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Gauge Freedom and the Potentials as the Working Variables

Different scalar and vector potentials can give the same electric and magnetic fields. A gauge condition restricts this freedom. Lorenz gauge gives sourced wave equations for the potentials; Coulomb gauge gives a Poisson equation for the scalar potential.

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  • Shows that a gauge transformation leaves the measured fields unchanged
  • Picks a gauge condition and states what it simplifies
  • Writes the sourced wave equation for the potentials in a chosen gauge

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For classical electromagnetism in vacuum with sources, use SI units: E = -grad phi - ∂t A and B = curl A. phi is the scalar potential and A the vector potential; ∂t is a time derivative. Different potentials can describe the same fields. E and B, and measurable predictions made from them, are gauge independent.

For a smooth function chi(x,t), set A' = A + grad chi and phi' = phi - ∂t chi. The prime labels a new potential description in the same frame, not a Lorentz boost. The two shifts must be matched. An identity transformation may leave the potentials unchanged.

A' = A + grad chiphi' = phi - dt chiB' = BE' = E
Here dt chi means ∂t chi. A gauge change leaves field components unchanged in the same observer frame.

The curl of a gradient is zero, so B' = curl(A + grad chi) = B. Also E' = -grad(phi - ∂t chi) - ∂t(A + grad chi) = E: grad ∂t chi cancels ∂t grad chi because smooth mixed derivatives commute. A spatially uniform vector potential has zero curl.

The Lorenz condition is div A + (1/c²)∂t phi = 0. Define L = ∇² - (1/c²)∂t², where ∇² is the spatial Laplacian. Maxwell's equations give L phi = -rho/eps0 and L A = -mu0 j. Here rho and j are charge and current densities; eps0 and mu0 are vacuum constants, with 1/c² = mu0 eps0. These are sourced wave equations.

L = ∇² - ∂t²/c²L phi = -rho/eps0L A = -mu0 j
L is the wave operator. Charge density sources the scalar equation; current density sources the vector equation.

A gauge condition need not select unique potentials. Let G = div A + (1/c²)∂t phi. A matched gauge shift gives G' = G + L chi. Thus L chi = 0 preserves Lorenz gauge. These are residual gauge transformations; initial and boundary conditions can restrict them further.

The Coulomb condition is div A = 0. Since div E = -∇²phi - ∂t(div A), Gauss's law gives ∇²phi = -rho/eps0. This Poisson equation applies to static or time-varying sources. It gives a direct route to phi for static charges, although Lorenz gauge is valid for statics too. With time-varying sources, its scalar solution can look instantaneous.

Scalar: +grad dt chiVector: -grad dt chiSum of changes: 0E' = E
These are changes in the two contributions to E. Neither contribution alone is gauge independent.

For localized source changes with causal initial and boundary data, physical field changes reach distant points no faster than c. In Coulomb gauge, both -grad phi and -∂t A contribute to E. Their sum has the same causal response as in Lorenz gauge. The scalar potential alone is not the full time-dependent electric field.

A physical answer can be a measured force or an energy transfer, as well as E or B. Such predictions must be gauge independent. Potential values may change with gauge; an absolute potential-energy value can also depend on its reference. Matching E and B from two gauges checks their agreement, not correctness. Check Maxwell's equations for the given sources and the initial and boundary conditions too.

Matched potential shifts leave E and B unchanged. Lorenz gauge gives sourced wave equations. Coulomb gauge gives a scalar Poisson equation, not faster physical signals.

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Gauge Freedom and the Potentials as the Working Variables · Science, ages 22 to 24 · LightMySky