Maxwell-Boltzmann distribution — full derivation: Worked examples

2.4.13 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Maxwell-Boltzmann distribution — full derivation

Yeh page Maxwell-Boltzmann distribution ke liye exhaustive drill sheet hai. Parent note ne curve derive kiya tha; yahan hum har tarah ke question hit karte hain jo yeh distribution throw kar sakti hai — har limiting case, har degenerate input, har sign, plus ek word problem aur ek exam twist.

Shuru karne se pehle, do cheezein yaad kar lo jo hum baar baar use karenge.

Yahan $m$ = ek molecule ka mass (kg), $k_{B} = 1.38 \times 1 0^{- 23} J/K$ (Boltzmann's constant, yani "energy per kelvin per molecule"), $T$ = temperature (K). Neeche symbol N₂ ka matlab sirf nitrogen molecule hai (do nitrogen atoms bonded, molar mass 28 g/mol); O₂ oxygen molecule hai (32 g/mol). Teen landmark speeds hain:

$v_{p} = \frac{2 k _{B} T}{m}, ⟨ v ⟩ = \frac{8 k _{B} T}{π m}, v_{r m s} = \frac{3 k _{B} T}{m} .$

Scenario matrix

Is distribution ke baare mein har question exactly inhi case classes mein se kisi ek mein aata hai. Neeche ke examples label kiye gaye hain ki woh kis cell ko fill karte hain.

Cell	Case class	Isme kya "extreme" hai	Example
A	Plug-and-chug landmark speed	ordinary numbers	Ex 1
B	Ek component ka sign	$v_{x} < 0$ vs $v_{x} > 0$ , dono halves	Ex 2
C	Degenerate input $v = 0$	speed exactly zero	Ex 3
D	Limiting behaviour $v \to \infty$	door wali tail	Ex 4
E	Limit $T \to 0$ aur $T \to \infty$	freezing / infinitely hot	Ex 5
F	Usi $T$ par heavy vs light molecule	mass extremes	Ex 6
G	Real-world word problem	atmosphere se escape	Ex 7
H	Exam twist — dimensionless ratio	temperature/mass cancel	Ex 8
I	Finite band mein fraction	ek integral jo tumhe set up karni hai	Ex 9

Ab hum cells A→I chalte hain.

Example 1 — Cell A: plug-and-chug landmark speed

Worked example Oxygen (O₂) ki mean speed

$T = 300 K$ par Ek O₂ molecule ka mass $m = 32 \times 1.66 \times 1 0^{- 27} = 5.31 \times 1 0^{- 26} kg$ hai. Mean speed $⟨ v ⟩$ nikalo.

Forecast: Ek nitrogen molecule (N₂, molar mass 28 g/mol) ki 300 K par most-probable speed ~420 m/s ke paas hoti hai. O₂, N₂ se heavier hai, toh andaaza lagao thodi slow — kuch sau m/s. Apna guess likh lo.

Step 1. Sahi formula chuno. Hume mean chahiye, toh use karo $⟨ v ⟩ = 8 k_{B} T / (π m)$ . Yeh step kyun? Teen landmark speeds exist karti hain; "mean" specifically factor $8/ π$ carry karta hai, 2 ya 3 nahi.

Step 2. Root ke andar numerator assemble karo: $8 k_{B} T = 8 (1.38 \times 1 0^{- 23}) (300) = 3.312 \times 1 0^{- 20}$ . Yeh step kyun? Divide karne se pehle saare constants group karo, taaki units clean rahen (J = kg·m²/s²).

Step 3. Divide karo aur root lo: $⟨ v ⟩ = \frac{3.312 \times 1 0 ^{- 20}}{π ( 5.31 \times 1 0 ^{- 26} )} = 1.986 \times 1 0^{5} \approx 446 m/s .$ Yeh step kyun? Root (m²/s²) ko wapas (m/s) mein convert karta hai.

Verify: Units: $J / kg = m^{2} / s^{2} = m/s$ ✓. Value ~446 m/s, "kuch sau" wali range mein jo humne forecast ki thi ✓.

Example 2 — Cell B: ek component ka sign (dono halves)

Worked example Wo molecules ka fraction jo

$v_{p}$ se tez leftward move kar rahe hain $x$ ke saath Single-component Gaussian $g (v_{x})$ use karke, kitne fraction ka $v_{x} < - v_{p}$ hai? (Yahan $- v_{p}$ ka matlab hai $- x$ direction mein move karna jisme component speed most-probable speed magnitude se zyada ho.)

Forecast: $g$ 0 ke baare mein symmetric hai, isliye leftward aur rightward tails barabar hain. Andaaza: "utna hi jitna fraction $v_{x} > + v_{p}$ ke saath hai."

Step 1. Symmetry recognize karo: $g (- v_{x}) = g (v_{x})$ kyunki yeh sirf $v_{x}^{2}$ par depend karta hai. Yeh step kyun? Yeh leftward question ko equivalent rightward question mein convert karta hai — koi naya integral nahi. $P (v_{x} < - v_{p}) = P (v_{x} > + v_{p}) .$

Step 2. $u = v_{x} / v_{p}$ substitute karke rescale karo. Do cheezein saath mein change hoti hain:

Exponent: kyunki $v_{p}^{2} = 2 k_{B} T / m$ hai, hume milta hai $\frac{m v _{x}^{2}}{2 k _{B} T} = \frac{v _{x}^{2}}{v _{p}^{2}} = u^{2}$ .
Measure (Jacobian): $v_{x} = u v_{p}$ isliye $d v_{x} = v_{p} d u$ — width $d v_{x}$ ka har slice width $v_{p} d u$ ka slice ban jaata hai.

Ab prefactor track karo. $g (v_{x}) d v_{x}$ likhte hain aur dono pieces substitute karte hain: $g (v_{x}) d v_{x} = (\frac{m}{2 π k _{B} T})^{1/2} e^{- u^{2}} (v_{p} d u) .$ Prefactor times Jacobian hai $(\frac{m}{2 π k _{B} T})^{1/2} v_{p}$ . Kyunki $v_{p} = 2 k_{B} T / m$ hai, woh product hai $\frac{m}{2 π k _{B} T} \cdot \frac{2 k _{B} T}{m} = \frac{1}{π} = \frac{1}{π}$ . Toh $P (v_{x} > v_{p}) = \frac{1}{π} \int_{1}^{\infty} e^{- u^{2}} d u = \frac{1}{2} erfc (1) .$ Yeh step kyun? Hume Jacobian $d v_{x} = v_{p} d u$ carry karna hi hoga, warna integral $v_{p}$ ke factor se off ho jaata. Jacobian exactly wahi hai jo prefactor ke dimensional part ko cancel karta hai, chhod ke clean dimensionless number $1/ π$ — standard normalised Gaussian tail (jo upar define kiya gaya $erfc$ integral hai).

Step 3. Evaluate karo: $erfc (1) = 0.1573$ , toh $P = \frac{1}{2} (0.1573) = 0.0786 \approx 7.9%$ . Yeh step kyun? $erfc (z)$ hamara " $z$ ke aage area" function hai; yahan $z = 1$ hai kyunki $u = v_{p}$ 1 pe map hota hai.

Verify: Figure s01 mein do red shaded tails (symmetric bell $g$ ki leftward tail $u < - 1$ aur rightward tail $u > 1$ ) mirror images hain, har ek ~7.9% area carry karta hai, dashed vertical lines $\pm v_{p}$ mark karti hain. Dono tails mein total $= 15.7%$ , baaki $84%$ ka $∣ v_{x} ∣ < v_{p}$ hai, jo ek Gaussian ke "zyaadatar mass centre ke paas" se match karta hai ✓.

Figure s01: Symmetric single-component Gaussian $g (v_{x})$ ko $v_{x} / v_{p}$ ke against plot kiya gaya hai; $\pm 1$ ke baad ke do red-shaded regions equal tails hain, har ek ~7.9% total area ka.

Example 3 — Cell C: degenerate input $v = 0$

Worked example

$f (0)$ kya hai, aur exactly kitne molecules rest par hain?

Forecast: Boltzmann factor $e^{0} = 1$ $v = 0$ par maximum hai. Naive guess: "zyaadatar molecules rest ke paas hain." Kya yeh sahi hai?

Step 1. $v = 0$ ko $f (v) = 4 π (\dots) v^{2} e^{- \dots}$ mein plug karo. $v^{2}$ factor $0$ deta hai. Yeh step kyun? $f$ ek product hai; agar koi bhi factor 0 hai toh puri cheez 0 hai, chahe exponential kitna bhi bada ho. $f (0) = 4 π (\dots) \cdot 0 \cdot 1 = 0.$

Step 2. Interpret karo. "Exactly rest par" ke liye $v_{x} = v_{y} = v_{z} = 0$ simultaneously chahiye — 3D velocity space mein ek single point, jiska zero volume hai. Yeh step kyun? Shell $4 π v^{2} d v$ kuch nahi ban jaati jab uski radius $\to 0$ hoti hai.

Verify: Figure s02 mein red speed curve origin par zero touch karti hai phir apne hump tak chadhti hai. Compare karo dashed single-component Gaussian se, jo 0 par maximal hai. Difference purely geometric $v^{2}$ shell factor hai ✓.

Figure s02: Red = speed distribution $f (v)$ , jo $f (0) = 0$ se shuru hoti hai; dashed black = ek velocity-component Gaussian, jo origin par peak karti hai. Dono ke beech ka gap $v^{2}$ shell factor hai.

Example 4 — Cell D: door wali tail $v \to \infty$

Worked example Kya

$f (v)$ kabhi decrease karna band karta hai, aur tail kitni tezi se marti hai?

Forecast: Andaaza lagao ki kya girta exponential eventually hamesha badhte $v^{2}$ par jeetega.

Step 1. Bade $v$ ke liye do factors compare karo. $v^{2}$ polynomially badhta hai; $e^{- α v^{2}}$ ( $α = m /2 k_{B} T > 0$ ke saath) super-exponentially decay karta hai. Yeh step kyun? Ek key limit: $lim_{v \to \infty} v^{2} e^{- α v^{2}} = 0$ , kyunki $v^{2}$ ka exponential $v$ ki kisi bhi power ko beat karta hai.

Step 2. Curve kahan turn over karta hai, yahan isi page par nikalo. $v^{2} e^{- α v^{2}}$ ki derivative zero karo: $\frac{d}{d v} (v^{2} e^{- α v^{2}}) = (2 v - 2 α v^{3}) e^{- α v^{2}} = 0 \Rightarrow 1 = α v^{2} \Rightarrow v = \frac{1}{α} = \frac{2 k _{B} T}{m} = v_{p} .$ Toh single peak $v_{p}$ par baithta hai; curve $v < v_{p}$ ke liye rise karta hai aur $v > v_{p}$ ke liye monotonically girta hai. Yeh step kyun? Yeh ek hump locate karta hai kisi aur page se borrow kiye bina — jo tail hume chahiye woh $v_{p}$ ke right ki sab cheez hai.

Step 3. Tail ko, maan lo, $v = 3 v_{p}$ par estimate karo: exponent $= α v^{2} = α (9/ α) = 9$ , toh factor $e^{- 9} \approx 1.23 \times 1 0^{- 4}$ . Peak ke relative, height ratio $f (3 v_{p}) / f (v_{p}) = 9 e^{- 9} / (1 \cdot e^{- 1}) = 9 e^{- 8} \approx 3.0 \times 1 0^{- 3}$ . Yeh step kyun? Numbers "tail tiny hai lekin nonzero" ko concrete banate hain — peak height ka 0.3%.

Verify: $9 e^{- 8} = 9 (3.355 \times 1 0^{- 4}) = 3.02 \times 1 0^{- 3}$ ✓. Tail finite $v$ ke liye kabhi zero nahi hoti lekin negligible hai — isliye kuch fast molecules hamesha escape kar sakti hain (dekho Ex 7).

Example 5 — Cell E: temperature limits $T \to 0$ aur $T \to \infty$

Worked example

$f (v)$ kaisi shape approach karta hai jab gas freeze hoti hai ( $T \to 0$ ) ya infinitely hot hoti hai ( $T \to \infty$ )?

Forecast: Andaaza lagao ki peak $v_{p} = 2 k_{B} T / m$ har limit mein kahan jaati hai.

Step 1 ( $T \to 0$ ). $v_{p} \to 0$ aur exponent $- m v^{2} /2 k_{B} T \to - \infty$ kisi bhi $v > 0$ ke liye, isliye curve everywhere zero pe push ho jaati hai sirf origin par chhod ke. Lekin height prefactor $(\frac{m}{2 π k _{B} T})^{3/2} \to \infty$ jab $T \to 0$ . Yeh do effects conflict mein nahi hain — yeh cooperate karte hain:

surviving region ki width $v_{p} \propto T \to 0$ ki tarah shrink hoti hai,
prefactor $T^{- 3/2} \to \infty$ ki tarah blow up karta hai,

aur yeh exactly usi rate par blow up/shrink karte hain jo har temperature par $\int_{0}^{\infty} f (v) d v = 1$ rakhta hai. Isliye limiting object infinitely tall, infinitely thin spike of unit area at $v = 0$ hai — ek Dirac delta $δ (v)$ (ek "spike of area 1"). Yeh step kyun? Diverging prefactor ke bina total area 0 ho jaata aur $f$ probability distribution rehna band kar deti. Normalisation ( $\int f = 1$ ) forced hai har $T$ par, aur yahi reason hai ki height infinity pe jaani chahiye jab width zero pe jaati hai. Physically: frozen gas mein har molecule essentially rest par hai.

Step 2 ( $T \to \infty$ ). $v_{p} \to \infty$ ; peak bina bound ke right slide karta hai aur curve flatten hoti hai (uska area 1 rehna chahiye, isliye wider curve lower hogi). Yeh step kyun? Hotter = faster aur broader, jaisa parent ki Forecast-then-Verify ne predict kiya tha.

Step 3. Sanity number: $T$ double karne par har landmark speed $2 \approx 1.414$ se multiply hoti hai. Yeh step kyun? $v \propto T$ ka matlab hai ki $4 \times$ temperature jump sirf speeds double karta hai — heating molecules ko speed karna inefficient tarika hai.

Verify: $2 = 1.41421$ ✓. Figure s03 mein red curve ( $2 T$ ) black ( $T$ ) curve se right aur neeche baithti hai; dono unit area 1 enclose karte hain, "wider means lower" trade-off confirm karta hai ✓.

Figure s03: Do speed distributions temperature $T$ (black) aur $2 T$ (red) par. Heating peak ko right push karta hai aur use lower karta hai; dono curves unit area rakhte hain.

Example 6 — Cell F: usi $T$ par heavy vs light

Worked example

$300 K$ par helium (He, $m = 4$ u) ki mean speed xenon (Xe, $m = 131$ u) se kitni zyada tez hai?

Forecast: Dono usi temperature feel karte hain, isliye same average energy — lekin halka wala use carry karne ke liye bahut tezi se move karna chahiye. Ratio guess karo.

Step 1. $⟨ v ⟩ = 8 k_{B} T / π m$ mein sirf $m$ gases ke beech different hai. Toh $\frac{⟨ v ⟩ _{H e}}{⟨ v ⟩ _{X e}} = \frac{m _{X e}}{m _{H e}} .$ Yeh step kyun? $T$ aur saare constants cancel ho jaate hain — ratio sirf mass ratio par depend karta hai.

Step 2. $131/4 = 32.75 = 5.72$ . Yeh step kyun? Speed $1/ m$ ke scale hoti hai: 33× heavier ⇒ ~5.7× slower.

Verify: $32.75 = 5.7227$ ✓. Physically, equipartition dono ko same $\frac{3}{2} k_{B} T$ energy deti hai, toh $v \propto 1/ m$ — helium balloons se xenon se kahin zyada tezi se leak karta hai. (Dekho Effusion and Graham's law.)

Example 7 — Cell G: real-world word problem

Worked example Kya hydrogen Earth ke atmosphere se escape kar sakti hai, lekin oxygen nahi?

Earth ki escape speed $v_{esc} = 1.12 \times 1 0^{4} m/s$ hai. Upper-atmosphere temperature $\approx 1000 K$ . H₂ ( $m = 2$ u) aur O₂ ( $m = 32$ u) ke liye $v_{r m s}$ compare karo. Ek gas geological time par leak hoti hai jab $v_{esc} ≲ 6 v_{r m s}$ .

Forecast: Halka H₂ fast hai — guess karo yeh leak kar sakta hai. Heavy O₂ slow hai — guess karo yeh ruk jaayega.

Step 1. H₂: $m = 2 \times 1.66 \times 1 0^{- 27} = 3.32 \times 1 0^{- 27}$ kg. $v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 1000 )}{3.32 \times 1 0 ^{- 27}} = 1.247 \times 1 0^{7} \approx 3530 m/s .$ Yeh step kyun? $v_{r m s}$ (factor 3) escape argument ke liye sabse fair "energy speed" hai.

Step 2. O₂: $m = 16 \times$ heavier, toh $v_{r m s}$ $16 = 4$ × smaller hai: $3530/4 \approx 883 m/s$ . Yeh step kyun? Scratch se recompute karne ki bajay $1/ m$ scaling reuse karo.

Step 3. Threshold $v_{esc} /6 = 1.12 \times 1 0^{4} /6 = 1867$ m/s se compare karo.

H₂: $3530 > 1867$ ⇒ uski fast tail escape karta hai. Leak ho jaata hai. ✓ forecast
O₂: $883 < 1867$ ⇒ $v_{esc}$ par tail negligible. Retained. ✓ forecast Yeh step kyun? Escape tail se govern hota hai (Ex 4), aur tail $v_{r m s}$ ke fixed multiple par baithti hai.

Verify: $1.247 \times 1 0^{7} = 3531$ m/s ✓; $3531/4 = 883$ m/s ✓. Yahi exactly reason hai ki Earth ne apna O₂ aur N₂ rakha lekin apna almost saara free hydrogen khо diya.

Example 8 — Cell H: exam twist, dimensionless ratio

Example 9 — Cell I: finite speed band mein fraction

Worked example Kitne molecules ka speed

$v_{p}$ aur $2 v_{p}$ ke beech hai?

Forecast: Yeh band hump ke peak aur girte side ka kuch hissa cover karta hai — guess karo ek bada chunk, shayad aadha.

Step 1. $u = v / v_{p}$ se non-dimensionalise karo, toh $v = u v_{p}$ aur $d v = v_{p} d u$ , aur exponent $m v^{2} /2 k_{B} T = u^{2}$ ho jaata hai. Phir $f (v) d v \propto u^{2} e^{- u^{2}} d u$ aur, normalisation ke liye $\int_{0}^{\infty} u^{2} e^{- u^{2}} d u = π /4$ use karke, $fraction = \frac{\int _{1}^{2} u ^{2} e ^{- u^{2}} d u}{\int _{0}^{\infty} u ^{2} e ^{- u^{2}} d u} = \frac{4}{π} \int_{1}^{2} u^{2} e^{- u^{2}} d u .$ Yeh step kyun? Scaling $m, T$ remove karta hai; answer universal hai (kisi bhi gas ke liye kisi bhi $T$ par hold karta hai). Dekho Gaussian integrals.

Step 2. Antiderivative $\int u^{2} e^{- u^{2}} d u = \frac{π}{4} erf (u) - \frac{u}{2} e^{- u^{2}}$ use karke finite integral evaluate karo ( $erf$ upar define ki gayi hai): $\int_{1}^{2} u^{2} e^{- u^{2}} d u = [\frac{π}{4} erf (u) - \frac{u}{2} e^{- u^{2}}]_{1}^{2} = 0.2332.$ Yeh step kyun? Closed antiderivative numerical guessing avoid karta hai; $erf$ Gaussian part handle karta hai.

Step 3. Final fraction paane ke liye normalisation prefactor se multiply karo: $fraction = \frac{4}{π} (0.2332) = \frac{4}{1.7725} (0.2332) = 0.526 \approx 53%.$ Yeh step kyun? Prefactor $4/ π$ normalising integral $π /4$ ka reciprocal hai; multiply karne se raw integral ek fraction of all molecules mein convert ho jaata hai.

Verify: Figure s04 mein $v_{p}$ se $2 v_{p}$ tak ka red shaded band visibly curve ke neeche total area ka lagbhag aadha cover karta hai, $0.53$ se match karta hai ✓. (Hamara "shayad aadha" forecast bilkul sahi nikla — band peak ko bracket karta hai, isliye zyaadatar molecules capture ho jaati hain.)

Figure s04: Speed distribution vs $v / v_{p}$ ; $v_{p}$ aur $2 v_{p}$ ke beech red-shaded region total area ka ~53% hai, $u = 1$ aur $u = 2$ par dashed lines mark karti hain.

Recall Har example kaun si cell hai?

A→Ex1 (plug-in) · B→Ex2 (signs, dono tails) · C→Ex3 ( $v = 0$ ) · D→Ex4 ( $v \to \infty$ ) · E→Ex5 ( $T \to 0, \infty$ ) · F→Ex6 (mass extremes) · G→Ex7 (word problem) · H→Ex8 (ratio) · I→Ex9 (finite band).

Active Recall

f (0) = 0

kyun hai jabki Boltzmann factor wahan maximal hai?

Geometric shell factor

v^{2} \to 0

: radius 0 ka sphere ka koi volume nahi hota, isliye koi bhi velocity vectors ka magnitude exactly zero nahi hota.

Usi

T

par do gases ki mean speeds kaise compare hoti hain?

⟨ v ⟩ \propto 1/ m

— mass ratio ka inverse square root;

T

cancel ho jaata hai.

Kya

⟨ v ⟩ / v_{p}

gas-dependent hai?

Nahi — yeh har ideal gas ke liye har temperature par

2/ π \approx 1.128

ke barabar hai.

T \to 0

par

f (v)

ka kya hota hai?

Yeh

v = 0

par Dirac delta spike mein collapse ho jaata hai; diverging prefactor

\propto T^{- 3/2}

area ko 1 ke barabar rakhta hai jab width

\propto T

shrink hoti hai.

erfc (z)

kya measure karta hai?

Ek Gaussian ke area ka woh fraction jo

z

ke aage tail mein hai, yani

erfc (z) = 1 - erf (z)