The Ionosphere as a Mirror

The problem

Why does shortwave radio travel around the world at night, while FM barely reaches the next town and GPS punches straight through? One dispersion relation answers all three.

The physics in one line

\[\omega^2 = \omega_{pe}^2 + c^2k^2 \qquad\Longrightarrow\qquad n_{\rm ref}^2 = 1 - \frac{\omega_{pe}^2}{\omega^2}\]

Below \(\omega_{pe}\) the refractive index is imaginary: the wave cannot propagate and is reflected. Above it, the wave passes.

Step 1: the critical frequency

The F2 layer peaks around 300 km altitude with a daytime density of about \(n_e = 10^{12}\) m⁻³ \(= 10^6\) cm⁻³:

\[f_{pe} = 8980\sqrt{n_e[\text{cm}^{-3}]} = 8980\sqrt{10^6} = 8.98\times10^6\ \text{Hz} \approx 9\ \text{MHz}\]

So the vertical critical frequency is about 9 MHz. Anything below bounces; anything above escapes. This is measured hourly worldwide by ionosondes, which sweep frequency and time the echo — the resulting ionogram is a direct plot of density against altitude.

Step 2: the classification

Service	Frequency	vs. \(f_{pe} \approx 9\) MHz	Behaviour
AM broadcast	0.5–1.7 MHz	far below	reflects
Shortwave	3–30 MHz	straddles	reflects at the low end
FM broadcast	88–108 MHz	far above	passes through
TV / mobile	0.5–3 GHz	far above	passes through
GPS	1.2 / 1.5 GHz	far above	passes through

Shortwave broadcasting occupies exactly the band that straddles the critical frequency, and it is not a coincidence: the band was chosen because it bounces.

Step 3: oblique incidence buys headroom

A wave arriving at angle \(\theta\) from vertical reflects when

\[f = f_{pe}\sec\theta\]

At a grazing \(\theta = 80°\), \(\sec\theta = 5.8\), so a 9 MHz critical frequency reflects signals up to about 52 MHz. This is why long-distance shortwave uses low takeoff angles: skipping at a shallow angle both raises the usable frequency and covers more ground per hop. Multi-hop paths bouncing between ionosphere and ocean circle the planet.

Step 4: why night changes everything

The D layer (~60–90 km) is dense enough to absorb — not reflect — medium-wave signals during the day, because collisions there are frequent and the wave energy is dissipated. After sunset the D layer's ionisation source is removed and it recombines within about an hour, while the F layer, far thinner and slower to recombine, persists all night.

Result: after dark, AM signals reach the F layer intact and reflect. A 50 kW AM station audible for 100 km at noon can be heard 2000 km away at midnight. This is why the FCC requires many US stations to reduce power or shut down at sunset, and why the phenomenon has a name — "skywave" — in broadcast regulation.

Step 5: what GPS pays anyway

Well above cutoff the ionosphere is nearly transparent — but \(n_{\rm ref} < 1\) still, so the group delay is non-zero and depends on the integrated electron content along the path:

\[\Delta t \propto \frac{1}{f^2}\int n_e\,dl\]

That delay is tens of nanoseconds, worth 5–15 m of position error — far too much for precision work. Since it scales as \(f^{-2}\), transmitting on two frequencies (L1 at 1575 MHz and L2 at 1228 MHz) allows the delay to be solved for and removed. Dual-frequency GPS exists because of this dispersion relation, and the correction is a direct measurement of the ionosphere's total electron content — GPS receivers are now one of the largest ionospheric sensor networks in existence.

Worked check

An ionosonde returns an echo at 8 MHz but not 10 MHz. What is the peak density?

\[\sqrt{n_e[\text{cm}^{-3}]} = \frac{8\times10^6}{8980} = 891 \qquad n_e = 7.9\times10^5\ \text{cm}^{-3} = 7.9\times10^{11}\ \text{m}^{-3}\]

EM waves in plasma · Plasma frequency · Dielectric tensor — add Earth's field and get Faraday rotation too · Cold plasma wave explorer · Debye shielding lab