How to Derive the Speed of Light from Maxwell's Equations

Begin with Maxwell's Equations in vacuum. In vacuum, charge density ρ = 0 {\displaystyle \rho =0} \rho =0 and current density J = 0. {\displaystyle \mathbf {J} =0.} {\mathbf {J}}=0. ∇ ⋅ E = 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 ϵ 0 ∂ E ∂ t {\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &=0\\\nabla \cdot \mathbf {B} &=0\\\nabla \times \mathbf {E} &=-{\frac {\partial \mathbf {B} }{\partial t}}\\\nabla \times \mathbf {B} &=\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\end{aligned}}} {\begin{aligned}\nabla \cdot {\mathbf {E}}&=0\\\nabla \cdot {\mathbf {B}}&=0\\\nabla \times {\mathbf {E}}&=-{\frac {\partial {\mathbf {B}}}{\partial t}}\\\nabla \times {\mathbf {B}}&=\mu _{0}\epsilon _{0}{\frac {\partial {\mathbf {E}}}{\partial t}}\end{aligned}} where μ 0 {\displaystyle \mu _{0}} \mu _{0} is the magnetic permeability constant and ϵ 0 {\displaystyle \epsilon _{0}} \epsilon _{0} is the electric permittivity constant. The intertwining between the electric and magnetic fields is on full display here.

Take the curl of both sides of Faraday's Law. ∇ × ( ∇ × E ) = ∇ × − ∂ B ∂ t = − ∂ ∂ t ( ∇ × B ) {\displaystyle {\begin{aligned}\nabla \times (\nabla \times \mathbf {E} )&=\nabla \times -{\frac {\partial \mathbf {B} }{\partial t}}\\&=-{\frac {\partial }{\partial t}}(\nabla \times \mathbf {B} )\end{aligned}}} {\begin{aligned}\nabla \times (\nabla \times {\mathbf {E}})&=\nabla \times -{\frac {\partial {\mathbf {B}}}{\partial t}}\\&=-{\frac {\partial }{\partial t}}(\nabla \times {\mathbf {B}})\end{aligned}} Note that partial derivatives commute with each other, given well-behaved functions.

Substitute the Ampere-Maxwell Law. Using the BAC-CAB identity ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E {\displaystyle \nabla \times (\nabla \times \mathbf {E} )=\nabla (\nabla \cdot \mathbf {E} )-\nabla ^{2}\mathbf {E} } \nabla \times (\nabla \times {\mathbf {E}})=\nabla (\nabla \cdot {\mathbf {E}})-\nabla ^{{2}}{\mathbf {E}} on the left side and recognizing that ∇ ⋅ E = 0 , {\displaystyle \nabla \cdot \mathbf {E} =0,} \nabla \cdot {\mathbf {E}}=0, ∇ ( ∇ ⋅ E ) − ∇ 2 E = − μ 0 ϵ 0 ∂ 2 E ∂ t 2 ∇ 2 E = μ 0 ϵ 0 ∂ 2 E ∂ t 2 . {\displaystyle {\begin{aligned}\nabla (\nabla \cdot \mathbf {E} )-\nabla ^{2}\mathbf {E} &=-\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}\\\nabla ^{2}\mathbf {E} &=\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}.\end{aligned}}} {\begin{aligned}\nabla (\nabla \cdot {\mathbf {E}})-\nabla ^{{2}}{\mathbf {E}}&=-\mu _{0}\epsilon _{0}{\frac {\partial ^{{2}}{\mathbf {E}}}{\partial t^{{2}}}}\\\nabla ^{{2}}{\mathbf {E}}&=\mu _{0}\epsilon _{0}{\frac {\partial ^{{2}}{\mathbf {E}}}{\partial t^{{2}}}}.\end{aligned}} The above equation is the wave equation in three dimensions.

Rewrite the wave equation in one dimension. ∂ 2 E ∂ x 2 = μ 0 ϵ 0 ∂ 2 E ∂ t 2 . {\displaystyle {\frac {\partial ^{2}E}{\partial x^{2}}}=\mu _{0}\epsilon _{0}{\frac {\partial ^{2}E}{\partial t^{2}}}.} {\frac {\partial ^{{2}}E}{\partial x^{{2}}}}=\mu _{0}\epsilon _{0}{\frac {\partial ^{{2}}E}{\partial t^{{2}}}}. The general solution to this equation is f ( x − v t ) + g ( x + v t ) , {\displaystyle f(x-vt)+g(x+vt),} f(x-vt)+g(x+vt), where v {\displaystyle v} v is the velocity and λ {\displaystyle \lambda } \lambda is the wavelength. Here, f {\displaystyle f} f and g {\displaystyle g} g are two arbitrary functions that describe a wave propagating in the positive and negative directions, respectively. Since this is quite general, we can opt for the most common solution of just a sinusoidal function traveling in the direction of propagation. So we can write the solution as E = E 0 sin ⁡ ( 2 π λ ( x − v t ) ) , {\displaystyle E=E_{0}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right),} E=E_{0}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right), where E 0 {\displaystyle E_{0}} E_{0} is the amplitude of the electric field (this quantity will cancel out later).

Twice differentiate the solution with respect to x {\displaystyle x} x and t {\displaystyle t} t. ∂ 2 E ∂ x 2 = − E 0 ( 2 π λ ) 2 sin ⁡ ( 2 π λ ( x − v t ) ) ∂ 2 E ∂ t 2 = − E 0 ( 2 π v λ ) 2 sin ⁡ ( 2 π λ ( x − v t ) ) {\displaystyle {\begin{aligned}{\frac {\partial ^{2}E}{\partial x^{2}}}&=-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{2}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right)\\{\frac {\partial ^{2}E}{\partial t^{2}}}&=-E_{0}\left({\frac {2\pi v}{\lambda }}\right)^{2}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right)\end{aligned}}} {\begin{aligned}{\frac {\partial ^{{2}}E}{\partial x^{{2}}}}&=-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{{2}}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right)\\{\frac {\partial ^{{2}}E}{\partial t^{{2}}}}&=-E_{0}\left({\frac {2\pi v}{\lambda }}\right)^{{2}}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right)\end{aligned}}

Substitute these equations back into the wave equation. Note that the sin {\displaystyle \sin } \sin expressions cancel out. − E 0 ( 2 π λ ) 2 = μ 0 ϵ 0 [ − E 0 ( 2 π v λ ) 2 ] 1 = μ 0 ϵ 0 v 2 {\displaystyle {\begin{aligned}-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{2}&=\mu _{0}\epsilon _{0}\left[-E_{0}\left({\frac {2\pi v}{\lambda }}\right)^{2}\right]\\1&=\mu _{0}\epsilon _{0}v^{2}\end{aligned}}} {\begin{aligned}-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{{2}}&=\mu _{0}\epsilon _{0}\left[-E_{0}\left({\frac {2\pi v}{\lambda }}\right)^{{2}}\right]\\1&=\mu _{0}\epsilon _{0}v^{{2}}\end{aligned}}

Arrive at the answer. v = 1 μ 0 ϵ 0 ≈ 3 × 10 8 m s − 1 . {\displaystyle v={\sqrt {\frac {1}{\mu _{0}\epsilon _{0}}}}\approx 3\times 10^{8}{\text{ m s}}^{-1}.} v={\sqrt {{\frac {1}{\mu _{0}\epsilon _{0}}}}}\approx 3\times 10^{{8}}{\text{ m s}}^{{-1}}. The expression on the right happens to equal the speed of light. In fact, light does not only travel at the speed of electromagnetic waves, it is an electromagnetic wave.