15 Jun 2025
Plane Electromagnetic Waves in Vacuum and Dielectrics
Plane-wave propagation, transversality, refractive index, dielectric constant, and wave impedance.
An unbounded medium has no interface to select a reflected wave or a discrete transverse mode. In a homogeneous, linear, isotropic, source-free dielectric,
\[\rho_{\mathrm f}=0,\qquad \mathbf J=0,\qquad \mathbf D=\epsilon\mathbf E,\qquad \mathbf B=\mu\mathbf H.\]Maxwell’s equations give
\[\nabla^2\mathbf E-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}=0, \qquad \nabla^2\mathbf H-\mu\epsilon\frac{\partial^2\mathbf H}{\partial t^2}=0.\]Harmonic plane-wave solution
Use the real part of the phasor
\[\mathbf E(\mathbf r,t)=\Re\!\left\{\mathbf E_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}\right\}.\]Every spatial derivative contributes $i\mathbf k$ and every time derivative contributes $-i\omega$. The wave equation becomes
\[(-k^2+\mu\epsilon\omega^2)\mathbf E_0=\mathbf0,\]so a nonzero field requires
\[\boxed{k=\omega\sqrt{\mu\epsilon}}, \qquad \boxed{v=\frac{\omega}{k}=\frac1{\sqrt{\mu\epsilon}}}.\]$k$ is in $\mathrm{rad\,m^{-1}}$, $\omega$ in $\mathrm{rad\,s^{-1}}$, and $v$ in $\mathrm{m\,s^{-1}}$. Constant phase means $\mathbf k\cdot\mathbf r-\omega t=\text{constant}$; differentiating it along $\hat{\mathbf k}$ confirms that the phase advances at $v=\omega/k$.
Transverse character from Gauss’s laws
In a homogeneous medium, $\nabla\cdot\mathbf D=0$ implies
\[i\epsilon\mathbf k\cdot\mathbf E_0=0 \quad\Longrightarrow\quad \boxed{\mathbf k\cdot\mathbf E_0=0}.\]Similarly, $\nabla\cdot\mathbf B=0$ gives
\[\boxed{\mathbf k\cdot\mathbf H_0=0}.\]The curl equations give the remaining geometry:
\[\boxed{\mathbf k\times\mathbf E_0=\omega\mu\mathbf H_0}, \qquad \boxed{\mathbf k\times\mathbf H_0=-\omega\epsilon\mathbf E_0}.\]Thus $\mathbf E_0$, $\mathbf H_0$, and $\mathbf k$ form a mutually perpendicular right-handed triad. In particular,
\[\mathbf H_0=\frac1\eta\hat{\mathbf k}\times\mathbf E_0, \qquad \boxed{\eta=\frac{E_0}{H_0}=\sqrt{\frac\mu\epsilon}}.\]The intrinsic wave impedance $\eta$ has unit ohm. For propagation in $+z$ with $\mathbf E$ along $+x$, $\mathbf H$ points along $+y$, since $\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z}$.
Vacuum and isotropic dielectric
In vacuum,
\[c=\frac1{\sqrt{\mu_0\epsilon_0}}, \qquad \eta_0=\sqrt{\frac{\mu_0}{\epsilon_0}}\approx376.73\ \Omega.\]Write the material constants as
\[\epsilon=\epsilon_r\epsilon_0, \qquad \mu=\mu_r\mu_0.\]The refractive index is the ratio of vacuum phase speed to material phase speed:
\[\boxed{n=\frac cv=\sqrt{\epsilon_r\mu_r}}.\]For most transparent optical dielectrics $\mu_r\simeq1$, so
\[\boxed{n\simeq\sqrt{\epsilon_r}}, \qquad \boxed{\epsilon_r\simeq n^2}.\]This relation uses the dielectric response at the wave frequency; a low-frequency static dielectric constant need not equal the optical value. The wavelength in the medium is
\[\lambda=\frac{2\pi}{k}=\frac{v}{f}=\frac{\lambda_0}{n},\]while the frequency $f=\omega/(2\pi)$ is fixed by the source and does not change on entering a stationary medium.
The material impedance can also be written
\[\boxed{\eta=\eta_0\sqrt{\frac{\mu_r}{\epsilon_r}}}.\]For a nonmagnetic dielectric, $\eta\simeq\eta_0/n$: a larger refractive index means a smaller ratio $E/H$.
Energy transported by a harmonic plane wave
For peak phasors, the cycle-averaged Poynting vector is
\[\langle\mathbf S\rangle =\frac12\Re(\mathbf E_0\times\mathbf H_0^{\ast}) =\boxed{\frac{\lvert E_0\rvert^2}{2\eta}\hat{\mathbf k}}.\]It has unit $\mathrm{W\,m^{-2}}$. In a lossless dielectric, $\mathbf E_0$ and $\mathbf H_0$ are in phase, so the average power flows in the phase-propagation direction.
Solved Problems
1. Reconstructing a plane wave from its frequency and electric amplitude
A $100\ \mathrm{MHz}$ plane wave travels along $+z$ through a nonmagnetic dielectric with $\epsilon_r=4.00$. Its peak electric field is $100\ \mathrm{V\,m^{-1}}$ along $+x$. Find $v$, $\lambda$, $k$, $\mathbf H_0$, and $\langle\mathbf S\rangle$.
Solution. The propagation speed and wavelength are
\[v=\frac{c}{\sqrt{\epsilon_r}}=1.499\times10^8\ \mathrm{m\,s^{-1}},\] \[\lambda=\frac vf=1.499\ \mathrm m, \qquad k=\frac{2\pi}{\lambda}=4.192\ \mathrm{rad\,m^{-1}}.\]The impedance is $\eta=\eta_0/2=188.37\ \Omega$. To make $\mathbf E\times\mathbf H$ point along $+z$, $\mathbf H$ must point along $+y$:
\[\mathbf H_0=\frac{100}{188.37}\hat{\mathbf y} =0.5309\hat{\mathbf y}\ \mathrm{A\,m^{-1}}.\]The mean power flux is
\[\langle\mathbf S\rangle =\frac{E_0^2}{2\eta}\hat{\mathbf z} =26.54\hat{\mathbf z}\ \mathrm{W\,m^{-2}}.\]Thus the phase convention $\cos(kz-\omega t)$ and the vector triad give
\[\boxed{v=1.499\times10^8\ \mathrm{m\,s^{-1}},\quad \lambda=1.499\ \mathrm m,\quad k=4.192\ \mathrm{rad\,m^{-1}},\quad \mathbf H_0=0.5309\hat{\mathbf y}\ \mathrm{A\,m^{-1}}}.\]$k\lambda=2\pi$ is dimensionless, and $E_0/H_0$ has unit ohm. If $\epsilon_r\to1$, both $v$ and $\eta$ approach their vacuum values.
2. Field amplitudes from a specified energy flux
A nonmagnetic dielectric of refractive index $n=1.50$ carries a $+z$-directed plane-wave intensity of $20.0\ \mathrm{W\,m^{-2}}$. Find the peak $E$ and $H$ amplitudes and the average total energy density.
Solution. The impedance is
\[\eta=\frac{\eta_0}{n}=251.15\ \Omega.\]Using peak phasors and the positive-power convention,
\[E_0=\sqrt{2\eta\langle S\rangle}=100.23\ \mathrm{V\,m^{-1}},\] \[H_0=\frac{E_0}{\eta}=0.3991\ \mathrm{A\,m^{-1}}.\]Energy travels at $v=c/n$, so
\[\langle u\rangle=\frac{\langle S\rangle}{v} =1.001\times10^{-7}\ \mathrm{J\,m^{-3}}.\]Hence
\[\boxed{E_0=100.23\ \mathrm{V\,m^{-1}},\quad H_0=0.3991\ \mathrm{A\,m^{-1}},\quad \langle u\rangle=1.001\times10^{-7}\ \mathrm{J\,m^{-3}}}.\]The relation $\langle S\rangle/\langle u\rangle=v$ supplies both the physical interpretation and a dimensional check. All three quantities vanish consistently as the specified intensity tends to zero.
Descriptive Questions
- Use the phasor form of Maxwell’s equations to prove that a uniform plane electromagnetic wave is transverse.
- Explain why frequency is unchanged but wavelength changes when a monochromatic wave enters a stationary dielectric.
- Distinguish phase velocity, intrinsic impedance, and refractive index, including the material assumptions behind $n\simeq\sqrt{\epsilon_r}$.
- Why must the electric field, magnetic field, and propagation vector form a right-handed triad for positive energy flow?
Numerical Problems
- Light of vacuum wavelength $600\ \mathrm{nm}$ enters a nonmagnetic dielectric of index $1.50$. Find its wavelength in the medium. Final answer: $\boxed{\lambda=400\ \mathrm{nm}}$.
- A transparent nonmagnetic material has $n=2.20$ at the operating frequency. Estimate its relative permittivity. Final answer: $\boxed{\epsilon_r=4.84}$.
- Find the intrinsic impedance of a nonmagnetic dielectric with $\epsilon_r=2.25$. Final answer: $\boxed{\eta=251.15\ \Omega}$.
- A plane wave with $E_0=90.0\ \mathrm{V\,m^{-1}}$ travels in a nonmagnetic medium of index $1.50$. Find its peak magnetic flux density. Final answer: $\boxed{B_0=4.503\times10^{-7}\ \mathrm T}$.
Every worked and numerical result is checked in the MJ-8 problem-verification worksheet; every printed check is zero.
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