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advanced29 min studyLesson 243 in Physics

Vector potential and gauge choice

Construct different potentials for the same magnetic field and preserve measurable quantities under a consistent gauge transformation.

Before you begin
On this page

The question for today

What you’ll understand

  • Recover a magnetic field by taking the curl of its potential.
  • Relate two vector potentials using an explicit gauge function.
  • Distinguish a gauge-dependent local potential from gauge-invariant fields and closed-loop circulation.

A vector potential offers a convenient representation of a divergence-free magnetic field, but the representation has redundancy. Different arrows drawn for the potential can describe exactly the same physical magnetic field. Gauge choice removes some of this freedom to simplify a calculation. The practical skill is to compare curls, boundary conditions, and closed-loop quantities instead of interpreting every change in the potential's local components as a new magnetic experiment.

A curl representation with room to choose

Writing the magnetic field as a curl automatically enforces zero divergence wherever the potential is sufficiently smooth. The potential has units of tesla metre, equivalent to weber per metre. Adding the gradient of a smooth scalar changes its components but leaves its curl unchanged. This local freedom is the origin of gauge redundancy. Global constructions need care in domains containing excluded flux tubes or other topological obstructions; a locally defined gradient need not supply one globally single-valued gauge function.

For a uniform field along z, one possible potential treats x and y symmetrically, while another has only a y component. Both give the same z curl. A uniform physical field therefore does not require a uniform potential. Conversely, a large local potential does not by itself establish a large local field. Derivatives and gauge-invariant combinations determine the observable classical field, so comparing potential magnitudes across arbitrary gauges is not a physical diagnostic.

B=∇×A,A′=A+∇χ,∇×∇χ=0\mathbf B=\nabla\times\mathbf A,\qquad \mathbf A'=\mathbf A+\nabla\chi,\qquad \nabla\times\nabla\chi=\mathbf0
A is in T m and the smooth gauge function χ is in T m². These relations express local classical gauge freedom; global claims require compatible domains, regularity, and single-valued gauge functions.↔ Scroll wide equations horizontally.

Gauge fixing simplifies an equation without adding a force

In magnetostatics the Coulomb condition sets the divergence of the potential to zero. Combining it with Ampère's law gives a vector Poisson equation, component by component in Cartesian coordinates. With localized steady currents and the appropriate decay condition, the usual current integral supplies a solution. The gauge condition does not force the physical magnetic field to vanish. It specifies a representative potential whose derivatives are easier to manipulate.

Residual gauge functions satisfying Laplace's equation can preserve the Coulomb condition, so boundary information is still needed to discuss uniqueness. In time-dependent problems the scalar potential must transform as well. Changing the vector potential while leaving an incompatible scalar potential fixed can change the electric field and is then no longer a gauge transformation of the same electromagnetic experiment. The cancellation of the two derivative contributions is the direct check.

E=−∇ϕ−∂tA,ϕ′=ϕ−∂tχ,∇⋅A=0 ⟹ ∇2A=−μ0J (magnetostatics)\mathbf E=-\nabla\phi-\partial_t\mathbf A,\qquad \phi'=\phi-\partial_t\chi,\qquad \nabla\cdot\mathbf A=0\ \Longrightarrow\ \nabla^2\mathbf A=-\mu_0\mathbf J\ \text{(magnetostatics)}
φ is in volts, t is in seconds, and J is in A/m². The simultaneous transformations preserve E and B. The final Poisson equation additionally assumes vacuum steady currents and Coulomb gauge; it is not the general radiating-field equation.↔ Scroll wide equations horizontally.

Closed-loop circulation survives the redundancy

For a surface on which the potential is smooth, Stokes' theorem equates its boundary circulation to magnetic flux. Adding a globally single-valued gradient contributes zero around that boundary. The resulting circulation has units of webers and is independent of the particular spanning surface when the magnetic field has no monopole sources and the surfaces belong to an admissible domain. An open-path integral lacks this gauge invariance unless combined with appropriate endpoint information.

This distinction also prevents an overstatement about the potential being meaningless. Its pointwise components depend on gauge, but gauge-invariant phase and flux information built from potentials can be physically significant. Here the setting remains classical: the calculation checks flux and fields without claiming to simulate a quantum interference experiment. Choose a gauge to exploit the geometry, then verify a measurable field or circulation before attaching physical meaning to a plotted potential.

Work it through

Two descriptions of one uniform field

For B=0.40z^ T\mathbf B=0.40\hat{\mathbf z}\,\mathrm T, compare As=(−By/2,Bx/2,0)\mathbf A_s=(-By/2,Bx/2,0) and Al=(0,Bx,0)\mathbf A_l=(0,Bx,0) at x=0.30 m, y=0.20 m. Find a gauge function connecting them and the flux through a 0.30 m by 0.20 m rectangle in the xy plane.

  1. Evaluate the representatives: As=(−0.040,0.060,0) T m\mathbf A_s=(-0.040,0.060,0)\,\mathrm{T\,m} and Al=(0,0.120,0) T m\mathbf A_l=(0,0.120,0)\,\mathrm{T\,m}. The unequal components do not yet establish unequal magnetic fields; their derivatives must be compared.

  2. Their difference is (By/2,Bx/2,0)=∇(Bxy/2)(By/2,Bx/2,0)=\nabla(Bxy/2). Thus χ=Bxy/2\chi=Bxy/2 connects them, and its value at this point is 0.012 T m². Taking either curl gives the same +0.40 T z component.

  3. The oriented area is 0.30(0.20)=0.060 m20.30(0.20)=0.060\,\mathrm{m^2} along +z. Flux and counterclockwise potential circulation are therefore BA=0.024 WbB A=0.024\,\mathrm{Wb} in either gauge; the closed-loop integral of the added gradient is zero.

The result

The local potentials differ by the gradient of Bxy/2, while both describe the same +0.40 T field and the same +0.024 Wb rectangle flux.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

A curl-free addition to an illustrative field

Reset and set both controls to Source strength q=0 and Rotation Ω=1. For a fresh comparison, Reset and set both controls to Source strength q=1 and Rotation Ω=1. The added softened radial field is the gradient of q ln(r²+ε)/2, so predict a changed arrow pattern with the same planar curl 2Ω. This is an analogy for adding a gradient, not a magnetic-field computation.

A curl-free addition to an illustrative fieldThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA field picture can change after adding a gradient while its curl remains fixed. The softened planar model is not an electromagnetic potential solver and supplies no gauge-invariant flux readout.

Pause & reason

Check your understanding

Which simultaneous change preserves both electric and magnetic fields for a smooth gauge function χ?

Your turn

Try it yourself

A uniform 0.30 T field along +z is represented by A1=(−By,0,0)\mathbf A_1=(-By,0,0) or A2=(0,Bx,0)\mathbf A_2=(0,Bx,0). Find χ with A2=A1+∇χ\mathbf A_2=\mathbf A_1+\nabla\chi and compare the two potentials at x=0.20 m, y=−0.10 m.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Subtract the two vectors first. A scalar proportional to xy has derivatives proportional to y and x in the required positions.

Compare with the solution

The difference is (By,Bx,0)(By,Bx,0), so χ=Bxy\chi=Bxy works and has units T m². At the specified point its gradient is (−0.030,0.060,0) T m. The first potential is (0.030,0,0) T m and the second is (0,0.060,0) T m. Their curls both give +0.30 T along z because ∂xAy−∂yAx\partial_x A_y-\partial_y A_x equals B in each representation. The negative value of χ here has no effect on that physical field.

Keep the idea

Take it with you

  • Local components of the vector potential depend on gauge while B does not.
  • A time-dependent gauge transformation must change both scalar and vector potentials consistently.
  • Closed-loop potential circulation equals magnetic flux under the domain assumptions of Stokes' theorem.

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