Many potentials, same fields

Science · Electricity & Magnetism · ages 22-24

Name ______________________ Date ____________

Use SI units and L=∇²-(1/c²)∂t². Which condition and resulting sourced equations describe Lorenz gauge?
- div A - (1/c²)∂t phi = 0; L phi = -rho/eps0; L A = -mu0 j
- div A + (1/c²)∂t phi = 0; L phi = +rho/eps0; L A = -mu0 j
- div A + (1/c²)∂t phi = 0; L phi = -rho/eps0; L A = -mu0 j
Under a matched gauge transformation in one observer frame, which quantities must remain unchanged?
- The electric and magnetic fields
- The scalar and vector potentials
- The condition used to select a gauge
A physical answer must always be written as E or B, never as a measured force or an energy transfer.
Circle one: True False
You want Gauss's law to give a scalar Poisson equation for general time-varying sources. Which gauge condition and resulting equation provide this simplification in SI units?
- div A = 0; ∇²phi = +rho/eps0
- div A + (1/c²)∂t phi = 0; ∇²phi = -rho/eps0
- div A = 0; ∇²phi = -rho/eps0
Start with A=0 and phi=0. Choose chi=axt, where a is a nonzero constant and xhat is the unit vector along +x. The matched shift gives A'=at xhat and phi'=-ax. What are the new fields?
- E' = a xhat and B' = 0
- E' = 0 and B' = 0
- E' = 2a xhat and B' = 0
Let L=∇²-(1/c²)∂t². Which statement about residual freedom in Lorenz gauge is correct?
- L chi = 0 permits a residual gauge shift
- Only chi = 0 can preserve the Lorenz condition
- Every smooth chi preserves the Lorenz condition
Two gauge calculations give identical E and B. Is this agreement alone enough to prove they solve the stated source problem?
- Yes; gauge agreement proves every required equation
- No; source equations and boundary data still need checks
- Yes; two gauges also fix every initial condition
With causal initial and boundary data, a student claims the instantaneous-looking Coulomb-gauge scalar potential makes a localized source change affect distant E instantly. What is missing?
- Coulomb gauge cannot be used with time-varying sources
- The term -∂t A; the complete field response is causal
- A gauge change alters measured signal arrival times

Answer key

For grown-ups. Fold this page away before handing over the rest.

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div A + (1/c²)∂t phi = 0; L phi = -rho/eps0; L A = -mu0 j · Lorenz gauge imposes div A+(1/c²)∂t phi=0. With the stated L, the scalar and vector potentials obey sourced wave equations with sources -rho/eps0 and -mu0 j.
The electric and magnetic fields · A matched gauge shift leaves E and B unchanged. Potentials may change, and a general shift need not preserve a chosen gauge condition.
False · A measured force or an energy transfer can also be a physical answer. The prediction must be gauge independent; it need not be written as a field.
div A = 0; ∇²phi = -rho/eps0 · Choose the Coulomb condition div A=0. It removes ∂t(div A) from div E, so Gauss's law becomes the scalar Poisson equation ∇²phi=-rho/eps0, including for time-varying sources.
E' = 0 and B' = 0 · -grad phi'=a xhat and -∂t A'=-a xhat cancel, so E'=0. A' is spatially uniform, so its curl and B' are zero.
L chi = 0 permits a residual gauge shift · The Lorenz expression changes by L chi. A shift with L chi=0 preserves that condition; initial and boundary data may restrict it further.
No; source equations and boundary data still need checks · Agreement alone does not prove correctness. Both calculations could share an error; the fields must satisfy Maxwell's equations and the given initial and boundary conditions.
The term -∂t A; the complete field response is causal · E includes both -grad phi and -∂t A. With causal data, their combined response does not transmit localized source changes faster than c.