Equipartition theorem — ½k_BT per quadratic degree of freedom

2.4.14 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

EK "quadratic degree of freedom" HOTA KYA HAI?

YE SACH KYU HAI? — Scratch se Derivation

Hum classical canonical (Boltzmann) distribution use karte hain: energy $E$ ke microstate ki probability $\propto e^{- E / k_{B} T}$ hoti hai. Hum chahte hain ke EK quadratic term $E_{q} = α q^{2}$ mein store average energy nikalen.

Step 1 — Average ko Boltzmann-weighted integral ke roop mein likhein. $⟨ α q^{2} ⟩ = \frac{\int _{- \infty}^{\infty} α q ^{2} e ^{- α q^{2} / k_{B} T} d q}{\int _{- \infty}^{\infty} e ^{- α q^{2} / k_{B} T} d q}$

Ye step kyun? Numerator har $q^{2}$ ki value ko us $q$ ki probability se weight karta hai; denominator normalize karta hai. Baaki saare coordinates ke factors $e^{- (\dots) / k_{B} T}$ numerator aur denominator mein identical hote hain aur cancel ho jaate hain, isliye hum sirf ek variable $q$ se deal karte hain. Yahi crucial simplification hai.

Step 2 — Dimensionless banane ke liye substitute karein. Maano $β = 1/ k_{B} T$ aur define karein $Z (β) = \int_{- \infty}^{\infty} e^{- β α q^{2}} d q .$ Ye step kyun? Dhyan do ke numerator exactly $- \frac{\partial}{\partial β} (\frac{1}{α} \cdot α ...)$ hai. Zyada cleanly: $\int α q^{2} e^{- β α q^{2}} d q = - \frac{\partial}{\partial β} \int e^{- β α q^{2}} d q = - Z^{'} (β) .$ Toh $⟨ α q^{2} ⟩ = - \frac{Z ^{'} ( β )}{Z ( β )} = - \frac{d}{d β} ln Z$ . Kyun? Exponential ko differentiate karne par $- α q^{2}$ neeche aa jaata hai, jo humein chahiye wala factor bana deta hai — ek standard trick jo ek awkward integral ko derivative mein badal deti hai.

Step 3 — Gaussian evaluate karein. $\int_{- \infty}^{\infty} e^{- a x^{2}} d x = π / a$ use karke, $a = β α$ ke saath: $Z (β) = \frac{π}{β α} = \frac{π}{α} β^{- 1/2} .$

Step 4 — Log-derivative lein. $ln Z = const - \frac{1}{2} ln β ⟹ - \frac{d l n Z}{d β} = \frac{1}{2 β} = \frac{1}{2} k_{B} T .$

Figure — Equipartition theorem — ½k_BT per quadratic degree of freedom

Real gases ke liye DOF count karna

Common mistake Common errors ko steel-man karke samjhein

Galti 1: "Har degree of freedom $\frac{1}{2} k_{B} T$ deta hai, toh ek vibrating bond $\frac{1}{2} k_{B} T$ deta hai." Kyun sahi lagta hai: "ek bond = ek motion." Fix: equipartition quadratic energy terms count karta hai, motions nahi. Ek vibration energy DONO kinetic ( $\overset{r}{˙}^{2}$ ) aur potential ( $r^{2}$ ) forms mein store karta hai → do terms → total $k_{B} T$ .

Galti 2: "Stiffer/heavier molecules har mode mein zyada thermal energy store karte hain." Kyun sahi lagta hai: ek stiff spring "zyada energy rakhta hai." Fix: derivation ne dikhaaya ke $α$ cancel hota hai. Fixed $T$ par, har quadratic slot exactly $\frac{1}{2} k_{B} T$ rakhta hai, mass ya stiffness se independent.

Galti 3: "Ye hamesha kaam karta hai." Kyun sahi lagta hai: ye khubsoorat taur par universal hai. Fix: ye ek classical result hai. Jab quantum levels ke beech energy gap $≳ k_{B} T$ ho (jaise room temp par vibration, low T par H₂ ka rotation), toh woh mode "frozen out" ho jaata hai aur kuch contribute nahi karta. Equipartition tabhi kaam karta hai jab $k_{B} T ≫$ quantum level spacing ho.

Galti 4: "Non-quadratic energy par bhi apply hota hai." Kyun sahi lagta hai: tum ek universal rule chahte ho. Fix: $\frac{1}{2} k_{B} T$ Gaussian integral se aata hai, jiske liye $E \propto q^{2}$ chahiye. $E \propto q^{4}$ ke liye tumhe $\frac{1}{4} k_{B} T$ milega (generalized theorem try karo!).

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho ek bade kamre mein uchhalti hue balls hain, aur "temperature" waise hai jaise har bachche ke paas pocket money hoti hai. Equipartition kehta hai: har alag tarike se ek ball jo bhi hilaaye use SAME amount ka pocket money milta hai, matlab $\frac{1}{2} k_{B} T$ . Ek ball jo left-right, up-down, aur front-back move kar sakti hai ke teen tarike hain → teen lots of money. Ek spring se connected ball ko spring stretch karne ke liye bhi money milti hai. Koi farak nahi padta ke spring strong hai ya weak, ya ball heavy hai ya light — rule energy fairly share karta hai. (Sirf ek catch hai: agar koi hilaav "start karna bahut expensive" ho — jaise ek quantum jump jo ek baar mein ek bada chunk maange — woh hilaav skip ho jaata hai, aur equipartition use kuch nahi deta.)

Active-recall checkpoint

Flashcards

Har quadratic degree of freedom average par kitni energy carry karta hai?

\frac{1}{2} k_{B} T

Ek degree of freedom "quadratic" kya banata hai?

Uska energy term kisi coordinate ya momentum mein

α q^{2}

form ka hota hai.

Result stiffness/mass constant

α

se independent kyun hai?

α

ln Z

mein sirf ek additive constant ke roop mein aata hai, jo

- d / d β

ke neeche vanish ho jaata hai; physically ek stiffer well

⟨ q^{2} ⟩

ko itna narrow karta hai ke

α ⟨ q^{2} ⟩ = \frac{1}{2} k_{B} T

bana rahe.

Monatomic ideal gas ki mean energy per atom?

\frac{3}{2} k_{B} T

(3 translational DOF), isliye

C_{V} = \frac{3}{2} R

Diatomic gas mein room temperature par

C_{V} = \frac{5}{2} R

kyun hota hai?

3 translation + 2 rotation = 5 quadratic DOF; 3rd rotation aur vibration frozen out hain.

Ek vibrational mode 2 quadratic DOF kyun count hota hai?

Ye dono kinetic (

\frac{1}{2} μ \overset{r}{˙}^{2}

) aur potential (

\frac{1}{2} κ r^{2}

) energy store karta hai, dono quadratic hain.

Classical solid ki heat capacity (Dulong–Petit)?

C_{V} = 3 R

, har atom ke liye 3 KE + 3 PE quadratic DOF se =

3 k_{B} T

Equipartition kab FAIL karta hai?

Jab kisi mode ki quantum energy-level spacing

≳ k_{B} T

ho, toh woh mode freeze out ho jaata hai (jaise room T par vibrations).

Equipartition derive karne ke liye starting formula?

⟨ α q^{2} ⟩ = \frac{\int α q ^{2} e ^{- β α q^{2}} d q}{\int e ^{- β α q^{2}} d q} = - \frac{d ln Z}{d β}

with

Z = π / (β α)

E = α ∣ q ∣^{n}

ke liye generalized result?

⟨ E ⟩ = \frac{1}{n} k_{B} T

; equipartition ka

\frac{1}{2}

sirf

n = 2

case hai.

Air ke liye

γ = 1.4

equipartition se kyun explain hota hai?

Diatomic 5 DOF →

C_{V} = \frac{5}{2} R

C_{P} = \frac{7}{2} R

γ = 7/5 = 1.4

Connections

Boltzmann distribution — poore proof ka base woh $e^{- E / k_{B} T}$ weighting provide karta hai.
Partition function Z — $Z (β)$ aur $- d ln Z / d β$ trick har jagah reuse hoti hai.
Maxwell–Boltzmann speed distribution — 3 translational DOF dete hain $⟨ \frac{1}{2} m v^{2} ⟩ = \frac{3}{2} k_{B} T$ .
Heat capacity of gases / Dulong–Petit law — direct applications.
Quantum freezing of degrees of freedom — equipartition kahan aur kyun break down karta hai.
Ideal gas law PV=NkT — translational equipartition + kinetic theory isse derive karta hai.
Brownian motion — equipartition ek suspended particle ki kinetic energy fix karta hai.