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PU-204 / L07 · lecture

Scalar and Vector Potentials, Gauge Freedom

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You must be comfortable with vector calculus identities, specifically the curl of a gradient and the divergence of a curl, and you should recall the four Maxwell equations in their full time-dependent form.

Why this matters

Potentials simplify the mathematics of electromagnetism by reducing four vector equations to two scalar/vector equations, while gauge freedom reveals the true physical content of the theory.

In the previous lecture, we established that Maxwell’s equations are consistent and complete. However, solving for the electric and magnetic fields directly is often algebraically cumbersome. By introducing potentials, we exploit the mathematical structure of the homogeneous Maxwell equations to define variables that automatically satisfy the constraints. This not only makes solving for fields easier but also exposes a profound symmetry—gauge invariance—that is central to modern physics, including quantum electrodynamics and the Standard Model.

The Idea in Plain Language

Imagine trying to describe the flow of water in a complex pipe system. Instead of tracking the velocity of every water molecule (the field), you might describe the pressure difference between points (the potential). The pressure is a simpler quantity; the flow is derived from how the pressure changes in space and time. Similarly, the electric and magnetic fields are the "flow" of the electromagnetic interaction. The scalar potential \(\phi\) and vector potential \(\mathbf{A}\) are the underlying "pressure" variables. Crucially, adding a constant to the pressure everywhere does not change the flow. In electromagnetism, we can add specific mathematical functions to the potentials without changing the physical fields at all. This redundancy is called gauge freedom. It tells us that the potentials are not directly observable; only the fields \(\mathbf{E}\) and \(\mathbf{B}\) are physical. This distinction is vital because it means the potentials are mathematical tools, not physical entities in the classical sense, though they play a crucial role in quantum mechanics.

Defining the Potentials

The homogeneous Maxwell equations are \(\nabla \cdot \mathbf{B} = 0\) and \(\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}\). The first equation states that the magnetic field has no sources or sinks; it is solenoidal. A fundamental theorem of vector calculus states that if the divergence of a vector field is zero, that field can be expressed as the curl of another vector field. Therefore, we define the vector potential \(\mathbf{A}\) such that:

\[ \mathbf{B} = \nabla \times \mathbf{A} \]

Substituting this into the second homogeneous equation, we get \(\nabla \times \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{A})\). Rearranging, we find \(\nabla \times (\mathbf{E} + \frac{\partial \mathbf{A}}{\partial t}) = 0\). A vector field with zero curl is conservative and can be written as the negative gradient of a scalar function. Thus, we define the scalar potential \(\phi\):

\[ \mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t} \]

These definitions automatically satisfy the two homogeneous Maxwell equations. The inhomogeneous equations, \(\nabla \cdot \mathbf{E} = \rho/\epsilon_0\) and \(\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\), now serve as the equations of motion for the potentials \(\phi\) and \(\mathbf{A}\).

Gauge Freedom

The potentials \(\phi\) and \(\mathbf{A}\) are not unique. Suppose we have a valid pair \((\phi, \mathbf{A})\). Consider a new pair defined by an arbitrary smooth scalar function \(\chi(\mathbf{r}, t)\):

\[ \mathbf{A}' = \mathbf{A} + \nabla \chi \] \[ \phi' = \phi - \frac{\partial \chi}{\partial t} \]

Let us verify that the physical fields remain unchanged. For the magnetic field:

\[ \mathbf{B}' = \nabla \times \mathbf{A}' = \nabla \times (\mathbf{A} + \nabla \chi) = \nabla \times \mathbf{A} + \nabla \times (\nabla \chi) \]

Since the curl of a gradient is always zero (\(\nabla \times \nabla \chi = 0\)), we have \(\mathbf{B}' = \mathbf{B}\). For the electric field:

\[ \mathbf{E}' = -\nabla \phi' - \frac{\partial \mathbf{A}'}{\partial t} = -\nabla \left( \phi - \frac{\partial \chi}{\partial t} \right) - \frac{\partial}{\partial t} \left( \mathbf{A} + \nabla \chi \right) \]

Expanding the terms:

\[ \mathbf{E}' = -\nabla \phi + \nabla \left( \frac{\partial \chi}{\partial t} \right) - \frac{\partial \mathbf{A}}{\partial t} - \frac{\partial}{\partial t} (\nabla \chi) \]

Assuming the spatial and temporal derivatives commute (which is true for smooth functions), \(\nabla (\frac{\partial \chi}{\partial t}) = \frac{\partial}{\partial t} (\nabla \chi)\). These two terms cancel exactly, leaving \(\mathbf{E}' = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t} = \mathbf{E}\). Thus, the transformation leaves the physics invariant. This is the gauge freedom.

Choosing a Gauge

While the potentials are not unique, we can choose a specific "gauge" to simplify the equations of motion. The most common choice is the Coulomb gauge, defined by \(\nabla \cdot \mathbf{A} = 0\). In this gauge, the scalar potential satisfies Poisson's equation, \(\nabla^2 \phi = -\rho/\epsilon_0\), which is instantaneous and resembles electrostatics. The vector potential satisfies a wave equation with a source term. Another important choice is the Lorenz gauge (note the 'z'), defined by:

\[ \nabla \cdot \mathbf{A} + \mu_0 \epsilon_0 \frac{\partial \phi}{\partial t} = 0 \]

In the Lorenz gauge, both \(\phi\) and \(\mathbf{A}\) satisfy the same inhomogeneous wave equation:

\[ \nabla^2 \phi - \mu_0 \epsilon_0 \frac{\partial^2 \phi}{\partial t^2} = -\frac{\rho}{\epsilon_0} \] \[ \nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J} \]

This symmetry is crucial for special relativity, as it treats space and time on equal footing. The choice of gauge does not change the physical predictions, but it changes the mathematical form of the solutions.

The derivation

◆ DERIVATION EMBED · D-163 · not re-derived here

Potentials and Gauge Freedom

Because the two homogeneous Maxwell equations, ∇·B = 0 and ∇×E = −∂B/∂t, are constraints of pure structure, the fields can be written as B = ∇×A and E = −∇φ − ∂A/∂t in terms of a vector potential A(r,t) and a scalar potential φ(r,t). These potentials are not unique: the simultaneous replacement A → A + ∇χ, φ → φ − ∂χ/∂t, for any smooth scalar field χ(r,t), leaves both E and B unchanged. This is the gauge freedom of electromagnetism.

B = ∇×A,   E = −∇φ − ∂A/∂t
Open full derivation →
Tier 3 · watch out

A common misconception is that the gauge transformation \(\chi\) must be zero at infinity for the potentials to vanish. While this is often assumed for localized sources, the gauge freedom is local. In quantum mechanics, the phase of the wavefunction is also gauge-dependent, and the Aharonov-Bohm effect demonstrates that the potentials have physical significance even in regions where \(\mathbf{E}\) and \(\mathbf{B}\) are zero, provided the gauge transformation is not globally trivial. C

Worked example

Problem. Consider a scalar potential \(\phi(\mathbf{r}, t) = \alpha x^2 t\) with \(\alpha = 2.0\ \mathrm{V\,m^{-2}\,s^{-1}}\) and a vector potential \(\mathbf{A}(\mathbf{r}) = \beta y^2 \hat{\mathbf{x}}\) with \(\beta = 2.0\ \mathrm{T\,m^{-1}}\) (two separate constants, because \(\phi\) and \(\mathbf{A}\) have different units). Determine the electric field \(\mathbf{E}\) and magnetic field \(\mathbf{B}\) at the point \(\mathbf{r} = (1.0, 1.0, 0.0) \ \mathrm{m}\) at time \(t = 0.5 \ \mathrm{s}\).

  1. Calculate the gradient of the scalar potential: \(\nabla \phi = \frac{\partial \phi}{\partial x} \hat{\mathbf{x}} + \frac{\partial \phi}{\partial y} \hat{\mathbf{y}} + \frac{\partial \phi}{\partial z} \hat{\mathbf{z}}\). Since \(\phi = k x^2 t\), \(\frac{\partial \phi}{\partial x} = 2\alpha xt\), \(\frac{\partial \phi}{\partial y} = 0\), \(\frac{\partial \phi}{\partial z} = 0\).
  2. Calculate the time derivative of the vector potential: \(\frac{\partial \mathbf{A}}{\partial t} = 0\) because \(\mathbf{A}\) does not depend explicitly on \(t\).
  3. Compute the electric field: \(\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t} = -(2\alpha xt) \hat{\mathbf{x}}\).
  4. Calculate the curl of the vector potential: \(\mathbf{B} = \nabla \times \mathbf{A}\). Since \(\mathbf{A} = A_x \hat{\mathbf{x}}\) with \(A_x = \beta y^2\), \(\mathbf{B} = (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) \hat{\mathbf{x}} + (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) \hat{\mathbf{y}} + (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}) \hat{\mathbf{z}}\). Here \(A_y = A_z = 0\), so \(\mathbf{B} = -\frac{\partial A_x}{\partial y} \hat{\mathbf{z}} = -2\beta y \hat{\mathbf{z}}\).
  5. Substitute the values: \(x=1.0\), \(y=1.0\), \(t=0.5\), \(\alpha = \beta = 2.0\). \(\mathbf{E} = -(2)(2.0)(1.0)(0.5) \hat{\mathbf{x}} = -2.0 \hat{\mathbf{x}} \ \mathrm{V/m}\). \(\mathbf{B} = -2(2.0)(1.0) \hat{\mathbf{z}} = -4.0 \hat{\mathbf{z}} \ \mathrm{T}\).
\[ \mathbf{E} = -2.0\ \hat{\mathbf{x}}\ \mathrm{V/m}, \quad \mathbf{B} = -4.0\ \hat{\mathbf{z}}\ \mathrm{T} \]

Units check. \(\alpha x^2 t\) has units \(\mathrm{V\,m^{-2}\,s^{-1}} \times \mathrm{m^2} \times \mathrm{s} = \mathrm{V}\), and \(\beta y^2\) has units \(\mathrm{T\,m^{-1}} \times \mathrm{m^2} = \mathrm{T\,m}\), the units of a vector potential. \(\nabla \phi\) has units \(\mathrm{V/m}\), which is correct for \(\mathbf{E}\). \(\nabla \times \mathbf{A}\) has units \(\mathrm{T}\), which is correct for \(\mathbf{B}\).

CHECK YOURSELF

Three quick tests

  1. Why does the curl of a gradient vanish identically?
    AnswerIt is a fundamental identity in vector calculus arising from the commutativity of partial derivatives for smooth functions.
  2. If \(\nabla \cdot \mathbf{A} = 0\), what is the name of this gauge?
    AnswerCoulomb gauge.
  3. Does the gauge transformation change the Lorentz force on a charge?
    AnswerNo, because the electric and magnetic fields are invariant under gauge transformations.
PROBLEMS

Problem set for PU-204 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-163 · Threads: force, energy, light