Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

The potentials are not unique. Let $ψ (r, t)$ be any scalar function and define

A^{'} = A + \nabla ψ, V^{'} = V - \frac{\partial ψ}{\partial t} . (5.5.45)

Since the curl of a gradient vanishes, $\nabla \times A^{'} = \nabla \times A + \nabla \times \nabla ψ = B$ . For the electric field, using that $\nabla$ and $\partial / \partial t$ commute,

- \nabla V^{'} - \frac{\partial A ^{'}}{\partial t} = - \nabla V + \nabla \frac{\partial ψ}{\partial t} - \frac{\partial A}{\partial t} - \frac{\partial}{\partial t} \nabla ψ = - \nabla V - \frac{\partial A}{\partial t} = E . (5.5.46)

Definition · Gauge transformation

The change Eq. (5.5.45) of the potentials, with an arbitrary $ψ$ , is a gauge transformation. It leaves $E$ and $B$ unchanged. A condition that selects among the equivalent pairs, such as Eq. (5.5.41), is a gauge.

Only $E$ and $B$ are measurable (through the force on a charge), so $(A^{'}, V^{'})$ and $(A, V)$ describe the same physics (Figure 5.5.8). The Lorenz gauge uses this freedom. Suppose a pair $(A, V)$ does not satisfy it, so that $Λ = \nabla \cdot A + μ ε \partial V / \partial t \neq = 0$ . Then the new pair has $\nabla \cdot A^{'} + μ ε \partial V^{'} / \partial t = Λ + \nabla^{2} ψ - μ ε \partial^{2} ψ / \partial t^{2}$ , and choosing $ψ$ as a solution of

\nabla^{2} ψ - μ ε \frac{\partial ^{2} ψ}{\partial t ^{2}} = - (\nabla \cdot A + μ ε \frac{\partial V}{\partial t}) (5.5.47)

makes it zero: the new pair satisfies the Lorenz gauge. Such a $ψ$ always exists, since Eq. (5.5.47) is a wave equation with a known source, of the kind solved in Section 5.5.10.

**Figure 5.5.8.** Gauge freedom. Any two pairs of potentials related by Eq. (5.5.45) give the same measurable fields. A gauge picks one pair. The Coulomb gauge is the one of magnetostatics, the Lorenz gauge the one of waves, because it gives two *uncoupled* wave equations of the same form.

Two common choices. The Coulomb gauge $\nabla \cdot A = 0$ is the one of magnetostatics (Section 4.1). In it $V$ obeys Poisson's equation at every instant, as if the charges were static, and all the retardation sits in $A$ . The Lorenz gauge Eq. (5.5.41) treats $V$ and $A$ alike and is the one used for waves. Even the Lorenz gauge does not fix the potentials completely: any $ψ$ that satisfies the homogeneous wave equation can still be added (Exercise 5.5.6).

Consistency. The wave equations Eq. (5.5.44) were derived assuming the gauge. Home exercise 5.5.3 shows that this costs nothing: the Lorenz gauge is consistent with the conservation of charge $\nabla \cdot J + \partial ρ_{v} / \partial t = 0$ , which Maxwell's equations contain anyway. Problem 5.5.5 shows the other side of the freedom: a pair of potentials that looks like a travelling wave but describes no field at all.

Common mistake · Taking the potentials for the physics

Concluding from a nonzero $V$ or $A$ that a field is present, or reading $V (a) - V (b)$ as the voltage between $a$ and $b$ . A gauge transformation changes both without changing anything measurable. A potential difference is a voltage only in a description in which $\partial A / \partial t$ contributes nothing along the path (Exercises 5.5.6 and 5.5.7). Always derive $E$ and $B$ before interpreting.

Quick check

What do the Lorenz gauge and the equations for the potentials become when nothing depends on time?

With $\partial / \partial t = 0$ the Lorenz gauge becomes $\nabla \cdot A = 0$ , the Coulomb gauge of magnetostatics, the wave equations become Poisson's equations $\nabla^{2} V = - ρ_{v} / ε$ and $\nabla^{2} A = - μ J$ , and $E = - \nabla V$ : exactly the left column of Table 5.5.2.

5.5.9 Gauge freedom