Exercises — Plasma sheath — communications blackout

What we do: apply the engineering shortcut directly — no derivation needed, this is a recognition drill. $f_p=8.98\sqrt{n_e}=8.98\sqrt{10^{16}}=8.98\times 10^8\text{Hz}\approx 0.90\text{GHz}.$ What it means: the square root of $10^{16}$ is $10^8$ (halve the exponent). Multiply by $8.98$. Answer $\approx 900\text{MHz}$.

What we do: compare $f_{\text{radio}}$ to $f_p$ — that is the entire blackout test. Since $0.40\text{GHz}<0.90\text{GHz}$, we have $f_{\text{radio}}<f_p$ → the wave is reflected → blackout. Look back at the s01 figure: $0.40$ GHz sits in the shaded reflect region, to the left of the red line.

Step 1 — build the fraction inside the root. $\frac{n_e{e}^2}{{\epsilon }_0{m}_e}=\frac{(5\times 10^{17})(1.6\times 10^{-19}{)}^2}{(8.85\times 10^{-12})(9.11\times 10^{-31})}.$ Numerator: $(5\times 10^{17})(2.56\times 10^{-38})=1.28\times 10^{-20}$. Denominator: $(8.85\times 10^{-12})(9.11\times 10^{-31})=8.06\times 10^{-42}$. Ratio: $1.28\times 10^{-20}/8.06\times 10^{-42}=1.588\times 10^{21}{\text{s}}^{-2}$.

Step 2 — square root gives ${\omega }_p$. ${\omega }_p=\sqrt{1.588\times 10^{21}}=3.98\times 10^{10}\text{rad/s}.$

Step 3 — divide by $2\pi$ to get $f_p$ in Hz. $f_p=\frac{3.98\times 10^{10}}{2\pi }=6.34\times 10^9\text{Hz}\approx 6.3\text{GHz}.$

Cross-check: the shortcut gives $8.98\sqrt{5\times 10^{17}}=8.98\times (7.07\times 10^8)=6.35\times 10^9$ Hz. ✓ The full formula and the shortcut agree, as they must — the shortcut is $\frac{1}{2\pi }\sqrt{e^2/({\epsilon }_0{m}_e)}$ baked into one constant.

What we do: blackout begins exactly when $f_p=f_{\text{radio}}$, so set the shortcut equal to the radio frequency and solve for $n_e$. $f=8.98\sqrt{n_{\text{crit}}}\Rightarrow n_{\text{crit}}={\left(\frac{f}{8.98}\right)}^2.$ $n_{\text{crit}}={\left(\frac{2.2\times 10^9}{8.98}\right)}^2=(2.45\times 10^8{)}^2=6.0\times 10^{16}{\text{m}}^{-3}.$ What it means: any sheath denser than $6.0\times 10^{16}{\text{m}}^{-3}$ kills the 2.2 GHz link. Peak re-entry densities ($10^{18}$–$10^{19}$) blow past this — guaranteed blackout.

What we do: we do NOT need any constants. Because $f_p\propto \sqrt{n_e}$, we have $n_{\text{crit}}\propto f^2$, so the ratio of critical densities is the square of the frequency ratio. $\frac{n_{\text{crit},2}}{n_{\text{crit},1}}={\left(\frac{f_2}{f_1}\right)}^2={\left(\frac{26}{2.2}\right)}^2=(11.82{)}^2=139.7\approx 140.$ What it means: Ka-band tolerates a plasma about 140× denser before it too gets blocked. That is the whole reason "just use a higher frequency" is a real engineering fix — the tolerance grows as the square of frequency, not linearly.

Step 1 — form the ratio ${\omega }_p^2/{\omega }^2$. Since $\omega =1.5{\omega }_p$, we get ${\omega }_p^2/{\omega }^2=1/1.5^2=1/2.25=0.4444$.

Step 2 — subtract and root. $n=\sqrt{1-0.4444}=\sqrt{0.5556}=\mathrm{0.745.}$

Step 3 — interpret. $n$ is a real number between 0 and 1 → the wave propagates (it is above cutoff). Had $\omega$ been below ${\omega }_p$, the quantity under the root would go negative and $n$ would be imaginary → total reflection. Notice $n<1$: the phase speed $c/n$ exceeds $c$, which is fine — no information travels faster than light, only the phase pattern does.

Step 1 — evaluate the density at $d=6$ cm. $n_e(6)=10^{19}{e}^{-6/2}=10^{19}{e}^{-3}=10^{19}\times 0.0498=4.98\times 10^{17}{\text{m}}^{-3}.$ The exponential drop of the wake is exactly why relay-through-the-wake works: three decay lengths cut the density by $e^{-3}\approx 20\times$.

Step 2 — plasma frequency there. $f_p=8.98\sqrt{4.98\times 10^{17}}=8.98\times (7.06\times 10^8)=6.34\times 10^9\text{Hz}\approx 6.3\text{GHz}.$

Step 3 — compare with the link. $f_{\text{radio}}=26\text{GHz}>f_p=6.3\text{GHz}$ → the wave gets through. ✓

What it means: two fixes stacked — going to Ka-band and looking at the thinned wake — together clear the link. Either alone might have failed at the dense stagnation point ($n_0=10^{19}$ gives $f_p\approx 28.4$ GHz, which even Ka-band cannot beat).

Step 1 — target density is the critical density for 2.2 GHz (from L2-Q2): $n_{\text{crit}}=6.0\times 10^{16}{\text{m}}^{-3}$.

Step 2 — required reduction factor. $\frac{n_0}{n_{\text{crit}}}=\frac{1\times 10^{19}}{6.0\times 10^{16}}=167.$ What it means: you must destroy about 99.4% of the free electrons ($1-1/167=0.994$) to reopen S-band. That is a brutal target — which is why quenchant injection is hard, and why relay-through-the-wake or higher frequency are usually preferred.

(a) Wavenumber. Set $\omega ={\omega }_p$: ${\omega }_p^2={\omega }_p^2+c^2{k}^2\Rightarrow c^2{k}^2=0\Rightarrow k=0.$ (b) Refractive index. $n=\sqrt{1-{\omega }_p^2/{\omega }^2}=\sqrt{1-1}=0.$ (c) Physical picture. $k=0$ means the spatial wavelength $\lambda =2\pi /k$ is infinite — the field oscillates in time everywhere in phase, with no propagation in space. The wave neither travels forward nor decays exponentially; it sits at the exact knife-edge between the propagating region ($\omega >{\omega }_p$, $k$ real) and the evanescent region ($\omega <{\omega }_p$, $k$ imaginary). This is the boundary case that separates the two halves of the s01 figure — the red line itself.

Step 1 — convert to angular frequencies. $\omega =2\pi (1\times 10^9)=6.283\times 10^9\text{rad/s},{\omega }_p=2\pi (2\times 10^9)=1.2566\times 10^{10}\text{rad/s}.$

Step 2 — form $\sqrt{{\omega }_p^2-{\omega }^2}$. ${\omega }_p^2=1.579\times 10^{20},{\omega }^2=3.948\times 10^{19},{\omega }_p^2-{\omega }^2=1.184\times 10^{20}.$ $\sqrt{1.184\times 10^{20}}=1.088\times 10^{10}{\text{s}}^{-1}.$

Step 3 — divide by $c$ to get $|k|$, then invert for $\delta$. $|k|=\frac{1.088\times 10^{10}}{3\times 10^8}=36.3{\text{m}}^{-1},\delta =\frac{1}{|k|}=0.0276\text{m}\approx 2.8\text{cm}.$ What it means: the blocked wave dies within a few centimetres of penetration — comparable to the sheath thickness itself, which is exactly why the capsule is sealed off. Below cutoff the plasma is not merely a poor conductor; it is an efficient exponential wall.

Plasma frequency: $f_p=8.98\sqrt{n_e}\to 8.98\sqrt{0}=0$. With no free electrons there is no restoring sheet-charge (Step 1 of the parent derivation gives $\sigma =n_{e}ex=0$), so nothing sloshes — the natural frequency collapses to zero. Critical density: for any positive $f_{\text{radio}}$, $n_{\text{crit}}=(f/8.98{)}^2>0$, so the vanishing $n_e=0$ is always below it. Blackout condition: $f_{\text{radio}}<f_p=0$ is impossible for a real radio, so no blackout — every frequency propagates. Why it must be true: a plasma with no free electrons is just neutral gas, and neutral gas is transparent to radio. The formula correctly reproduces "empty space lets all radio through." A physical law that failed this limit would be wrong.

Blackout test in one inequality ::: $f_{\text{radio}}<f_p$, i.e. $n_e>n_{\text{crit}}$. Scaling of critical density with radio frequency ::: $n_{\text{crit}}\propto f^2$ (square-law, not linear). Refractive index at exact cutoff $\omega ={\omega }_p$ ::: $n=0$ and $k=0$ — infinite wavelength, no propagation. Skin depth below cutoff ::: $\delta =1/|k|=c/\sqrt{{\omega }_p^2-{\omega }^2}$. Zero-density limit ::: $f_p\to 0$, no blackout — neutral gas is transparent. Is a below-cutoff wave absorbed or reflected? ::: Reflected (in a lossless plasma); it decays as $e^{-x/\delta }$ but carries no net power inward.