The electromagnetic wave equation

The e&M wave equation

Lets recall Maxwell's equations in differential form $${\displaystyle \begin{array}{c}\vec{\nabla }\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}\\ \vec{\nabla }\times \vec{B}={\mu }_0\left(\vec{J}+{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\right)\\ \vec{\nabla }\cdot \vec{E}=\frac{\rho }{{\epsilon }_0}\\ \vec{\nabla }\cdot \vec{B}=0\end{array}}$$ In free space there are no charges or currents these become: $${\displaystyle \begin{array}{c}\vec{\nabla }\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}\\ \vec{\nabla }\times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\\ \vec{\nabla }\cdot \vec{E}=0\\ \vec{\nabla }\cdot \vec{B}=0\end{array}}$$ Lets take the time derivative of $$\vec{\nabla }\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$ $$\vec{\nabla }\times \frac{\partial \vec{E}}{\partial t}=-\frac{{\partial }^2\vec{B}}{\partial t^2}$$ $$\vec{\nabla }\times \vec{\nabla }\times \vec{B}=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}$$ but recall $$\vec{\nabla }\times \vec{\nabla }\times \vec{C}=\vec{\nabla }(\vec{\nabla }\cdot \vec{C})-(\vec{\nabla }\cdot \vec{\nabla })\vec{C}$$ so using that and $\vec{\nabla }\cdot \vec{B}=0$ we get $${\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}$$ This is the 3d wave equation! Note that is a second time derivative on one side and a second space derivative on the other sideIt is left as an exercise to show that $${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$$ we also see from this equation that the speed of light in vacuum is $$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$$