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Wyatt's Notes — University

Maxwell's Equations

1.1 The Four Equations

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)}}

∮SB⋅dA=0(Gauss′s Law for Magnetism)\oint_S \mathbf{B} \cdot d\mathbf{A} = 0 \quad \mathrm{(Gauss's\ Law\ for\ Magnetism)}

∮CE⋅dl=−dΦBdt(Faraday′s Law)\oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} \quad \mathrm{(Faraday's\ Law)}

∮CB⋅dl=μ0Ienc+μ0ε0dΦEdt(Ampere−Maxwell Law)\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)}}

Differential Form:

∇⋅E=ρε0(Gauss′s Law)\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \quad \mathrm{(Gauss's\ Law)}

∇⋅B=0(Gauss′s Law for Magnetism)\nabla \cdot \mathbf{B} = 0 \quad \mathrm{(Gauss's\ Law\ for\ Magnetism)}

∇×E=−∂B∂t(Faraday′s Law)\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \quad \mathrm{(Faraday's\ Law)}

∇×B=μ0J+μ0ε0∂E∂t(Ampere−Maxwell Law)\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} \quad \mathrm{(Ampere{-}Maxwell\ Law)}

Where ρ\rho is the charge density, J\mathbf{J} is the current density, ε0\varepsilon_0 is the permittivity Of free space, and μ0\mu_0 is the permeability of free space.

1.2 Derivation from Integral to Differential Form

Gauss’s Law. Apply the divergence theorem to the integral form:

∮SE⋅dA=∫V(∇⋅E) dV=1ε0∫Vρ dV\oint_S \mathbf{E} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{E})\, dV = \frac{1}{\varepsilon_0}\int_V \rho\, dV

Since this holds for any volume VV: ∇⋅E=ρ/ε0\nabla \cdot \mathbf{E} = \rho / \varepsilon_0.

Faraday’s Law. Apply Stokes’ theorem:

∮CE⋅dl=∫S(∇×E)⋅dA=−∫S∂B∂t⋅dA\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}

Since this holds for any surface SS: ∇×E=−∂B/∂t\nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t.

Gauss’s Law for Magnetism. By the divergence theorem:

∮SB⋅dA=∫V(∇⋅B) dV=0\oint_S \mathbf{B} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{B})\, dV = 0

Since VV is arbitrary: ∇⋅B=0\nabla \cdot \mathbf{B} = 0. This expresses the absence of magnetic monopoles.

Ampere-Maxwell Law. Apply Stokes’ theorem:

∮CB⋅dl=∫S(∇×B)⋅dA=μ0∫SJ⋅dA+μ0ε0ddt∫SE⋅dA\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}

Since SS is arbitrary: ∇×B=μ0J+μ0ε0 ∂E/∂t\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0\, \partial \mathbf{E}/\partial t.

1.3 Continuity Equation

Taking the divergence of the Ampere-Maxwell law:

∇⋅(∇×B)=0=μ0∇⋅J+μ0ε0∂∂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})

Using Gauss’s law: ∇⋅J+∂ρ∂t=0\nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0.

This is the continuity equation, expressing conservation of charge.

1.4 Boundary Conditions at Interfaces

At an interface between two linear media (labelled 1 and 2) with surface normal n^\hat{\mathbf{n}} Pointing from 2 into 1, Maxwell’s equations impose four boundary conditions.

Normal component of D\mathbf{D}. Apply Gauss’s law for D\mathbf{D} to a thin pillbox Straddling the interface:

∮D⋅dA=σfA  ⟹  D1n−D2n=σf\oint \mathbf{D} \cdot d\mathbf{A} = \sigma_f A \implies D_{1n} - D_{2n} = \sigma_f

Tangential component of E\mathbf{E}. Apply Faraday’s law to a rectangular loop Perpendicular to the interface. As the loop height Δh→0\Delta h \to 0The flux through the Loop vanishes:

∮E⋅dl=0  ⟹  E1t=E2t\oint \mathbf{E} \cdot d\mathbf{l} = 0 \implies E_{1t} = E_{2t}

In vector form: n^×(E1−E2)=0\hat{\mathbf{n}} \times (\mathbf{E}_1 - \mathbf{E}_2) = \mathbf{0}.

Normal component of B\mathbf{B}. Apply Gauss’s law for B\mathbf{B} to a pillbox:

B1n=B2nB_{1n} = B_{2n}

Tangential component of H\mathbf{H}. Apply Ampere’s law for H\mathbf{H} to a loop Perpendicular to the interface:

n^×(H1−H2)=Kf\hat{\mathbf{n}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{K}_f

Where Kf\mathbf{K}_f is the free surface current density.

Summary (no free charges or currents, σf=0\sigma_f = 0, Kf=0\mathbf{K}_f = \mathbf{0}):

FieldNormal componentTangential component
E\mathbf{E}ε1E1n=ε2E2n\varepsilon_1 E_{1n} = \varepsilon_2 E_{2n}E1t=E2tE_{1t} = E_{2t}
D\mathbf{D}D1n=D2nD_{1n} = D_{2n}D1t/ε1=D2t/ε2D_{1t}/\varepsilon_1 = D_{2t}/\varepsilon_2
B\mathbf{B}μ1B1n=μ2B2n\mu_1 B_{1n} = \mu_2 B_{2n}B1t/μ1=B2t/μ2B_{1t}/\mu_1 = B_{2t}/\mu_2
H\mathbf{H}μ2H1n=μ1H2n\mu_2 H_{1n} = \mu_1 H_{2n}H1t=H2tH_{1t} = H_{2t}

1.5 Worked Example: Deriving the Electromagnetic Wave Equation

Problem. Starting from Maxwell’s equations in free space (ρ=0\rho = 0, J=0\mathbf{J} = \mathbf{0}), Derive the wave equations for E\mathbf{E} and B\mathbf{B}And show that the wave speed is c=1/μ0ε0c = 1/\sqrt{\mu_0 \varepsilon_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=−∂B∂t,∇×B=μ0ε0∂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}

Take the curl of Faraday’s law:

∇×(∇×E)=−∂∂t(∇×B)=−μ0ε0∂2E∂t2\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}

Apply the vector identity ∇×(∇×E)=∇(∇⋅E)−∇2E\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}. Since ∇⋅E=0\nabla \cdot \mathbf{E} = 0:

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

∇2E=μ0ε0∂2E∂t2\boxed{\nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}}

An identical calculation, taking the curl of the Ampere-Maxwell law, yields:

∇2B=μ0ε0∂2B∂t2\boxed{\nabla^2 \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2}}

Comparing with the standard wave equation ∇2F=1v2∂2F∂t2\nabla^2 \mathbf{F} = \frac{1}{v^2}\frac{\partial^2 \mathbf{F}}{\partial t^2} The wave speed is:

c=1μ0ε0≈2.998×108 m/sc = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 2.998 \times 10^8\ \mathrm{m}/s

■\blacksquare