills Lagrangian $\frac{1}{2} \int F^{\alpha \beta} F_{\alpha \beta}$ works just fine, providing the four of the eight Maxwell equations not coming from curvature forms being closed. Dec 27, 2020 at 4:15
• @KonstantinosKanakoglou: In this post I explicitly suppose that there are no charges, i.e. no term $j_{\alpha}A^{\alpha}$.
– makt
Dec 27, 2020 at 6:44
• @TerryTao: Your first sentence sounds to be in the right direction. However I do not understand the following example. If one considers the standard Lagrangian $\frac{1}{2}\int F^{\mu\nu}F_{\mu\nu}$ where $F_{\mu\nu}$ is any (not necessarily closed) 2-form, then the EL-equation is 6 equations $\vec E=\vec B=0$ (equivalently $F_{\mu\nu}=0$). It is much stronger than 8 Maxwell equations.
– makt
Dec 27, 2020 at 6:51

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.

• I think the main point of my question is that $L=\frac{1}{2}(E^2-B^2)$ does NOT give Maxwell equations. It does not contain derivatives of $\vec E,\vec B$ at all. Hence if one varies this $L$ one gets 0 order equation $\vec E=\vec B=0$. To get Maxwell equtions one has to assume that $\vec E,\vec B$ are not arbitrary, but are first derivatives of another field called potential.
– makt
Dec 26, 2020 at 11:36
• @makt. It is true that $L$ does not contain derivatives. But remark that if it did, then the E.-L. equations would be second order, while the Maxwell's E. are first-order only. The point is that the constraint $d\Omega=0$ does involve derivatives. Please do the calculations by yourself. Dec 26, 2020 at 12:14
• Indeed, at the level of the classical $\vec{E}$ and $\vec{B}$, there is no strict need for the potentials - one merely has to properly take into account the constraints on the dynamics, very nice. Only when one couples a quantum particle to the electromagnetic field, the Bohm-Aharonov effect appears - and suddenly the potentials appear to be more than just an artifice. Dec 26, 2020 at 15:27
• @DenisSerre: There is an abuse of terminilogy. For my $d\Omega=0$ is not a constrain but a part of Maxwell equations as it is common in physics. It encodes, in particular, the Faraday induction law. For me electromagnetic field is a pair $\vec E,\vec B)$ without any constrains. Thus I am looking for a Lagrangian suth that its EL equation would imply the equation $d\Omega=0$.
– makt
Dec 26, 2020 at 15:48
• @makt - div $\vec{E} =$ div $\vec{B} =0$ are manifestly not dynamical equations, but constraint equations - they do not contain time derivatives. As such, they can of course not be derived from the Euler-Lagrange equations of an action functional of the form $S=\int L dt$. Dec 26, 2020 at 16:06

There is a trivial sense in which the answer is "yes": the solutions to the Maxwell equations are (formally, at least) the global minimisers to the functional

$$\int\int |\mathrm{div} E|^2 + |\mathrm{div} B|^2 + |\mathrm{rot} E + \frac{1}{c} \dot{B} |^2 + |\mathrm{rot} B - \frac{1}{c} \dot{E} |^2\ dx dt.$$

I would suppose you consider this as cheating, but in order to rule this sort of degenerate answer out you would need to formulate your question with more mathematical precision.

• Is this example ruled out by requiring Lorentz invariance? Dec 27, 2020 at 19:09
• Hmm, good question. There are no positive definite Lorentz-invariant inner products on 4-vectors, so this construction wouldn't work directly, but it's hard to see how to rule out some more sophisticated example of this form (involving various derivatives of the Maxwell equations) that would be both Lorentz invariant and minimized by the solutions to Maxwell. Dec 27, 2020 at 19:28
• @TerryTao You're almost cheating in the spirit of Feynman's universal law of nature ;-) Take the sum of the squares of the absolute value of every equation in the universe and set it equal to zero. Dec 27, 2020 at 19:38
• @TerryTao: At the moment I care less about Lorentz invariance: after all, the Lagrangian $m\dot x^2/2$ of a free non-relativistic particle is not Galileo invariant, only up to total derivatives, although in relativity Lagrangians are usually Lorentz invariant as far as I know. The problem is that EL equation for this Lagrangian is of second order. I do not see immediately if it's equivalent to Maxwell equations. Probably not: how a system of linear second order equations could be equivalent to a system of linear first order equations?
– makt
Dec 27, 2020 at 19:53
• There are analytic reasons (in this case, elliptic rigidity) that allow a second-order system to be equivalent to a first-order system given suitable boundary conditions. The prototypical example of this is Liouville's theorem for harmonic functions, which shows that the second-order equation $\Delta u = 0$ is equivalent to the first order system $\nabla u = 0$ if $u$ is bounded. More generally, the Euler-Lagrange equations for $\int \sum_i |L_i u|^2$ is $\sum_i L_i^* L_i u = 0$, which is equivalent to the system $L_i u = 0$ if $u$ does not grow too fast at infinity. Dec 27, 2020 at 20:46

The nature of the Maxwell equations (including their Lagrangian aspects) can be clarified by viewing these equations as part of a larger family of classical field theories in physics. For simplicity I will focus on just four theories here:

1. Vacuum classical Yang-Mills theory;
2. Abelian vacuum classical Yang-Mills theory with gauge group $$U(1)$$, aka the vacuum Maxwell equations;
3. The vacuum Einstein equations;
4. The vacuum linearised Einstein equations.

However one could vastly generalise the discussion if desired, by adding a source matter term to any of these equations, or by coupling with other field theories to obtain such classical field theories as Maxwell-Klein-Gordon, Maxwell-Chern-Simons, Einstein-Maxwell, Yang-Mills-Higgs, Einstein-Vlasov, nonlinear sigma fields (wave maps), etc.. One can also similarly discuss quantum field theories but I will not attempt to do so here.

These equations all describe the dynamics of one or more underlying geometric objects, expressible in local coordinates by tensors (which I will call the primary fields):

1. For Yang-Mills the primary field is the connection one-form $$A_\alpha$$ (whose components take values in a Lie algebra $${\mathfrak g}$$).
2. For Maxwell the primary field is the vector potential $$A_\alpha$$ (whose components take values in $${\bf R}$$).
3. For Einstein the primary field is the metric $$g_{\alpha \beta}$$.
4. For linearised Einstein the primary field is the metric perturbation $$h_{\alpha \beta}$$.

For coupled systems one could have more than one primary field, but here we focus on simpler models in which there is just one primary field in each theory. From this primary field one can then create some secondary fields:

1. For Yang-Mills one can form the curvature tensor $$F_{\alpha \beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha + [A_\alpha, A_\beta]$$ (normalisation conventions can vary from author to author here).
2. For Maxwell one can form the electromagnetic field $$F_{\alpha \beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha$$.
3. For Einstein one can form the Christoffel symbols $$\Gamma_{\alpha \beta}^\gamma = \frac{1}{2} g^{\gamma \delta} (\partial_\beta g_{\gamma \alpha} + \partial_\alpha g_{\gamma \beta} - \partial_\gamma g_{\alpha \beta})$$ and the Riemann curvature tensor $$R_{\alpha \beta \gamma}^\delta = \partial_\beta \Gamma_{\alpha \gamma}^\delta - \partial_\alpha \Gamma_{\beta \gamma}^\delta + O(\Gamma^2)$$ (where I omit the precise nature of the quadratic term $$O(\Gamma^2)$$ for simplicity).
4. For linearised Einstein one can form the linearised Christoffel symbol $$\Gamma_{\alpha \beta}^\gamma = \frac{1}{2} \eta^{\gamma \delta} (\partial_\beta h_{\gamma \alpha} + \partial_\alpha h_{\gamma \beta} - \partial_\gamma h_{\alpha \beta})$$ and the linearised Riemann curvature tensor $$R_{\alpha \beta \gamma}^\delta = \partial_\beta \Gamma_{\alpha \gamma}^\delta - \partial_\alpha \Gamma_{\beta \gamma}^\delta$$.

These secondary fields are not unconstrained, but by their definition obey various Bianchi-type identities (as well as some symmetry and antisymmetry identities not mentioned here):

1. For Yang-Mills one has the Bianchi identity $$D_\alpha F_{\beta \gamma} + D_\beta F_{\gamma \alpha} + D_\gamma F_{\alpha \beta} = 0$$, where $$D_\alpha = \partial_\alpha + [A_\alpha,]$$ is the covariant derivative.
2. For Maxwell one has the Gauss-Faraday law $$\partial_\alpha F_{\beta \gamma} + \partial_\beta F_{\gamma \alpha} + \partial_\gamma F_{\alpha \beta} = 0$$.
3. For Einstein one has the second Bianchi identity $$\nabla_\alpha R_{\beta \gamma \delta}^\sigma + \nabla_\beta R_{\gamma \alpha \delta}^\sigma + \nabla_\gamma R_{\alpha \beta \delta}^\sigma = 0$$.
4. For linearised Einstein one has the linearised second Bianchi identity $$\partial_\alpha R_{\beta \gamma \delta}^\sigma + \partial_\beta R_{\gamma \alpha \delta}^\sigma + \partial_\gamma R_{\alpha \beta \delta}^\sigma = 0$$.

These identities are best thought of as geometric identities rather than true laws of motion, despite the fact that some components of these equations would involve a time derivative when viewed in standard coordinate systems. (All of these Bianchi identities can be derived mathematically from the Jacobi identity $$[\nabla_\alpha,[\nabla_\beta, \nabla_\gamma]] + [\nabla_\beta,[\nabla_\gamma,\nabla_\alpha]] + [\nabla_\gamma,[\nabla_\alpha,\nabla_\beta]]=0$$ applied to the covariant derivative $$\nabla_\alpha$$ that is naturally associated to the field theory.) In the case of Maxwell and linearised Einstein, the Bianchi-type identities can be expressed purely in terms of secondary fields, but in general we see that the Bianchi identities involve both primary and secondary fields.

To each of these field theories one can associate a Lagrangian which is typically expressed in terms of both primary and secondary fields:

1. For Yang-Mills the functional is $$\frac{1}{2} \int F^{\alpha \beta} \cdot F_{\alpha \beta}\ d\eta$$ (using an invariant inner product on $${\mathfrak g}$$, and with $$d\eta$$ denoting the Minkowski volume form).
2. For Maxwell the functional is $$\frac{1}{2} \int F^{\alpha \beta} F_{\alpha \beta}\ d\eta$$.
3. For Einstein the functional is the Einstein-Hilbert action $$\frac{1}{2} \int R\ dg$$ ($$R = g^{\alpha \beta} R_{\alpha \gamma \beta}^\gamma$$ being the scalar curvature).
4. For linearised Einstein the functional is the quadratic component of the Einstein-Hilbert action applied to an infinitesimal perturbation $$g = \eta + \varepsilon h$$; it has a moderately complicated form, see e.g., equations (6), (7) of this paper of Menon.

One can express these Lagrangians purely in terms of the primary field by expanding every appearance of a secondary field in terms of primary fields. In the case of Yang-Mills and Maxwell one can also express the functional purely in terms of secondary fields, but this is a fluke that is not expected to occur for more general field equations.

By varying these functionals with respect to the primary fields (or with respect to the secondary fields using the Bianchi-type identities as constraints that produce Lagrange multipliers, in the cases where those Bianchi identities can be expressed purely in terms of secondary fields), one obtains the equations of motion as Euler-Lagrange equations:

1. For Yang-Mills the equations of motion are $$D^\alpha F_{\alpha \beta} = 0$$.
2. For Maxwell the equations of motion are the Gauss-Ampere law $$\partial^\alpha F_{\alpha \beta} = 0$$.
3. For Einstein the equations of motion are $$R_{\alpha \beta \gamma}^\beta - \frac{1}{2} R g_{\alpha \gamma} = 0$$.
4. For linearised Einstein the equations of motion are $$R_{\alpha \beta \gamma}^\beta - \frac{1}{2} R \eta_{\alpha \gamma} = 0$$ (with $$R = \eta^{\alpha \beta} R_{\alpha \gamma \beta}^\gamma$$ being the linearised scalar curvature).

In the latter three cases one can express the equations of motion purely in terms of secondary fields, but again this is something of a fluke that should not be expected in general (particularly when working with nonlinear theories). Note that in all four cases the insertion of matter into the theory would add a matter term to the Lagrangian and thus a source term to the equations of motion, but would not affect the geometric Bianchi-type identities, which highlights a key difference between the equations of motion and the Bianchi-type identities.

In all of these cases, there is also a gauge invariance that prevents one from observing the primary fields directly.

1. For Yang-Mills, one can replace the connection $$A_\alpha$$ by the gauge-equivalent connection $$U A_\alpha U^{-1} - (\partial_\alpha U) U^{-1}$$ without affecting the physical observables.
2. For Maxwell, one can similarly replace the vector potential $$A_\alpha$$ by the gauge-equivalent potential $$A_\alpha + \partial_\alpha \lambda$$ without affecting the physical observables.
3. For Einstein, one can apply a diffeomorphic change of coordinates to spacetime and to the metric $$g^{\alpha \beta}$$ without affecting the physical observables (the general principle of relativity).
4. For linearised Einstein, one can replace the metric perturbation $$h_{\alpha \beta}$$ by $$h_{\alpha \beta} + \partial_\alpha X_\beta + \partial_\beta X_\alpha$$ for any vector field $$X^\alpha$$ (which can be viewed as a linearised (or infinitesimal) diffeomorphism) without affecting the physical observables.

Because of this gauge invariance, it is initially tempting in all four cases to try to ignore the primary field as much as possible and try to formulate the physics in terms of the secondary fields (which tend to be much more directly physically measurable, being less affected by gauge transforms). However, despite the fact that in some cases the primary field can indeed be expunged from the equations of motion and/or the Lagrangian, there is no a priori reason why the geometric Bianchi-type identities should then be subsumed into the Euler-Lagrange equations (though one can always do so artificially, as explained in my other answer). The fact that the Bianchi identities and the equations of motion look so similar in the case of Maxwell's equations (when specialised to 3+1 spacetime dimensions) is a misleading red herring; once one sees how things work for all the other field equations one sees that these two systems of equations should really be treated as being of a very different nature.

• Thanks for the interesting answer, Sorry for being boring, but I think that here the same point I do not understand appears in an equivalent form. "For Maxwell the primary field is the vector potential $A_\alpha$..." In Maxwell case existence of the potential is not given in advance, its existence is not assumed (in the contrary to further Yang-Mills generalizations) but it is equivalent to what you called Gauss-Faraday law $div(\vec B)=0, rot(\vec E)=-\frac{1}{c}\dot{\vec B}$.
– makt
Dec 29, 2020 at 6:34
• This law is not a mathematical identity like Bianchi identity, but a law of nature established experimentally. For this reason vector potential $A_\alpha$ cannot be considered as an a priori given field, it is a consequence of (Maxwell) equations of motion. The fact that $A_\alpha$ is defined up to a gauge is not the main difficulty, I think. Thus one comes back to the same issue: part of equaitons of motion is replaced by something else mathematically equivalent to a constrain.
– makt
Dec 29, 2020 at 6:44
• It's a question of interpretation. You are using the nineteenth century interpretation of classical electromagnetism in which the electromagnetic field is taken to be primary and the vector potential is secondary. I am using the post-Yang-Mills interpretation in which the vector potential is primary and the electromagnetic field, by definition, is the curl of the vector potential. The subsequent development of the laws of physics (e.g., the electroweak theory linking electromagnetism to the U(2) Yang-Mills theory of the weak force) has shown this is the more natural interpretation. Dec 29, 2020 at 16:26
• You are confusing historical order of discovery of theories with physical causation. Kepler's laws of planetary motion were deep experimental facts that directly led to Newton's laws of universal gravitation, but it is the latter that explains the former, not the other way around. Now, previously unexplained empirical facts such as Kepler's second law are mathematical consequences of deeper facts, in this case rotational covariance (and Noether's theorem). This is completely analogous to how the Gauss-Faraday law is now a mathematical consequence of the deeper fact of the U(1) nature of EM. Dec 30, 2020 at 17:14
• To continue the gravitational analogy, your original question is analogous to the question "Can Kepler's laws be expressed as an Euler-Lagrange equation without the gravitational potential?". Perhaps this is mathematically possible by some artificial trick, such as the sum-of-squares trick in my other answer, but it is hard to see the point of asking such a question in view of the completely satisfying success of Newtonian gravitation in explaining Kepler's laws. Similarly with Maxwell's equations and U(1) Yang-Mills theory. Dec 30, 2020 at 17:32

Here is a very simple Lagrangian that yields the Maxwell equations upon variation, taken from this post (also by me) on physics.SE. The trick is to introduce Lagrange multipliers.

Let $$\begin{equation} S[\chi,\tilde\chi,F,\tilde F]\overset{\mathrm{def}}=\int \chi_\nu(\partial_\mu F^{\mu \nu} - J^\nu)+ \tilde{\chi}_\nu(\partial_\mu \tilde{F}^{\mu \nu} - \tilde{J}^\nu)\mathrm dx \end{equation}$$

Variations with respect to $$\chi$$ and $$\tilde \chi$$ give you the Maxwell equations. Variation with respect to $$F$$ and $$\tilde F$$ give you $$\begin{equation} \partial_{[\mu}\chi_{\nu]}=\partial_{[\mu}\tilde\chi_{\nu]}=0 \end{equation}$$

The equations for $$F,\tilde F$$ and those for $$\chi,\tilde \chi$$ are decoupled so you can forget about the latter. The standard Maxwell equations are obtained by also taking the magnetic current to vanish, $$\tilde J=0$$. The general case with $$\tilde J\neq0$$ also allows the existence of magnetic monopoles, a fascinating subject itself.

This trick (and variations thereof) is used in proving $$S$$-duality of Maxwell's theory, and generalizations e.g. in Seiberg-Witten theory.

• Thanks. It seems however that if one allows auxiliary fields, like your $\chi_\mu,\tilde\chi_\mu$, one can write any system of PDE as EL-equations. Maxwell equations are not unique in this sense.
– makt
Dec 27, 2020 at 7:14
• @makt Indeed, it's a common technique for making an apparently first-order PDE an ELE; often, we add a total derivative, thereby unsubtly disguising what we did (although the motive may be e.g. ensuring hermiticity). When complex fields are involved, we can often use the conjugate, adjoint etc. as the multiplier, for much the same reason as $z^\ast$ can be treated as "independent" of $z$.
– J.G.
Dec 27, 2020 at 20:51
• @J.G. : Do you have examples of physical interest? An example I am aware of is the (first order) Dirac equation say for a free particle. Its Lagrangian has no auxiliary fields or constrains.
– makt
Dec 28, 2020 at 7:11
• @makt The Dirac & Schrödinger Lagrangians (the latter Galilean-invariant) use this conjugate-instead-of-new-auxiliary technique. I imagine this has also been done with Rarita–Schwinger. The Batalin–Vilkovisky formalism is an example with novel multipliers.
– J.G.
Dec 28, 2020 at 8:27

The actual number of degrees of freedom of the electromagnetic field is 2, per point in 3-dimensional space. One can see this starting from the formulation in terms of potentials, which feature 4 components per point. Gauge invariance allows one to choose, for example, the axial gauge $$A_3 =0$$, reducing the number of components to 3; then, one furthermore sees that $$A_0$$ is not in fact a dynamical degree of freedom, since the conjugate momentum vanishes (there is no time derivative of $$A_0$$ in the Lagrangean). In effect, $$A_0$$ acts as a Lagrange multiplier enforcing the constraint $$\vec{\nabla } \cdot \vec{E} =0$$. Indeed, one physically sees electromagnetic waves with only 2 polarization degrees of freedom.

Now, the formulation in terms of electromagnetic fields $$\vec{E}$$, $$\vec{B}$$ formally features 6 components per point, so these must be viewed as constrained variables, with 4 constraints. And indeed, stating that the electromagnetic fields are components of a closed 2-form is the most economical and symmetric way of stating such 4 constraints, as discussed by Denis Serre.

If we insist on keeping all 6 variables, as opposed to eliminating 4 of them by choosing a minimal, unconstrained set of generalized coordinates, then we need to implement the constraints using 4 Lagrange multiplier fields. In that respect, electromagnetism isn't quite that exceptional, compared to classical mechanics. Even if we tend to discuss mechanical examples in the Lagrange formalism of the second kind, since that is the most economical, of course also the Lagrange formalism of the first kind is available and instructive. So, while we don't need the full 8 auxiliary fields introduced by AccidentalFourierTransform, we do need 4 Lagrange multiplier fields for the electromagnetic case.

And that is the point where we perhaps shouldn't let aesthetics get in the way: Returning for a moment to the formulation in terms of vector fields, one actually only has to introduce one new Lagrange multiplier field in addition to the 4 vector field components $$A_{\mu }$$, namely, a Lagrange multiplier field, let's call it $$\lambda$$, implementing the gauge choice. The other Lagrange multiplier comes from among the original fields themselves - it is $$A_0$$, as discussed above! Now, is there really a good reason to allow for the variable $$A_0$$, but not the variable $$\lambda$$? What is the difference, beyond aesthetics? What if we renamed $$A_0$$ into $$\kappa$$?

Analogously, in the case of the formulation in terms of $$\vec{E}$$ and $$\vec{B}$$, is it reasonable to object to the introduction of additional Lagrange multiplier fields? After all, there really are redundant components - electromagnetic waves have only 2 polarizations. However, if we accept that Lagrange multiplier fields are natural and necessary in such a situation, then the answer is the standard electromagnetic Lagrangean supplemented with the 2-form constraint introduced via Lagrange multipliers.

• Well, I would on the contrary say that, from the physical point of view, the electromagnetic field has 2 degrees of freedom. That is what is actually seen in electromagnetic waves! We're the ones who are inflating the number of components to 6. Of course, we have reasons to do so, related to the different ways we are able to observe electromagnetic phenomena. I don't think this is that exceptional - think of a mechanical object constrained to move on a circle of radius $R$ in a plane. We can and do choose at times to consider that in terms of variables $r$, $\phi$ with a constraint $r=R$ ... Dec 27, 2020 at 16:35
• @makt "I have never seen it in other examples of constructing Lagrangians in classical mechanics"... actually general relativity provides a good example. If one wished to formulate GR in terms of the Riemann curvature tensor instead of the metric (much as Maxwell's equations are formulated in terms of the EM field rather than the vector potential), one would see a "mysterious and artificial" division into the "constraint" equations (Bianchi identities) and "dynamic" equations (Einstein), with only the latter having a Lagrangian interpretation (Einstein-Hilbert action). Dec 28, 2020 at 2:16
• Admittedly the nonlinear aspects of general relativity weaken the analogy with the linear equations of Maxwellian electromagnetism. But one can easily repair this aspect of the analogy by coimparing Maxwell to linearised gravity (en.wikipedia.org/wiki/Linearized_gravity) instead. Now the Bianchi identities use the Minkowski metric instead of a variable coefficient metric, and similarly the Einstein-Hilbert functional will use the Minkowski volume element instead of one generated by a dynamic metric. Dec 28, 2020 at 16:34
• One can imagine as a thought experiment that the equations for linearised gravity (Bianchi+Einstein, expressed in terms of a tensor field $R_{\alpha \beta \gamma}^\delta$) were discovered before the full Einstein equations. It was observed that it was technically convenient to express this tensor $R_{\alpha \beta \gamma}^\delta$ as the Riemann curvature of a "metric potential" $h_{\alpha \beta}$, but the equations could be written without reference to such a potential, so one might be led to the belief that $R_{\alpha \beta \gamma}^\delta$ could be treated as an unconstrained tensor. Dec 28, 2020 at 16:38
• However, such a mindset would have hindered the ability to discover the full Einstein equations, or to discover more general equations of motion (e.g., Einstein-Maxwell, Einstein-Vlasov, etc.) in which gravitation is coupled with other physical fields. As I said, the ability for Maxwell's equations to be expressible purely in terms of the curvature tensor $F_{\alpha \beta}$ (analogous to $R_{\alpha \beta \gamma}^\delta$) is a fluke (only shared by a few other linear field models such as linearised gravity) and is somewhat misleading as a guide to more general physical field theories. Dec 28, 2020 at 16:41