10.02.02 · em-sr / magnetostatics

Magnetic vector potential, gauge freedom, and the Coulomb gauge

shipped3 tiersLean: none

Anchor (Master): Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 5-6; Cohen-Tannoudji, Dupont-Roc, & Grynberg, Photons and Atoms (1997), Ch. I

Intuition Beginner

Since the magnetic field $B$ has no sources or sinks (no magnetic monopoles), we can always write $B$ as the curl of another vector field:

B = curl (A)

The vector field $A$ is called the magnetic vector potential. It is not unique: adding the gradient (slope) of any scalar function $λ$ to $A$ gives the same $B$ , because the curl of a gradient is always zero: $curl (A + grad λ) = curl (A)$ . This freedom to change $A$ without changing $B$ is called gauge freedom, and the transformation $A \to A + grad λ$ is a gauge transformation.

The vector potential is useful because it simplifies many calculations. Instead of computing $B$ directly from the Biot-Savart law (which involves a cross product), we can compute $A$ from a simpler integral and then take the curl. For a steady current $I$ in a wire loop, the vector potential is:

A (r) = \frac{μ _{0} I}{4 π} \oint \frac{d l ^{'}}{∣ r - r ^{'} ∣}

This is simpler than the Biot-Savart law because the integrand is a vector divided by a scalar, with no cross product.

Visual Beginner

Quantity	Definition	Units
Magnetic field $B$	curl of $A$	Tesla (T)
Vector potential $A$	Potential for $B$	T $\cdot$ m
Gauge function $λ$	$A \to A + grad λ$	T $\cdot$ m
Coulomb gauge	divergence of $A$ equals zero	(dimensionless condition)

Worked example Beginner

Compute $A$ for an infinite straight wire carrying current $I$ in the $z$ -direction.

By symmetry, $A$ points in the $z$ -direction and depends only on the distance $s$ from the wire: $A = A_{z} (s) \hat{z}$ . The integral formula gives $A_{z} (s) = - \frac{μ _{0} I}{2 π} ln (s / s_{0})$ where $s_{0}$ is a reference distance.

Now take the curl: $B = curl (A)$ . In cylindrical coordinates, $B = - \frac{d A _{z}}{d s} \hat{ϕ} = \frac{μ _{0} I}{2 π s} \hat{ϕ}$ . This is the familiar result from Ampere's law.

Notice that $A$ is not unique. We can add any gradient $grad λ$ to $A$ without changing $B$ . For example, adding $- μ_{0} I z / (2 π s_{0}) \hat{z}$ would shift the reference point. This is gauge freedom.

Check your understanding Beginner

Formal definition Intermediate+

The magnetic vector potential $A$ is defined by the relation $B = \nabla \times A$ , which is guaranteed to exist by the divergence-free condition $\nabla \cdot B = 0$ and the Helmholtz decomposition theorem.

Gauge transformations. For any smooth scalar function $λ (r, t)$ :

A \to A^{'} = A + \nabla λ, V \to V^{'} = V - \frac{\partial λ}{\partial t}

leave the physical fields $E = - \nabla V - \partial A / \partial t$ and $B = \nabla \times A$ unchanged.

Coulomb gauge. The choice $\nabla \cdot A = 0$ . In this gauge, the magnetostatic equation becomes:

\nabla^{2} A = - μ_{0} J

which is three independent Poisson equations (one for each component of $A$ ). The solution is:

A (r) = \frac{μ _{0}}{4 π} \int \frac{J ( r ^{'} )}{∣ r - r ^{'} ∣} d^{3} r^{'}

Lorenz gauge. The choice $\nabla \cdot A + μ_{0} ϵ_{0} \partial V / \partial t = 0$ . This gauge is Lorentz-invariant and decouples the equations for $V$ and $A$ into wave equations:

□ V = - ρ / ϵ_{0}, □ A = - μ_{0} J

where $□ = μ_{0} ϵ_{0} \partial^{2} / \partial t^{2} - \nabla^{2}$ is the d'Alembertian.

In the language of differential forms. $B$ is a 2-form and $A$ is a 1-form with $B = d A$ . The gauge transformation is $A \to A + d λ$ . The Coulomb gauge corresponds to the codifferential condition $δ A = 0$ (where $δ = \astd *$ is the Hodge codifferential).

Key derivation Intermediate+

Derivation (The Coulomb gauge: existence, uniqueness, and the Poisson equation).

Theorem. For any vector field $A$ , there exists a gauge transformation $A \to A + \nabla λ$ such that $\nabla \cdot (A + \nabla λ) = 0$ . The function $λ$ satisfies $\nabla^{2} λ = - \nabla \cdot A$ .

Proof. We seek $λ$ such that $\nabla \cdot (A + \nabla λ) = 0$ , i.e., $\nabla^{2} λ = - \nabla \cdot A$ . This is a Poisson equation with source $- \nabla \cdot A$ . The solution is:

λ (r) = \frac{1}{4 π} \int \frac{\nabla ^{'} \cdot A ( r ^{'} )}{∣ r - r ^{'} ∣} d^{3} r^{'}

which is well-defined provided $\nabla \cdot A$ falls off faster than $1/ r^{2}$ at infinity. Under this gauge transformation, $\nabla \cdot A^{'} = 0$ . $□$

Derivation (Poisson equation for $A$ in the Coulomb gauge). Starting from Ampere's law $\nabla \times B = μ_{0} J$ and substituting $B = \nabla \times A$ :

\nabla \times (\nabla \times A) = μ_{0} J

Using the vector identity $\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A$ :

\nabla (\nabla \cdot A) - \nabla^{2} A = μ_{0} J

In the Coulomb gauge ( $\nabla \cdot A = 0$ ), the first term vanishes:

\nabla^{2} A = - μ_{0} J

This is the vector Poisson equation, with solution:

A (r) = \frac{μ _{0}}{4 π} \int \frac{J ( r ^{'} )}{∣ r - r ^{'} ∣} d^{3} r^{'}

The Coulomb gauge $\nabla \cdot A = 0$ is satisfied automatically when $\nabla \cdot J = 0$ (steady currents, continuity equation). In this case, $\nabla \cdot A = \frac{μ _{0}}{4 π} \int \nabla \cdot \frac{J ( r ^{'} )}{∣ r - r ^{'} ∣} d^{3} r^{'}$ . Integrating by parts and using $\nabla \cdot J = 0$ plus the boundary condition $J \to 0$ at infinity gives $\nabla \cdot A = 0$ .

Bridge. The vector potential and gauge freedom are the 3D precursors of the 4-potential $A^{μ} = (V / c, A)$ and the gauge transformation $A^{μ} \to A^{μ} + \partial^{μ} λ$ that appears in the covariant formulation 10.06.01. The foundational insight is that gauge freedom is not a mere mathematical convenience but a deep structural property of electromagnetism: the physical fields ( $E$ , $B$ ) are gauge-invariant, while the potentials ( $V$ , $A$ ) are gauge-dependent. The choice of gauge (Coulomb for statics, Lorenz for radiation) selects the most convenient representation. The central message is that in the Coulomb gauge, the vector potential satisfies a simple Poisson equation, making it as easy to compute as the scalar potential in electrostatics. This simplicity is why the Coulomb gauge is preferred for magnetostatic problems. The Lorenz gauge, by contrast, makes the Lorentz invariance manifest and decouples the equations for $V$ and $A$ into wave equations, which is essential for radiation problems 10.07.02 and the retarded potential formalism 10.08.01.

Exercises Intermediate+

Exercise 1 (easy, symbolic).

Compute $A$ and $B$ for a circular current loop of radius $R$ carrying current $I$ , on the axis of symmetry (the $z$ -axis).

Hint

By symmetry, $A$ has only a $ϕ$ -component on axis. But on the $z$ -axis, $A_{ϕ} = 0$ by symmetry. Use $A = \frac{μ _{0} I}{4 π} \oint d l^{'} /∣ r - r^{'} ∣$ directly.

Answer

For a point on the $z$ -axis at distance $z$ from the center, $∣ r - r^{'} ∣ = R^{2} + z^{2}$ for all points on the loop. The integral $\oint d l^{'}$ around a closed loop is zero (the tangent vector integrates to zero around a closed curve). So $A = 0$ on axis. Then $B = \nabla \times A = 0$ on axis? No — $A = 0$ only on the $z$ -axis, and the curl depends on the off-axis behaviour. Direct calculation gives $B_{z} = μ_{0} I R^{2} / [2 (R^{2} + z^{2})^{3/2}]$ from the Biot-Savart law. This shows that $A$ must be computed off-axis to get $B$ by taking the curl.

Exercise 2 (easy, symbolic).

Show that the Coulomb gauge $\nabla \cdot A = 0$ is achieved for the vector potential $A = \frac{μ _{0}}{4 π} \int \frac{J ( r ^{'} )}{∣ r - r ^{'} ∣} d^{3} r^{'}$ when $\nabla \cdot J = 0$ .

Hint

Move the divergence inside the integral. $\nabla \cdot \frac{J ( r ^{'} )}{∣ r - r ^{'} ∣} = J (r^{'}) \cdot \nabla \frac{1}{∣ r - r ^{'} ∣}$ . Integrate by parts.

Answer

\nabla \cdot A = \frac{μ _{0}}{4 π} \int \nabla \cdot \frac{J ( r ^{'} )}{∣ r - r ^{'} ∣} d^{3} r^{'} = \frac{μ _{0}}{4 π} \int J (r^{'}) \cdot \nabla \frac{1}{∣ r - r ^{'} ∣} d^{3} r^{'}

Since $\nabla (1/∣ r - r^{'} ∣) = - \nabla^{'} (1/∣ r - r^{'} ∣)$ (the derivative switches from unprimed to primed with a sign change), we get:

\nabla \cdot A = - \frac{μ _{0}}{4 π} \int J (r^{'}) \cdot \nabla^{'} \frac{1}{∣ r - r ^{'} ∣} d^{3} r^{'}

Integrating by parts: $= - \frac{μ _{0}}{4 π} [\oint \frac{J}{∣ r - r ^{'} ∣} \cdot d S^{'} - \int \frac{\nabla ^{'} \cdot J}{∣ r - r ^{'} ∣} d^{3} r^{'}]$ . The surface integral vanishes (J = 0 at infinity) and $\nabla^{'} \cdot J = 0$ for steady currents. So $\nabla \cdot A = 0$ .

Exercise 4 (medium, symbolic).

Show that in the Lorenz gauge ( $\nabla \cdot A + \frac{1}{c ^{2}} \frac{\partial V}{\partial t} = 0$ ), both $V$ and $A$ satisfy the wave equation with sources.

Hint

Start from Maxwell's equations in terms of potentials. $E = - \nabla V - \partial A / \partial t$ , $B = \nabla \times A$ . Substitute into Gauss's law $\nabla \cdot E = ρ / ϵ_{0}$ and Ampere-Maxwell $\nabla \times B = μ_{0} J + μ_{0} ϵ_{0} \partial E / \partial t$ .

Answer

From Gauss's law: $\nabla \cdot (- \nabla V - \partial A / \partial t) = ρ / ϵ_{0}$ , giving $- \nabla^{2} V - \frac{\partial}{\partial t} (\nabla \cdot A) = ρ / ϵ_{0}$ . Using Lorenz gauge $\nabla \cdot A = - \frac{1}{c ^{2}} \frac{\partial V}{\partial t}$ : $- \nabla^{2} V + \frac{1}{c ^{2}} \frac{\partial ^{2} V}{\partial t ^{2}} = ρ / ϵ_{0}$ , i.e., $□ V = ρ / ϵ_{0}$ .

From Ampere-Maxwell: $\nabla \times (\nabla \times A) = μ_{0} J + μ_{0} ϵ_{0} \frac{\partial}{\partial t} (- \nabla V - \partial A / \partial t)$ . Expanding the LHS: $\nabla (\nabla \cdot A) - \nabla^{2} A$ . Substituting Lorenz gauge: $- \frac{1}{c ^{2}} \nabla \frac{\partial V}{\partial t} - \nabla^{2} A = μ_{0} J - \frac{1}{c ^{2}} \nabla \frac{\partial V}{\partial t} - \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}}$ . The gradient terms cancel, giving $- \nabla^{2} A + \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}} = μ_{0} J$ , i.e., $□ A = μ_{0} J$ .

Exercise 5 (hard, symbolic).

The Aharonov-Bohm effect demonstrates that $A$ (not just $B$ ) has physical significance. Describe the setup and explain why it implies that $A$ is not merely a mathematical convenience.

Hint

Electrons pass on both sides of a solenoid (where $B = 0$ but $A \neq = 0$ ). The interference pattern shifts by an amount proportional to the flux, even though the electrons never enter the region of nonzero $B$ .

Answer

Setup. An infinite solenoid carries current, producing a magnetic field $B$ confined entirely inside the solenoid. Outside, $B = 0$ but $A \neq = 0$ : $A_{ϕ} = Φ/ (2 π s)$ where $Φ$ is the magnetic flux through the solenoid and $s$ is the distance from the axis.

Electrons are directed around the solenoid in a double-slit arrangement, passing through the region where $B = 0$ but $A \neq = 0$ . The phase of each electron's wavefunction is shifted by $e \int A \cdot d l /ℏ$ . The two paths acquire a phase difference:

Δ ϕ = \frac{e}{ℏ} \oint A \cdot d l = \frac{e Φ}{ℏ}

by Stokes' theorem (the loop integral of $A$ equals the flux). This shifts the interference pattern.

Significance. The electrons never enter the region of nonzero $B$ , yet their behaviour depends on $A$ (or, more precisely, on the gauge-invariant loop integral $\oint A \cdot d l$ ). This shows that $A$ has direct physical significance in quantum mechanics, not just in the classical $B = \nabla \times A$ sense. The gauge-invariant quantity is the Wilson loop $exp (i e \oint A \cdot d l /ℏ)$ .

Exercise 6 (hard, symbolic).

Show that the magnetic flux through a surface $S$ equals the line integral of $A$ around the boundary: $Φ = \int_{S} B \cdot d a = \oint_{\partial S} A \cdot d l$ (Stokes' theorem). Use this to compute the flux through a circle of radius $s$ around an infinite wire.

Hint

For the wire, $A = - \frac{μ _{0} I}{2 π} ln (s) \hat{z}$ (in one gauge). Compute $\oint A \cdot d l$ around the circle. Or use $B = μ_{0} I / (2 π s) \hat{ϕ}$ directly.

Answer

By Stokes' theorem: $Φ = \int_{S} B \cdot d a = \int_{S} (\nabla \times A) \cdot d a = \oint_{\partial S} A \cdot d l$ .

For the infinite wire: $B = \frac{μ _{0} I}{2 π s} \hat{ϕ}$ . The flux through a circle of radius $s$ (in a plane perpendicular to the wire): $Φ = \int_{0}^{s} \frac{μ _{0} I}{2 π s ^{'}} 2 π s^{'} d s^{'} = μ_{0} I \int_{0}^{s} \frac{d s ^{'}}{s ^{'}}$ . This integral diverges at the lower limit — the flux through a circle around an infinite wire is infinite. This reflects the fact that the total flux of $B$ from an infinite line source is unbounded.

For the line integral approach: $\oint A \cdot d l$ around a circle of radius $s$ . Using $A = - \frac{μ _{0} I}{2 π} ln (s) \hat{z}$ , the line integral around a circle (in the $\hat{ϕ}$ direction) of a purely $z$ -directed vector is zero. This is consistent: the flux through a bounded surface (like a disk) requires $A$ to have an azimuthal component. In a different gauge ( $A = \frac{μ _{0} I}{2 π} ln (s_{0} / s) \hat{ϕ}$ , which is a valid gauge in 2D), the line integral gives $μ_{0} I ln (s_{0} / s) \cdot 2 π / (2 π) = μ_{0} I ln (s_{0} / s)$ , which is gauge-dependent (as expected for an open surface with unbounded flux).

Exercise 7 (medium, symbolic).

Compute $A$ inside and outside an infinite solenoid of radius $R$ carrying current $I$ with $n$ turns per unit length. Verify that $\nabla \times A$ gives the expected $B$ .

Hint

By symmetry, $A = A_{ϕ} (s) \hat{ϕ}$ . Inside: $B = μ_{0} n I \hat{z}$ . Outside: $B = 0$ . Use $Φ = \int B \cdot d a = \oint A \cdot d l$ to find $A_{ϕ}$ .

Answer

Inside ( $s < R$ ): $Φ = B π s^{2} = μ_{0} n I π s^{2}$ . By Stokes: $\oint A \cdot d l = 2 π s A_{ϕ} = μ_{0} n I π s^{2}$ . So $A_{ϕ} = μ_{0} n I s /2$ . Check: $B = \nabla \times A = \frac{1}{s} \frac{\partial}{\partial s} (s A_{ϕ}) \hat{z} = \frac{1}{s} \frac{\partial}{\partial s} (μ_{0} n I s^{2} /2) \hat{z} = μ_{0} n I \hat{z}$ . Correct.

Outside ( $s > R$ ): $Φ = μ_{0} n I π R^{2}$ (total flux through the solenoid). $2 π s A_{ϕ} = μ_{0} n I π R^{2}$ . So $A_{ϕ} = μ_{0} n I R^{2} / (2 s)$ . Check: $B = \frac{1}{s} \frac{\partial}{\partial s} (s \cdot μ_{0} n I R^{2} / (2 s)) \hat{z} = \frac{1}{s} \frac{\partial}{\partial s} (μ_{0} n I R^{2} /2) \hat{z} = 0$ . Correct: $B = 0$ outside the solenoid.

Note: $\nabla \cdot A = 0$ in both regions (Coulomb gauge satisfied). The vector potential is continuous at $s = R$ : $A_{ϕ} (R) = μ_{0} n I R /2$ from both expressions.

Exercise 8 (medium, symbolic).

Explain why the Coulomb gauge is convenient for magnetostatics but inconvenient for radiation problems.

Hint

In the Coulomb gauge, $V$ satisfies the Poisson equation (instantaneous), but $A$ satisfies a complicated equation with $V$ as a source. For radiation, we want retarded solutions.

Answer

Coulomb gauge for statics: $\nabla^{2} A = - μ_{0} J$ (simple Poisson equation) and $\nabla^{2} V = - ρ / ϵ_{0}$ (also Poisson). Both are elliptic equations with instantaneous (non-retarded) solutions. For static or quasi-static problems, this is natural.

Coulomb gauge for time-dependent problems: The equations become $\nabla^{2} V = - ρ / ϵ_{0}$ (still instantaneous — $V$ responds instantly to changes in $ρ$ , which violates causality in appearance though not in the physical fields) and $\nabla^{2} A - \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}} = - μ_{0} J + \frac{1}{c ^{2}} \nabla \frac{\partial V}{\partial t}$ . The second equation couples $A$ to the time derivative of $V$ , making it harder to solve.

Lorenz gauge for radiation: Both $V$ and $A$ satisfy decoupled wave equations $□ V = - ρ / ϵ_{0}$ and $□ A = - μ_{0} J$ . These have retarded solutions that automatically satisfy causality. The Lorenz gauge is Lorentz-invariant, making it natural for relativistic problems 10.06.01.

Lean formalization Intermediate+

Mathlib has differential forms and the exterior derivative, providing the abstract setting for $B = d A$ . The Helmholtz decomposition exists. However, the gauge transformation $A \to A + d λ$ , the Coulomb/Lorenz gauge conditions, and the Green's function solution of the vector Poisson equation are all absent. The physical boundary conditions and the existence/uniqueness theory are also missing. lean_status: none.

Advanced results Master

The Helmholtz decomposition. Any sufficiently smooth vector field $F$ that vanishes at infinity can be uniquely decomposed as $F = - \nabla U + \nabla \times W$ where $U$ is a scalar potential and $W$ is a vector potential. The scalar potential $U$ satisfies $\nabla^{2} U = - \nabla \cdot F$ (Poisson equation for the divergence part), and the vector potential $W$ satisfies $\nabla^{2} W = - \nabla \times F$ (Poisson equation for the curl part, in the Coulomb gauge). This decomposition is the mathematical foundation for the existence of the vector potential.

The Lorenz gauge and Lorentz invariance. The Lorenz gauge condition $\nabla \cdot A + \frac{1}{c ^{2}} \frac{\partial V}{\partial t} = 0$ can be written in covariant form as $\partial_{μ} A^{μ} = 0$ . This is manifestly Lorentz-invariant (a scalar equation), and the gauge transformation $A^{μ} \to A^{μ} + \partial^{μ} λ$ preserves the Lorenz condition if $□ λ = 0$ . The four-potential $A^{μ} = (V / c, A)$ transforms as a four-vector under Lorentz transformations, and the Lorenz gauge condition selects a particular representative from each gauge equivalence class.

Gauge fixing as a constrained optimization. The Coulomb gauge minimises $\int ∣ A ∣^{2} d^{3} r$ over all gauge-equivalent potentials. The Lorenz gauge minimises a similar functional involving the four-potential. This variational characterisation provides an alternative proof of the existence of the gauge fixing.

Synthesis. The vector potential $A$ is the bridge between the magnetic field $B$ (a physical, measurable quantity) and the mathematical framework that simplifies calculations. The foundational insight is that gauge freedom — the ability to change $A$ without changing $B$ — is not a deficiency but a feature: it reflects a redundancy in the description that can be exploited to simplify different problems. The central message is that the Coulomb gauge ( $\nabla \cdot A = 0$ ) makes magnetostatic problems tractable (Poisson equation for each component), while the Lorenz gauge ( $\partial_{μ} A^{μ} = 0$ ) makes radiation and relativistic problems tractable (decoupled wave equations). Putting these together, the vector potential provides the unified description of magnetostatics, electrodynamics, and radiation 10.07.02, and generalises to the four-potential 10.06.01 in the covariant formulation.

Full proof set Master

Proposition (Uniqueness of $A$ in the Coulomb gauge). If $A_{1}$ and $A_{2}$ both satisfy $\nabla \times A = B$ , $\nabla \cdot A = 0$ , and $A \to 0$ at infinity, then $A_{1} = A_{2}$ .

Proof. Let $D = A_{1} - A_{2}$ . Then $\nabla \times D = 0$ and $\nabla \cdot D = 0$ . Since $\nabla \times D = 0$ , there exists a scalar $ϕ$ with $D = \nabla ϕ$ . Then $0 = \nabla \cdot D = \nabla^{2} ϕ$ . Since $ϕ$ satisfies Laplace's equation and $\nabla ϕ \to 0$ at infinity, we must have $ϕ =$ const (the only solution to Laplace's equation that does not grow at infinity and whose gradient vanishes at infinity). Therefore $D = 0$ and $A_{1} = A_{2}$ . $□$

Connections Master

Faraday tensor 10.06.01 is the 4-dimensional formulation of the electromagnetic field; the 4-potential $A^{μ} = (V / c, A)$ is the covariant generalisation of the vector potential.
Biot-Savart/Ampere 10.02.01 is the direct computation of $B$ ; the vector potential provides an alternative route.
Magnetic scalar potential 10.02.05 is an alternative potential that works only in current-free regions; it complements the vector potential.
Radiation 10.07.02 uses the retarded vector potential in the Lorenz gauge to compute radiation fields.
Green's function methods 10.08.01 provide the general framework for solving the Poisson and wave equations for the potentials.
Differential forms 10.04.01 give the coordinate-free formulation: $B = d A$ , gauge transformation $A \to A + d λ$ .

Historical & philosophical context Master

The vector potential was introduced by Franz Ernst Neumann in 1845 and developed by Wilhelm Weber and James Clerk Maxwell. Maxwell's original formulation of electromagnetism (the "20 equations in 20 unknowns" of the 1865 paper) used the vector potential as a fundamental quantity; the field-based formulation (using $E$ and $B$ directly) was popularised by Heaviside and Hertz in the 1880s.

The gauge principle — that the potentials are not uniquely determined by the fields — was initially regarded as a mathematical nuisance. Its physical significance became apparent only with quantum mechanics: the Aharonov-Bohm effect (1959) showed that the vector potential has observable consequences even in regions where $B = 0$ . This established gauge freedom as a fundamental principle of physics, not merely a computational device.

The modern view (developed by Yang and Mills in the 1950s and standard in particle physics) is that gauge invariance is the defining principle of the electromagnetic interaction: the Lagrangian is invariant under local gauge transformations, and this invariance requires the existence of the electromagnetic field. This perspective inverts the historical one: instead of the field implying the potential, the gauge principle implies the field.

Bibliography Master

Maxwell, J. C., "A Dynamical Theory of the Electromagnetic Field," Phil. Trans. R. Soc. 155, 459-512 (1865).
Aharonov, Y. and Bohm, D., "Significance of electromagnetic potentials in the quantum theory," Phys. Rev. 115, 485-491 (1959).
Jackson, J. D., Classical Electrodynamics, 3rd ed. (Wiley, 1999).
Griffiths, D. J., Introduction to Electrodynamics, 4th ed. (Cambridge, 2017).
Cohen-Tannoudji, C., Dupont-Roc, J., and Grynberg, G., Photons and Atoms: Introduction to Quantum Electrodynamics (Wiley, 1997).

Prerequisites

10.02.01
10.01.01

Used in

10.02.05
10.03.02
10.06.01

Tier anchors

beginner: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 5.4
intermediate: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 5.4; Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 5.1-5.4
master: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 5-6; Cohen-Tannoudji, Dupont-Roc, & Grynberg, Photons and Atoms (1997), Ch. I

References

Griffiths — Introduction to Electrodynamics, 4th ed. (Cambridge, 2017) · Ch. 5.4 The vector potential; Ch. 10.1.2 Gauge transformations
Jackson — Classical Electrodynamics, 3rd ed. (Wiley, 1999) · Ch. 5.1-5.4 The vector potential; gauge transformations; Coulomb and Lorenz gauges; magnetostatic boundary-value problems
Cohen-Tannoudji, Dupont-Roc, & Grynberg — Photons and Atoms (Wiley, 1997) · Ch. I.B The potentials; gauge invariance; Coulomb and Lorenz gauges

Estimated time

beginner: 15m
intermediate: 35m
master: 55m