Wave Propagation in Lossy and Conducting Media
In free space a wave glides forever, its amplitude untouched. Real media bleed energy: every wiggle of the field drives a conduction current, and that current heats the medium. The bookkeeping for that loss was hiding all along inside the propagation constant \(\gamma=\sqrt{j\omega\mu(\sigma+j\omega\varepsilon)}\). Splitting it into \(\alpha+j\beta\) now gives a wave that not only oscillates but decays — and in a metal it dies so fast it survives only within a thin skin. This chapter turns \(\gamma\) into numbers, from a low-loss dielectric to a perfect conductor.
- The exact formulas for the attenuation constant \(\alpha\) and phase constant \(\beta\) in any medium.
- The attenuated wave \(E_0e^{-\alpha z}\cos(\omega t-\beta z)\) and the neper-to-decibel conversion.
- The complex intrinsic impedance and the resulting \(\vec{E}\!-\!\vec{H}\) phase lag.
- The low-loss dielectric limit \(\sigma/\omega\varepsilon\ll1\) and its simple approximations.
- The good-conductor limit \(\sigma/\omega\varepsilon\gg1\), where \(\alpha=\beta=\sqrt{\pi f\mu\sigma}\).
- Skin depth \(\delta=1/\alpha\) and why fields hug the surface of a metal.
The General Propagation Constant
Chapter 17 wrote the propagation constant as \(\gamma=\sqrt{j\omega\mu(\sigma+j\omega\varepsilon)}=\alpha+j\beta\). Squaring and matching real and imaginary parts gives two equations for the two unknowns; solving them yields closed forms that hold for every linear medium, lossless to highly conducting:
Everything is governed by the lone dimensionless ratio \(\sigma/\omega\varepsilon\) — the loss tangent from Chapter 17. When it is tiny the bracket for \(\alpha\) collapses toward zero (almost no loss); when it is huge, \(\alpha\) and \(\beta\) grow together and become equal. The phase velocity and wavelength keep their definitions but now read off the loss-aware \(\beta\):
The same material can sit at either extreme depending on frequency, because \(\omega\) lives in the denominator. The two limits below are not different physics — just the general formulas evaluated where the bracket simplifies.
The Attenuated Travelling Wave
With \(\gamma=\alpha+j\beta\), the forward wave from Chapter 18 picks up a real exponential factor. The phasor \(E_0e^{-\gamma z}=E_0e^{-\alpha z}e^{-j\beta z}\) gives a field that oscillates and shrinks as it advances:
Attenuation is quoted in nepers per metre (the natural-log measure) or, more often in engineering, in decibels. The two scales are fixed multiples of one another:
Complex Intrinsic Impedance
The intrinsic impedance from Chapter 18 generalises by carrying the loss term inside the square root. It is now a complex number, with a magnitude and a phase angle:
The phase angle \(\theta_\eta\) is the price of loss. In free space it was zero and the fields rose and fell together; here \(\vec{H}\) lags \(\vec{E}\) by \(\theta_\eta\), which ranges from \(0^\circ\) (perfect dielectric) up to a hard ceiling of \(45^\circ\) (perfect conductor). That lag is exactly why a conductor stores and dissipates power differently from vacuum — a theme Chapter 20 makes quantitative.
Low-Loss Dielectrics
When \(\sigma/\omega\varepsilon\ll1\) — a good insulator, or any dielectric well above its crossover frequency — expand the brackets to first order. The loss appears only as a small \(\alpha\); the phase behaviour is essentially that of the lossless wave:
The striking feature is that \(\alpha\) is (to leading order) independent of frequency: a low-loss dielectric attenuates a 1 MHz and a 1 GHz wave by nearly the same nepers per metre, because the rise of \(\sigma\)-driven loss with \(\omega\) is cancelled by the \(\omega\) in the denominator of the loss tangent. Velocity, wavelength, and impedance stay within a whisker of their lossless values.
Good Conductors
At the opposite extreme \(\sigma/\omega\varepsilon\gg1\): the conduction current dwarfs the displacement current, so \(\sigma+j\omega\varepsilon\approx\sigma\) and \(\gamma\approx\sqrt{j\omega\mu\sigma}\). The square root of \(j\) contributes \(45^\circ\), splitting equally between \(\alpha\) and \(\beta\):
Three facts fall out at once. The attenuation \(\alpha\) now grows as \(\sqrt{f}\) — higher frequencies die faster. The impedance angle is locked at exactly \(45^\circ\), so \(\vec{H}\) lags \(\vec{E}\) by an eighth of a cycle. And because \(\alpha=\beta\), the wave loses most of its amplitude within a single wavelength. The table below collects all three regimes side by side:
| Regime | Attenuation α | Phase β | Impedance η |
|---|---|---|---|
| Lossless (σ = 0) | \(0\) | \(\omega\sqrt{\mu\varepsilon}\) | \(\sqrt{\mu/\varepsilon}\ \angle\,0^\circ\) |
| Low-loss dielectric | \(\tfrac{\sigma}{2}\sqrt{\mu/\varepsilon}\) | \(\approx\omega\sqrt{\mu\varepsilon}\) | \(\approx\sqrt{\mu/\varepsilon}\ \angle\,\text{small}\) |
| Good conductor | \(\sqrt{\pi f\mu\sigma}\) | \(\sqrt{\pi f\mu\sigma}\) | \(\sqrt{\omega\mu/\sigma}\ \angle\,45^\circ\) |
Skin Depth
The distance over which the wave's amplitude falls to \(1/e\approx 37\%\) of its surface value is the skin depth (or depth of penetration) \(\delta\). It is simply the reciprocal of the attenuation constant, and in a good conductor it takes a famously compact form:
Because \(\delta\propto 1/\sqrt{f}\), raising the frequency confines the current to an ever-thinner shell at the surface — the skin effect that drives up the AC resistance of wires and lets a thin metal screen shield a room.
The numbers are dramatic. The same copper that a 60 Hz wave penetrates a few millimetres into admits a microwave only a couple of micrometres — which is why a thin plating of silver works as well as solid metal at GHz frequencies, and why seawater (a modest conductor) still swallows radio so thoroughly that submarines must use extremely low frequencies:
| Medium | Conductivity σ (S/m) | Frequency | Skin depth δ |
|---|---|---|---|
| Copper | \(5.8\times10^{7}\) | 60 Hz | ≈ 8.5 mm |
| Copper | \(5.8\times10^{7}\) | 1 MHz | ≈ 66 μm |
| Copper | \(5.8\times10^{7}\) | 1 GHz | ≈ 2.1 μm |
| Seawater | ≈ 4 | 1 kHz | ≈ 8 m |
Worked Examples
Problem. A medium has \(\sigma=0.1\ \text{S/m}\), \(\varepsilon_r=4\), \(\mu_r=1\), at \(f=100\ \text{MHz}\). Find \(\alpha\) and \(\beta\).
Solution. First the loss tangent: \(\sigma/\omega\varepsilon=0.1/[(2\pi\cdot10^{8})(4\varepsilon_0)]\approx 4.49\) — neither limit applies, so use the exact formulas with \(\omega\sqrt{\mu\varepsilon}\approx 4.19\):
Problem. For the wave above (\(\alpha\approx 5.6\ \text{Np/m}\)), how much is it attenuated over \(0.5\ \text{m}\)?
Solution. Compute \(\alpha z\) in nepers, then convert:
Problem. Find the skin depth of copper (\(\sigma=5.8\times10^{7}\)) at \(1\ \text{MHz}\), and the field fraction surviving at \(3\delta\).
Solution. Use \(\delta=1/\sqrt{\pi f\mu_0\sigma}\):
Problem. For the medium of Example 1, find \(|\eta|\) and the angle by which \(\vec{H}\) lags \(\vec{E}\).
Solution. With \(\sqrt{\mu/\varepsilon}=\eta_0/2=188.5\ \Omega\) and \(\sigma/\omega\varepsilon=4.49\):
Problem. A dielectric has \(\varepsilon_r=2.25\), \(\mu_r=1\), and \(\tan\delta=\sigma/\omega\varepsilon=10^{-3}\) at \(3\ \text{GHz}\). Find \(\alpha\), \(\beta\), and \(\eta\).
Solution. Use the low-loss forms; \(\sqrt{\mu/\varepsilon}=\eta_0/1.5\approx 251\ \Omega\), \(\beta\approx(\omega/c)\sqrt{\varepsilon_r}\):
Problem. Seawater has \(\sigma=4\ \text{S/m}\). Find the skin depth at \(1\ \text{kHz}\) and the depth at which a signal falls to \(1\%\).
Solution. At \(1\ \text{kHz}\) seawater is a good conductor, so \(\delta=1/\sqrt{\pi f\mu_0\sigma}\); \(1\%\) means \(e^{-\alpha z}=0.01\Rightarrow \alpha z\approx 4.6\):
Chapter Summary
Exact \(\alpha,\beta\) from \(\gamma=\sqrt{j\omega\mu(\sigma+j\omega\varepsilon)}\); both set by \(\sigma/\omega\varepsilon\).
\(E_0e^{-\alpha z}\cos(\omega t-\beta z)\); loss (dB) \(=8.686\,\alpha z\).
\(\eta=|\eta|\angle\theta_\eta\); \(\vec{H}\) lags \(\vec{E}\) by \(0\!-\!45^\circ\).
\(\alpha\approx\tfrac{\sigma}{2}\sqrt{\mu/\varepsilon}\) (≈ frequency-independent); \(\beta\approx\omega\sqrt{\mu\varepsilon}\).
\(\alpha=\beta=\sqrt{\pi f\mu\sigma}\); \(\eta=\sqrt{\omega\mu/\sigma}\angle45^\circ\).
\(\delta=1/\alpha=1/\sqrt{\pi f\mu\sigma}\); fields hug the surface as \(f\) rises.
Problems
For each item, first compute \(\sigma/\omega\varepsilon\) to pick the regime, then reach for the matching formula — exact, low-loss, or good-conductor. Difficulty rises down the list.
- Show that the general \(\alpha\) and \(\beta\) reduce to \(0\) and \(\omega\sqrt{\mu\varepsilon}\) when \(\sigma=0\).
- A wave with \(\alpha=0.2\ \text{Np/m}\) travels \(10\ \text{m}\). Find the loss in dB and the surviving amplitude fraction.
- A medium has \(\varepsilon_r=9\), \(\mu_r=1\), \(\sigma=0\). Find \(\beta\), \(\lambda\), and \(\eta\) at \(500\ \text{MHz}\).
- For \(\sigma=5\ \text{S/m}\), \(\varepsilon_r=80\) at \(1\ \text{MHz}\), compute \(\sigma/\omega\varepsilon\) and classify the medium.
- Using the result of Problem 4, find \(\alpha\), \(\beta\), and \(\delta\) with the good-conductor formulas.
- Find the skin depth of aluminium (\(\sigma=3.5\times10^{7}\)) at \(1\ \text{MHz}\).
- By what factor does copper's skin depth change between \(1\ \text{kHz}\) and \(1\ \text{GHz}\)?
- A good conductor has \(\eta=(1+j)\,\Omega\) at some frequency. Find \(|\eta|\) and \(\theta_\eta\), and confirm the \(45^\circ\) rule.
- Show that in a good conductor the wavelength inside the metal is \(\lambda=2\pi\delta\).
- A \(1\ \text{GHz}\) wave enters copper. How many skin depths reduce the field to \(0.1\%\) of its surface value?
- For the low-loss limit, show that \(\alpha\) is independent of frequency while \(\beta\) is nearly linear in \(\omega\).
- Explain physically why \(\theta_\eta\) cannot exceed \(45^\circ\), no matter how large \(\sigma/\omega\varepsilon\) becomes.