Exercises — Plasma sheath — communications blackout

Kya karein: engineering shortcut directly apply karo — koi derivation nahi chahiye, yeh ek recognition drill hai. $f_p=8.98\sqrt{n_e}=8.98\sqrt{10^{16}}=8.98\times 10^8\text{Hz}\approx 0.90\text{GHz}.$ Matlab kya hai: $10^{16}$ ka square root $10^8$ hota hai (exponent ko aadha karo). $8.98$ se multiply karo. Answer $\approx 900\text{MHz}$.

Kya karein: $f_{\text{radio}}$ ko $f_p$ se compare karo — yahi poora blackout test hai. Kyunki $0.40\text{GHz}<0.90\text{GHz}$ hai, hamare paas $f_{\text{radio}}<f_p$ hai → wave reflect ho jaati hai → blackout. s01 figure dekho: $0.40$ GHz shaded reflect region mein hai, red line ke left mein.

Step 1 — root ke andar fraction banao. $\frac{n_e{e}^2}{{\epsilon }_0{m}_e}=\frac{(5\times 10^{17})(1.6\times 10^{-19}{)}^2}{(8.85\times 10^{-12})(9.11\times 10^{-31})}.$ 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 se ${\omega }_p$ milta hai. ${\omega }_p=\sqrt{1.588\times 10^{21}}=3.98\times 10^{10}\text{rad/s}.$

Step 3 — $2\pi$ se divide karke $f_p$ Hz mein nikalo. $f_p=\frac{3.98\times 10^{10}}{2\pi }=6.34\times 10^9\text{Hz}\approx 6.3\text{GHz}.$

Cross-check: shortcut deta hai $8.98\sqrt{5\times 10^{17}}=8.98\times (7.07\times 10^8)=6.35\times 10^9$ Hz. ✓ Full formula aur shortcut agree karte hain, jaisa hona chahiye — shortcut hi hai $\frac{1}{2\pi }\sqrt{e^2/({\epsilon }_0{m}_e)}$ ek constant mein baked.

Kya karein: blackout tab shuru hota hai jab exactly $f_p=f_{\text{radio}}$, toh shortcut ko radio frequency ke barabar set karo aur $n_e$ solve karo. $f=8.98\sqrt{n_{\text{crit}}}\Rightarrow n_{\text{crit}}={\left(\frac{f}{8.98}\right)}^2.$ $n_{\text{crit}}={\left(\frac{2.2\times 10^9}{8.98}\right)}^2=(2.45\times 10^8{)}^2=6.0\times 10^{16}{\text{m}}^{-3}.$ Matlab kya hai: $6.0\times 10^{16}{\text{m}}^{-3}$ se zyada dense koi bhi sheath 2.2 GHz link ko khatam kar deti hai. Peak re-entry densities ($10^{18}$–$10^{19}$) is se kaafi aage hain — guaranteed blackout.

Kya karein: hume koi constants ki zaroorat nahi. Kyunki $f_p\propto \sqrt{n_e}$ hai, hamare paas $n_{\text{crit}}\propto f^2$ hai, toh critical densities ka ratio frequency ratio ka square hai. $\frac{n_{\text{crit},2}}{n_{\text{crit},1}}={\left(\frac{f_2}{f_1}\right)}^2={\left(\frac{26}{2.2}\right)}^2=(11.82{)}^2=139.7\approx 140.$ Matlab kya hai: Ka-band ek plasma ko roughly 140× denser tolerate karta hai block hone se pehle. Yahi reason hai ki "bas higher frequency use karo" ek real engineering fix hai — tolerance frequency ke square ki tarah badhti hai, linearly nahi.

Step 1 — ratio ${\omega }_p^2/{\omega }^2$ banao. Kyunki $\omega =1.5{\omega }_p$ hai, humein milta hai ${\omega }_p^2/{\omega }^2=1/1.5^2=1/2.25=0.4444$.

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

Step 3 — interpret karo. $n$ ek real number hai 0 aur 1 ke beech → wave propagate karti hai (yeh cutoff se upar hai). Agar $\omega$ neeche hota ${\omega }_p$ se, toh root ke andar quantity negative ho jaati aur $n$ imaginary hoti → total reflection. Note karo $n<1$: phase speed $c/n$ ka $c$ se zyada hona theek hai — koi information $c$ se tez nahi jaati, sirf phase pattern jaata hai.

Step 1 — $d=6$ cm pe density evaluate karo. $n_e(6)=10^{19}{e}^{-6/2}=10^{19}{e}^{-3}=10^{19}\times 0.0498=4.98\times 10^{17}{\text{m}}^{-3}.$ Wake ka exponential drop exactly yahi reason hai ki relay-through-the-wake kaam karta hai: teen decay lengths density ko $e^{-3}\approx 20\times$ kaat dete hain.

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

Step 3 — link se compare karo. $f_{\text{radio}}=26\text{GHz}>f_p=6.3\text{GHz}$ → wave nikal jaati hai. ✓

Matlab kya hai: do fixes stack hue — Ka-band pe jaana aur thinned wake dekhna — milkar link clear kar dete hain. Akele mein se koi bhi dense stagnation point pe fail ho sakta tha ($n_0=10^{19}$ deta hai $f_p\approx 28.4$ GHz, jise Ka-band bhi beat nahi kar sakta).

Step 1 — target density 2.2 GHz ke liye critical density hai (L2-Q2 se): $n_{\text{crit}}=6.0\times 10^{16}{\text{m}}^{-3}$.

Step 2 — required reduction factor. $\frac{n_0}{n_{\text{crit}}}=\frac{1\times 10^{19}}{6.0\times 10^{16}}=167.$ Matlab kya hai: S-band reopen karne ke liye tumhe roughly 99.4% free electrons destroy karne honge ($1-1/167=0.994$). Yeh ek brutal target hai — yahi reason hai ki quenchant injection mushkil hai, aur relay-through-the-wake ya higher frequency usually prefer ki jaati hai.

(a) Wavenumber. $\omega ={\omega }_p$ set karo: ${\omega }_p^2={\omega }_p^2+c^2{k}^2\Rightarrow c^2{k}^2=0\Rightarrow k=0.$ (b) Refractive index. $n=\sqrt{1-{\omega }_p^2/{\omega }^2}=\sqrt{1-1}=0.$ (c) Physical picture. $k=0$ ka matlab hai spatial wavelength $\lambda =2\pi /k$ infinite hai — field har jagah time mein oscillate karta hai same phase mein, space mein koi propagation nahi. Wave na toh aage jaati hai na exponentially decay karti hai; yeh exactly propagating region ($\omega >{\omega }_p$, $k$ real) aur evanescent region ($\omega <{\omega }_p$, $k$ imaginary) ke beech knife-edge pe hai. Yeh boundary case hai jo s01 figure ke do halfon ko alag karta hai — red line khud.

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

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

Step 3 — $c$ se divide karke $|k|$ nikalo, phir $\delta$ ke liye invert karo. $|k|=\frac{1.088\times 10^{10}}{3\times 10^8}=36.3{\text{m}}^{-1},\delta =\frac{1}{|k|}=0.0276\text{m}\approx 2.8\text{cm}.$ Matlab kya hai: blocked wave kuch centimetres ke andar hi khatam ho jaati hai — sheath thickness se comparable, yahi reason hai ki capsule sealed off ho jaata hai. Cutoff ke neeche plasma sirf poor conductor nahi hai; yeh ek efficient exponential wall hai.

Plasma frequency: $f_p=8.98\sqrt{n_e}\to 8.98\sqrt{0}=0$. Koi free electrons nahi toh koi restoring sheet-charge nahi (parent derivation ka Step 1 deta hai $\sigma =n_{e}ex=0$), toh kuch bhi slosh nahi karta — natural frequency zero pe collapse kar jaati hai. Critical density: kisi bhi positive $f_{\text{radio}}$ ke liye, $n_{\text{crit}}=(f/8.98{)}^2>0$, toh vanishing $n_e=0$ hamesha iske neeche hai. Blackout condition: $f_{\text{radio}}<f_p=0$ real radio ke liye impossible hai, toh koi blackout nahi — har frequency propagate karti hai. Kyun sahi hona chahiye: koi free electrons nahi wala plasma sirf neutral gas hai, aur neutral gas radio ke liye transparent hai. Formula sahi se reproduce karta hai "empty space sab radio ko nikaalne deta hai." Koi bhi physical law jo is limit mein fail ho woh galat hoga.

Blackout test ek inequality mein ::: $f_{\text{radio}}<f_p$, yaani $n_e>n_{\text{crit}}$. Scaling of critical density with radio frequency ::: $n_{\text{crit}}\propto f^2$ (square-law, linear nahi). Refractive index at exact cutoff $\omega ={\omega }_p$ ::: $n=0$ aur $k=0$ — infinite wavelength, koi propagation nahi. Skin depth below cutoff ::: $\delta =1/|k|=c/\sqrt{{\omega }_p^2-{\omega }^2}$. Zero-density limit ::: $f_p\to 0$, koi blackout nahi — neutral gas transparent hai. Is a below-cutoff wave absorbed or reflected? ::: Reflected (lossless plasma mein); yeh $e^{-x/\delta }$ ki tarah decay karti hai lekin andar koi net power carry nahi karti.