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Potentials and Gauge Transformations

6.1 Scalar and Vector Potentials

Section titled “6.1 Scalar and Vector Potentials”

We can express the fields in terms of potentials:

E=−∇V−∂A∂t,B=∇×A\mathbf{E} = -\nabla V - \frac{\partial \mathbf{A}}{\partial t}, \quad \mathbf{B} = \nabla \times \mathbf{A}

In electrostatics, A=0\mathbf{A} = \mathbf{0} and E=−∇V\mathbf{E} = -\nabla V.

6.2 Gauge Transformations

Section titled “6.2 Gauge Transformations”

The potentials are not unique. The transformation

V"=V−∂χ∂t,A′=A+∇χV" = V - \frac{\partial \chi}{\partial t}, \quad \mathbf{A}' = \mathbf{A} + \nabla \chi

For any scalar function χ(r,t)\chi(\mathbf{r}, t) leaves E\mathbf{E} and B\mathbf{B} unchanged. This is a gauge transformation.

Common gauges:

  • Coulomb gauge: ∇⋅A=0\nabla \cdot \mathbf{A} = 0. Useful in magnetostatics.
  • Lorenz gauge: ∇⋅A+μ0ε0∂V∂t=0\nabla \cdot \mathbf{A} + \mu_0 \varepsilon_0 \frac{\partial V}{\partial t} = 0. Simplifies the wave equations for VV and A\mathbf{A}:

∇2V−μ0ε0∂2V∂t2=−ρε0\nabla^2 V - \mu_0 \varepsilon_0 \frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\varepsilon_0}

∇2A−μ0ε0∂2A∂t2=−μ0J\nabla^2 \mathbf{A} - \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}

:::caution Common Pitfall The Lorenz gauge (with one “r”) is named after Ludvig Lorenz, not Hendrik Lorentz. It is frequently Misspelled “Lorentz gauge.” The two are different people, and the correct spelling is “Lorenz gauge.”

6.3 Derivation of the Lorenz Gauge Condition

Section titled “6.3 Derivation of the Lorenz Gauge Condition”

Starting from the definitions E=−∇V−∂A/∂t\mathbf{E} = -\nabla V - \partial\mathbf{A}/\partial t and B=∇×A\mathbf{B} = \nabla \times \mathbf{A}Substitute into Gauss’s law:

∇⋅E=−∇2V−∂∂t(∇⋅A)=ρε0\nabla \cdot \mathbf{E} = -\nabla^2 V - \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = \frac{\rho}{\varepsilon_0}

\nabla^2 V + \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = -\frac{\rho}{\varepsilon_0} \tag{6.1}

Substitute into the Ampere-Maxwell law:

∇×B=∇(∇⋅A)−∇2A=μ0J+μ0ε0∂∂t(−∇V−∂A∂t)\nabla \times \mathbf{B} = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} = \mu_0\mathbf{J} + \mu_0\varepsilon_0\frac{\partial}{\partial t}(-\nabla V - \frac{\partial\mathbf{A}}{\partial t})

\nabla^2 \mathbf{A} - \mu_0\varepsilon_0\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{J} + \nabla\!\left(\nabla \cdot \mathbf{A} + \mu_0\varepsilon_0\frac{\partial V}{\partial t}\right) \tag{6.2}

Equations (6.1) and (6.2) are coupled through the term ∇⋅A+μ0ε0 ∂V/∂t\nabla \cdot \mathbf{A} + \mu_0\varepsilon_0\,\partial V/\partial t.

The Lorenz gauge sets this term to zero:

∇⋅A+μ0ε0∂V∂t=0\nabla \cdot \mathbf{A} + \mu_0\varepsilon_0\frac{\partial V}{\partial t} = 0

This is always achievable. If the current potentials do not satisfy this condition, perform a Gauge transformation with χ\chi satisfying:

∇2χ−μ0ε0∂2χ∂t2=−(∇⋅A+μ0ε0∂V∂t)\nabla^2\chi - \mu_0\varepsilon_0\frac{\partial^2\chi}{\partial t^2} = -\left(\nabla \cdot \mathbf{A} + \mu_0\varepsilon_0\frac{\partial V}{\partial t}\right)

In the Lorenz gauge, (6.1) and (6.2) decouple into inhomogeneous wave equations:

∇2V−μ0ε0∂2V∂t2=−ρε0\nabla^2 V - \mu_0\varepsilon_0\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\varepsilon_0}

∇2A−μ0ε0∂2A∂t2=−μ0J\nabla^2 \mathbf{A} - \mu_0\varepsilon_0\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{J}

Both VV and A\mathbf{A} satisfy wave equations with sources ρ/ε0\rho/\varepsilon_0 and μ0J\mu_0\mathbf{J} And both propagate at speed cc. The Lorenz gauge makes manifest the relativistic covariance Of the theory (Section 7).

6.4 Retarded Potentials

Section titled “6.4 Retarded Potentials”

The inhomogeneous wave equations in the Lorenz gauge have causal solutions --- the potentials At (r,t)(\mathbf{r}, t) depend on the sources at the retarded time tr=t−R/ct_r = t - R/c where R=∣r−r′∣R = \lvert\mathbf{r} - \mathbf{r}'\rvert:

V(r,t)=14πε0∫ρ(r′,tr)R d3r′V(\mathbf{r}, t) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}', t_r)}{R}\,d^3\mathbf{r}'

A(r,t)=μ04π∫J(r′,tr)R d3r′\mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}', t_r)}{R}\,d^3\mathbf{r}'

Physical interpretation. Information about changes in the source travels outward at speed cc. The field at point r\mathbf{r} and time tt is determined by the source configuration at the Earlier time trt_r when a light signal would have left r′\mathbf{r}' to arrive at r\mathbf{r} at Time tt.

Verification that retarded potentials satisfy the wave equation

We verify for VV; the argument for A\mathbf{A} is identical. Define

V(r,t)=14πε0∫[ρ]R d3r′V(\mathbf{r}, t) = \frac{1}{4\pi\varepsilon_0}\int \frac{[\rho]}{R}\,d^3\mathbf{r}'

Where [ρ]=ρ(r′,t−R/c)[\rho] = \rho(\mathbf{r}', t - R/c). Applying the d’Alembertian operator □2=∇2−1c2∂2∂t2\square^2 = \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} and carefully Differentiating under the integral (the derivatives act on both the explicit 1/R1/R and the Implicit RR in [ρ][\rho] through trt_r):

□2V=14πε0∫[□2[ρ]R]d3r′\square^2 V = \frac{1}{4\pi\varepsilon_0}\int \left[\frac{\square^2[\rho]}{R}\right] d^3\mathbf{r}'

The key identity is □2(f(tr)/R)=−4πf(t) δ3(r−r′)\square^2(f(t_r)/R) = -4\pi f(t)\,\delta^3(\mathbf{r} - \mathbf{r}')Which Follows from the fact that ∇2(1/R)=−4πδ3(r−r′)\nabla^2(1/R) = -4\pi\delta^3(\mathbf{r} - \mathbf{r}') and that the Time derivatives cancel the 1/R1/R propagation effects. Therefore:

□2V=−1ε0∫ρ(r′,t) δ3(r−r′) d3r′=−ρ(r,t)ε0\square^2 V = -\frac{1}{\varepsilon_0}\int \rho(\mathbf{r}', t)\,\delta^3(\mathbf{r} - \mathbf{r}')\,d^3\mathbf{r}' = -\frac{\rho(\mathbf{r}, t)}{\varepsilon_0}

This confirms that VV satisfies the wave equation. ■\blacksquare

6.5 Lienard-Wiechert Potentials

Section titled “6.5 Lienard-Wiechert Potentials”

For a moving point charge qq following trajectory rs(t)\mathbf{r}_s(t)The retarded potentials Cannot be evaluated naively because the retarded time trt_r satisfies a non-trivial equation:

c(t−tr)=∣r−rs(tr)∣c(t - t_r) = \lvert\mathbf{r} - \mathbf{r}_s(t_r)\rvert

The Lienard-Wiechert potentials are the exact solutions:

V(r,t)=q4πε01(κR)∣tr,A(r,t)=qv4πε0c21(κR)∣trV(\mathbf{r}, t) = \frac{q}{4\pi\varepsilon_0}\frac{1}{(\kappa R)}\bigg|_{t_r}, \quad \mathbf{A}(\mathbf{r}, t) = \frac{q\mathbf{v}}{4\pi\varepsilon_0 c^2}\frac{1}{(\kappa R)}\bigg|_{t_r}

Where R=r−rs(tr)\mathbf{R} = \mathbf{r} - \mathbf{r}_s(t_r), R=∣R∣R = \lvert\mathbf{R}\rvert v=r˙s(tr)\mathbf{v} = \dot{\mathbf{r}}_s(t_r)And κ=1−R^⋅v/c\kappa = 1 - \hat{\mathbf{R}} \cdot \mathbf{v}/c.

The factor κ\kappa corrects for the Doppler effect: when the charge moves toward the Observation point, the radiation is compressed (higher density of field lines).

Fields of a moving charge. The electric field splits into two parts:

E=Evel+Eacc\mathbf{E} = \mathbf{E}_{\mathrm{vel} + \mathbf{E}_{\mathrm{acc}}}

The velocity field (Coulomb-like, falls off as 1/R21/R^2):

Evel=q4πε0(1−β2)(R^−β)κ3R2∣tr\mathbf{E}_{\mathrm{vel} = \frac{q}{4\pi\varepsilon_0}\frac{(1-\beta^2)(\hat{\mathbf{R}} - \boldsymbol{\beta})}{\kappa^3 R^2}\bigg|_{t_r}}

Where β=v/c\boldsymbol{\beta} = \mathbf{v}/c.

The acceleration field (radiation, falls off as 1/R1/R):

Eacc=q4πε0cR^×[(R^−β)×β˙]κ3R∣tr\mathbf{E}_{\mathrm{acc} = \frac{q}{4\pi\varepsilon_0 c}\frac{\hat{\mathbf{R}} \times [(\hat{\mathbf{R}} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}]}{\kappa^3 R}\bigg|_{t_r}}

Only the acceleration field contributes to radiation at large distances. The magnetic field Is always:

B=1cR^×E\mathbf{B} = \frac{1}{c}\hat{\mathbf{R}} \times \mathbf{E}

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