Electromagnetism – Making Symmetry Between E and B Fields Manifest in Lagrangian

electromagnetismlagrangian-formalism

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?

Best Answer

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.

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