Maxwell-Boltzmann distribution — full derivation: Visual walkthrough

2.4.13 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Maxwell-Boltzmann distribution — full derivation

Step 0 — "Velocity" kya hai aur hum kya count kar rahe hain?

KYA. Ek molecule ka kisi ek instant mein ek velocity hoti hai: ek arrow jo dikhata hai woh kidhar ja raha hai, jiska length uski speed hai. 3D mein us arrow ki teen walls par teen shadows hoti hain — uske components $v_{x}, v_{y}, v_{z}$ (woh kitni tezi se left-right, forward-back, up-down move karta hai).

KYUN. Kyunki speed — arrow ki length — andar teen independent numbers chhupa ke rakhti hai. Speed samajhne ke liye pehle har shadow ko samjhte hain, phir reassemble karte hain.

PICTURE. Blue arrow ek molecule ki velocity hai. Uske teen coloured shadows components hain. Arrow ki length hai $v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ (3D mein Pythagoras).

Figure — Maxwell-Boltzmann distribution — full derivation

Step 1 — Har component ek bell curve follow karta hai, aur kyun zaroori hai

KYA. Hum claim karte hain ki ek molecule ka sideways speed $v_{x}$ ke paas hone ka probability ek Gaussian (zero par centred symmetric bell) hai:

g (v_{x}) = A e^{- b v_{x}^{2}} .

KYUN yahi shape aur koi nahin? Do physical facts ise pin karte hain:

Independence — left-right speed aur up-down speed ke beech koi link nahi. Toh teeno ka chance product hai $g (v_{x}) g (v_{y}) g (v_{z})$ .
Isotropy — space ka koi favourite direction nahi, toh joint chance sirf arrow ki length par depend kar sakta hai, yaani $v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ par.

Left side par product jo right side par sum of squares par depend kare — yeh exponential ki pehchaan hai. $ln$ lene se product sum ban jaata hai: $ln g (v_{x}) + ln g (v_{y}) + ln g (v_{z})$ ko $v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ ka function hona chahiye. Iska ek hi tarika hai: $ln g (v_{x}) = a - b v_{x}^{2}$ har ek ke liye. $ln$ undo karo aur Gaussian milta hai.

PICTURE. Bell $v_{x} = 0$ par peak karti hai (zyaadatar molecules sirf gently sideways drift karte hain) aur bade $∣ v_{x} ∣$ ke liye khatam ho jaati hai. Yeh symmetric hai: ek molecule ka left ya right jaane ka equal chance hai.

Step 2 — Constants fix karna: normalization + temperature

KYA. Do unknowns bacha hain: $A$ aur $b$ . Inhe hum do facts se fix karte hain.

KYUN (fact 1: total = 1). Saari probability ek mein add honi chahiye. Ek Gaussian ke neeche ka area ek classic Gaussian integral hai:

\int_{- \infty}^{\infty} A e^{- b v_{x}^{2}} d v_{x} = A \frac{π}{b} = 1 \Rightarrow A = \frac{b}{π} .

KYUN (fact 2: temperature). $b$ jaanne ke liye humein pata hona chahiye gas kitni hot hai. Equipartition theorem kehta hai ki har sideways motion average energy $\frac{1}{2} k_{B} T$ carry karta hai:

⟨ \frac{1}{2} m v_{x}^{2} ⟩ = \frac{1}{2} k_{B} T \Rightarrow ⟨ v_{x}^{2} ⟩ = \frac{k _{B} T}{m} .

Ek Gaussian ke liye, spread (variance) hai $⟨ v_{x}^{2} ⟩ = \frac{1}{2 b}$ . Inhe equal set karte hain:

\frac{1}{2 b} = \frac{k _{B} T}{m} \Rightarrow b = \frac{m}{2 k _{B} T} .

PICTURE. Do temperatures. Hot bell choti aur chaudi hai (bada spread, chhota $b$ ); cold bell lambi aur patli hai. Dono ka exactly same area = 1 hai.

Step 3 — Teen bells stack karo: velocity space mein cloud

KYA. Full velocity distribution teen identical bells ka product hai:

F (v) = (\frac{m}{2 π k _{B} T})^{3/2} e^{- m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2}) /2 k_{B} T} .

KYUN. Independence phir se — joint chance product hai. Teen squared terms recombine hokar $v^{2}$ ban jaate hain, toh $F$ sirf origin se distance par depend karta hai. Iska matlab hai probability cloud spherically symmetric hai: centre par dense, barabar sabhi directions mein bahar ki taraf fade hota hua.

PICTURE. Velocity space mein dots ka ek fuzzy ball. Centre ke paas dense (slow molecules common), bahar patla (fast molecules rare). Koi direction favoured nahi — ek perfect fuzzy sphere.

Step 4 — Woh trick jo $v^{2}$ ko janam deti hai: shell count karna

KYA. Hum speed chahte hain, arrow nahi. Bahut saare different arrows ka same length $v$ hota hai — woh sab ek sphere of radius $v$ par baithte hain. Speed $v$ ke paas hone ka chance paane ke liye humein us poori sphere par cloud density add karni hogi.

KYUN. Density $F (v)$ sphere par har jagah same hai (spherical symmetry). Toh total sirf density × shell ka surface area hai:

f (v) d v = density F (v) \times shell volume 4 π v^{2} d v .

Radius $v$ ke sphere ka surface area $4 π v^{2}$ hai. Yahan $v^{2}$ paida hota hai — yeh pure geometry hai, "speed $v$ par kitni jagah hai."

PICTURE. Radius $v$ , thickness $d v$ ka ek thin spherical shell. Uska area $4 π v^{2}$ shell bade hone par badhta hai. $v = 0$ par shell point ban jaata hai — zero area, zero jagah.

Step 5 — Tug-of-war: do curves ladte hain, ek jeetta hai

KYA. Sab kuch collect karke Maxwell-Boltzmann speed distribution milti hai:

f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- m v^{2} /2 k_{B} T} .

KYUN hill shape? Do competing factors:

$v^{2}$ — geometric factor, 0 se uthta hua, bada $v$ chahta hai.
$e^{- m v^{2} /2 k_{B} T}$ — energy factor (Boltzmann factor se), bade $v$ ko crush karta hai.

Ek uthti parabola ko ek girte bell se multiply karo aur ek aisi curve milti hai jo 0 se shuru hoti hai, chadhti hai, peak karti hai, phir khatam — ek lopsided hill.

PICTURE. Teen curves overlaid: uthta $v^{2}$ (green), girta exponential (red), aur unka product $f (v)$ (blue). Dekho kaise product dono ends par zero ho jaata hai lekin beech mein bulge karta hai.

Step 6 — Edge cases: curve ke do ends

KYA. Woh boundaries check karo jo reader ko milenge.

Case $v = 0$ (ruka hua molecule). $f (0) = 4 π (\dots)^{3/2} \cdot 0^{2} \cdot 1 = 0$ . Bhaale exponential maximum ho ( $e^{0} = 1$ ), phir bhi $v^{2}$ shell ka zero area hai. Koi molecule exactly frozen nahi hai — energetically forbidden hai isliye nahi, balki $v = 0$ par koi jagah nahi hai.

Case $v \to \infty$ (superfast molecule). Yahan $v^{2}$ unbounded badhta hai lekin $e^{- m v^{2} /2 k_{B} T}$ kaafi faster girta hai (exponential hamesha kisi bhi power ko beat karta hai). Toh $f (v) \to 0$ . Koi molecule infinitely fast nahi hai.

PICTURE. Dono tails par zoom: left par parabola curve ko zero pin karta hai; right par exponential use wapas neeche kheenchta hai. Curve dono ends par zero ko chhuti hai.

Step 7 — Peak kahan baitta hai (balance point)

KYA. Peak $v_{p}$ wahan hai jahan uthta $v^{2}$ aur girta $e^{- α v^{2}}$ (jahan $α = m /2 k_{B} T$ ) exactly balance karte hain. Derivative zero set karo:

\frac{d}{d v} (v^{2} e^{- α v^{2}}) = 2 v e^{- α v^{2}} - 2 α v^{3} e^{- α v^{2}} = 0 \Rightarrow 1 = α v^{2} \Rightarrow v_{p} = \frac{2 k _{B} T}{m} .

KYUN derivative? Peak ek flat spot hai — jahan curve uthna band karke girna shuru karta hai, uski slope zero hoti hai. Derivative hi slope hai. Yeh "geometry right push kar raha hai" ( $2 v$ ) minus "energy left pull kar rahi hai" ( $2 α v^{3}$ ) mein split hoti hai; woh exactly $v_{p}$ par cancel karte hain.

PICTURE. Teen points par arrows ke roop mein do forces: $v_{p}$ se neeche geometry jeetta hai (curve uthta hai), $v_{p}$ par barabar (flat top), $v_{p}$ ke upar energy jeetti hai (curve girta hai).

Ek-picture summary

Sab kuch ek canvas par: fuzzy velocity cloud (Step 3) → uska spherical shell (Step 4) → $v^{2}$ vs exponential tug-of-war (Step 5) → final hill jisme $v_{p} < ⟨ v ⟩ < v_{r m s}$ marked hai (Steps 6–7).

Recall Feynman retelling — poora walkthrough simple words mein

Har molecule ko ek 3D graph mein ek arrow imagine karo. Kyunki left-right ko up-down se koi link nahi, aur koi direction special nahi, arrow-tips ka cloud ek fuzzy ball hai — beech mein dense, edges par patla, aur teeno shadows ek hi simple bell curve follow karte hain. Bell ko finite banana use decay karne par majboor karta hai; ise temperature se match karna exactly fix karta hai ki woh kitni tezi se decay karta hai (andar ka $m /2 k_{B} T$ ).

Ab mujhe parwah nahi ki arrow kidhar point karta hai, sirf uski length — speed. Same length ke saare arrows ek sphere par hote hain, aur bade spheres ka zyaada surface hota hai, $v^{2}$ ki tarah badhta hua. Toh ek speed ka chance hai (wahan cloud kitna dense hai) × (sphere kitni jagah deta hai). Density exponentially girti hai; room $v^{2}$ ki tarah badhta hai. Dono ladte hain: choti speeds par room jeetta hai (toh zero speed impossible hai — jagah nahi), badi speeds par energy jeetti hai (toh infinite speed impossible hai). Woh exactly peak $v_{p} = 2 k_{B} T / m$ par barabar hote hain. Kyunki right tail lamba hai, mean aur rms speeds peak se thodi aage baithti hain. Gas ko heat karo aur poori hill right slide karke spread ho jaati hai. Woh hill Maxwell-Boltzmann distribution hai.

Recall Self-test

$f (0) = 0$ kyun hai bhaale $e^{0} = 1$ ho? ::: $v^{2}$ shell-area factor $v = 0$ par zero hai — jagah nahi, bhaale energy cost sabse kam ho. Humne kaun sa ek external fact import kiya? ::: Boltzmann factor: high energy exponentially unlikely hai, jo $e^{- m v^{2} /2 k_{B} T}$ deta hai. $4 π v^{2}$ kahan se aata hai? ::: Velocity space mein radius $v$ ke sphere ka surface area (Step 4). $b > 0$ kyun hona chahiye? ::: Warna Gaussian unbounded badhta hai aur normalize nahi ho sakta.

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