## Electromagnetic Wave Equation Taking the vector identity $ \vec{\nabla} \times(\vec{\nabla} \times \vec{A})= \vec{\nabla}(\vec{\nabla}\cdot \vec{A})-\vec{\nabla}^{2}\vec{A} $ and substituting $\vec{E}$, we find we can simplify using Maxwell's equations: $ \vec{\nabla} \times\underbrace{ (\vec{\nabla} \times \vec{E}) }_{ -\frac{ \partial B }{ \partial t } }= \vec{\nabla}\underbrace{ (\vec{\nabla}\cdot \vec{E}) }_{ =0 \text{ in vacuum} }-\vec{\nabla}^{2}\vec{E} $ $ -\frac{ \partial }{ \partial t } \underbrace{ (\vec{\nabla} \times \vec{B}) }_{ \mu_{0}\epsilon_{0} \frac{ \partial \vec{E} }{ \partial t } }=-\vec{\nabla}^{2}\vec{E} $ This can be rearranged to form a 3D wave equation: $ \vec{\nabla}^{2}\vec{E}=\mu_{0}\epsilon_{0} \frac{ \partial^{2} \vec{E} }{ \partial t^{2} } $ We can do the same thing but plugging in $\vec{B}$ instead of $\vec{E}$ and get the equation $ \vec{\nabla}^{2}\vec{B}=\mu_{0}\epsilon_{0} \frac{ \partial^{2} \vec{B} }{ \partial t^{2} } $ Uniform plane waves in lossless medium are in-phase as illustrated below: ![[Pasted image 20260913172058.png|300]] ### Uniform Plane Wave Solution $ \vec{E}(\vec{r},t) = \mathrm{Re}(\vec{E}(\vec{r})e^{j\omega t}) $ ${} \omega =\frac{2\pi}{T}= {}$ angular frequency $[\text{rad/s}]$ $k = \frac{2\pi}{\lambda} =$ wavenumber $[1/\text{m}]$ $\lambda =$ wavelength $[\text{m}]$ Plugging this solution into the wave equation gives us the **dispersion relation**: $k^{2}=\omega^{2}\mu_{0}\epsilon_{0}$ If we let $\omega t-kz$ be constant, then the speed of propagation is $v = \frac{dz}{dt} = \frac{\omega}{k}=f\lambda=c$ Combined with the dispersion relation, we get the speed of light: $ c = \frac{1}{\sqrt{ \mu_{0}\epsilon_{0}}} \approx 3\cdot 10^{8} \text{ m/s} $ ### Helmholz Equation If we assume one frequency $\omega$, then we can simplify the wave equation to $ \nabla^{2}\vec{E}+k^{2}\vec{E}=0 $ ### Phase and Group Velocity ![[Pasted image 20260922202257.png|367]]