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

Last Updated: March 1, 2021 Approved

Maxwell's Equations, along with describing how the electric field $\mathbf {E}$ and magnetic field $\mathbf {B}$ interact, also predict the speed of light, for light is an electromagnetic wave. Thus, the end goal here is to obtain a wave equation.

Steps

1
Begin with Maxwell's Equations in vacuum. In vacuum, charge density $\rho =0$ $\rho =0$ and current density $\mathbf {J} =0.$ ${\mathbf {J}}=0.$
- ${\begin{aligned}\nabla \cdot \mathbf {E} &=0\\\nabla \cdot \mathbf {B} &=0\\\nabla \times \mathbf {E} &=-{\frac {\partial \mathbf {B} }{\partial t}}\\\nabla \times \mathbf {B} &=\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\end{aligned}}$
- where $\mu _{0}$ is the magnetic permeability constant and $\epsilon _{0}$ is the electric permittivity constant. The intertwining between the electric and magnetic fields is on full display here.
2
Take the curl of both sides of Faraday's Law.
- ${\begin{aligned}\nabla \times (\nabla \times \mathbf {E} )&=\nabla \times -{\frac {\partial \mathbf {B} }{\partial t}}\\&=-{\frac {\partial }{\partial t}}(\nabla \times \mathbf {B} )\end{aligned}}$
- Note that partial derivatives commute with each other, given well-behaved functions.
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3
Substitute the Ampere-Maxwell Law.
- Using the BAC-CAB identity $\nabla \times (\nabla \times \mathbf {E} )=\nabla (\nabla \cdot \mathbf {E} )-\nabla ^{2}\mathbf {E}$ on the left side and recognizing that $\nabla \cdot \mathbf {E} =0,$
- ${\begin{aligned}\nabla (\nabla \cdot \mathbf {E} )-\nabla ^{2}\mathbf {E} &=-\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}\\\nabla ^{2}\mathbf {E} &=\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}.\end{aligned}}$
- The above equation is the wave equation in three dimensions.
4
Rewrite the wave equation in one dimension.
- ${\frac {\partial ^{2}E}{\partial x^{2}}}=\mu _{0}\epsilon _{0}{\frac {\partial ^{2}E}{\partial t^{2}}}.$
- The general solution to this equation is $f(x-vt)+g(x+vt),$ where $v$ is the velocity and $\lambda$ is the wavelength. Here, $f$ and $g$ 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 $E=E_{0}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right),$ where $E_{0}$ is the amplitude of the electric field (this quantity will cancel out later).
5
Twice differentiate the solution with respect to $x$ and $t$ .
- ${\begin{aligned}{\frac {\partial ^{2}E}{\partial x^{2}}}&=-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{2}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right)\\{\frac {\partial ^{2}E}{\partial t^{2}}}&=-E_{0}\left({\frac {2\pi v}{\lambda }}\right)^{2}\sin \left({\frac {2\pi }{\lambda }}(x-vt)\right)\end{aligned}}$
6
Substitute these equations back into the wave equation. Note that the $\sin$ $\sin$ expressions cancel out.
- ${\begin{aligned}-E_{0}\left({\frac {2\pi }{\lambda }}\right)^{2}&=\mu _{0}\epsilon _{0}\left[-E_{0}\left({\frac {2\pi v}{\lambda }}\right)^{2}\right]\\1&=\mu _{0}\epsilon _{0}v^{2}\end{aligned}}$
7
Arrive at the answer.
- $v={\sqrt {\frac {1}{\mu _{0}\epsilon _{0}}}}\approx 3\times 10^{8}{\text{ m s}}^{-1}.$
- 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.