Maxwell-Boltzmann distribution — full derivation: Exercises

2.4.13 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Maxwell-Boltzmann distribution — full derivation

Poore note mein hum parent note ke boxed results use karte hain. Main toolkit dobara state karta hoon taaki koi bhi symbol unexplained na rahe:

Level 1 — Recognition

L1.1 — High-speed tail ko kaun sa factor khatam karta hai?

$f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- m v^{2} /2 k_{B} T}$ mein, us factor ka naam batao jo $v \to \infty$ pe $f (v) \to 0$ karta hai, aur us factor ka naam batao jo $f (0) = 0$ karta hai.

Recall Solution

KYA: Do competing factors shape ko control karne ke liye ladte hain.

Jab $v \to \infty$ : Boltzmann factor $e^{- m v^{2} /2 k_{B} T}$ kisi bhi power se tez decay karta hai, isliye yeh jeet jaata hai aur $f (v) \to 0$ ho jaata hai.
$v = 0$ pe: geometric factor $v^{2}$ (shell volume $4 π v^{2}$ ) zero hai, isliye $f (0) = 0$ .

YEH AISA KYU DIKHTA HAI: ek bell jo zero se shuru hoti hai, $v^{2}$ ki wajah se upar uthti hai, phir exponential se crush ho jaati hai. $e^{- E / k_{B} T}$ kahan se aata hai, yeh jaanne ke liye Boltzmann factor and partition function dekho.

L1.2 — Teeno speeds ka order

Kuch bhi compute kiye bina, $v_{p}, ⟨ v ⟩, v_{r m s}$ ko increasing order mein likho aur numeric ratio do.

Recall Solution

$v_{p} : ⟨ v ⟩ : v_{r m s} = 2 : 8/ π : 3 \approx 1.414 : 1.596 : 1.732.$ Toh $== v_{p} < ⟨ v ⟩ < v_{r m s} ==$ hai. Lamba high-speed tail mean aur rms ko peak ke dahiney kheenchta hai. Mnemonic: Please Make Room (Peak, Mean, RMS).

Level 2 — Application

L2.1 — 300 K pe helium ki most probable speed

Helium atom mass $m = 4 u$ . $T = 300$ K pe $v_{p}$ calculate karo.

Recall Solution

KYA: $v_{p} = 2 k_{B} T / m$ mein directly dalo. $m = 4 \times 1.66 \times 1 0^{- 27} = 6.64 \times 1 0^{- 27} kg .$ $v_{p} = \frac{2 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{6.64 \times 1 0 ^{- 27}} = 1.247 \times 1 0^{6} \approx 1117 m/s .$ KYU: halke atoms tez chalte hain — He, N₂ se $28/4 = 7 \approx 2.6 \times$ zyada tez hai same $T$ pe.

L2.2 — Mean aur most-probable speed ka ratio

Dikhao ki $⟨ v ⟩ / v_{p}$ ek pure number hai aur iska evaluation karo.

Recall Solution

$\frac{⟨ v ⟩}{v _{p}} = \frac{8 k _{B} T / π m}{2 k _{B} T / m} = \frac{8}{2 π} = \frac{4}{π} = \frac{2}{π} \approx 1.128.$ YEH UNIVERSAL KYU HAI: $k_{B} T$ aur $m$ completely cancel ho jaate hain, isliye yeh ratio har ideal gas ke liye har temperature pe same hai — yeh MB shape ka ek fingerprint hai.

L2.3 — $v_{p}$ double karne ke liye kitna temperature chahiye

Ek gas ka $v_{p} = 400$ m/s hai $T_{1} = 300$ K pe. Kis temperature pe $v_{p} = 800$ m/s hoga?

Recall Solution

KYA: $v_{p} \propto T$ use karo (mass fixed hai). $\frac{v _{p, 2}}{v _{p, 1}} = \frac{T _{2}}{T _{1}} \Rightarrow \frac{800}{400} = 2 = \frac{T _{2}}{300} .$ Square karo: $T_{2} = 4 \times 300 = 1200$ K. KYU: speed double karne ke liye $4 \times$ temperature chahiye kyunki kinetic energy $\propto v^{2} \propto T$ hoti hai. Equipartition theorem se link hai.

Level 3 — Analysis

L3.1 — Jab $T$ badhta hai toh peak height kyun girta hai?

$f (v)$ ke neeche total area hamesha $1$ hai. $v_{p} \propto T$ use karke explain karo ki gas ko garm karne se $f (v)$ ki peak kyun neeche aati hai. Figure dekho.

Figure — Maxwell-Boltzmann distribution — full derivation

Recall Solution

FIGURE KYA DIKHATA HAI: do MB curves, cooler (lavender) aur hotter (coral). Hotter wali dahiney zyada peak karti hai aur neeche bhi hoti hai. KYU: area $\int_{0}^{\infty} f (v) d v = 1$ conserved hai (yeh ek probability hai). Jab peak dahiney khisak ke bade $v_{p} \propto T$ pe jaati hai, toh curve wider bhi hoti hai (iska spread bhi $T$ ke saath scale karta hai). Ek fixed area jo wider base pe phailti hai, toh zaroor choti hogi. Toh: hotter = tez aur flat, kabhi taller nahi. Yeh exactly Kinetic theory of gases ka picture hai jisme molecules zyada energetic aur zyada spread-out hain.

L3.2 — $+ x$ hemisphere mein chalte molecules ka fraction

Kitne fraction molecules ka $v_{x} > 0$ hai? Aur kitne fraction ka dono $v_{x} > 0$ aur $v_{y} > 0$ hai?

Recall Solution

KYA: har component $g (v_{x})$ ek even Gaussian hai jo $0$ pe centred hai, isliye $\int_{0}^{\infty} g (v_{x}) d v_{x} = \frac{1}{2} .$ $v_{x} > 0$ wale molecules ka fraction $1/2$ hai. Components independent hain, isliye joint probability multiply hoti hai: $P (v_{x} > 0 and v_{y} > 0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} .$ KYU: isotropy koi bhi preferred direction forbid karta hai, isliye $⟨ v_{x} ⟩ = 0$ aur $+ x$ aur $- x$ ke beech perfect symmetry hoti hai.

L3.3 — Energy distribution ka peak kahan hai?

$f (v)$ ko energy distribution mein convert karo aur most probable kinetic energy $E_{p}$ dhundho. Kya $E_{p} = \frac{1}{2} m v_{p}^{2}$ hai?

Recall Solution

HUM KYA KARTE HAIN: variable change karo $E = \frac{1}{2} m v^{2}$ , toh $v = 2 E / m$ aur $d v = \frac{d E}{2 m E}$ hoga. $f (v) d v$ mein substitute karo: $f (v) d v \propto v^{2} e^{- m v^{2} /2 k_{B} T} d v \propto \frac{2 E}{m} e^{- E / k_{B} T} \cdot \frac{d E}{2 m E} \propto E e^{- E / k_{B} T} d E .$ Toh $F (E) \propto E^{1/2} e^{- E / k_{B} T}$ hai. Maximize karo: $\frac{d}{d E} (E^{1/2} e^{- E / k_{B} T}) = 0$ deta hai $\frac{1}{2} E^{- 1/2} e^{- E / k_{B} T} - \frac{1}{k _{B} T} E^{1/2} e^{- E / k_{B} T} = 0 \Rightarrow \frac{1}{2} = \frac{E}{k _{B} T} \Rightarrow E_{p} = \frac{1}{2} k_{B} T .$ Twist yeh hai: $\frac{1}{2} m v_{p}^{2} = \frac{1}{2} m \cdot \frac{2 k _{B} T}{m} = k_{B} T$ compare karo. Toh $E_{p} = \frac{1}{2} k_{B} T \neq = k_{B} T$ ! YEH KYUN DIFFERENT HAIN: Jacobian $d v / d E \propto E^{- 1/2}$ peak ko warp kar deta hai. Most-probable speed aur most-probable energy alag coordinate systems mein rehte hain — nonlinear change of variable peaks ko move kar deta hai. Ek beautiful, sneaky fact hai yeh.

Level 4 — Synthesis

L4.1 — Effusion fast molecules ko favour karta hai

Jo rate jis pe molecules ek tiny hole se escape karte hain woh $v f (v)$ ke proportional hai (tez molecules hole ko zyada baar hit karte hain). Escaping molecules ki most probable speed dhundho, aur andar ke $v_{p}$ se compare karo.

Recall Solution

KYA: effusion beam density $\propto v \cdot f (v) \propto v \cdot v^{2} e^{- α v^{2}} = v^{3} e^{- α v^{2}}$ hai, $α = m /2 k_{B} T$ ke saath. Maximize karo: $\frac{d}{d v} (v^{3} e^{- α v^{2}}) = (3 v^{2} - 2 α v^{4}) e^{- α v^{2}} = 0 \Rightarrow 3 = 2 α v^{2}$ , toh $v_{p, eff} = \frac{3}{2 α} = \frac{3 k _{B} T}{m} = v_{r m s} .$ KYU: "tez = zyada baar escape karta hai" wala extra $v$ factor escaping beam ki peak ko interior $v_{r m s}$ tak push kar deta hai — beam gas se hotter hai jo peeche reh jaati hai. Yeh Effusion and Graham's law aur isotope enrichment ke peeche ki physics hai.

L4.2 — Effusion flux se Graham's law

Do gases same $T$ pe same hole se effuse karte hain. Dhyan se dikhao ki unke effusion rates $\frac{R _{1}}{R _{2}} = \frac{m _{2}}{m _{1}}$ obey karte hain, phir H₂ ( $m = 2 u$ ) aur O₂ ( $m = 32 u$ ) ke liye rate ratio dhundho.

Recall Solution

Rate ko kaun sa moment govern karta hai? Area $A$ wale hole se crossing karte molecules ki number per unit time flux hai: un molecules ko count karo jinke paas hole ki taraf positive velocity component $v_{z}$ hai, weighted by kitni tezi se woh approach karte hain: $R = n A \int_{v_{z} > 0} v_{z} g (v_{z}) d v_{z},$ jahan $n$ number density hai aur $g (v_{z})$ one-component Gaussian hai. $v_{z}$ kyun aur $v_{r m s}$ kyun nahi? Sirf hole ke normal along motion molecules ko us se uthata hai; sideways motion ( $v_{x}, v_{y}$ ) matter nahi karta. Toh governing quantity $⟨ v_{z} ⟩_{+}$ hai, yani escaping half ke upar one-directional speed ka mean — nahi rms speed aur nahi full mean speed $⟨ v ⟩$ . Moment evaluate karo: $α = m /2 k_{B} T$ aur $g (v_{z}) = α / π e^{- α v_{z}^{2}}$ ke saath, $\int_{0}^{\infty} v_{z} g (v_{z}) d v_{z} = \frac{α}{π} \cdot \frac{1}{2 α} = \frac{1}{2} \frac{1}{π α} = \frac{1}{2} \frac{2 k _{B} T}{π m} .$ Mass dependence yahan sab kuch hai: fixed $T$ aur equal $n$ , $A$ pe, $R \propto \frac{1}{m} \Rightarrow \frac{R _{1}}{R _{2}} = \frac{m _{2}}{m _{1}} .$ Moment integral $1/ m$ kyun scale karta hai: prefactor $α \cdot α^{- 1} = α^{- 1/2} = 2 k_{B} T / m$ hi sirf mass dependence carry karta hai — toh poora flux $\propto m^{- 1/2}$ hai, bhaale ki specific moment directional mean ho, $v_{r m s}$ nahi. H₂ aur O₂ ke liye: $\frac{R _{H_{2}}}{R _{O_{2}}} = \frac{32}{2} = 16 = 4.$ KYU: hydrogen $4 \times$ tez effuse karta hai kyunki yeh $16 \times$ halka hai aur flux $1/ m$ ke saath scale karta hai. Yahi Graham's law hai.

L4.3 — Sirf Gaussian integral se $b$ recover karo

$g (v_{x}) = A e^{- b v_{x}^{2}}$ se shuru karo aur sirf constraint $⟨ v_{x}^{2} ⟩ = k_{B} T / m$ se, Gaussian integrals use karke $b$ rederive karo. Parent ka answer quote mat karo.

Recall Solution

HUM KYA CHAHIYE: do Gaussian integrals. $\int_{- \infty}^{\infty} e^{- b x^{2}} d x = \frac{π}{b}, \int_{- \infty}^{\infty} x^{2} e^{- b x^{2}} d x = \frac{1}{2} \frac{π}{b ^{3}} .$ Variance ratio hai ( $A$ aur $π$ cancel ho jaate hain): $⟨ v_{x}^{2} ⟩ = \frac{\frac{1}{2} π / b ^{3}}{π / b} = \frac{1}{2 b} .$ Physics input $\frac{k _{B} T}{m}$ ke equal set karo: $\frac{1}{2 b} = \frac{k _{B} T}{m} \Rightarrow b = \frac{m}{2 k _{B} T} .$ Doosra integral kyun? Yeh $\int e^{- b x^{2}} d x = π / b$ ko $b$ ke respect mein differentiate karke milta hai — ek slick trick jo "multiply by $x^{2}$ " ko " $- \partial_{b}$ " mein convert karta hai. Dekho Gaussian integrals.

Level 5 — Mastery

L5.1 — $⟨ v ⟩$ ka full normalization check

Proof karo ki $⟨ v ⟩ = \frac{8 k _{B} T}{π m}$ hai $\int_{0}^{\infty} v f (v) d v$ scratch se evaluate karke.

Recall Solution

KYA: $α = m /2 k_{B} T$ aur $C = 4 π (α / π)^{3/2}$ ke saath (toh $f (v) = C v^{2} e^{- α v^{2}}$ ), $⟨ v ⟩ = \int_{0}^{\infty} v \cdot C v^{2} e^{- α v^{2}} d v = C \int_{0}^{\infty} v^{3} e^{- α v^{2}} d v .$ Key integral: $\int_{0}^{\infty} v^{3} e^{- α v^{2}} d v = \frac{1}{2 α ^{2}}$ ( $u = v^{2}$ substitute karo). Toh $⟨ v ⟩ = 4 π (\frac{α}{π})^{3/2} \cdot \frac{1}{2 α ^{2}} = \frac{2}{π} α^{- 1/2} = \frac{2}{π} \frac{2 k _{B} T}{m} = \frac{8 k _{B} T}{π m} . ✓$ Odd power easy kyun tha: $[0, \infty)$ pe $v^{odd} e^{- α v^{2}}$ ke integrals elementary hote hain ( $u = v^{2}$ sub karo); sirf even powers ko $π$ chahiye.

L5.2 — $v_{p}$ se tez molecules ka fraction

Dikhao ki $v > v_{p}$ wale molecules ka fraction $\frac{2}{π} e^{- 1} + erfc (1)$ hai, aur numerically evaluate karo. (Yahan $erfc (x) = \frac{2}{π} \int_{x}^{\infty} e^{- t^{2}} d t$ "complementary error function" hai — Gaussian ka tail area, "fraction beyond a threshold" ke liye exact tool.)

Recall Solution

KYA: dimensionless speed $s = v / v_{p}$ use karo, toh $α v^{2} = s^{2}$ (kyunki $v_{p}^{2} = 1/ α$ ) aur $f (v) d v = \frac{4}{π} s^{2} e^{- s^{2}} d s$ hai. Humein chahiye $P (v > v_{p}) = \frac{4}{π} \int_{1}^{\infty} s^{2} e^{- s^{2}} d s .$ Integration by parts $u = s, d w = s e^{- s^{2}} d s$ ke saath (toh $w = - \frac{1}{2} e^{- s^{2}}$ ): $\int_{1}^{\infty} s^{2} e^{- s^{2}} d s = [- \frac{1}{2} s e^{- s^{2}}]_{1}^{\infty} + \frac{1}{2} \int_{1}^{\infty} e^{- s^{2}} d s = \frac{1}{2} e^{- 1} + \frac{1}{2} \cdot \frac{π}{2} erfc (1) .$ $4/ π$ se multiply karo: $P (v > v_{p}) = \frac{4}{π} \cdot \frac{1}{2} e^{- 1} + \frac{4}{π} \cdot \frac{π}{4} erfc (1) = \frac{2}{π} e^{- 1} + erfc (1) .$ Numbers: $\frac{2}{π} e^{- 1} = 1.1284 \times 0.3679 = 0.4151$ ; $erfc (1) = 0.1573$ . Sum $\approx 0.572$ . Erfc kyun? Gaussian ka "fraction beyond a threshold" exactly complementary error function ki definition hai; $s^{2}$ integral ke liye iske bina koi elementary closed form nahi hai. Roughly 57% molecules most-probable speed se tez chalte hain — kyunki peak ek asymmetric hump ka middle nahi hota.

L5.3 — 2D gas: speed distribution rederive karo

Ek two-dimensional gas mein, velocity space ek plane hai, toh shell ek sphere ki jagah ring hai jiska circumference $2 π v$ hai. $f_{2 D} (v)$ aur iska most probable speed derive karo. Figure dekho.

Recall Solution

KYA BADALATA HAI: 2D mein sirf do components $v_{x}, v_{y}$ hain, dono ab bhi Gaussian $e^{- α v_{i}^{2}}$ hain jisme $α = m /2 k_{B} T$ hai. Geometric factor ab ring area $2 π v d v$ ban jaata hai (circumference $\times$ thickness), $4 π v^{2} d v$ nahi. Joint density do identical Gaussians ka product hai, dono normalized hain taaki 2D prefactor $α / π$ ho: $f_{2 D} (v) d v = 2D norm \frac{α}{π} e^{- α (v_{x}^{2} + v_{y}^{2})} \cdot ring 2 π v d v = 2 α v e^{- α v^{2}} d v .$ Normalization check karo: $u = α v^{2}$ , $d u = 2 α v d v$ substitute karo: $\int_{0}^{\infty} 2 α v e^{- α v^{2}} d v = \int_{0}^{\infty} e^{- u} d u = 1. ✓$ Toh $f_{2 D} (v) = 2 α v e^{- α v^{2}}$ ek genuine probability density hai. Most probable speed: $v e^{- α v^{2}}$ maximize karo: $\frac{d}{d v} (v e^{- α v^{2}}) = (1 - 2 α v^{2}) e^{- α v^{2}} = 0 \Rightarrow 1 = 2 α v^{2} \Rightarrow v_{p}^{2 D} = \frac{1}{2 α} = \frac{k _{B} T}{m} .$ 3D se chota kyun hai ( $v_{p}^{3 D} = 2 k_{B} T / m$ ): geometric factor sirf $v^{1}$ ki tarah grow karta hai (ek ring ka circumference) $v^{2}$ ki jagah (ek sphere ka surface area). Fewer dimensions matlab speed hone ke liye kam "room", isliye falling exponential pehle jeet jaata hai aur peak lower speed pe hoti hai. Visual: figure mein 2D ring (left) ka area per unit $v$ bahut kam add hota hai 3D shell (right) ki comparison mein, isliye "more room" aur "energy cost" ka balance point andar move karta hai. Physically, ek plane pe confined molecule ke paas kinetic energy store karne ke liye ek axis kam hai, isliye same temperature pe uski typical speed lower hai — yeh Equipartition theorem ke saath consistent hai ( $\frac{1}{2} k_{B} T$ per degree of freedom, ab sirf do hain).

Active Recall

Recall Universal ratio

$⟨ v ⟩ / v_{p}$ ? $2/ π \approx 1.128$ — $T$ aur $m$ cancel ho jaate hain, har ideal gas ke liye same.

Recall Heating karne se

$f (v)$ ki peak kyun neeche aati hai? Fixed area $= 1$ ; peak dahiney khisak jaati hai ( $v_{p} \propto T$ ) aur curve wider ho jaati hai, isliye choti honi hi padegi.

Recall Most probable energy vs most probable speed ki energy?

$E_{p} = \frac{1}{2} k_{B} T$ , lekin $\frac{1}{2} m v_{p}^{2} = k_{B} T$ — yeh differ karte hain change of variable mein Jacobian ki wajah se.

Recall Effusion rate ko kaun sa moment govern karta hai, aur yeh mass ke saath kaise scale karta hai?

Directional flux $\int_{0}^{\infty} v_{z} g (v_{z}) d v_{z} = \frac{1}{2} 2 k_{B} T / π m$ — escaping half ke inward component ka mean, $v_{r m s}$ nahi; yeh $m^{- 1/2}$ ke saath scale karta hai, Graham's law deta hai.

Recall 2D most-probable speed?

$k_{B} T / m$ , $f_{2 D} (v) = 2 α v e^{- α v^{2}}$ se (ring factor $2 π v$ ).