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The Electromagnetic Wave Equation

Module by: Paul Padley

Summary: We derive the electromagnetic wave equation from Maxwell's equations.

The E&M Wave Equation

Lets recall Maxwell's equations in differential form ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × B ⃗ = μ 0 ( J ⃗ + ε 0 ∂ E ⃗ ∂ t ) ∇ ⃗ ⋅ E ⃗ = ρ ε 0 ∇ ⃗ ⋅ B ⃗ = 0 ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × B ⃗ = μ 0 ( J ⃗ + ε 0 ∂ E ⃗ ∂ t ) ∇ ⃗ ⋅ E ⃗ = ρ ε 0 ∇ ⃗ ⋅ B ⃗ = 0 In free space there are no charges or currents these become: ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × B ⃗ = μ 0 ε 0 ∂ E ⃗ ∂ t ∇ ⃗ ⋅ E ⃗ = 0 ∇ ⃗ ⋅ B ⃗ = 0 ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × B ⃗ = μ 0 ε 0 ∂ E ⃗ ∂ t ∇ ⃗ ⋅ E ⃗ = 0 ∇ ⃗ ⋅ B ⃗ = 0 Lets take the time derivative of ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × ∂ E ⃗ ∂ t = − ∂ 2 B ⃗ ∂ t 2 ∇ ⃗ × ∂ E ⃗ ∂ t = − ∂ 2 B ⃗ ∂ t 2 ∇ ⃗ × ∇ ⃗ × B ⃗ = − μ 0 ε 0 ∂ 2 B ⃗ ∂ t 2 ∇ ⃗ × ∇ ⃗ × B ⃗ = − μ 0 ε 0 ∂ 2 B ⃗ ∂ t 2 but recall ∇ ⃗ × ∇ ⃗ × C ⃗ = ∇ ⃗ ( ∇ ⃗ ⋅ C ⃗ ) − ( ∇ ⃗ ⋅ ∇ ⃗ ) C ⃗ ∇ ⃗ × ∇ ⃗ × C ⃗ = ∇ ⃗ ( ∇ ⃗ ⋅ C ⃗ ) − ( ∇ ⃗ ⋅ ∇ ⃗ ) C ⃗ so using that and ∇ ⃗ ⋅ B ⃗ = 0 ∇ ⃗ ⋅ B ⃗ = 0 we get ∇ 2 B ⃗ = μ 0 ε 0 ∂ 2 B ⃗ ∂ t 2 ∇ 2 B ⃗ = μ 0 ε 0 ∂ 2 B ⃗ ∂ t 2 This is the 3d wave equation! Note that is a second time derivative on one side and a second space derivative on the other side It is left as an exercise to show that ∇ 2 E ⃗ = μ 0 ε 0 ∂ 2 E ⃗ ∂ t 2 ∇ 2 E ⃗ = μ 0 ε 0 ∂ 2 E ⃗ ∂ t 2 we also see from this equation that the speed of light in vacuum is c = 1 μ 0 ε 0 c = 1 μ 0 ε 0

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