Under a gauge transformation, which of the following quantities changes?
AThe electric field E
BThe magnetic field B
CThe vector potential A
DThe current density J
Gauge transformations are defined precisely to leave E and B unchanged — they are the observable fields. The transformation A → A + ∇λ and φ → φ − ∂λ/∂t changes the potentials, but these changes cancel when computing E = −∇φ − ∂A/∂t and B = ∇×A. Source quantities (ρ, J) are physical and do not transform. Only the potentials (A and φ) change under a gauge transformation.
Question 2 Multiple Choice
In the Coulomb gauge, the scalar potential φ satisfies the instantaneous Poisson equation, appearing to change everywhere at once when charge moves. Why doesn't this violate special relativity?
AIt does violate relativity; the Coulomb gauge is only valid for non-relativistic problems
BThe instantaneous propagation is real but the effect is too small to detect at ordinary energies
Cφ alone is not observable; only the combination that produces E and B is physical, and that propagates causally
DIn the Coulomb gauge, the vector potential A cancels the instantaneous term before any observable effect reaches a test charge
The observable quantity is E = −∇φ − ∂A/∂t, not φ alone. The unphysical instantaneous part of φ in the Coulomb gauge is exactly canceled by a corresponding term in ∂A/∂t, so the total electric field propagates causally. This is a crucial conceptual point: gauge potentials are mathematical tools with no direct physical meaning individually — only the combinations that give E and B are observable. The apparent 'action at a distance' in φ is a gauge artifact, not a physical signal.
Question 3 True / False
The Lorentz gauge is preferred over the Coulomb gauge in relativistic treatments because the Lorentz condition preserves its form under Lorentz transformations.
TTrue
FFalse
Answer: True
The Lorentz gauge condition ∇·A + (1/c²)∂φ/∂t = 0 is Lorentz covariant — it holds in all inertial frames related by Lorentz boosts. This makes it natural for relativistic field theory and quantum electrodynamics, where maintaining spacetime symmetry is essential. The Coulomb gauge condition ∇·A = 0 is not Lorentz covariant; boosting to a new frame breaks the condition and requires a re-gauging. The Lorentz gauge manifests the relativistic unity of space and time directly in the potential equations.
Question 4 True / False
The Lorentz gauge condition substantially fixes the gauge — once you impose ∇·A + (1/c²)∂φ/∂t = 0, there is a unique pair (φ, A) describing the physical situation.
TTrue
FFalse
Answer: False
The Lorentz condition constrains gauge freedom but does not eliminate it. There remains residual gauge freedom: you can still perform additional gauge transformations with any scalar function λ that satisfies the wave equation □²λ = 0, since such λ preserves the Lorentz condition while changing the potentials. The Lorentz condition defines a family of gauges, not a single unique one. Complete gauge fixing — selecting a unique representative — requires additional constraints beyond the Lorentz condition.
Question 5 Short Answer
Explain why choosing between the Lorentz gauge and the Coulomb gauge does not change the physical predictions of a problem, even though the equations look very different in each gauge.
Think about your answer, then reveal below.
Model answer: Both gauges describe the same physical fields E and B — the observable quantities that govern all measurable forces and radiation. The potentials φ and A are not directly observable; only ∇×A = B and −∇φ − ∂A/∂t = E are physical. Different gauge choices change the potentials but cannot change these combinations. Like choosing a coordinate system, a gauge choice affects only the algebra, not the physics.
This is the deep meaning of gauge freedom: an entire family of mathematically different (φ, A) pairs all describe exactly the same physics. The Lorentz gauge makes radiation problems algebraically cleaner by decoupling the potential equations symmetrically. The Coulomb gauge makes static charge problems simpler by reducing φ to a Poisson equation. In both cases, computing E and B gives identical results. Gauge choice is a calculational strategy, and the freedom to choose is a consequence of the fact that potentials carry redundant mathematical information.