Plasma sheath — communications blackout: Worked examples

3.4.23 · D3 · Physics › Rocket Flight Mechanics › Plasma sheath — communications blackout

Ek bhi example touch karne se pehle, humein har symbol earn karna hoga jo hum use karenge. Parent note ne $ω$ , $ω_{p}$ , $k$ , $c$ aur $n$ freely use kiye — yahan hum inhe ek-ek karke ground up se banate hain, plain words aur pictures ke saath.

Definition Woh symbols jo hum use karenge — har ek pehle define hoga

$f$ — ordinary frequency: ek second mein wave kitne complete wobbles karti hai. Unit: hertz (Hz).
$ω$ (Greek "omega") — angular frequency: wahi wobble lekin radians per second mein measure ki gayi. Ek full wobble $2 π$ radians ka hota hai, isliye $ω = 2 π f$ . Socho ek dot circle mein ghoom raha hai: $f$ laps count karta hai, $ω$ radians count karta hai.
$f_{p}$ — plasma frequency Hz mein; $ω_{p} = 2 π f_{p}$ uska angular version hai. Same physical "sloshing pitch", do alag unit systems.
$c$ — speed of light, $c = 3 \times 1 0^{8} m/s$ : wave empty space mein kitni fast travel karti hai.
$k$ — wavenumber: wave ki phase ke kitne radians ek metre space mein fit hote hain. Unit: $m^{- 1}$ . Agar $k$ real number hai toh wave aage badhti hai; agar $k$ imaginary nikle toh wave travel karne ki bajaye wahi decay karti hai.
$n$ — refractive index: woh factor jitna plasma, wave ki phase ko vacuum ke comparison mein slow karta hai. Vacuum mein $n = 1$ .

Formula Dispersion relation aur refractive index kahan se aate hain

Plasma ke andar speed $v = ω / k$ se travel karne wali ek plane wave ko Maxwell's equations follow karni padti hain. Unhe $ω_{p}$ par oscillate karte free electrons ke cloud ke liye solve karne par milta hai dispersion relation — woh rule jo wave ki frequency ko uske wavenumber se connect karta hai: $ω^{2} = ω_{p}^{2} + c^{2} k^{2} ⟹ k = \frac{1}{c} ω^{2} - ω_{p}^{2}$ Aise padho: wave ke total "frequency budget" $ω^{2}$ mein se, plasma $ω_{p}^{2}$ sirf electrons ko slosh karwane mein spend karta hai; sirf bacha hua $ω^{2} - ω_{p}^{2}$ wave ko space mein $c^{2} k^{2}$ ke roop mein aage push karne ke liye available hai. Refractive index define hota hai $n = c k / ω$ (vacuum speed divided by wave ki actual phase speed). Upar se $k$ substitute karne par: $n = \frac{c k}{ω} = 1 - \frac{ω _{p}^{2}}{ω ^{2}} = 1 - \frac{f _{p}^{2}}{f ^{2}}$ ( $ω = 2 π f$ aur $ω_{p} = 2 π f_{p}$ mein $2 π$ top aur bottom cancel ho jaate hain, isliye plain $f$ se bhi kaam chalta hai). Yeh ek formula Examples 4, 6 aur 7 ko drive karta hai.

Ab woh do everyday workhorses jinmein hum numbers plug karenge:

Scenario matrix

Is topic ke har possible problem inn cells mein se ek mein aati hai. Har row ek "shape" ka question hai; neeche ke worked examples mein us cell ka label diya gaya hai jise woh cover karta hai.

Cell	Kya unknown / special hai	Typical question	Example
A — forward	$n_{e}$ given hai, $f_{p}$ find karo	"Plasma frequency kya hai?"	Ex 1
B — compare	$n_{e}$ aur $f_{radio}$ given hain, decide karo	"Blackout hai ya nahi?"	Ex 2
C — inverse	$f_{radio}$ given hai, $n_{crit}$ find karo	"Mujhe block karne wali density kya hai?"	Ex 3
D — boundary	$f_{radio} = f_{p}$ exactly	"Cutoff par kya hota hai?"	Ex 4
E — degenerate	$n_{e} = 0$ (ya un-ionized)	"Koi electron nahi — kuch block hoga?"	Ex 5
F — limiting	$f_{radio} ≫ f_{p}$	"Bahut badi frequency kitni slow hoti hai?"	Ex 6
G — depth/decay	cutoff ke neeche evanescent skin depth	"Kitna thick sheath actually block karta hai?"	Ex 7
H — real-world word problem	S-band capsule re-entry par	"Ek working link design karo"	Ex 8
I — exam twist	wrong tool trap ( $ω$ vs $f$ )	" $2 π$ ka factor pakdo"	Ex 9

Hum A, B, C, D, E, F, G, H, I — har cell ko cover karenge. Neeche ki picture inke sab ke liye map hai: yeh refractive index $n$ ko frequency ke against plot karti hai aur us cutoff ko mark karti hai jis par hum baar baar laute hain.

Figure — Plasma sheath — communications blackout

Worked examples

Worked example Example 1 — Cell A (forward): ek dense sheath ki plasma frequency

Ek peak re-entry sheath mein $n_{e} = 4 \times 1 0^{18} m^{- 3}$ hai. $f_{p}$ find karo.

Forecast: Guess karo — $f_{p}$ 1 GHz ke kareeb hogi ya 20 GHz ke kareeb?

Density ka square root lo: $4 \times 1 0^{18} = 2 \times 1 0^{9}$ . Yeh step kyun? Formula ko $n_{e}$ chahiye kyunki restoring field electron density ke saath grow karta hai, aur ek oscillator ki frequency stiffness ke square root ke saath badhti hai.
Constant se multiply karo: $f_{p} = 8.98 \times 2 \times 1 0^{9} = 1.796 \times 1 0^{10} Hz \approx 18 GHz$ . Yeh step kyun? $8.98$ ke andar $e$ , $ε_{0}$ , $m_{e}$ aur $1/2 π$ ek hi number mein bundle hain jo already hertz mein output deta hai.

Verify: Units — $m^{- 3} = m^{- 1.5}$ , aur constant baaki milaa kar Hz deta hai (neeche numerically checked). Parent ka Ex 1 ( $1 0^{18} \to 9$ GHz) se zyada dense sheath zyada bada $f_{p}$ dena chahiye, aur $18 > 9$ hai. ✓

Worked example Example 2 — Cell B (compare): kya ek X-band 10 GHz link survive karta hai?

Wahi sheath jaise Example 1 mein ( $f_{p} \approx 18 GHz$ ). Tum $f_{radio} = 10 GHz$ par transmit karte ho.

Forecast: Through hoga ya block hoga?

Compare karo: $f_{radio} = 10 GHz$ versus $f_{p} = 18 GHz$ . Yeh step kyun? Sirf yahi decision rule hai: inequality $f_{radio} ≶ f_{p}$ .
Kyunki $10 < 18$ hai, radio cutoff se neeche hai → blackout. Yeh step kyun? $f_{p}$ se neeche electrons wave ke saath chal lete hain aur use cancel kar dete hain (woh evanescent ho jaati hai). Figure s01 par yeh cutoff line ke baayein wala coral region hai.

Verify: Sanity check — X-band bhi 18 GHz sheath se haarta hai. Jeetnay ke liye Ka-band chahiye ( $\sim$ 26 GHz $> 18$ ). Inequality ki direction matter karti hai: zyada frequency help karti hai. ✓

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

Ek sheath mein $n_{e} = n_{crit}$ hai teri frequency ke liye, matlab $f_{radio} = f_{p}$ exactly (equivalently $ω = ω_{p}$ , kyunki $ω = 2 π f$ ). Wavenumber $k$ kya hai, aur physically kya hota hai?

Forecast: Wave travel karti hai, rukti hai, ya reflect hoti hai?

Dispersion relation yaad karo (upar build ki gayi) $ω^{2} = ω_{p}^{2} + c^{2} k^{2}$ , toh $k = \frac{1}{c} ω^{2} - ω_{p}^{2}$ . Yeh step kyun? $k$ (radians of phase per metre) batata hai ki wave aage chale ( $k$ real) ya wahi mare ( $k$ imaginary). Yeh woh single quantity hai jo decide karta hai "travel karna hai ya nahi".
$ω = ω_{p}$ par: $ω^{2} - ω_{p}^{2} = 0$ , isliye $k = \frac{1}{c} 0 = 0$ . Yeh step kyun? Hum equality ko relation mein substitute karte hain; frequency budget poora electrons ko slosh karwane mein spend ho jaata hai, wave ko aage push karne ke liye kuch nahi ( $c^{2} k^{2} = 0$ ) bachta.
$k = 0$ matlab infinite wavelength, zero forward progress — wave na plasma mein energy propagate karti hai aur na cleanly reflect hoti hai; yeh dono ke beech ka knife-edge hai. Yeh step kyun? Wavelength $λ = 2 π / k$ hai, aur jaise $k \to 0$ hota hai yeh $\infty$ tak blow up ho jaata hai: koi spatial oscillation bilkul nahi.

Verify: Refractive index $n = 1 - ω_{p}^{2} / ω^{2}$ use karo (upar $n = c k / ω$ se defined). $ω = ω_{p}$ par, $n = 1 - 1 = 0$ . Zero refractive index exactly boundary hai — $ω > ω_{p}$ ke liye real, $ω < ω_{p}$ ke liye imaginary. Yahi figure s01 mein dono curves ka pinch point hai. ✓

Worked example Example 5 — Cell E (degenerate): ek garam lekin un-ionized gas

Ek capsule 1500 K par hot air se guzarti hai jo abhi ionized nahi hui hai: $n_{e} = 0$ . Kya radio signal pass hoga?

Forecast: Yeh toh blistering hot hai — zaroor signal block hoga?

Calculate karo $f_{p} = 8.98 0 = 0 Hz$ . Yeh step kyun? Koi free electrons nahi matlab slosh karne ke liye kuch nahi — plasma frequency zero ho jaati hai.
Blackout condition hai $f_{radio} < f_{p} = 0$ , jo koi bhi real frequency satisfy nahi karti. Yeh step kyun? Har positive $f_{radio}$ zero se upar hai, isliye har frequency pass ho jaati hai.

Verify: Refractive index $n = 1 - 0/ ω^{2} = 1$ — bilkul vacuum jaisa. Yeh prove karta hai ki parent-note ki galat soch "heat radio jam karti hai" galat hai: heat sirf isliye matter karti hai kyunki woh electrons banati hai. ✓

Worked example Example 6 — Cell F (limiting): ek bahut badi frequency plasma ko barely notice karti hai

Ek Ka-band link $f_{radio} = 26 GHz$ par, $f_{p} = 9 GHz$ wali sheath se guzarti hai. Uska refractive index 1 se kitne factor neeche hai (woh kitni "slow" hoti hai)?

Forecast: Barely, ya bahut zyada?

Ratio banao $\frac{f _{p}}{f _{radio}} = \frac{9}{26} = 0.3462$ ( $ω_{p} / ω$ ke barabar, kyunki $2 π$ cancel ho jaate hain). Yeh step kyun? Upar ki derivation se, refractive index sirf is ek ratio par depend karta hai: $n = 1 - (f_{p} / f)^{2}$ . Wave ke baare mein aur kuch matter nahi karta.
Square karo: $0.346 2^{2} = 0.1198$ . Yeh step kyun? Formula ratio ka square subtract karta hai, kyunki dispersion relation $ω^{2}$ aur $ω_{p}^{2}$ se bana hai.
Refractive index: $n = 1 - 0.1198 = 0.8802 = 0.938$ . Yeh step kyun? Root lena frequency budget ko wapas ek actual phase-slowing factor mein convert karta hai.

Verify: $n = 0.938$ almost 1 ke kareeb hai — wave almost vacuum jaise sail kar jaati hai. Jaise $f_{radio} \to \infty$ , $(f_{p} / f)^{2} \to 0$ isliye $n \to 1$ (limiting behaviour): bahut badi frequencies electrons ko effectively frozen dekhti hain. Figure s01 par yeh mint curve hai jo far right par $n = 1$ ki taraf creep karti hai. ✓

Worked example Example 7 — Cell G (depth/decay): ek blocked wave kitni door penetrate karti hai?

Ek 300 MHz wave $f_{p} = 2 GHz$ wali sheath se takraati hai. Cutoff ke neeche, $k$ imaginary hai: likho $k = iκ$ , toh field $e^{ik z} = e^{- κ z}$ jaisi ho jaati hai — ek wave nahi, ek decay. Skin depth $δ = 1/ κ$ find karo, woh distance jis par field $1/ e$ (roughly 37%) tak drop ho jaati hai.

Forecast: Millimetres, centimetres, ya metres?

Cutoff ke neeche, $ω < ω_{p}$ se $ω^{2} - ω_{p}^{2}$ negative ho jaata hai, isliye $κ = \frac{1}{c} ω_{p}^{2} - ω^{2} = \frac{2 π}{c} f_{p}^{2} - f^{2}$ . Yeh step kyun? Jab $k = \frac{1}{c} ω^{2} - ω_{p}^{2}$ ke root ke andar ka number negative ho jaata hai, $k$ imaginary ban jaata hai. Root ke andar order flip karne se ek real decay rate $κ$ milti hai — yeh evanescent (decaying) field ki pehchaan hai, travelling field ki nahi.
Root evaluate karo: $(2 \times 1 0^{9})^{2} - (3 \times 1 0^{8})^{2} = 4 \times 1 0^{18} - 9 \times 1 0^{16} = 3.91 \times 1 0^{18} = 1.977 \times 1 0^{9}$ . Yeh step kyun? Hum $f_{p}$ aur $f$ ko leftover-budget $f_{p}^{2} - f^{2}$ mein plug karte hain; chota $f^{2}$ bade $f_{p}^{2}$ ko barely dent karta hai, yeh batata hai ki wave cutoff se bahut deep neeche hai aur jaldi maregi.
$2 π / c$ prefactor lagao: $κ = \frac{2 π}{3 \times 1 0 ^{8}} \times 1.977 \times 1 0^{9} = 41.4 m^{- 1}$ . Yeh step kyun? $2 π / c$ , Hz-mein-frequency ko radians-of-decay-per-metre mein convert karta hai — exactly wahi jo decay rate ki units honi chahiye ( $m^{- 1}$ ).
Depth ke liye invert karo: $δ = 1/ κ = 0.0241 m \approx 2.4 cm$ . Yeh step kyun? Skin depth, decay rate ka reciprocal hai — bada $κ$ matlab field thodi si distance mein hi mar jaati hai.

Verify: Ek real sheath centimetres-thick hoti hai, $δ$ se comparable, isliye wave roughly ek factor $e$ ya zyada per sheath knock down ho jaati hai — total blackout ke liye kaafi hai. Units: $[κ] = m^{- 1}$ isliye $δ$ metres mein hai. Figure s02 exactly yahi decay dikhata hai. ✓

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

Ek crewed capsule ki peak sheath $n_{e} = 1.5 \times 1 0^{18} m^{- 3}$ hai. Mission rules mein direct link chahiye (koi relay nahi). In bands mein se kaun kaam karega: UHF 0.4 GHz, S-band 2.2 GHz, X-band 10 GHz, Ka-band 32 GHz?

Forecast: Charon mein se kitne survive karenge?

$f_{p}$ find karo: $1.5 \times 1 0^{18} = 1.225 \times 1 0^{9}$ ; $f_{p} = 8.98 \times 1.225 \times 1 0^{9} = 1.100 \times 1 0^{10} Hz \approx 11 GHz$ . Yeh step kyun? Ek hi number sab kuch decide karta hai — har kisi ko isse beat karna hoga.
Har band ko 11 GHz se compare karo:
- UHF 0.4 GHz $< 11$ → blocked.
- S-band 2.2 GHz $< 11$ → blocked.
- X-band 10 GHz $< 11$ → blocked (sirf barely!).
- Ka-band 32 GHz $> 11$ → passes. ✓ Yeh step kyun? Inequality har band par apply hoti hai; note karo X-band sirf 1 GHz se haarta hai, yeh dikhata hai ki thoda zyada dense sheath answer ko flip kar sakta hai.
Decision: sirf Ka-band direct link mein survive karta hai.

Verify: Yeh parent Example 4 ki advice se match karta hai — " $f_{radio}$ ko $f_{p}$ (Ka-band) se upar raise karo." Aur yeh explain karta hai ki real capsules backup ke liye relay-via-wake option bhi kyun carry karti hain. ✓

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

$ω$ -versus- $f$ ka trap Ek exam mein likha hai: "Ek sheath ka $f_{p} = 3 GHz$ hai. Ek student angular plasma frequency $ω_{p}$ compute karta hai aur phir galti se blackout declare karta hai har radio angular frequency ke liye jo $3 \times 1 0^{9}$ se neeche ho." Sahi $ω_{p}$ kya hai, aur galti kahan hai?

Forecast: Kya $ω_{p}$ , $3 \times 1 0^{9}$ se bada hai ya chota?

Sahi se convert karo: $ω_{p} = 2 π f_{p} = 2 π \times 3 \times 1 0^{9} = 1.885 \times 1 0^{10} rad/s$ . Yeh step kyun? $ω = 2 π f$ hamesha — angular frequency radians count karta hai, ordinary frequency full cycles (yeh wahi pehla symbol hai jo humne define kiya tha).
Student ne radio $ω$ (rad/s) ko $3 \times 1 0^{9}$ number se compare kiya jo actually $f_{p}$ in Hz hai, $ω_{p}$ nahi. Unhone units mix kar diye $2 π$ ke factor se. Yeh step kyun? $8.98 n_{e}$ shortcut $f_{p}$ Hz mein output karta hai; tumhe ise angular frequency se compare nahi karna chahiye.
Sahi rule: like se like compare karo — ya toh $f_{radio}$ vs $f_{p}$ , ya $ω_{radio}$ vs $ω_{p}$ .

Verify: $ω_{p} / f_{p} = 1.885 \times 1 0^{10} /3 \times 1 0^{9} = 6.283 = 2 π$ . ✓ Exactly woh stray factor jiske baare mein parent-note ka warning hai.

Active recall

Recall Khud test karo

Angular frequency $ω$ aur ordinary frequency $f$ mein kya relation hai? ::: $ω = 2 π f$ — wahi wobble, cycles per second ki bajaye radians per second mein count ki gayi. Wavenumber $k$ kya batata hai, aur imaginary $k$ ka kya matlab hai? ::: $k$ radians of phase per metre hai; real $k$ = travelling wave, imaginary $k$ = evanescent (decaying) field jiska skin depth $δ = 1/ κ$ hai. $n = 1 - (f_{p} / f)^{2}$ kahan se aata hai? ::: $n = c k / ω$ ko dispersion relation $ω^{2} = ω_{p}^{2} + c^{2} k^{2}$ ke saath combine karne se. Exactly $f_{radio} = f_{p}$ par, $k$ aur refractive index kya hai? ::: $k = 0$ aur $n = 0$ — propagation aur reflection ke beech ka knife-edge. Agar $n_{e} = 0$ ho, toh $f_{p}$ kya hai aur kya kuch wave ko block karta hai? ::: $f_{p} = 0$ ; koi bhi frequency usse neeche nahi hai, isliye sab kuch pass hota hai ( $n = 1$ ). $8.98 n_{e}$ shortcut kaunsi quantity deta hai? ::: $f_{p}$ hertz mein — $ω_{p}$ NAHI. Stray $2 π$ mat lagao.