Maxwell-Boltzmann distribution — full derivation

2.4.13 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

Hum derive kya kar rahe hain?

Hum temperature $T$ par ek ideal gas ke liye $f (v)$ ki explicit form chahte hain.

Step 1 — Har velocity component mein Gaussian KYU hota hai

KAISE. Maano $v_{x}$ ko $d v_{x}$ mein paane ki probability $g (v_{x}) d v_{x}$ hai, similarly $y, z$ ke liye. Independence ki wajah se joint distribution factorize hoti hai:

F (v_{x}, v_{y}, v_{z}) = g (v_{x}) g (v_{y}) g (v_{z}) .

Isotropy ki wajah se, $F$ sirf magnitude $v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ par depend kar sakta hai, yaani $v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ par. To:

g (v_{x}) g (v_{y}) g (v_{z}) = Φ (v_{x}^{2} + v_{y}^{2} + v_{z}^{2}) .

To $ln g (v_{x}) = a - b v_{x}^{2}$ , jisse milta hai:

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

Hamein $b > 0$ chahiye (warna normalize nahi ho sakta — large speeds infinitely likely ho jaenge).

Step 2 — Normalization se constants fix karo

Normalization $\int_{- \infty}^{\infty} g (v_{x}) d v_{x} = 1$ , Gaussian integral $\int_{- \infty}^{\infty} e^{- b x^{2}} d x = π / b$ use karke:

A \frac{π}{b} = 1 \Rightarrow A = \frac{b}{π} .

$b$ ko physics se fix karo. $b$ ko temperature se relate karna hoga. Equipartition use karo: har translational degree of freedom 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} .

Gaussian $g (v_{x}) = A e^{- b v_{x}^{2}}$ ke liye, variance hai $⟨ v_{x}^{2} ⟩ = \frac{1}{2 b}$ .

$\frac{1}{2 b} = \frac{k _{B} T}{m}$ set karne par:

b = \frac{m}{2 k _{B} T}, g (v_{x}) = (\frac{m}{2 π k _{B} T})^{1/2} e^{- m v_{x}^{2} /2 k_{B} T} .

Step 3 — Components se speed tak

Joint density teen identical Gaussians ka product hai:

F (v) d^{3} 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} d v_{x} d v_{y} d v_{z} .

Spherical shells mein convert karo, $d^{3} v \to 4 π v^{2} d v$ :

Figure — Maxwell-Boltzmann distribution — full derivation

Step 4 — Teen characteristic speeds (derive karo, memorize mat karo)

Shorthand ke liye $α = \frac{m}{2 k _{B} T}$ lo, to $f (v) \propto v^{2} e^{- α v^{2}}$ .

Most probable speed $v_{p}$ — $f$ maximize karo: $\frac{d}{d v} (v^{2} e^{- α v^{2}}) = 0$ set karo:

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? Peak wahan hai jahan rising $v^{2}$ exactly falling exponential ko balance karta hai.

Mean speed $⟨ v ⟩$ — $\int_{0}^{\infty} v f (v) d v$ :

⟨ v ⟩ = \frac{8 k _{B} T}{π m} .

RMS speed $v_{r m s}$ — $⟨ v^{2} ⟩ = \int_{0}^{\infty} v^{2} f (v) d v = \frac{3 k _{B} T}{m}$ se ( $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ ke saath consistent):

v_{r m s} = \frac{3 k _{B} T}{m} .

$T$ badhne par $f (v)$ ka kya hoga? Pehle forecast karo, phir check karo. Forecast: peak right shift karega aur curve flatten/broaden hogi. Verify: $v_{p} \propto T$ → peak right move karta hai. Area 1 rehta hai, isliye higher peak position matlab lower, wider curve. Hotter gas = faster aur zyada spread-out speeds. ✓

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bachche ko samjhao

Ek bumper cars ki bheed imagine karo jo randomly crash kar rahi hain. Kuch time baad, kuch cars zoom kar rahi hain, kuch crawl kar rahi hain, aur zyaatar ek medium speed par hain. Agar inhe speed ke hisaab se sort karo, ek hill-shaped graph milta hai: almost koi frozen nahi hain (kuch to speed chahiye), almost koi super-fast nahi hain (usme huge energy lagti hai), aur beech ka bump "usual" speed hai. Floor ko heat karo taaki cars zyada push karein — poori hill faster speeds ki taraf slide karti hai aur spread out hoti hai. Woh hill Maxwell-Boltzmann distribution hai.

Active Recall

f (v)

ko shape karne wale do competing factors kaunse hain?

Geometric factor

4 π v^{2}

(high

v

par velocity-space mein zyada room) vs Boltzmann factor

e^{- m v^{2} /2 k_{B} T}

(energy cost).

f (0) = 0

kyun hai?

v^{2}

shell-volume factor

v = 0

par vanish karta hai — radius 0 ki shell mein koi velocity vector zero magnitude nahi rakhta.

Single-component distribution

g (v_{x})

ki form?

(\frac{m}{2 π k _{B} T})^{1/2} e^{- m v_{x}^{2} /2 k_{B} T}

, ek Gaussian 0 par centered.

Full MB speed distribution?

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

Most probable speed

v_{p}

2 k_{B} T / m

\frac{d}{d v} (v^{2} e^{- α v^{2}}) = 0

se.

Mean speed

⟨ v ⟩

8 k_{B} T / π m

RMS speed

v_{r m s}

3 k_{B} T / m

Teen speeds ka order?

v_{p} < ⟨ v ⟩ < v_{r m s}

(ratios

2 : 8/ π : 3

g (v_{x})

Gaussian (exponential of

- v_{x}^{2}

) kyun hona chahiye?

Independence joint density ko product mein factorize karta hai; isotropy use sirf

v_{x}^{2} + v_{y}^{2} + v_{z}^{2}

par dependent banata hai; sirf exponential products ko sums mein convert karta hai.

Constant

b = m /2 k_{B} T

kya fix karta hai?

Equipartition:

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

⇒

⟨ v_{x}^{2} ⟩ = 1/ (2 b) = k_{B} T / m

Higher

T

par curve kaise change hoti hai?

Peak right shift karta hai (

v_{p} \propto T

), curve broaden aur lower hoti hai (area conserved).

Connections

Boltzmann factor and partition function — derivation ke core mein $e^{- E / k_{B} T}$ weight
Equipartition theorem — $⟨ v_{x}^{2} ⟩ = k_{B} T / m$ deta hai $b$ fix karne ke liye
Kinetic theory of gases — $v_{r m s}$ pressure ko feed karta hai $P = \frac{1}{3} nm ⟨ v^{2} ⟩$
Gaussian integrals — normalization aur moments
Effusion and Graham's law — flux $\propto ⟨ v ⟩ \propto 1/ m$
Maxwell-Boltzmann vs Fermi-Dirac vs Bose-Einstein — quantum statistics ka classical limit

Maxwell-Boltzmann distribution — full derivation

Hum derive kya kar rahe hain?

Step 1 — Har velocity component mein Gaussian KYU hota hai

Step 2 — Normalization se constants fix karo

Step 3 — Components se speed tak

Step 4 — Teen characteristic speeds (derive karo, memorize mat karo)

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Active Recall

Connections

Concept Map