My goal is to derive Maxwell's equations of electromagnetism with almost no effort at all. As often in mathematics, things look simpler when there is less structure. Here, as in mechanics, we do not assume any prior metric, so the geometry of the space at hand is very simple.

polarized electromagnetic wave

The Lagrangian

In all generality, on any manifold $\Man$ (think for instance $\Man = \mathbf{R}^n$ in any dimension $n$), the Lagrangian for a generalized version of electromagnetism takes the general simple form

\[ \mathcal{L}(A, \dd A) = \langle G(\dd A), \dd A \rangle + \langle J, A \rangle \]

Let us review the terms appearing in the Lagrangian:

  • The electromagnetic potential $A$ is a one-form
  • The electromagnetic field $F = \dd A$ is thus a (closed) two-form
  • The (free) current $J$ is a one-current (an element in the dual of the space of one-forms); “Free” means that $J$ does not depend on $A$.
  • The displacement $G(F)$ is a two-current, which depends on the field $F$.

The bracket $\langle \cdot,\cdot \rangle$ is the duality bracket.

Analogy with Classical Mechanics

There is a loose analogy with classical mechanics as follows

  • $A$ is the position
  • $F = \dd A$ is the velocity
  • $G$ is the momentum

From this point of view, we see that the “potential” $\langle J, A \rangle$ is linear. This has the consequence that the corresponding force does not depend on $A$. An analogy in mechanics, is the gravitation potential of the (flat) earth, given by $mgh$, which gives a constant force $m g$, the weight of the mass $m$.

Affine Assumption

A further assumption is that $G$ is an affine function of $F$:

\[ G(F) = \frac{1}{2}μ(F) + {M} \]

The two-current ${M}$ is the magnetisation-polarisation.

We further assume that the linear part $μ$ is symmetric, i.e.:

\[ \langle μ(F_1), F_2 \rangle = \langle μ(F_2), F_1 \rangle \]

Euler–Lagrange Equations

The Euler–Lagrange equations are now easy to derive, and we obtain Maxwell's equations in weak form: \[ \langle μ(F), \dd B \rangle = \langle M, \dd B \rangle + \langle J, B \rangle \qquad \forall B \] where $B$ is an arbitrary one-form.

One defines the operator $∂$, dual to the exterior derivative by

\[ \langle ∂ G, A \rangle := \langle G, \dd A \rangle \]

Note that $∂ ∂ = 0$.

This gives Maxwell's equations in its familiar guise: \[ ∂ μ(F) = ∂ {M} + J \]

We see that to stand any chance to solve this equation, we need that

\[ ∂ J = 0 \]

This is the charge conservation condition.

The quantity $∂ {M}$ is called the bound current. Clearly, as $∂(∂{M})= 0$, the “bound charge” is conserved as well.

The equation $\dd F= 0$ contains both the magnetic Gauß's law, and the induction law. The equation $∂μ(F) = ∂{M} + J$ contains both the electric Gauß's law and the Ampère circuit law.

By writing the equation in $A$ instead, one obtains \[ \label{eq:generalwave} ∂μ(\dd A) = ∂ {M} + J \]

In the absence of current and of magnetisation or polarisation, the last equation becomes a kind of wave equation $∂μ(\dd A) = 0$.

Curvature and Gauge Invariance

curved shell

There is another fundamental symmetry of the equations, which stems from the observation that the final equations do not depend on $A$, only on $\dd A$. The fact that the final equations are invariant with respect to the transformation $A \to A + \dd φ$ is called gauge invariance.

Why is that?

A proper generalisation of Maxwell's equations, which leads to the Yang–Mills equations, is that the field $F$ represents a curvature. This is the subject of a future post, but let me simply explain the idea of the curvature in this case. The electromagnetic potential $A$ is (the potential for) a connection. What does it indicate in this case, and what is the role of the gauge invariance?

A field of matter is described as a function $ψ$ from $\Man$ to $\CC$, the complex numbers. A fundamental observation is that the phase of that function does not matter for the calculations. In particular, “keeping the phase fixed“ while moving in space has no intrinsic meaning. This is where the connection $θ$, which corresponds to the potential $A$, comes in.

The essential point is that this connection $θ$ is a form defined on an extended space (technically, a principal bundle), which in this case is roughly speaking $\Man \times U(1)$, where $U(1)$ is the group of unitary complex numbers. In general, there is no way to see this connection as a form on the space $\Man$. However, one can choose a gauge, which is a section of the bundle above, and this gauge pulls the connection $θ$ back to a form on $\Man$, the familiar potential!

Of course, the potential varies depending on the gauge choice, and that gauge choice has no bearing whatsoever on the solution of the equations.

What happens when one changes the gauge? Well, it turns out that changing the gauge, in this case, amounts to add a closed form to the form $A$. As a result, the Euler–Lagrange equations are ill-posed in the variable $A$.

Another point of view is that adding a connection with corresponding zero curvature does not change the final curvature. Indeed, changing the gauge is equivalent to adding a flat connection (a bit similar to the active/passive point of view in physics).

Anyhow, the magnetic field $F$ can be regarded as a curvature, which essentially tells us by how much the phase changes by following a small circle (and coming back to the same point).

Finally, we see that, in a sense, the Lagrangian penalises the curvature.

As a final comment (which is also to be a future note in this blog), there is a far reaching Theorem (called Noether's second Theorem) which gives very general conditions for the gauge invariance of a Lagrangian, in any number of variables, and of any order (here the order is just one, as there are only one derivative in the independent variables).

With a Metric

In the presence of a metric, there is a natural choice for $μ$, given by $\langle μ(F_1), F_2 \rangle := (F_1,F_2)$, where $(\cdot,\cdot)$ is the metric at hand. It is symmetric as required.

The operation $∂$ is related to the codifferential $\dd^{\star}$ by $∂ μ(F) = \dd^{\star} F$.

If ${M} = 0$, one thus obtains: \[ \dd F = 0 \\ \dd^{\star}F = J \]

Classical Maxwell Equations

In the case of a $(-,+,+,+)$ metric, we recover Maxwell's equations. The following table lists the classical names of the time and space components of the various terms used above.

Name Kind Time component Space component
$F$ Electromagnetic tensor 2-form $E$ Electric field $B$ Magnetic field
$G$ Displacement tensor 2-current $D$ Electric displacement field $H$ Magnetizing field
$A$ Four-Potential 1-form $φ$ Electric potential $A$ Magnetic potential
$J$ Four-Current 1-current $ρ$ Charge density $j$ Electric current
$M$ Magnetisation-polarisation tensor 2-current $P$ Polarisation $M$ Magnetisation
Say thanks!