Electromagnetic Potentials

The electromagnetic fields can be expressed in terms of scalar (φ) and vector (A) potentials. Gauge freedom allows transformation between equivalent potential descriptions without changing the physical fields E and B.

Key Concepts

B = ∇×A (vector potential)
E = -∇φ - ∂A/∂t (scalar potential)
Gauge freedom: physics unchanged by φ→φ-∂Λ/∂t, A→A+∇Λ
Coulomb gauge: ∇·A = 0
Lorenz gauge: ∇·A + (1/c²)∂φ/∂t = 0

Key Equations

Fields from potentials

\mathbf{E} = -\nabla\phi - \frac{\partial\mathbf{A}}{\partial t},\quad \mathbf{B} = \nabla\times\mathbf{A}

Gauge transformation

\phi\to\phi - \frac{\partial\Lambda}{\partial t},\quad \mathbf{A}\to\mathbf{A}+\nabla\Lambda

Lorenz gauge condition

\nabla\cdot\mathbf{A} + \frac{1}{c^2}\frac{\partial\phi}{\partial t} = 0

Wave equations (Lorenz)

\Box^2\phi = -\rho/\varepsilon_0,\quad \Box^2\mathbf{A} = -\mu_0\mathbf{J}

Worked Example

Example Problem

Problem

A static uniform E-field E₀ẑ can be written as φ = -E₀z, A=0. After gauge transformation with Λ=E₀zt, find the new potentials.

Solution

φ' = φ - ∂Λ/∂t = -E₀z - E₀z = -2E₀z? No: ∂Λ/∂t = E₀z → φ' = -E₀z - E₀z? Let me redo: Λ=E₀zt → ∂Λ/∂t = E₀z, ∇Λ = E₀t ẑ. So φ' = -E₀z - E₀z = -2E₀z, A' = E₀t ẑ. Check E': E' = -∇φ' - ∂A'/∂t = 2E₀ẑ - E₀ẑ = E₀ẑ. Correct.

Practice

Exercises

7 problems

1 of 7

For a static charge Q=1.0 nC at origin, the scalar potential at r=1.0 m is φ=Q/(4πε₀r). Find φ in V.

Your answer

2 of 7

For a long solenoid with B=B₀ẑ inside (radius R), the vector potential outside (r>R) is A_φ = B₀R²/(2r). For R=2 cm, B₀=0.1 T, r=4 cm, find A_φ in mT·m.

Your answer

mT·m

3 of 7

The Lorenz gauge condition: ∇·A + (1/c²)∂φ/∂t = 0. For a plane wave φ=φ₀cos(kx-ωt), A=Aₓcos(kx-ωt)x̂, find the Lorenz condition relating kAₓ and ωφ₀/c².

Unlock Exercise 3