Maxwell-Boltzmann speed distribution — derivation (key for propulsion): Visual walkthrough

1.7.12 · D2 · Physics › Thermodynamics › Maxwell-Boltzmann speed distribution — derivation (key for p

Hum ek cheez dhundh rahe hain — speed distribution $f (v)$ :

Step 1 — Ek molecule, teen arrows

KYA. Ek akela molecule space mein uda ja raha hai. Uski velocity ek arrow (vector) hai jiske teen shadows hain: $x$ direction mein kitna fast, $y$ mein, $z$ mein. Un teen numbers ko $v_{x}, v_{y}, v_{z}$ kaho.

KYU. Pehle hum speed (ek number) ki baat kar sakein, pehle velocity (teen numbers) ki baat karni padegi, kyunki collisions teen directions ko independently scramble karti hain. Speed phir 3D Pythagoras rule se milti hai: $v = along x v_{x}^{2} + along y v_{y}^{2} + along z v_{z}^{2} .$ Root ke andar har term ek shadow ka contribution hai; poora root arrow ki actual length hai.

PICTURE. Arrow dekho aur axes par uske teen coloured shadow-lengths dekho.

Step 2 — Ek direction at a time: bell curve

KYA. Sirf ek shadow $v_{x}$ par dhyan do. $g (v_{x})$ ko fraction maano un molecules ka jinki $x$ -velocity $v_{x}$ ke aaspaas hai. Hum claim karte hain iska shape zero par centred ek bell curve hai: $g (v_{x}) = A e^{- β v_{x}^{2}} .$

YEH SHAPE HI KYU, KUCH AUR KYU NAHI? Do demands isse force karti hain:

Teen shadows independent hain, isliye joint chance ek product hai: $F = g (v_{x}) g (v_{y}) g (v_{z})$ .
Isotropy kehta hai $F$ sirf $v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ par depend kar sakta hai.

Sirf woh function jahan teen copies multiply karna exponent mein sums ko bahar product mein badal deta hai, woh exponential hai. Har symbol dekho: $g (v_{x}) = height fixer A e^{- β width fixer v_{x}^{2}} .$

$A$ height scale karta hai taaki total probability $1$ ho.
$β$ (ek positive number) width set karta hai: bada $β$ = narrow spike, chhota $β$ = wide spread.
Minus sign probability ko bade $∣ v_{x} ∣$ ke liye shrink karta hai — warna infinitely fast molecules common ho jaate (jo absurd hai).
$v_{x}^{2}$ ( $v_{x}$ nahi) ise symmetric rakhta hai: left jaana utna hi likely hai jitna right jaana.

PICTURE. Bell, $v_{x} = 0$ ke baare mein symmetric, jisme $β$ control karta hai kitni fat hai.

Step 3 — $β$ ki width ko temperature se pin karna

KYA. Abstract $β$ ko real physics se connect karna hoga. Hum Equipartition Theorem use karte hain: motion ki har direction average par $\frac{1}{2} k_{B} T$ energy store karti hai.

KYU. Temperature define hoti hai average kinetic energy se. $T$ formula mein yahi ek darwaza hai jisse enter karta hai. Symbols mein: $⟨ \frac{1}{2} m v_{x}^{2} ⟩ = \frac{1}{2} k_{B} T .$

$⟨ ⟩$ matlab "saare molecules par average."
$\frac{1}{2} m v_{x}^{2}$ woh kinetic energy hai jo sirf $x$ -motion mein store hai.

Bell $g (v_{x}) \propto e^{- β v_{x}^{2}}$ ke liye, ek standard integral deta hai $⟨ v_{x}^{2} ⟩ = \frac{1}{2 β}$ (jitni wider bell, utna bada yeh spread). Substitute karo: $\frac{1}{2} m \cdot \frac{1}{2 β} = \frac{1}{2} k_{B} T ⟹ β = \frac{m}{2 k _{B} T} .$

Heavy molecule (bada $m$ ) ⟹ bada $β$ ⟹ narrow bell ⟹ sluggish.
Hot gas (bada $T$ ) ⟹ chhota $β$ ⟹ wide bell ⟹ fast.

PICTURE. Do temperatures par do bells — hot wali wider aur shorter hai.

Step 4 — Teen bells ko velocity-space mein stack karna

KYA. Ab teen bells ko wapas saath laao. Full velocity-space density — ek specific arrow $(v_{x}, v_{y}, v_{z})$ ki chance — yeh hai: $F (v_{x}, v_{y}, v_{z}) = A^{3} e^{- β v^{2}} = (\frac{m}{2 π k _{B} T})^{3/2} e^{- \frac{m v ^{2}}{2 k _{B} T}} .$

$(\frac{m}{2 π k _{B} T})^{3/2}$ yani $A^{3}$ — teen height-fixers multiply kiye.
$e^{- \frac{m v ^{2}}{2 k _{B} T}}$ woh exponential hai jisme $β$ ko uski physical value se replace kiya gaya hai.

KYU. Yeh $F$ density hai velocity-space ke unit volume per. Yeh origin par sabse bada hai (zero length ka arrow) kyunki wahan exponent zero hai aur exponential sabse bada hota hai.

PICTURE. Velocity-space mein ek cloud, centre par sabse dense, bahar jaate jaate fade hota hua — brightness = $F$ .

Step 5 — Spherical shell: $v^{2}$ kahan se aata hai

KYA. Ek hi speed $v$ wale saare arrows ki length equal hoti hai, isliye unki tips velocity-space mein radius $v$ ke sphere par hoti hain. Yeh count karne ke liye ki speed $v$ ke aaspaas kitne molecules hain, hum $F$ ko us poori thin shell par add karte hain.

KYU. Radius $v$ aur thickness $d v$ ki shell ka volume = (surface area) × (thickness): $shell surface 4 π v^{2} \times thickness d v .$

$4 π v^{2}$ radius $v$ ke sphere ka surface area hai.
Bada $v$ matlab bada sphere matlab us length ke arrows ke liye zyada room.

Shell ke constant density $F$ ko shell ke volume se multiply karo: $f (v) d v = kitne directions 4 π v^{2} F, density per arrow (\frac{m}{2 π k _{B} T})^{3/2} e^{- \frac{m v ^{2}}{2 k _{B} T}} d v .$

PICTURE. Do shells, ek chhoti ek badi — badi thin hai par vast hai, bahut zyada arrows hold karti hai.

Step 6 — Woh tug of war jo peak banata hai

KYA. Final law assemble karo: $f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- \frac{m v ^{2}}{2 k _{B} T}}$

KYU iska hump hai. Do factors ladte hain:

$v^{2}$ zero se badhta hai — $v$ badhne ke saath zyada directions.
$e^{- \frac{m v ^{2}}{2 k _{B} T}}$ girta hai — fast molecules energetically rare hote hain.

Chhote $v$ par badhta $v^{2}$ jeet jaata hai (curve chadhti hai). Bade $v$ par girta exponential jeet jaata hai (curve dub jaati hai). Dono ke beech ek peak baitha hai — most probable speed.

PICTURE. Badhta $v^{2}$ , girta exponential, aur unka product — lopsided MB curve.

Step 7 — Long tail aur teen speeds

KYA. Curve skewed hai: peak ke left taraf steeply girta hai par right taraf ek lamba tail kheenchta hai. Us tail mein teen landmark speeds hain, har ek alag sawaal ka jawaab deta hai:

$v_{p} = \frac{2 k _{B} T}{m}$ — peak (most common speed).
$\overset{v}{ˉ} = \frac{8 k _{B} T}{π m}$ — plain average.
$v_{r m s} = \frac{3 k _{B} T}{m}$ — energy-weighted speed.

KYU $v_{p} < \overset{v}{ˉ} < v_{r m s}$ ka order hai. Tail average ko peak se right kheenchta hai; $v_{r m s}$ ke andar squaring fast tail ko aur zyada weight deti hai, use sabse zyada right push karti hai. Numerically $2 : 8/ π : 3 \approx 1.000 : 1.128 : 1.225$ .

PICTURE. Curve par teen speeds mark kiye, tail shaded hai yeh dikhane ke liye ki fast, propulsion-critical molecules kahan rehte hain.

Ek picture mein summary

Upar sab kuch, compressed: teen bells → ek velocity cloud → ek shell se slice kiya jiska area $4 π v^{2}$ hai → deta hai skewed speed curve apni teen landmark speeds ke saath.

Recall Feynman retelling — poora walkthrough plain words mein

Ek buzzing molecule imagine karo. Uski motion ke teen independent shadows hain — sideways, forward, up. Kyunki box mein kuch bhi kisi direction ko prefer nahi karta, har shadow wahi gentle bell curve follow karta hai, zero ke paas likeliest aur speed badhne par rare. Teen bells multiply karo aur maths tumhe kuch deta hai jo sirf total speed par depend karta hai — "velocity land" mein ek soft glowing cloud, dead centre par sabse bright. Lekin hum ek exact arrow nahi chahte; hum kisi bhi arrow ki ek given length chahte hain. Woh saare arrows ek sphere par hain, aur bade spheres ki huge surface hoti hai — isliye faster speeds ke paas bahut zyada room hai chahe har individual fast arrow rare ho. Woh "more room" hi $v^{2}$ hai jo curve ko zero se upar kheenchta hai (kuch bhi kabhi perfectly still nahi hota). Saath hi exponential truly fast waalon ko crush karta hai. Badhta $v^{2}$ versus girta exponential: peak par exactly tie hote hain, phir exponential jeet jaata hai aur ek lamba fast tail chhod jaata hai. Woh tail hi rockets ke liye saari baat hai — usme speedsters rehte hain, aur lighter gas use aur fast stretch karti hai. Peak, average, RMS usi order mein lagte hain kyunki tail averages ko lagataar rightward kheenchti rehti hai.

Connections

Kinetic Theory of Gases — woh buzzing molecules jinka yeh sab describe karta hai
Equipartition Theorem — $\frac{1}{2} k_{B} T$ provide kiya jisne $β$ pin kiya (Step 3)
Boltzmann Distribution — hamare exponential ka energy-space parent
Mean Free Path — Step 7 ka $\overset{v}{ˉ}$ use karta hai
Rocket Propulsion & Specific Impulse — kyun fast tail aur low $m$ jeetat hain
Effusion and Graham's Law — escape rate $\propto \overset{v}{ˉ} \propto 1/ m$