Lorentz. It is frequently Misspelled “Lorentz gauge.” The two are different people, and the correct spelling is “Lorenz gauge.”
Starting from the definitions E = − ∇ V − ∂ A / ∂ t \mathbf{E} = -\nabla V - \partial\mathbf{A}/\partial t E = − ∇ V − ∂ A / ∂ t B = ∇ × A \mathbf{B} = \nabla \times \mathbf{A} B = ∇ × A
∇ ⋅ E = − ∇ 2 V − ∂ ∂ t ( ∇ ⋅ A ) = ρ ε 0 \nabla \cdot \mathbf{E} = -\nabla^2 V - \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = \frac{\rho}{\varepsilon_0} ∇ ⋅ E = − ∇ 2 V − ∂ t ∂ ( ∇ ⋅ A ) = ε 0 ρ ∇ 2 V + ∂ ∂ t ( ∇ ⋅ A ) = − ρ ε 0 (6.1) \nabla^2 V + \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = -\frac{\rho}{\varepsilon_0} \tag{6.1} ∇ 2 V + ∂ t ∂ ( ∇ ⋅ A ) = − ε 0 ρ ( 6.1 ) Substitute into the Ampere-Maxwell law:
∇ × B = ∇ ( ∇ ⋅ A ) − ∇ 2 A = μ 0 J + μ 0 ε 0 ∂ ∂ 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}) ∇ × B = ∇ ( ∇ ⋅ A ) − ∇ 2 A = μ 0 J + μ 0 ε 0 ∂ t ∂ ( − ∇ V − ∂ t ∂ A ) ∇ 2 A − μ 0 ε 0 ∂ 2 A ∂ t 2 = − μ 0 J + ∇ ( ∇ ⋅ A + μ 0 ε 0 ∂ V ∂ t ) (6.2) \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} ∇ 2 A − μ 0 ε 0 ∂ t 2 ∂ 2 A = − μ 0 J + ∇ ( ∇ ⋅ A + μ 0 ε 0 ∂ t ∂ V ) ( 6.2 ) Equations (6.1) and (6.2) are coupled through the term ∇ ⋅ A + μ 0 ε 0 ∂ V / ∂ t \nabla \cdot \mathbf{A} + \mu_0\varepsilon_0\,\partial V/\partial t ∇ ⋅ A + μ 0 ε 0 ∂ V / ∂ t
The Lorenz gauge sets this term to zero:
∇ ⋅ A + μ 0 ε 0 ∂ V ∂ t = 0 \nabla \cdot \mathbf{A} + \mu_0\varepsilon_0\frac{\partial V}{\partial t} = 0 ∇ ⋅ A + μ 0 ε 0 ∂ t ∂ V = 0 This is always achievable. If the current potentials do not satisfy this condition, perform a Gauge transformation with χ \chi χ
∇ 2 χ − μ 0 ε 0 ∂ 2 χ ∂ t 2 = − ( ∇ ⋅ A + μ 0 ε 0 ∂ 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) ∇ 2 χ − μ 0 ε 0 ∂ t 2 ∂ 2 χ = − ( ∇ ⋅ A + μ 0 ε 0 ∂ t ∂ V ) In the Lorenz gauge, (6.1) and (6.2) decouple into inhomogeneous wave equations:
∇ 2 V − μ 0 ε 0 ∂ 2 V ∂ t 2 = − ρ ε 0 \nabla^2 V - \mu_0\varepsilon_0\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\varepsilon_0} ∇ 2 V − μ 0 ε 0 ∂ t 2 ∂ 2 V = − ε 0 ρ ∇ 2 A − μ 0 ε 0 ∂ 2 A ∂ t 2 = − μ 0 J \nabla^2 \mathbf{A} - \mu_0\varepsilon_0\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{J} ∇ 2 A − μ 0 ε 0 ∂ t 2 ∂ 2 A = − μ 0 J Both V V V A \mathbf{A} A ρ / ε 0 \rho/\varepsilon_0 ρ / ε 0 μ 0 J \mu_0\mathbf{J} μ 0 J c c c
The inhomogeneous wave equations in the Lorenz gauge have causal solutions --- the potentials At ( r , t ) (\mathbf{r}, t) ( r , t ) retarded time t r = t − R / c t_r = t - R/c t r = t − R / c R = ∣ r − r ′ ∣ R = \lvert\mathbf{r} - \mathbf{r}'\rvert R = ∣ r − r ′ ∣
V ( r , t ) = 1 4 π ε 0 ∫ ρ ( r ′ , t r ) R d 3 r ′ V(\mathbf{r}, t) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}', t_r)}{R}\,d^3\mathbf{r}' V ( r , t ) = 4 π ε 0 1 ∫ R ρ ( r ′ , t r ) d 3 r ′ A ( r , t ) = μ 0 4 π ∫ J ( r ′ , t r ) R d 3 r ′ \mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}', t_r)}{R}\,d^3\mathbf{r}' A ( r , t ) = 4 π μ 0 ∫ R J ( r ′ , t r ) d 3 r ′ Physical interpretation. Information about changes in the source travels outward at speed c c c r \mathbf{r} r t t t t r t_r t r r ′ \mathbf{r}' r ′ r \mathbf{r} r t t t
Verification that retarded potentials satisfy the wave equation We verify for V V V A \mathbf{A} A
V ( r , t ) = 1 4 π ε 0 ∫ [ ρ ] R d 3 r ′ V(\mathbf{r}, t) = \frac{1}{4\pi\varepsilon_0}\int \frac{[\rho]}{R}\,d^3\mathbf{r}' V ( r , t ) = 4 π ε 0 1 ∫ R [ ρ ] d 3 r ′ Where [ ρ ] = ρ ( r ′ , t − R / c ) [\rho] = \rho(\mathbf{r}', t - R/c) [ ρ ] = ρ ( r ′ , t − R / c ) □ 2 = ∇ 2 − 1 c 2 ∂ 2 ∂ t 2 \square^2 = \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} □ 2 = ∇ 2 − c 2 1 ∂ t 2 ∂ 2 1 / R 1/R 1/ R R R R [ ρ ] [\rho] [ ρ ] t r t_r t r
□ 2 V = 1 4 π ε 0 ∫ [ □ 2 [ ρ ] R ] d 3 r ′ \square^2 V = \frac{1}{4\pi\varepsilon_0}\int \left[\frac{\square^2[\rho]}{R}\right] d^3\mathbf{r}' □ 2 V = 4 π ε 0 1 ∫ [ R □ 2 [ ρ ] ] d 3 r ′ The key identity is □ 2 ( f ( t r ) / R ) = − 4 π f ( t ) δ 3 ( r − r ′ ) \square^2(f(t_r)/R) = -4\pi f(t)\,\delta^3(\mathbf{r} - \mathbf{r}') □ 2 ( f ( t r ) / R ) = − 4 π f ( t ) δ 3 ( r − r ′ ) ∇ 2 ( 1 / R ) = − 4 π δ 3 ( r − r ′ ) \nabla^2(1/R) = -4\pi\delta^3(\mathbf{r} - \mathbf{r}') ∇ 2 ( 1/ R ) = − 4 π δ 3 ( r − r ′ ) 1 / R 1/R 1/ R
□ 2 V = − 1 ε 0 ∫ ρ ( r ′ , t ) δ 3 ( r − r ′ ) d 3 r ′ = − ρ ( r , t ) ε 0 \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} □ 2 V = − ε 0 1 ∫ ρ ( r ′ , t ) δ 3 ( r − r ′ ) d 3 r ′ = − ε 0 ρ ( r , t ) This confirms that V V V ■ \blacksquare ■
For a moving point charge q q q r s ( t ) \mathbf{r}_s(t) r s ( t ) t r t_r t r
c ( t − t r ) = ∣ r − r s ( t r ) ∣ c(t - t_r) = \lvert\mathbf{r} - \mathbf{r}_s(t_r)\rvert c ( t − t r ) = ∣ r − r s ( t r )∣ The Lienard-Wiechert potentials are the exact solutions:
V ( r , t ) = q 4 π ε 0 1 ( κ R ) ∣ t r , A ( r , t ) = q v 4 π ε 0 c 2 1 ( κ R ) ∣ t r V(\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} V ( r , t ) = 4 π ε 0 q ( κ R ) 1 t r , A ( r , t ) = 4 π ε 0 c 2 q v ( κ R ) 1 t r Where R = r − r s ( t r ) \mathbf{R} = \mathbf{r} - \mathbf{r}_s(t_r) R = r − r s ( t r ) R = ∣ R ∣ R = \lvert\mathbf{R}\rvert R = ∣ R ∣ v = r ˙ s ( t r ) \mathbf{v} = \dot{\mathbf{r}}_s(t_r) v = r ˙ s ( t r ) κ = 1 − R ^ ⋅ v / c \kappa = 1 - \hat{\mathbf{R}} \cdot \mathbf{v}/c κ = 1 − R ^ ⋅ v / c
The factor κ \kappa κ 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 = E v e l + E a c c \mathbf{E} = \mathbf{E}_{\mathrm{vel} + \mathbf{E}_{\mathrm{acc}}} E = E vel + E acc The velocity field (Coulomb-like, falls off as 1 / R 2 1/R^2 1/ R 2
E v e l = q 4 π ε 0 ( 1 − β 2 ) ( R ^ − β ) κ 3 R 2 ∣ t r \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}} E vel = 4 π ε 0 q κ 3 R 2 ( 1 − β 2 ) ( R ^ − β ) t r Where β = v / c \boldsymbol{\beta} = \mathbf{v}/c β = v / c
The acceleration field (radiation, falls off as 1 / R 1/R 1/ R
E a c c = q 4 π ε 0 c R ^ × [ ( R ^ − β ) × β ˙ ] κ 3 R ∣ t r \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}} E acc = 4 π ε 0 c q κ 3 R R ^ × [( R ^ − β ) × β ˙ ] t r Only the acceleration field contributes to radiation at large distances. The magnetic field Is always:
B = 1 c R ^ × E \mathbf{B} = \frac{1}{c}\hat{\mathbf{R}} \times \mathbf{E} B = c 1 R ^ × E 