Plane waves

Source-free, linear, homogeneous, isotropic medium; the same equation holds for ${\mathbf{H}}_s$. For a lossless medium ($\sigma =0$), $\gamma =j\beta$.

Time form ($E_0$ real): $\mathbf{E}=E_0\cos (\omega t-\beta z){\hat{a}}_x$ and $\mathbf{H}=(E_0/\eta )\cos (\omega t-\beta z){\hat{a}}_y$, in time phase. A wave in $-z$ has $e^{+j\beta z}$, i.e. $\cos (\omega t+\beta z)$.

In rad/m; $\lambda$ is the wavelength in the medium.

${\lambda }_0=c/f$ is the free-space wavelength. Refractive index $n=\sqrt{{\mu }_r{\epsilon }_r}=c/v_p$. The frequency does not change when a wave enters a new medium; the wavelength does.

Ratio $|\mathbf{E}|/|\mathbf{H}|$ for a single travelling wave. For a non-magnetic dielectric $\eta ={\eta }_0/n$.

Use these to get $\mathbf{H}$ from $\mathbf{E}$ (or back) for any propagation direction, including a reflected wave travelling in $-z$. They also hold in a lossy medium, with complex $\eta$.

Lossless medium; $\mathbf{r}=x{\hat{a}}_x+y{\hat{a}}_y+z{\hat{a}}_z$ and $\beta =\sqrt{k_x^2+k_y^2+k_z^2}=\omega \sqrt{\mu \epsilon }$. Time form (real ${\mathbf{E}}_0$): ${\mathbf{E}}_0\cos (\omega t-\mathbf{k}\cdot \mathbf{r})$. Example: $\cos (\omega t-3x-4y)$ with $x,y$ in metres has $\beta =5$ rad/m and ${\hat{a}}_k=0.6{\hat{a}}_x+0.8{\hat{a}}_y$. The wavelength measured along one axis, e.g. $2\pi /k_x$, is never shorter than $\lambda =2\pi /\beta$.

In W/m². Power through an area $A$ normal to ${\hat{a}}_k$ is $S_{\text{avg}}A$. With RMS amplitudes, drop the factor 1/2.

Single travelling wave with peak amplitudes $E_0$ and $H_0=E_0/\eta$, energy densities in J/m³. The electric and magnetic energy densities are equal, and the energy moves at $v_p$ (equal to $v_g$ in a non-dispersive medium): $v_p\cdot \frac{1}{2}\epsilon |E_0{|}^2=|E_0{|}^2/(2\eta )$.

Take the root with $\alpha \ge 0$ and $\beta \ge 0$ for a wave travelling in $+z$ as $e^{-\gamma z}$.

Valid for any $\sigma$. Use it when the loss tangent is neither very small nor very large.

Equals $|{\mathbf{J}}_c|/|{\mathbf{J}}_d|$. Writing ${\epsilon }^{\prime }=\epsilon$ and ${\epsilon }^{\prime \prime }=\sigma /\omega$ assumes all the loss is conduction loss. Good dielectric: $\tan \theta \ll 1$; good conductor: $\tan \theta \gg 1$. The same material can be either, depending on frequency.

$0\le {\theta }_{\eta }\le 45^{\circ }$ and $\theta =2{\theta }_{\eta }$. $\mathbf{E}$ leads $\mathbf{H}$ in time by ${\theta }_{\eta }$.

$E_0$ real. Both envelopes decay as $e^{-\alpha z}$; $\mathbf{H}$ lags $\mathbf{E}$ by ${\theta }_{\eta }$.

Power decays at $2\alpha$, twice the field rate, but the dB figure is the same for field and power: 1 Np = 8.686 dB.

For $\sigma /(\omega \epsilon )\ll 1$, to first order in the loss tangent. If $\sigma$ and $\epsilon$ do not depend on frequency, $\alpha$ is then independent of frequency; $\beta$, $v_p$ and $\eta$ are essentially the lossless values.

For $\sigma /(\omega \epsilon )\gg 1$. $|\eta |$ is tiny (a small fraction of an ohm for metals at radio frequencies), so a good conductor looks almost like a short circuit to an incident wave.

Distance over which the field amplitude falls to $1/e\approx 0.368$ of its value (power to $1/e^2\approx 0.135$). In a good conductor $\delta \propto 1/\sqrt{f}$; the last two results hold only there.

Good conductor, with thickness (or wire radius $a$) much greater than $\delta$: the current is treated as flowing uniformly in a surface layer of depth $\delta$. $\ell$ is the wire length.

Evaluate $d\beta /d\omega$ at the carrier frequency.

$\lambda$ is the wavelength in the medium. Normal dispersion ($d{v}_p/d\lambda >0$) gives $v_g<v_p$.

For $\cos ({\omega }_{1}t-{\beta }_{1}z)+\cos ({\omega }_{2}t-{\beta }_{2}z)$ the envelope $\cos \left(\frac{\Delta \omega }{2}t-\frac{\Delta \beta }{2}z\right)$ moves at $\Delta \omega /\Delta \beta$, while the carrier moves at $({\omega }_1+{\omega }_2)/({\beta }_1+{\beta }_2)$.

$N$ = electron density per m³, $e$ and $m$ the electron charge and mass. Propagation only for $\omega >{\omega }_p$; below $f_p$ the wave is evanescent and is reflected. Here $v_p=c/\sqrt{1-{\omega }_p^2/{\omega }^2}>c$ and $v_g=c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$.

$v_p{v}_g=c^2$ holds whenever $\beta =\sqrt{{\omega }^2-{\omega }_0^2}/c$ (plasma, air-filled waveguide above cut-off). In a good conductor $\beta =\sqrt{\omega \mu \sigma /2}$ gives $d\beta /d\omega =\beta /(2\omega )$.

$E_{x0},E_{y0}\ge 0$ and $-180^{\circ }<\varphi \le 180^{\circ }$. Phasor form: $\left(E_{x0}{\hat{a}}_x+E_{y0}{e}^{j\varphi }{\hat{a}}_y\right)e^{-j\beta z}$.

Reverse both for a wave travelling in $-z$ ($e^{+j\beta z}$). The optics convention names handedness the opposite way, so use the definition the question states.

AR = 1 for circular and $\infty$ for linear. $\psi$ is the angle of the major axis from the x-axis (take the root that lies along the major axis). For $\varphi =\pm 90^{\circ }$ the axes lie along x and y, and AR $=\max (E_{x0},E_{y0})/\min (E_{x0},E_{y0})$.

OA (with $+$) is the semi-major axis and OB (with $-$) the semi-minor axis, for any $\varphi$. Checks: ${\text{OA}}^2+{\text{OB}}^2=E_{x0}^2+E_{y0}^2$ and $\text{OA}\cdot \text{OB}=E_{x0}{E}_{y0}|\sin \varphi |$. Example: $E_{x0}=3$, $E_{y0}=4$, $\varphi =90^{\circ }$ gives OA = 4, OB = 3, AR = 4/3.

For real $\eta$: $-1\le \Gamma \le 1$ and $0\le \tau \le 2$. Perfect conductor: ${\eta }_2=0$, so $\Gamma =-1$ and $\tau =0$.

The reflected $\mathbf{H}$ has the opposite sign to the reflected $\mathbf{E}$ ratio because the propagation direction reverses.

Follows from $\eta ={\eta }_0/n$. Entering a denser medium ($n_2>n_1$) gives $\Gamma <0$.

These need only medium 1 to be lossless; ${\eta }_2$ may be complex. For a good conductor $T\approx 4R_s/{\eta }_1$, a tiny fraction. When both impedances are real, $T=({\eta }_1/{\eta }_2)|\tau {|}^2$; $T\ne |\tau {|}^2$ because the impedances on the two sides differ.

$1\le s\le \infty$. With lossless medium 1, $|E_1{|}_{\max }=(1+|\Gamma |)E_{i0}$ and $|E_1{|}_{\min }=(1-|\Gamma |)E_{i0}$.

Adjacent maximum and minimum are ${\lambda }_1/4$ apart. $\Gamma >0$ when ${\eta }_2>{\eta }_1$ (e.g. dielectric into air); $\Gamma <0$ when ${\eta }_2<{\eta }_1$ (e.g. air into a dielectric or conductor). $|{\mathbf{H}}_1|$ has its maxima where $|{\mathbf{E}}_1|$ has its minima. For complex $\Gamma =|\Gamma |\angle {\theta }_{\Gamma }$ (lossy medium 2, lossless medium 1), $|E_1|$ is maximum at $z=-({\theta }_{\Gamma }+2m\pi )/(2{\beta }_1)$, with the integers $m$ chosen so that $z\le 0$, and minimum ${\lambda }_1/4$ from each maximum.

$E_0$ real, conductor at $z=0$. $\mathbf{E}$ nulls at $z=-m{\lambda }_1/2$ (including the wall), $\mathbf{H}$ nulls at $z=-(2m+1){\lambda }_1/4$; no net power flows ($S_{\text{avg}}=0$).

Same form as a transmission-line input impedance. Quarter-wave slab ($d={\lambda }_2/4$, or any odd multiple): ${\eta }_{\text{in}}={\eta }_2^2/{\eta }_3$, so ${\eta }_2=\sqrt{{\eta }_1{\eta }_3}$ removes the reflection, but only at the design frequency. Half-wave slab ($d={\lambda }_2/2$): ${\eta }_{\text{in}}={\eta }_3$, so the reflection is the same as if medium 3 met medium 1 directly; it is zero only when ${\eta }_3={\eta }_1$ (e.g. a dielectric sheet in air).

$n=\sqrt{{\mu }_r{\epsilon }_r}$. Both follow from phase matching: the phase constant along the boundary is the same for all three waves.

Electric-field ratios. At ${\theta }_i=0$ they reduce to the normal-incidence $\Gamma$ and $\tau$.

Sign convention chosen so that ${\Gamma }_{\parallel }\to ({\eta }_2-{\eta }_1)/({\eta }_2+{\eta }_1)$ at normal incidence; some texts define ${\Gamma }_{\parallel }$ with the opposite sign. $|{\Gamma }_{\parallel }|$ and the Brewster condition do not depend on that choice.

Equivalently $\sin {\theta }_B=\sqrt{{\epsilon }_2/({\epsilon }_1+{\epsilon }_2)}$. The angle at which ${\Gamma }_{\parallel }=0$. Between non-magnetic media there is no Brewster angle for perpendicular polarisation, so an unpolarised wave incident at ${\theta }_B$ reflects as a purely perpendicular-polarised wave (the polarising angle).

Exists only going from the denser to the rarer medium. For ${\theta }_i\ge {\theta }_c$, $|{\Gamma }_{\perp }|=|{\Gamma }_{\parallel }|=1$ (total internal reflection) for both polarisations, and the reflected wave picks up a phase shift that differs between the two.

${\theta }_i>{\theta }_c$; $z>0$ is the distance into medium 2 from the boundary and $x$ runs along the boundary in the plane of incidence. The field in medium 2 is not zero, but it carries no time-average power across the boundary.