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

Step-by-Step Guide

1. 1

   Step 1: Begin with Maxwell's Equations in vacuum.

   In vacuum, charge density ρ=0{\displaystyle \rho =0} and current density J=0.{\displaystyle {\mathbf {J} }=0.} ∇⋅E=0∇⋅B=0∇×E=−∂B∂t∇×B=μ0ϵ0∂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}}} where μ0{\displaystyle \mu _{0}} is the magnetic permeability constant and ϵ0{\displaystyle \epsilon _{0}} is the electric permittivity constant.
   
   The intertwining between the electric and magnetic fields is on full display here.
2. 2

   Step 2: 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}}} Note that partial derivatives commute with each other, given well-behaved functions. , Using the BAC-CAB identity ∇×(∇×E)=∇(∇⋅E)−∇2E{\displaystyle \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,} ∇(∇⋅E)−∇2E=−μ0ϵ0∂2E∂t2∇2E=μ0ϵ0∂2E∂t2.{\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}}} The above equation is the wave equation in three dimensions. , ∂2E∂x2=μ0ϵ0∂2E∂t2.{\displaystyle {\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−vt)+g(x+vt),{\displaystyle f(x-vt)+g(x+vt),} where v{\displaystyle v} is the velocity and λ{\displaystyle \lambda } is the wavelength.
   
   Here, f{\displaystyle f} and g{\displaystyle 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=E0sin(2πλ(x−vt)),{\displaystyle E=E_{0}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right),} where E0{\displaystyle E_{0}} is the amplitude of the electric field (this quantity will cancel out later). , ∂2E∂x2=−E0(2πλ)2sin(2πλ(x−vt))∂2E∂t2=−E0(2πvλ)2sin(2πλ(x−vt)){\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}}} , Note that the sin{\displaystyle \sin } expressions cancel out. −E0(2πλ)2=μ0ϵ01=μ0ϵ0v2{\displaystyle {\begin{aligned}-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{2}&=\mu _{0}\epsilon _{0}\left\\1&=\mu _{0}\epsilon _{0}v^{2}\end{aligned}}} , v=1μ0ϵ0≈3×108 m s−1.{\displaystyle 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.
   
   Therefore, light does not only travel at the speed of electromagnetic waves, it is an electromagnetic wave.
3. 3

   Step 3: Substitute the Ampere-Maxwell Law.

4. 4

   Step 4: Rewrite the wave equation in one dimension.

5. 5

   Step 5: Twice differentiate the solution with respect to x{\displaystyle x} and t{\displaystyle t}.

6. 6

   Step 6: Substitute these equations back into the wave equation.

7. 7

   Step 7: Arrive at the answer.

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