Plasma sheath — communications blackout: Worked examples

Before we touch a single example, we must earn every symbol we'll use. The parent note used $ω$ , $ω_{p}$ , $k$ , $c$ and $n$ freely — here we build each from the ground up, in plain words and pictures.

Definition The symbols we will use — each defined before use

$f$ — ordinary frequency: how many full wobbles per second a wave makes. Unit: hertz (Hz).
$ω$ (Greek "omega") — angular frequency: the same wobble measured in radians per second. One full wobble is $2 π$ radians, so $ω = 2 π f$ . Think of a dot going round a circle: $f$ counts laps, $ω$ counts radians swept.
$f_{p}$ — plasma frequency in Hz; $ω_{p} = 2 π f_{p}$ is its angular version. Same physical "sloshing pitch", two unit systems.
$c$ — the speed of light, $c = 3 \times 1 0^{8} m/s$ : how fast the wave would travel in empty space.
$k$ — the wavenumber: how many radians of the wave's phase fit into one metre of space. Unit: $m^{- 1}$ . If $k$ is a real number the wave marches forward; if $k$ turns out imaginary the wave decays in place instead of travelling.
$n$ — the refractive index: the factor by which the plasma slows the wave's phase compared to vacuum. In vacuum $n = 1$ .

Formula Where the dispersion relation and refractive index come from

A plane wave travelling at speed $v = ω / k$ inside the plasma must obey Maxwell's equations. Working them through for a cloud of free electrons oscillating at $ω_{p}$ gives the dispersion relation — the rule linking a wave's frequency to its wavenumber: $ω^{2} = ω_{p}^{2} + c^{2} k^{2} ⟹ k = \frac{1}{c} ω^{2} - ω_{p}^{2}$ Read it as: of the wave's total "frequency budget" $ω^{2}$ , the plasma spends $ω_{p}^{2}$ just making electrons slosh; only the leftover $ω^{2} - ω_{p}^{2}$ is available to push the wave forward through space as $c^{2} k^{2}$ . The refractive index is defined as $n = c k / ω$ (vacuum speed over the wave's actual phase speed). Substituting $k$ from above: $n = \frac{c k}{ω} = 1 - \frac{ω _{p}^{2}}{ω ^{2}} = 1 - \frac{f _{p}^{2}}{f ^{2}}$ (the $2 π$ 's in $ω = 2 π f$ and $ω_{p} = 2 π f_{p}$ cancel top and bottom, so it works with plain $f$ too). This single formula drives Examples 4, 6 and 7.

Now the two everyday workhorses we'll plug numbers into:

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. Each row is one "shape" of question; the worked examples below are labelled with the cell they cover.

Cell	What is unknown / special	Typical question	Example
A — forward	given $n_{e}$ , find $f_{p}$	"What's the plasma frequency?"	Ex 1
B — compare	given $n_{e}$ and $f_{radio}$ , decide	"Blackout or not?"	Ex 2
C — inverse	given $f_{radio}$ , find $n_{crit}$	"What density blocks me?"	Ex 3
D — boundary	$f_{radio} = f_{p}$ exactly	"What happens at cutoff?"	Ex 4
E — degenerate	$n_{e} = 0$ (or un-ionized)	"No electrons — does anything block?"	Ex 5
F — limiting	$f_{radio} ≫ f_{p}$	"How much does a huge frequency slow?"	Ex 6
G — depth/decay	evanescent skin depth below cutoff	"How thick a sheath actually blocks?"	Ex 7
H — real-world word problem	S-band capsule at re-entry	"Design a working link"	Ex 8
I — exam twist	wrong tool trap ( $ω$ vs $f$ )	"Spot the factor of $2 π$ "	Ex 9

We hit A, B, C, D, E, F, G, H, I — every cell. The picture below is the map for all of them: it plots the refractive index $n$ against frequency and marks the cutoff we keep returning to.

Figure — Plasma sheath — communications blackout

Worked examples

Worked example Example 1 — Cell A (forward): plasma frequency of a dense sheath

A peak re-entry sheath has $n_{e} = 4 \times 1 0^{18} m^{- 3}$ . Find $f_{p}$ .

Forecast: Guess — will $f_{p}$ be closer to 1 GHz or 20 GHz?

Take the square root of the density: $4 \times 1 0^{18} = 2 \times 1 0^{9}$ . Why this step? The formula wants $n_{e}$ because the restoring field grows with electron density, and frequency of an oscillator grows with the square root of stiffness.
Multiply by the constant: $f_{p} = 8.98 \times 2 \times 1 0^{9} = 1.796 \times 1 0^{10} Hz \approx 18 GHz$ . Why this step? $8.98$ bundles $e$ , $ε_{0}$ , $m_{e}$ and the $1/2 π$ into one number that already outputs hertz.

Verify: Units — $m^{- 3} = m^{- 1.5}$ , and the constant carries the rest to give Hz (checked numerically below). A denser sheath than Ex 1 of the parent ( $1 0^{18} \to 9$ GHz) should give a higher $f_{p}$ , and $18 > 9$ . ✓

Worked example Example 4 — Cell D (boundary): exactly at cutoff

A sheath happens to have $n_{e} = n_{crit}$ for your frequency, i.e. $f_{radio} = f_{p}$ exactly (equivalently $ω = ω_{p}$ , since $ω = 2 π f$ ). What is the wavenumber $k$ , and what physically happens?

Forecast: Does the wave travel, stop, or reflect?

Recall the dispersion relation (built above) $ω^{2} = ω_{p}^{2} + c^{2} k^{2}$ , so $k = \frac{1}{c} ω^{2} - ω_{p}^{2}$ . Why this step? $k$ (radians of phase per metre) tells us whether a wave marches forward ( $k$ real) or dies in place ( $k$ imaginary). It is the single quantity that decides "travel or not".
At $ω = ω_{p}$ : $ω^{2} - ω_{p}^{2} = 0$ , so $k = \frac{1}{c} 0 = 0$ . Why this step? We substitute the equality into the relation; the frequency budget is entirely spent on sloshing electrons, leaving nothing ( $c^{2} k^{2} = 0$ ) to push the wave forward.
$k = 0$ means infinite wavelength, zero forward progress — the wave neither propagates energy into the plasma nor cleanly reflects; it's the knife-edge between the two. Why this step? Wavelength is $λ = 2 π / k$ , and as $k \to 0$ this blows up to $\infty$ : no spatial oscillation at all.

Verify: Use the refractive index $n = 1 - ω_{p}^{2} / ω^{2}$ (defined as $n = c k / ω$ above). At $ω = ω_{p}$ , $n = 1 - 1 = 0$ . A zero refractive index is exactly the boundary — real for $ω > ω_{p}$ , imaginary for $ω < ω_{p}$ . This is the pinch point of both curves in figure s01. ✓

Worked example Example 6 — Cell F (limiting): a huge frequency barely notices the plasma

A Ka-band link at $f_{radio} = 26 GHz$ passes through a sheath with $f_{p} = 9 GHz$ . By what factor is its refractive index below 1 (how much is it "slowed")?

Forecast: Barely, or a lot?

Form the ratio $\frac{f _{p}}{f _{radio}} = \frac{9}{26} = 0.3462$ (equal to $ω_{p} / ω$ , since the $2 π$ 's cancel). Why this step? From the derivation above, the refractive index depends only on this one ratio: $n = 1 - (f_{p} / f)^{2}$ . Nothing else about the wave matters.
Square it: $0.346 2^{2} = 0.1198$ . Why this step? The formula subtracts the square of the ratio, because the dispersion relation is built from $ω^{2}$ and $ω_{p}^{2}$ .
Refractive index: $n = 1 - 0.1198 = 0.8802 = 0.938$ . Why this step? Taking the root converts the frequency budget back into an actual phase-slowing factor.

Verify: $n = 0.938$ is close to 1 — the wave sails through almost as if in vacuum. As $f_{radio} \to \infty$ , $(f_{p} / f)^{2} \to 0$ so $n \to 1$ (limiting behaviour): huge frequencies see the electrons as effectively frozen. On figure s01 this is the mint curve creeping up toward $n = 1$ on the far right. ✓

Worked example Example 7 — Cell G (depth/decay): how far does a blocked wave penetrate?

A 300 MHz wave hits a sheath with $f_{p} = 2 GHz$ . Below cutoff, $k$ is imaginary: write $k = iκ$ , so the field goes as $e^{ik z} = e^{- κ z}$ — a decay, not a wave. Find the skin depth $δ = 1/ κ$ , the distance over which the field drops to $1/ e$ (about 37%).

Forecast: Millimetres, centimetres, or metres?

Below cutoff, $ω < ω_{p}$ makes $ω^{2} - ω_{p}^{2}$ negative, so $κ = \frac{1}{c} ω_{p}^{2} - ω^{2} = \frac{2 π}{c} f_{p}^{2} - f^{2}$ . Why this step? When the thing under the root in $k = \frac{1}{c} ω^{2} - ω_{p}^{2}$ goes negative, $k$ becomes imaginary. Flipping the order under the root gives a real decay rate $κ$ — the fingerprint of an evanescent (decaying) field rather than a travelling one.
Evaluate the root: $(2 \times 1 0^{9})^{2} - (3 \times 1 0^{8})^{2} = 4 \times 1 0^{18} - 9 \times 1 0^{16} = 3.91 \times 1 0^{18} = 1.977 \times 1 0^{9}$ . Why this step? We plug $f_{p}$ and $f$ into the leftover-budget $f_{p}^{2} - f^{2}$ ; the small $f^{2}$ barely dents the big $f_{p}^{2}$ , telling us the wave is deep below cutoff and will die fast.
Attach the $2 π / c$ prefactor: $κ = \frac{2 π}{3 \times 1 0 ^{8}} \times 1.977 \times 1 0^{9} = 41.4 m^{- 1}$ . Why this step? The $2 π / c$ converts a frequency-in-Hz into radians-of-decay-per-metre — that is exactly what a decay rate must have units of ( $m^{- 1}$ ).
Invert to get depth: $δ = 1/ κ = 0.0241 m \approx 2.4 cm$ . Why this step? Skin depth is the reciprocal of the decay rate — a big $κ$ means the field dies within a short distance.

Verify: A real sheath is centimetres-thick, comparable to $δ$ , so the wave is knocked down by roughly a factor $e$ or more per sheath — enough for total blackout. Units: $[κ] = m^{- 1}$ so $δ$ is in metres. Figure s02 shows exactly this decay. ✓

Worked example Example 8 — Cell H (real-world word problem): design a survivable link

A crewed capsule's peak sheath is $n_{e} = 1.5 \times 1 0^{18} m^{- 3}$ . Mission rules require a direct link (no relay). Which of these bands works: UHF 0.4 GHz, S-band 2.2 GHz, X-band 10 GHz, Ka-band 32 GHz?

Forecast: How many of the four survive?

Find $f_{p}$ : $1.5 \times 1 0^{18} = 1.225 \times 1 0^{9}$ ; $f_{p} = 8.98 \times 1.225 \times 1 0^{9} = 1.100 \times 1 0^{10} Hz \approx 11 GHz$ . Why this step? One number decides everything — everyone must beat it.
Compare each band to 11 GHz:
- UHF 0.4 GHz $< 11$ → blocked.
- S-band 2.2 GHz $< 11$ → blocked.
- X-band 10 GHz $< 11$ → blocked (just barely!).
- Ka-band 32 GHz $> 11$ → passes. ✓ Why this step? The inequality is applied band by band; note X-band loses by only 1 GHz, showing how a slightly denser sheath flips the answer.
Decision: only Ka-band survives a direct link.

Verify: This matches parent Example 4's advice — "raise $f_{radio}$ above $f_{p}$ (Ka-band)." And it explains why real capsules also carry a relay-via-wake option as backup. ✓

Worked example Example 9 — Cell I (exam twist): the

$ω$ -versus- $f$ trap An exam says: "A sheath has $f_{p} = 3 GHz$ . A student computes the angular plasma frequency $ω_{p}$ and then wrongly declares blackout for any radio angular frequency below $3 \times 1 0^{9}$ ." What is the true $ω_{p}$ , and where's the mistake?

Forecast: Is $ω_{p}$ bigger or smaller than $3 \times 1 0^{9}$ ?

Convert correctly: $ω_{p} = 2 π f_{p} = 2 π \times 3 \times 1 0^{9} = 1.885 \times 1 0^{10} rad/s$ . Why this step? $ω = 2 π f$ always — angular frequency counts radians, ordinary frequency counts full cycles (this is the very first symbol we defined).
The student compared a radio $ω$ (rad/s) against the number $3 \times 1 0^{9}$ which is actually $f_{p}$ in Hz, not $ω_{p}$ . They mixed units by a factor of $2 π$ . Why this step? The $8.98 n_{e}$ shortcut outputs $f_{p}$ in Hz; you must not compare it to an angular frequency.
Correct rule: compare like with like — either $f_{radio}$ vs $f_{p}$ , or $ω_{radio}$ vs $ω_{p}$ .

Verify: $ω_{p} / f_{p} = 1.885 \times 1 0^{10} /3 \times 1 0^{9} = 6.283 = 2 π$ . ✓ Exactly the stray factor the parent-note mistake warns about.

Active recall

Recall Test yourself

How is angular frequency $ω$ related to ordinary frequency $f$ ? ::: $ω = 2 π f$ — same wobble, counted in radians per second instead of cycles per second. What does the wavenumber $k$ tell you, and what does an imaginary $k$ mean? ::: $k$ is radians of phase per metre; real $k$ = travelling wave, imaginary $k$ = evanescent (decaying) field with skin depth $δ = 1/ κ$ . Where does $n = 1 - (f_{p} / f)^{2}$ come from? ::: From $n = c k / ω$ combined with the dispersion relation $ω^{2} = ω_{p}^{2} + c^{2} k^{2}$ . At exactly $f_{radio} = f_{p}$ , what is $k$ and the refractive index? ::: $k = 0$ and $n = 0$ — the knife-edge between propagation and reflection. If $n_{e} = 0$ , what is $f_{p}$ and does anything block the wave? ::: $f_{p} = 0$ ; no frequency is below it, so everything passes ( $n = 1$ ). The $8.98 n_{e}$ shortcut gives which quantity? ::: $f_{p}$ in hertz — NOT $ω_{p}$ . Don't add a stray $2 π$ .