Maxwell’s equations are the foundation of classical electromagnetism. In SI units:
Integral Form:
\oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\mathrm{enc}}{\varepsilon_0} \quad \mathrm{(Gauss's\ Law)}}
∮ S B ⋅ d A = 0 ( G a u s s ′ s L a w f o r M a g n e t i s m ) \oint_S \mathbf{B} \cdot d\mathbf{A} = 0 \quad \mathrm{(Gauss's\ Law\ for\ Magnetism)} ∮ S B ⋅ d A = 0 ( Gaus s ′ s Law for Magnetism )
∮ C E ⋅ d l = − d Φ B d t ( F a r a d a y ′ s L a w ) \oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} \quad \mathrm{(Faraday's\ Law)} ∮ C E ⋅ d l = − d t d Φ B ( Farada y ′ s Law )
∮ C B ⋅ d l = μ 0 I e n c + μ 0 ε 0 d Φ E d t ( A m p e r e − M a x w e l l L a w ) \oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\mathrm{enc} + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \quad \mathrm{(Ampere{-}Maxwell\ Law)}} ∮ C B ⋅ d l = μ 0 I enc + μ 0 ε 0 d t d Φ E ( Ampere − Maxwell Law )
Differential Form:
∇ ⋅ E = ρ ε 0 ( G a u s s ′ s L a w ) \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \quad \mathrm{(Gauss's\ Law)} ∇ ⋅ E = ε 0 ρ ( Gaus s ′ s Law )
∇ ⋅ B = 0 ( G a u s s ′ s L a w f o r M a g n e t i s m ) \nabla \cdot \mathbf{B} = 0 \quad \mathrm{(Gauss's\ Law\ for\ Magnetism)} ∇ ⋅ B = 0 ( Gaus s ′ s Law for Magnetism )
∇ × E = − ∂ B ∂ t ( F a r a d a y ′ s L a w ) \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \quad \mathrm{(Faraday's\ Law)} ∇ × E = − ∂ t ∂ B ( Farada y ′ s Law )
∇ × B = μ 0 J + μ 0 ε 0 ∂ E ∂ t ( A m p e r e − M a x w e l l L a w ) \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} \quad \mathrm{(Ampere{-}Maxwell\ Law)} ∇ × B = μ 0 J + μ 0 ε 0 ∂ t ∂ E ( Ampere − Maxwell Law )
Where ρ \rho ρ J \mathbf{J} J ε 0 \varepsilon_0 ε 0 μ 0 \mu_0 μ 0
Gauss’s Law. Apply the divergence theorem to the integral form:
∮ S E ⋅ d A = ∫ V ( ∇ ⋅ E ) d V = 1 ε 0 ∫ V ρ d V \oint_S \mathbf{E} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{E})\, dV = \frac{1}{\varepsilon_0}\int_V \rho\, dV ∮ S E ⋅ d A = ∫ V ( ∇ ⋅ E ) d V = ε 0 1 ∫ V ρ d V
Since this holds for any volume V V V ∇ ⋅ E = ρ / ε 0 \nabla \cdot \mathbf{E} = \rho / \varepsilon_0 ∇ ⋅ E = ρ / ε 0
Faraday’s Law. Apply Stokes’ theorem:
∮ C E ⋅ d l = ∫ S ( ∇ × E ) ⋅ d A = − ∫ S ∂ B ∂ t ⋅ d A \oint_C \mathbf{E} \cdot d\mathbf{l} = \int_S (\nabla \times \mathbf{E}) \cdot d\mathbf{A} = -\int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A} ∮ C E ⋅ d l = ∫ S ( ∇ × E ) ⋅ d A = − ∫ S ∂ t ∂ B ⋅ d A
Since this holds for any surface S S S ∇ × E = − ∂ B / ∂ t \nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t ∇ × E = − ∂ B / ∂ t
Gauss’s Law for Magnetism. By the divergence theorem:
∮ S B ⋅ d A = ∫ V ( ∇ ⋅ B ) d V = 0 \oint_S \mathbf{B} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{B})\, dV = 0 ∮ S B ⋅ d A = ∫ V ( ∇ ⋅ B ) d V = 0
Since V V V ∇ ⋅ B = 0 \nabla \cdot \mathbf{B} = 0 ∇ ⋅ B = 0
Ampere-Maxwell Law. Apply Stokes’ theorem:
∮ C B ⋅ d l = ∫ S ( ∇ × B ) ⋅ d A = μ 0 ∫ S J ⋅ d A + μ 0 ε 0 d d t ∫ S E ⋅ d A \oint_C \mathbf{B} \cdot d\mathbf{l} = \int_S (\nabla \times \mathbf{B}) \cdot d\mathbf{A} = \mu_0 \int_S \mathbf{J} \cdot d\mathbf{A} + \mu_0 \varepsilon_0 \frac{d}{dt}\int_S \mathbf{E} \cdot d\mathbf{A} ∮ C B ⋅ d l = ∫ S ( ∇ × B ) ⋅ d A = μ 0 ∫ S J ⋅ d A + μ 0 ε 0 d t d ∫ S E ⋅ d A
Since S S S ∇ × B = μ 0 J + μ 0 ε 0 ∂ E / ∂ t \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0\, \partial \mathbf{E}/\partial t ∇ × B = μ 0 J + μ 0 ε 0 ∂ E / ∂ t
Taking the divergence of the Ampere-Maxwell law:
∇ ⋅ ( ∇ × B ) = 0 = μ 0 ∇ ⋅ J + μ 0 ε 0 ∂ ∂ t ( ∇ ⋅ E ) \nabla \cdot (\nabla \times \mathbf{B}) = 0 = \mu_0 \nabla \cdot \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial}{\partial t}(\nabla \cdot \mathbf{E}) ∇ ⋅ ( ∇ × B ) = 0 = μ 0 ∇ ⋅ J + μ 0 ε 0 ∂ t ∂ ( ∇ ⋅ E )
Using Gauss’s law: ∇ ⋅ J + ∂ ρ ∂ t = 0 \nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0 ∇ ⋅ J + ∂ t ∂ ρ = 0
This is the continuity equation , expressing conservation of charge.
At an interface between two linear media (labelled 1 and 2) with surface normal n ^ \hat{\mathbf{n}} n ^
Normal component of D \mathbf{D} D Apply Gauss’s law for D \mathbf{D} D
∮ D ⋅ d A = σ f A ⟹ D 1 n − D 2 n = σ f \oint \mathbf{D} \cdot d\mathbf{A} = \sigma_f A \implies D_{1n} - D_{2n} = \sigma_f ∮ D ⋅ d A = σ f A ⟹ D 1 n − D 2 n = σ f
Tangential component of E \mathbf{E} E Apply Faraday’s law to a rectangular loop Perpendicular to the interface. As the loop height Δ h → 0 \Delta h \to 0 Δ h → 0
∮ E ⋅ d l = 0 ⟹ E 1 t = E 2 t \oint \mathbf{E} \cdot d\mathbf{l} = 0 \implies E_{1t} = E_{2t} ∮ E ⋅ d l = 0 ⟹ E 1 t = E 2 t
In vector form: n ^ × ( E 1 − E 2 ) = 0 \hat{\mathbf{n}} \times (\mathbf{E}_1 - \mathbf{E}_2) = \mathbf{0} n ^ × ( E 1 − E 2 ) = 0
Normal component of B \mathbf{B} B Apply Gauss’s law for B \mathbf{B} B
B 1 n = B 2 n B_{1n} = B_{2n} B 1 n = B 2 n
Tangential component of H \mathbf{H} H Apply Ampere’s law for H \mathbf{H} H
n ^ × ( H 1 − H 2 ) = K f \hat{\mathbf{n}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{K}_f n ^ × ( H 1 − H 2 ) = K f
Where K f \mathbf{K}_f K f
Summary (no free charges or currents, σ f = 0 \sigma_f = 0 σ f = 0 K f = 0 \mathbf{K}_f = \mathbf{0} K f = 0
Field Normal component Tangential component E \mathbf{E} E ε 1 E 1 n = ε 2 E 2 n \varepsilon_1 E_{1n} = \varepsilon_2 E_{2n} ε 1 E 1 n = ε 2 E 2 n E 1 t = E 2 t E_{1t} = E_{2t} E 1 t = E 2 t D \mathbf{D} D D 1 n = D 2 n D_{1n} = D_{2n} D 1 n = D 2 n D 1 t / ε 1 = D 2 t / ε 2 D_{1t}/\varepsilon_1 = D_{2t}/\varepsilon_2 D 1 t / ε 1 = D 2 t / ε 2 B \mathbf{B} B μ 1 B 1 n = μ 2 B 2 n \mu_1 B_{1n} = \mu_2 B_{2n} μ 1 B 1 n = μ 2 B 2 n B 1 t / μ 1 = B 2 t / μ 2 B_{1t}/\mu_1 = B_{2t}/\mu_2 B 1 t / μ 1 = B 2 t / μ 2 H \mathbf{H} H μ 2 H 1 n = μ 1 H 2 n \mu_2 H_{1n} = \mu_1 H_{2n} μ 2 H 1 n = μ 1 H 2 n H 1 t = H 2 t H_{1t} = H_{2t} H 1 t = H 2 t
Problem. Starting from Maxwell’s equations in free space (ρ = 0 \rho = 0 ρ = 0 J = 0 \mathbf{J} = \mathbf{0} J = 0 E \mathbf{E} E B \mathbf{B} B c = 1 / μ 0 ε 0 c = 1/\sqrt{\mu_0 \varepsilon_0} c = 1/ μ 0 ε 0
Solution In free space, Maxwell’s equations reduce to:
∇ ⋅ E = 0 , ∇ ⋅ B = 0 \nabla \cdot \mathbf{E} = 0, \quad \nabla \cdot \mathbf{B} = 0 ∇ ⋅ E = 0 , ∇ ⋅ B = 0
∇ × E = − ∂ B ∂ t , ∇ × B = μ 0 ε 0 ∂ E ∂ t \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} ∇ × E = − ∂ t ∂ B , ∇ × B = μ 0 ε 0 ∂ t ∂ E
Take the curl of Faraday’s law:
∇ × ( ∇ × E ) = − ∂ ∂ t ( ∇ × B ) = − μ 0 ε 0 ∂ 2 E ∂ t 2 \nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} ∇ × ( ∇ × E ) = − ∂ t ∂ ( ∇ × B ) = − μ 0 ε 0 ∂ t 2 ∂ 2 E
Apply the vector identity ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E \nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E ∇ ⋅ E = 0 \nabla \cdot \mathbf{E} = 0 ∇ ⋅ E = 0
− ∇ 2 E = − μ 0 ε 0 ∂ 2 E ∂ t 2 -\nabla^2 \mathbf{E} = -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} − ∇ 2 E = − μ 0 ε 0 ∂ t 2 ∂ 2 E
∇ 2 E = μ 0 ε 0 ∂ 2 E ∂ t 2 \boxed{\nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}} ∇ 2 E = μ 0 ε 0 ∂ t 2 ∂ 2 E
An identical calculation, taking the curl of the Ampere-Maxwell law, yields:
∇ 2 B = μ 0 ε 0 ∂ 2 B ∂ t 2 \boxed{\nabla^2 \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2}} ∇ 2 B = μ 0 ε 0 ∂ t 2 ∂ 2 B
Comparing with the standard wave equation ∇ 2 F = 1 v 2 ∂ 2 F ∂ t 2 \nabla^2 \mathbf{F} = \frac{1}{v^2}\frac{\partial^2 \mathbf{F}}{\partial t^2} ∇ 2 F = v 2 1 ∂ t 2 ∂ 2 F
c = 1 μ 0 ε 0 ≈ 2.998 × 10 8 m / s c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 2.998 \times 10^8\ \mathrm{m}/s c = μ 0 ε 0 1 ≈ 2.998 × 1 0 8 m / s
■ \blacksquare ■