Making symmetry between E and B fields manifest in Lagrangian

Question

Maxwell's equations are nearly symmetric between $E$ and $B$. If we add magnetic monopoles, or of course if we restrict ourselves to the sourceless case, then this symmetry is exact. This is not just a discrete symmetry of exchange. A "duality rotation" of $(E, B) \rightarrow (E \cos \theta - B \sin \theta, B \cos \theta + E \sin \theta)$ preserves the form of the equations. (The charges and currents, if any, would have to similarly mix as well. The upshot is that if all particles have the same ratio between their electric and magnetic charges, it's indistinguishable from just having electric charges. See e.g. Jackson 6.12.)

This shows up quite nicely in the Hamiltonian formulation as the invariance of the free Hamiltonian density $\mathcal{H} = E^2 + B^2$. The free Lagrangian density, however, is the Lorentz invariant scalar $\mathcal{L} = E^2 - B^2$, which does not appear to have the full continuous symmetry, though a quarter rotation leads to $\mathcal{L} \rightarrow - \mathcal{L}$, leaving the equations of motion unchanged. Is there an alternate free Lagrangian density that makes the full continuous symmetry manifest?

EDIT:

The normal Lagrangian expression is dependent on a gauge field A, whose existence depends on the non-existence of monopoles. As Peter Morgan points out, this obviously breaks the symmetry and explains why this Lagrangian needn't have this symmetry. However, that leaves the following questions:

1. Is there another Lagrangian giving the same equations of motion that does have this symmetry? If yes, can it be demonstrated? If not, is there a proof?
2. Even if the answer to 1 is no, is there a family of Lagrangians related by this symmetry that each give the same equations of motion? If yes, can it be demonstrated? If not, is there a proof?

Answer 1

Let us rephrase the question as follows.

Is there a Lagrangian density ${\cal L}$ that (1) implements the global $SO(2)$ symmetry manifestly, and (2) whose Euler-Lagrange equations yield Maxwell's equations in vacuum where there are no sources?

The answer is Yes. Below we will show this. Let the speed of light be $c=1$ from now on.

1) Field variables. The model has $2\times 3=6$ gauge potential fields ${\cal A}^a_i(\vec{x},t)$. Here $i=1,2,3$ are three spatial directions, and $a=1,2$ is an internal $SO(2)$ index. The gauge potential transforms

$${\cal A}^a_i\to \sum_{b=1}^2M^a{}_b {\cal A}^b_i $$

in the 2-dimensional fundamental representation of $SO(2)$, where

$$M^a{}_b =\left[\begin{array}{cc} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) &\cos(\theta)\end{array}\right] \in SO(2), \qquad \sum_{b,c=1}^2 (M^{t})_a{}^b g_{bc} M^c{}_d = g_{ad}, \qquad g_{ab}\equiv \delta_{ab}.$$

The magnetic field $\vec{B}$ and electric field $\vec{E}$ are given by the curl of the gauge potential $\vec{\cal A}^a$,

$$\vec{\cal B}^a := \vec{\nabla} \times \vec{\cal A}^a, \qquad a=1,2, $$

where

$$ \vec{B}\equiv\vec{\cal B}^1 \qquad \mathrm{and} \qquad \vec{E}\equiv \vec{\cal B}^2.$$

It is easy to check that

$${\cal B}_i^a\to \sum_{b=1}^2 M^a{}_b {\cal B}_i^b$$

implements the sought-for $SO(2)$ transformation on the magnetic and electric fields $\vec{B}$ and $\vec{E}$. The two scalar-valued Maxwell equations

$$\vec{\nabla}\cdot\vec{B}=0 \qquad \mathrm{and} \qquad \vec{\nabla}\cdot\vec{E}=0 $$

are identically satisfied, because

$$\vec{\nabla}\cdot\vec{\cal B}^a=\vec{\nabla}\cdot(\vec{\nabla}\times\vec{\cal A}^a)=0, \qquad a=1,2.$$

2) Lagrangian density. The Lagrangian density ${\cal L}$ is

$$ {\cal L} = \frac{1}{2}\sum_{a,b=1}^2 \vec{\cal B}^a \cdot\left( \epsilon_{ab} \frac{\partial \vec{\cal A}^b }{\partial t} - g_{ab} \vec{\cal B}^b\right), $$

where $\epsilon_{ab} = - \epsilon_{ba}$ is the Levi-Civita tensor in 2 dimensions (with $\epsilon_{12} = 1$). It is easy to check that the Lagrangian density ${\cal L}$ is manifestly invariant under global $SO(2)$ transformations, because both $\epsilon_{ab}$ and $g_{ab}\equiv \delta_{ab}$ are invariant tensors for $SO(2)$. The action is by definition

$$ S[{\cal A}]=\int d^4x \ {\cal L}. $$

Extremizing the action with respect to $2\times 3=6$ gauge potential fields ${\cal A}^a_i$ produces $6$ Euler-Lagrange equations. In detail, an arbitrary infinitesimal variation ${\cal A}^a_i\to {\cal A}^a_i+\delta{\cal A}^a_i$ induces a change $\delta{\cal L}$ in the Lagrangian density

$$ \delta{\cal L} ~\sim~ \sum_{a,b=1}^2\delta \vec{\cal B}^a \cdot\left( \epsilon_{ab} \frac{\partial \vec{\cal A}^b }{\partial t} - g_{ab} \vec{\cal B}^b\right) ~\sim~ \sum_{a,b=1}^2\delta \vec{\cal A}^a\cdot \left( \epsilon_{ab} \frac{\partial \vec{\cal B}^b }{\partial t} - g_{ab} \vec{\nabla} \times \vec{\cal B}^b\right), $$

where the "$\sim$" sign means equality modulo divergence terms. The variation $\delta S=0$ of the action vanishes iff the last parenthesis is zero. This yield precisely the two vector-valued Maxwell equations

$$ \frac{\partial \vec{B}}{\partial t} + \vec{\nabla} \times \vec{E}=\vec{0} \qquad \mathrm{and} \qquad \frac{\partial \vec{E}}{\partial t} = \vec{\nabla} \times \vec{B}. $$

References:

1) S. Deser and C. Teitelboim, "Duality Transformations Of Abelian And Nonabelian Gauge Fields", Phys.Rev.D 13 (1976) 1592.

2) C. Bunster and M. Henneaux, "Can (Electric-Magnetic) Duality Be Gauged?", Phys.Rev.D83 (2011) 045031, arXiv:1011.5889.

3) S. Deser, "No local Maxwell duality invariance", Class.Quant.Grav.28 (2011) 085009, arXiv:1012.5109.

Answer 2

No, the Lagrangian of electrodynamics indeed fails to be invariant under the particular continuous $SO(2)$ symmetry that you described. You can't change this fact by any field redefinition etc. because physically harmless field redefinitions don't change the action.

However, the Lagrangian is invariant under $$ (E,B) \to (E\cosh \eta + B \sinh \eta, B\cosh \eta + E \sinh \eta) $$ which corresponds to changing the metric on the $(E,B)$ space from the $(++)$ Euclidean signature to the $(+-)$ Lorentzian signature (like in special relativity). On the other hand, the energy density (with the positively definite quadratic invariant) obviously fails to be invariant under this modified transformation.

However, it's the latter, indefinite "Lorentzian" transformation that is closer to the actual continuous symmetries that may exist between electricity and magnetism. In $N=4$ gauge theory or type IIB string theory, it's easy to see that the most natural continuous symmetry is $SL(2,R)$ which is, for two possible "types of charges", equivalent to $Sp(2,R)$.

The quantization of electric charges breaks it to $SL(2,Z)$ which is the actual symmetry group - S-duality - of $N=4$ gauge theory or type IIB string theory. It is isomorphic neither to $SO(2)$ nor to $SO(1,1)$ because its bilinear invariant is antisymmetric, not symmetric.

This $SL(2,Z)$ group contains the "quarter rotation" you mentioned but it doesn't contain the full continuous group. Even the $SL(2,R)$ doesn't. At the level of the Lagrangian or quantum theory, one may see why the continuous group mixing $E,B$ is morally wrong. The action is not invariant, as you noticed. Also, the Dirac quantization rule leads one to consider the product of the elementary electric charge and elementary magnetic charge, which should be a multiple of $2\pi$, and this "symplectic" structure has to be preserved as well.

Answer 3

a somehow hidden assumption of your question is that you consider a theory in which the material response is the same to a magnetic as well as to an electric field. We have to elaborate on this a little now to answer your question from an electrodynamic point of view in addition to the nice general answer given by Luboš.

I suppose you know the linear constitutive relations $$\mathbf{D} = \epsilon \mathbf{E}\quad \text{and}\quad \mathbf{B}=\mu \mathbf{H}\ .$$ These link the electric and magnetic field to their "dual" counterparts using the permittivity $\epsilon = \epsilon_0 \epsilon_r$ and permeability $\mu = \mu_0 \mu_r$ assumed to be not depending on the fields itself.

In this case you will have a Lagrangian of the form $$\mathcal{L} \propto F_{\mu\nu}M^{\mu\nu}$$ where now the tensor $M$ is a quantity that is linearly linked to $\epsilon$ and $\mu$ and $F$ via $M^{\mu\nu} = \epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$. Generally, $M$ will depend on $F$ in some arbitrary complex way.

Now, if you assume linear materials with the same response to electric and magnetic excitation, $$\epsilon_r \equiv \mu_r\ ,$$

you find that you can write $$\mathcal{L} \propto F\wedge\star F$$ as in the case of vacuum where $\epsilon_r=\mu_r = 1$.

These materials are called impedance matched and only a special subclass of all materials as describes. The symmetry you refer to can then be used to interprete $\epsilon$ as a metric which is done in a field called transformation optics, see eg. Leonhardt, Philbin: Transformation Optics and the Geometry of Light where all the concepts are discussed in greater detail along with spectacular applications as cloaking:

So, to sum up, in general (macroscopic) electrodynamics, the symmetry between electric and magnetic fields breaks down. In e.g. optical frequencies, materials tend to respond only in an electric way, $$\mu^{\text{opt}}_r = \mu_0 \quad ,\ \epsilon^{\text{opt}}_r \text{ arbitrary}\ .$$

Answer 4

But the electric and magnetic field are fundamentally different so handling them as symmetric just obscures their elementarry role in the Pointcaré algebra.

The fields tensor expressed in Lorentz group generator form with $\mathsf{J}$ and $\mathsf{K}$ as the generators of rotations and boosts for (axial-)vector fields.

$ F^\mu_{~\nu} ~~=~~ \mathsf{B}\cdot\mathsf{J} ~+~ \mathsf{E}\cdot\mathsf{K} ~~=~~ \left( \begin{array}{rrrr} \ 0\ \ & ~~\mathsf{E}_x & ~~\mathsf{E}_y & ~~\mathsf{E}_z \ \\ \mathsf{E}_x & \ 0\ \ & ~~\mathsf{B}_z & - \mathsf{B}_y \ \\ \mathsf{E}_y & - \mathsf{B}_z & \ 0\ \ & ~~\mathsf{B}_x \ \\ \mathsf{E}_z & ~~\mathsf{B}_y & - \mathsf{B}_x & \ 0\ \ \ \end{array} \right) $

Here you see the fundamental difference: The magnetic field rotates while the electric field boosts. If you simply look at the signs then you'll recognize the $\mathsf{J}$ and $\mathsf{K}$ in the field tensor.

The field tensor is the boost/rotate generator and so it defines the changes in (proper) time of vector currents and axial currents.

$ \begin{aligned} &\frac{\partial}{\partial \tau}\Big(\bar{\psi}\gamma^\mu\psi\Big) &=~~ &\tfrac{q}{m}\,F^\mu_{~\nu}\Big(\bar{\psi}\gamma^\nu\psi\Big) \\ \\ &\frac{\partial}{\partial \tau}\Big(\bar{\psi}\gamma^5\gamma^\mu\psi\Big) &=~~ &\tfrac{q}{m}\,F^\mu_{~\nu}\Big(\bar{\psi}\gamma^5\gamma^\nu\psi\Big) \end{aligned} $

These are just the QFT equivalents of the classical expressions. See for instance Jackson section 11.11. where the axial expression is just the first term in the Thomas-Bargmann-Michel-Telegdi equation. The equivalent Lorentz generator ${\cal F}^\mu_{~\nu}$ for chiral spinors is.

$ \mathcal{F}^\mu_{~\nu} ~~=~~ \mathsf{B}\cdot J ~+~ \mathsf{E}\cdot K ~~=~~ -\tfrac12\left( \begin{array}{c} (i\mathsf{B}^i+\mathsf{E}^i)\sigma_i ~~~~~~0~~~~~~ \\ \\ ~~~~~~0~~~~~~ (i\mathsf{B}^i-\mathsf{E}^i)\sigma_i \\ \end{array} \right) $

These act on the spinors and, as it should be, the result for the vector and axial current is just the same: Let $\psi_{\,\tau}$ represent the chiral spinor field $\psi_o$ after the corresponding operator has acted on it for a total (proper) time $\tau$.

$ \psi_{\,\tau} ~~=~~ \exp\left(\tau\tfrac{q}{m}\,{\cal F}^\mu_{~\nu}\right)\psi_o $

Then we can write.

$ \begin{aligned} &\exp\left(\tau\tfrac{q}{m}\,F^\mu_{~\nu}\right)~\bar{\psi}_o\gamma^\nu\psi_o &=~~ &\bar{\psi}_{\,\tau}\gamma^\nu\psi_{\,\tau} \end{aligned} $

This means that operating with the Field tensor on the vector current is just the same as operating with the equivalent spinor boost/rotate operator on the spinor field.

Regards, Hans

Answer 5

An alternative way to look at this:

The Hamiltonian and the associated Lagrangian formulation that you give presupposes that the field is the electromagnetic potential, with the electric field as the momentum, which is only possible if we do not have monopoles in the mix. The electromagnetic field has to satisfy Gauss's law for magnetism and Faraday's law of induction(chosen from the first page of a Google search page because I didn't like the Wikipedia page much) in order for the six components of the electromagnetic field to be expressible as the exterior derivative of the four components of the electromagnetic potential.

Without the reduction of the EM field to the EM potential because of the no monopole constraint, one way to proceed to a Hamiltonian or Lagrangian formalism would be to introduce the time-like derivatives of the electromagnetic field as the momentum components, which, however, would change the EM field equations into second order equations that are not the Maxwell equations. I'm sure this must have been done somewhere in the literature, and various possible constrained dynamics will have been considered, because people have investigated the possibility of monopoles, but I regret that I don't know a good reference. The people who have investigated the possibility of monopoles have apparently not so far managed to make the phenomenology of such a dynamics work out nicely.

If we deal with the source-free Maxwell's equations, we could equally well express the electromagnetic field in terms of a potential as $\mathbf{B}(x)=\nabla\phi'(x)$ and $\mathbf{E}=\nabla\times\mathbf{A}'(x)$ instead of the usual prescription, or indeed in terms of any linear transform, $\mathbf{B}(x)=\cos(\alpha)\nabla\phi_\alpha(x)+\sin(\alpha)\nabla\times\mathbf{A}_\alpha(x)$ and $\mathbf{E}(x)=-\sin(\alpha)\nabla\phi_\alpha(x)+\cos(\alpha)\nabla\times\mathbf{A}_\alpha(x)$, although we have to keep sight of the fact that this breaks down as soon as we introduce electric charges or monopoles. I suggest you try seeing, for the free field case, what the effect is of changing the definition of the electromagnetic field in terms of the electromagnetic potential as you transform the electromagnetic field.