16 May 2025
Electromagnetic Boundaries, Dielectric Waves, and Energy Flow
Boundary conditions, source-free wave equations, dielectric plane waves, and Poynting's theorem.
Let a smooth interface separate medium 1 from medium 2, and define $\hat{\mathbf n}$ to point from 1 to 2. The limiting forms of Maxwell’s integral equations determine which field components can jump.
Normal boundary conditions
Place a thin pillbox of face area $\Delta A$ across the interface. As its height tends to zero, the side flux vanishes. Gauss’s electric law leaves
\[\mathbf D_2\cdot\hat{\mathbf n}\,\Delta A -\mathbf D_1\cdot\hat{\mathbf n}\,\Delta A =\rho_s\Delta A,\]so
\[\boxed{\hat{\mathbf n}\cdot(\mathbf D_2-\mathbf D_1)=\rho_s}.\]$\rho_s$ is free surface-charge density in $\mathrm{C\,m^{-2}}$. Applying the same pillbox to magnetic flux gives
\[\boxed{\hat{\mathbf n}\cdot(\mathbf B_2-\mathbf B_1)=0}.\]Thus $B_n$ is always continuous, while $D_n$ jumps only by the free surface charge.
Tangential boundary conditions
Use a narrow rectangular loop of length $\Delta\ell$ parallel to the interface. Its area tends to zero with its height. Faraday’s flux term then vanishes for finite $\partial\mathbf B/\partial t$, leaving
\[\boxed{\hat{\mathbf n}\times(\mathbf E_2-\mathbf E_1)=\mathbf0}.\]For the Ampere-Maxwell law, a free sheet current $\mathbf K$ can remain finite as the loop height tends to zero. With $\mathbf K$ in $\mathrm{A\,m^{-1}}$,
\[\boxed{\hat{\mathbf n}\times(\mathbf H_2-\mathbf H_1)=\mathbf K}.\]If $\rho_s=0$ and $\mathbf K=\mathbf0$, the continuous combinations are $D_n$, $B_n$, $\mathbf E_t$, and $\mathbf H_t$. The fields $E_n=D_n/\epsilon$ and $B_t=\mu H_t$ may still change when the material constants change.
Wave equations in a source-free dielectric
In a homogeneous, source-free, nonconducting dielectric,
\[\rho_{\mathrm f}=0,\qquad \mathbf J=0,\qquad \mathbf D=\epsilon\mathbf E,\qquad \mathbf B=\mu\mathbf H,\]with constant $\epsilon$ and $\mu$. Maxwell’s curl equations reduce to
\[\nabla\times\mathbf E=-\mu\frac{\partial\mathbf H}{\partial t}, \qquad \nabla\times\mathbf H=\epsilon\frac{\partial\mathbf E}{\partial t}.\]Take the curl of the first equation:
\[\nabla\times(\nabla\times\mathbf E) =-\mu\frac{\partial}{\partial t}(\nabla\times\mathbf H) =-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}.\]The identity $\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$ and Gauss’s law $\nabla\cdot\mathbf E=0$ give
\[\boxed{\nabla^2\mathbf E-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}=0}.\]The same steps yield
\[\boxed{\nabla^2\mathbf H-\mu\epsilon\frac{\partial^2\mathbf H}{\partial t^2}=0}.\]Both have wave speed
\[\boxed{v=\frac1{\sqrt{\mu\epsilon}}}.\]Plane wave in a dielectric
For propagation along $+z$, take
\[\mathbf E(z,t)=E_0\cos(kz-\omega t)\,\hat{\mathbf x}.\]Substitution in the wave equation gives
\[-k^2+\mu\epsilon\omega^2=0, \qquad \boxed{k=\omega\sqrt{\mu\epsilon}}.\]For a general plane-wave phasor $\mathbf E_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}$, Gauss’s law gives
\[i\mathbf k\cdot\mathbf E_0=0.\]Therefore $\mathbf k\cdot\mathbf E_0=0$: the electric field is transverse. Faraday’s law gives
\[\mathbf k\times\mathbf E_0=\omega\mu\mathbf H_0,\]so $\mathbf H_0$ is transverse to both $\mathbf k$ and $\mathbf E_0$. Its magnitude is fixed by the intrinsic impedance
\[\boxed{\eta=\frac{E_0}{H_0}=\sqrt{\frac{\mu}{\epsilon}}},\]measured in ohms. For the chosen axes,
\[\mathbf H(z,t)=\frac{E_0}{\eta}\cos(kz-\omega t)\,\hat{\mathbf y},\]and $\mathbf E\times\mathbf H$ points along $+z$.
Poynting theorem from Maxwell’s equations
Dot the Ampere-Maxwell equation with $\mathbf E$ and Faraday’s law with $\mathbf H$:
\[\mathbf E\cdot(\nabla\times\mathbf H) =\mathbf J\cdot\mathbf E+\mathbf E\cdot\frac{\partial\mathbf D}{\partial t},\] \[\mathbf H\cdot(\nabla\times\mathbf E) =-\mathbf H\cdot\frac{\partial\mathbf B}{\partial t}.\]Subtract the first relation from the second and use
\[\nabla\cdot(\mathbf E\times\mathbf H) =\mathbf H\cdot(\nabla\times\mathbf E) -\mathbf E\cdot(\nabla\times\mathbf H).\]For a linear, nondispersive medium,
\[\mathbf E\cdot\frac{\partial\mathbf D}{\partial t} =\frac{\partial}{\partial t}\left(\frac12\mathbf E\cdot\mathbf D\right), \qquad \mathbf H\cdot\frac{\partial\mathbf B}{\partial t} =\frac{\partial}{\partial t}\left(\frac12\mathbf B\cdot\mathbf H\right).\]Therefore
\[\boxed{\frac{\partial u}{\partial t}+\nabla\cdot\mathbf S+\mathbf J\cdot\mathbf E=0},\]where
\[\boxed{u=\frac12(\mathbf E\cdot\mathbf D+\mathbf B\cdot\mathbf H)}, \qquad \boxed{\mathbf S=\mathbf E\times\mathbf H}.\]$u$ is electromagnetic energy density in $\mathrm{J\,m^{-3}}$, $\mathbf S$ is energy flux in $\mathrm{W\,m^{-2}}$, and $\mathbf J\cdot\mathbf E$ is the rate per unit volume at which field energy becomes material energy. Integration over a fixed volume $V$ gives
\[\boxed{\frac{d}{dt}\int_Vu\,dV =-\oint_{\partial V}\mathbf S\cdot d\mathbf a -\int_V\mathbf J\cdot\mathbf E\,dV}.\]The outward surface normal fixes the signs: positive outward Poynting flux decreases the stored field energy.
For the dielectric plane wave, $H=E/\eta$ and $\eta^2=\mu/\epsilon$, so
\[u_E=\frac12\epsilon E^2, \qquad u_B=\frac12\mu H^2 =\frac12\mu\frac{E^2}{\eta^2} =\frac12\epsilon E^2.\]The electric and magnetic contributions are equal instantaneously. Averaging $\cos^2(kz-\omega t)$ over one cycle gives
\[\boxed{\langle\mathbf S\rangle =\frac{E_0H_0}{2}\hat{\mathbf z} =\frac{E_0^2}{2\eta}\hat{\mathbf z}}.\]The dispersion, wave-speed, impedance, equal-energy, and Poynting identities are independently checked in the Maxima worksheet; every printed residual is zero.
Discussion