## 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]] We define the vector potential $\vec{A}$ and the scalar potential $\Phi$ with the following: $\vec{B}=\nabla \times \vec{A}$ $E=-\frac{ \partial A }{ \partial t }-\nabla \Phi$ This leaves $\nabla\cdot \vec{A}$ undefined, so we can choose the Lorentz Gauge $\nabla\cdot A + \mu_{0}\epsilon \frac{ \partial \Phi }{ \partial t }=0$ This choice allows us to derive a wave equation for $\Phi$ and $\vec{A}$: $ \nabla^{2}\vec{A}-\mu_{0}\epsilon \frac{ \partial^{2}A }{ \partial t^{2} } =0 $ $ \nabla^{2}\Phi-\mu_{0}\epsilon \frac{ \partial^{2}\Phi }{ \partial t^{2} } =0 $ ### Uniform Plane Wave Solution $ \vec{E}(\vec{r},t) =E(\vec{r})e^{j\omega t-j\vec{k}\cdot \vec{r}} $ $\omega =$ 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**: $k^{2}=\omega^{2}\mu_{0}\epsilon_{0}$ To follow a point of constant phase, we let $\omega t-kz$ be constant. Then the speed of propagation is given by $v = \frac{dz}{dt} = \frac{\omega}{k}=f\lambda=c$ 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]] ### Polarization For a wave propagating in the $\hat{z}$ direction, the direction of the $\vec{E}$ field determines its polarization. A wave with $E_{y}=0$ is said to be $\hat{x}$-polarized and waves with $E_{x}=0$ are considered $\hat{y}$-polarized. More generally, the wave can follow a polarization ellipse defined by $ \vec{E}(z)=E_{x}e^{-jkz}(\hat{x}+Ae^{j\phi}\hat{y}) $ resulting in $x=\cos\omega t$ and $y=A\cos(\omega t+\phi)$ For a uniform plane wave in direction $\hat k$: $ \vec H=\frac{1}{\eta}\,\hat k\times\vec E,\qquad \eta=\sqrt{\frac{\mu}{\epsilon}} $ $\vec E$, $\vec H$ and $\hat k$ form a right-handed set. **Polarization handedness (IEEE):** point your right thumb along $\vec k$. If your fingers curl the way $\vec E$ rotates in time, the wave is right-handed. With $e^{+j\omega t}$ and propagation along $+\hat z$: - $\hat x-j\hat y$ gives $\hat x\cos\omega t+\hat y\sin\omega t$, rotating $x\to y$: **RHC**. - $\hat x+j\hat y$: **LHC**. - Traditional optics names (looking back at the source) are the **opposite**. Reversing the time convention also flips the sign. See [[conventions]] §6. **General complex vector** $\vec A=\vec A_r+j\vec A_i$: - $\mathrm{Re}\{\vec Ae^{j\omega t}\}=\vec A_r\cos\omega t-\vec A_i\sin\omega t$ traces an **ellipse**. - $\vec A_r$ and $\vec A_i$ are conjugate semi-diameters. At $t=0$ the field is at $\vec A_r$, and it moves toward $-\vec A_i$. - The polarization is linear if $\vec A_r\parallel\vec A_i$ (or one is zero), and circular if $|\vec A_r|=|\vec A_i|$ and $\vec A_r\perp\vec A_i$. - A complex vector takes 6 real numbers: 2 for the plane of the ellipse, 1 for its orientation within that plane, 2 for the axis lengths, and 1 for the time phase (Haus 1.6). Polarization transformations (Jones calculus): [[optical operators]]. ### Transversality comes from Gauss, not the wave equation For $\vec E=\vec E_0e^{-j\vec k\cdot\vec r}$, $\nabla\to-j\vec k$. Gauss's law ($\nabla\cdot\vec E=0$) gives $\vec k\cdot\vec E_0=0$. A longitudinal field $\vec E_0\parallel\vec k$ **does** satisfy the vector wave equation, because the wave equation was derived by *using* $\nabla\cdot\vec E=0$ and no longer carries it. Solving the wave equation is not the same as solving Maxwell's equations (pset 1 P1d). ### Complex wave vector and evanescent waves $ \vec k\cdot\vec k=\omega^2\mu\epsilon\qquad\text{(unconjugated dot product)} $ Write $\vec k=\vec k'-j\vec k''$. Then $ \vec E=\vec E_0\,e^{-\vec k''\cdot\vec r}\,e^{-j\vec k'\cdot\vec r},\qquad k'^2-k''^2=\omega^2\mu\epsilon,\qquad \vec k'\cdot\vec k''=0\ \ (\text{lossless}) $ - In a **lossless** medium the decay direction is forced to be perpendicular to the phase direction. That is the evanescent wave of total internal reflection ([[reflection]]). - For decay *along* the propagation direction, $\omega^2\mu\epsilon$ must be complex, i.e. the medium must be lossy (Haus 1.49–1.50). ### Phase and group velocity - Phase fronts move at $v_p=\dfrac{\omega}{k}$. - A wave packet's envelope moves at $v_g=\dfrac{d\omega}{dk}$. With $k=\dfrac{n\omega}{c}$, $v_g=\dfrac{c}{n_g}$ where $n_g=n+\omega\dfrac{dn}{d\omega}=n-\lambda\dfrac{dn}{d\lambda}$. - $v_p$ can exceed $c$ (plasma, waveguide); $v_g$ normally doesn't. - For pulses and group velocity dispersion: [[dispersion]]. ### Shape-preserving solutions (1D) Changing variables to $a=z-vt$, $b=z+vt$ turns the 1D wave equation into $\dfrac{\partial^2E}{\partial a\,\partial b}=0$. So $ E(z,t)=f(z-vt)+g(z+vt) $ Any shape travels unchanged at $v$ in $\pm z$. This breaks when $v$ depends on $\omega$ (dispersion).