Finding the Speed of Electromagnetic Light from Maxwell's Equations
Formulas
Vacuum Maxwell’s Equations
$$ \begin{align} \nabla \cdot \mathbf{E} &= 0 \\[1em] \nabla \cdot \mathbf{B} &= 0 \\[1em] \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\[1em] \nabla \times \mathbf{B} &= \mu_{0}\epsilon_{0}\frac{\partial \mathbf{E}}{\partial t} \end{align} $$
$$ \frac{\partial^2 f}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} $$
Three-dimensional wave equation
$$ \nabla ^2 f = \frac{1}{v^2}\frac{\partial ^2 f}{\partial t^2} $$
Derivation
The goal is to derive a wave equation form from Maxwell’s equations, regarding $\mathbf{E}$ and $\mathbf{B}$. By taking the curl of $(3)$,
$$ \begin{align*} \nabla \times (\nabla \times \mathbf{E}) &= \nabla \times \left( -\frac{\partial \mathbf{B}}{\partial t} \right) \\ &= -\frac{\partial }{\partial t} (\nabla \times \mathbf{B}) \\ &= -\frac{\partial}{\partial t} \left( \mu_{0}\epsilon_{0} \frac{\partial \mathbf{E}}{\partial t} \right) \\ &= -\mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{E}}{\partial t^2} \end{align*} $$
The third equality holds by $(4)$. Similarly, by taking the curl of $(4)$,
$$ \begin{align*} \nabla \times (\nabla \times \mathbf{B}) &= \nabla \times \left( \mu_{0} \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t} \right) \\ &= \mu_{0}\epsilon_{0} \frac{\partial }{\partial t}(\nabla \times \mathbf{E}) \\ &= -\mu_{0}\epsilon_{0} \frac {\partial ^2 \mathbf{B} }{\partial t^2} \end{align*} $$
And since $\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A})-\nabla^2\mathbf{A}$,
$$ \nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E})-\nabla^2\mathbf{E}=-\mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{E}}{\partial t^2} $$
$$ \nabla \times (\nabla \times \mathbf{B}) = \nabla(\nabla \cdot \mathbf{B})-\nabla^2\mathbf{B}=-\mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{B}}{\partial t^2} $$
Finally, because of $\nabla \cdot \mathbf{E}=0$, $\nabla \cdot \mathbf{B}=0$,
$$ \nabla ^2 \mathbf{E} = \mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{E}}{\partial t^2} $$
$$ \nabla ^2 \mathbf{B} = \mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{B}}{\partial t^2} $$
Now, from Maxwell’s equations, $\mathbf{E}$ and $\mathbf{B}$ are separated. It can be seen that it takes the form of a three-dimensional wave equation. Remarkably, the speed of electromagnetic waves is determined by $\dfrac{1}{v^2}=\mu_{0}\epsilon_{0}$ to be $v=\dfrac{1}{\sqrt{\mu_{0}\epsilon_{0}}}=3.00\times 10^8 m/s$, which is the same as the speed of light. Thus, it can be speculated that light is a type of electromagnetic wave, and its speed is constant.