Equipartition theorem — ½k_BT per quadratic degree of freedom: Exercises

2.4.14 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Equipartition theorem — ½k_BT per quadratic degree of freedo

Poore note mein, parent note se yeh core statement yaad rakho: har quadratic energy term average mein $\frac{1}{2} k_{B} T$ carry karti hai, jahan $k_{B}$ Boltzmann's constant hai aur $T$ absolute temperature hai (kelvin mein). Hum $k_{B} = 1.380649 \times 1 0^{- 23} J/K$ aur gas constant $R = N_{A} k_{B} = 8.314 J/(mol\cdotK)$ lete hain jahan $N_{A} = 6.022 \times 1 0^{23}$ Avogadro's number hai (atoms per mole).

Level 1 — Recognition

L1.1 — Quadratic slots count karo

Neeche di gayi har energy expression mein kitne quadratic degrees of freedom (DOF) hain, yeh batao. Ek quadratic DOF ek aisa additive term hota hai jiska form $α q^{2}$ ho (ek constant aur ek squared coordinate ya momentum ka product). Part (c) mein, $μ$ do bonded atoms ki reduced mass hai — ek single effective mass jo hume do atoms ki vibration ko ek particle of mass $μ$ ki motion ki tarah treat karne deti hai jo bond ke saath move karta hai; ise "vibration ki effective mass" samjho. Symbol $\overset{r}{˙}$ bond length $r$ ke change ki rate hai (uski stretching speed), aur $κ$ bond ka spring constant hai.

(a) $E = \frac{1}{2} m v_{x}^{2} + \frac{1}{2} m v_{y}^{2} + \frac{1}{2} m v_{z}^{2}$ (b) $E = \frac{1}{2} I ω^{2}$ (c) $E = \frac{1}{2} μ \overset{r}{˙}^{2} + \frac{1}{2} κ r^{2}$ (d) $E = \frac{1}{2} m v_{x}^{2} + a x^{4}$ (yeh $x^{4}$ term quadratic nahi hai)

Recall Solution

Sirf $α q^{2}$ shape wale additive terms count karo.

(a) teen aise terms ( $v_{x}^{2}, v_{y}^{2}, v_{z}^{2}$ ) → $== 3 ==$ quadratic DOF.
(b) ek term → $== 1 ==$ quadratic DOF.
(c) do terms, ek $\overset{r}{˙}^{2}$ mein (kinetic) aur ek $r^{2}$ mein (potential) → $== 2 ==$ quadratic DOF. Yahi vibration ka double-count hai.
(d) $v_{x}^{2}$ term quadratic hai → $== 1 ==$ quadratic DOF. $a x^{4}$ term quartic hai (coordinate fourth power tak raised hai), yeh $α q^{2}$ form ka nahi hai, isliye yeh quadratic DOF nahi hai aur $\frac{1}{2} k_{B} T$ contribute nahi karta. (Yeh actually kitna contribute karta hai, yeh L3.2/L5.2 mein neeche work out kiya gaya hai, tools build hone ke baad — yahaan assume mat karo.) Toh: $== 1 ==$ quadratic DOF plus ek non-quadratic term.

L1.2 — Mean energy state karo

Ek system mein temperature $T$ par exactly $f$ quadratic DOF hain. Uski mean total energy likho.

Recall Solution

$⟨ E ⟩ = f \times \frac{1}{2} k_{B} T = \frac{f}{2} k_{B} T .$ Mass, stiffness, ya geometry kuch bhi matter nahi karta — sirf count $f$ aur temperature matter karti hai.

Level 2 — Application

L2.1 — Ek molecule ki mean translational energy

$T = 300 K$ par ek single air molecule ki mean translational kinetic energy compute karo. Answer joules mein do.

Recall Solution

Translation mein 3 quadratic DOF hain ( $x, y, z$ direction mein motion), toh $⟨ E_{trans} ⟩ = \frac{3}{2} k_{B} T = \frac{3}{2} (1.380649 \times 1 0^{- 23}) (300) .$ $⟨ E_{trans} ⟩ \approx 6.21 \times 1 0^{- 21} J .$

L2.2 — Diatomic gas ki heat capacities

Nitrogen ( $N_{2}$ ) ke liye room temperature par, translation aur rotation active hain lekin vibration frozen hai. Per mole $C_{V}$ , $C_{P}$ , aur $γ = C_{P} / C_{V}$ nikalo.

Recall Solution

Active DOF: 3 translation + 2 rotation $= 5$ . Toh $U = \frac{5}{2} R T, C_{V} = (\frac{\partial U}{\partial T})_{V} = \frac{5}{2} R .$ Ideal gas ke liye $C_{P} = C_{V} + R = \frac{7}{2} R$ (yeh " $+ R$ " constant pressure par expansion ke kaam se aata hai $P V = N k T$ ke through, dekho Ideal gas law PV=NkT). $C_{V} = \frac{5}{2} (8.314) \approx 20.79 J/(mol\cdotK), C_{P} = \frac{7}{2} (8.314) \approx 29.10 J/(mol\cdotK) .$ $γ = \frac{C _{P}}{C _{V}} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4.$

L2.3 — Dulong–Petit prediction

Ek classical solid mein har atom ko ek 3D harmonic well mein baitha mana jaata hai. Molar heat capacity predict karo aur Dulong–Petit law ki value se compare karo.

Recall Solution

Har atom: 3 kinetic + 3 potential quadratic terms $= 6$ DOF. $U = 6 \times \frac{1}{2} R T = 3 R T ⟹ C_{V} = 3 R \approx 24.94 J/(mol\cdotK) .$ Yahi Dulong–Petit law hai, jo room temperature ke paas kaafi solids ke liye confirm hoti hai.

Level 3 — Analysis

L3.1 — Stiffness cancel kyun hoti hai? (numerical demonstration)

Do 1-D oscillators same temperature $T = 300 K$ par hain: oscillator A ka spring constant $κ_{A} = 1 N/m$ hai, oscillator B ka $κ_{B} = 100 N/m$ hai. Dono ke liye mean potential energy $⟨ \frac{1}{2} κ x^{2} ⟩ = \frac{1}{2} k_{B} T$ hai. Har ek ke liye mean squared displacement $⟨ x^{2} ⟩$ compute karo aur confirm karo ki mean potential energies equal hain.

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

Yeh figure har oscillator ke liye probability density $P (x)$ plot karta hai — yeh kitna likely hai ki mass equilibrium se $x$ displacement par mile (Boltzmann weight $\propto e^{- κ x^{2} /2 k_{B} T}$ , normalize kiya hua taaki har curve ke neeche area 1 ho). Horizontal axis displacement $x$ hai; vertical axis woh probability density hai. Cyan curve soft spring (A) ki hai, jo mass ko door tak jaane deta hai — ek choda bell. Amber curve stiff spring (B) ki hai, jo mass ko $x = 0$ ke paas pin karta hai — ek sankra, oouncha bell. Stiff bell $100 = 10 \times$ sankra hai. Figure ka point yeh hai: haalaanki dono bells ki widths bahut alag hain, phir bhi stored energy $\frac{1}{2} κ ⟨ x^{2} ⟩$ (width $^{2} \times$ stiffness) same aati hai.

Recall Solution

$\frac{1}{2} κ ⟨ x^{2} ⟩ = \frac{1}{2} k_{B} T$ set karo, toh $⟨ x^{2} ⟩ = k_{B} T / κ$ . $⟨ x^{2} ⟩_{A} = \frac{( 1.380649 \times 1 0 ^{- 23} ) ( 300 )}{1} \approx 4.14 \times 1 0^{- 21} m^{2},$ $⟨ x^{2} ⟩_{B} = \frac{( 1.380649 \times 1 0 ^{- 23} ) ( 300 )}{100} \approx 4.14 \times 1 0^{- 23} m^{2} .$ Stiff spring (B) mass ko 100× chhote $⟨ x^{2} ⟩$ mein confine karta hai — uska Gaussian sankra hai (figure mein tall thin amber curve dekho). Lekin product $\frac{1}{2} κ_{A} ⟨ x^{2} ⟩_{A} = \frac{1}{2} κ_{B} ⟨ x^{2} ⟩_{B} = \frac{1}{2} k_{B} T \approx 2.07 \times 1 0^{- 21} J$ identical hai. Stiffness $κ$ spread ko exactly utna hi narrow karti hai jitna stored energy ko fixed rakhne ke liye zaroori hai — isi liye $α$ (yahaan $κ /2$ ) general derivation mein cancel ho jaata hai.

L3.2 — Ek quartic potential

Ek particle 1-D mein energy $E = \frac{1}{2} m v^{2} + b x^{4}$ ( $b > 0$ ) ke saath move karta hai. Generalised equipartition rule $⟨ α q^{n} ⟩ = \frac{1}{n} k_{B} T$ use karke temperature $T$ par uski total mean energy nikalo (yeh rule L5.2 mein scratch se prove ki jaati hai; yahaan hum sirf ise apply kar rahe hain — yeh kehti hai: ek energy term jahan ek coordinate $n$ th power par hai, average mein $\frac{1}{n} k_{B} T$ carry karti hai).

Recall Solution

Kinetic term $\frac{1}{2} m v^{2}$ quadratic hai ( $n = 2$ ) → $\frac{1}{2} k_{B} T$ contribute karta hai.
Potential term $b x^{4}$ mein $n = 4$ hai → $\frac{1}{4} k_{B} T$ contribute karta hai. $⟨ E ⟩ = \frac{1}{2} k_{B} T + \frac{1}{4} k_{B} T = \frac{3}{4} k_{B} T .$ Ek quartic well large $x$ par ek spring se "stiffer" hai, toh woh quadratic potential ke $\frac{1}{2} k_{B} T$ se kam energy store karta hai — generalised rule ise cleanly capture karta hai.

Level 4 — Synthesis

L4.1 — Vibration turn on hoti hai

Ek diatomic gas ko room temperature se high temperature tak garam karo, jahan bond vibration unfreeze ho jaati hai. High- $T$ regime mein $C_{V}$ aur $γ$ nikalo, aur batao ki $C_{V}$ room temperature ki value se kitna jump hua.

Recall Solution

Room temp: 3 translation + 2 rotation = 5 DOF → $C_{V} = \frac{5}{2} R$ . High temp: vibration 2 DOF add karti hai (KE + PE) → total $5 + 2 = 7$ DOF. $C_{V} = \frac{7}{2} R \approx 29.10 J/(mol\cdotK), C_{P} = \frac{9}{2} R, γ = \frac{9/2}{7/2} = \frac{9}{7} \approx 1.286.$ Jump hai $Δ C_{V} = \frac{7}{2} R - \frac{5}{2} R = R \approx 8.314 J/(mol\cdotK)$ — ek poora $R$ , kyunki vibration do half- $R$ slots contribute karti hai. Dekho Quantum freezing of degrees of freedom taaki samjho ki yeh unfreezing gradual kyun hoti hai, ek sharp step nahi.

L4.2 — Energy se speed

Equipartition result $\frac{3}{2} k_{B} T = \frac{1}{2} m ⟨ v^{2} ⟩$ use karke, $T = 300 K$ par nitrogen molecule ( $m = 4.65 \times 1 0^{- 26} kg$ ) ki root-mean-square speed $v_{rms} = ⟨ v^{2} ⟩$ nikalo.

Recall Solution

Total translational KE ko $\frac{3}{2} k_{B} T$ ke barabar rakho: $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T ⟹ ⟨ v^{2} ⟩ = \frac{3 k _{B} T}{m} .$ $⟨ v^{2} ⟩ = \frac{3 ( 1.380649 \times 1 0 ^{- 23} ) ( 300 )}{4.65 \times 1 0 ^{- 26}} \approx 2.672 \times 1 0^{5} m^{2} / s^{2} .$ $v_{rms} = ⟨ v^{2} ⟩ \approx 517 m/s .$ Yahi $v_{rms}$ Maxwell–Boltzmann speed distribution mein appear karta hai — equipartition us distribution ki width fix karta hai.

Level 5 — Mastery

L5.1 — Ek quadratic term ke liye equipartition scratch se derive karo

Boltzmann distribution se shuru karo (kisi value $q$ ki probability $\propto e^{- α q^{2} / k_{B} T}$ hai), aur prove karo ki $⟨ α q^{2} ⟩ = \frac{1}{2} k_{B} T$ theorem quote kiye bina. Tum Gaussian integral $\int_{- \infty}^{\infty} e^{- a x^{2}} d x = π / a$ for $a > 0$ use kar sakte ho.

Recall Solution

Step 1 — weighted integral ke roop mein average. $β \equiv 1/ k_{B} T$ likho. Boltzmann-weighted average hai $⟨ α q^{2} ⟩ = \frac{\int _{- \infty}^{\infty} α q ^{2} e ^{- β α q^{2}} d q}{\int _{- \infty}^{\infty} e ^{- β α q^{2}} d q} .$ Kyun: numerator har $α q^{2}$ ko uski probability se weight karta hai; denominator normalize karta hai. Baaki sab coordinates top aur bottom mein cancel ho jaate hain.

Step 2 — numerator ko ek derivative mein badlo. Define karo $Z (β) = \int_{- \infty}^{\infty} e^{- β α q^{2}} d q$ . Integral ke andar differentiate karne par ek factor $- α q^{2}$ pull down hota hai: $\frac{\partial Z}{\partial β} = \int_{- \infty}^{\infty} (- α q^{2}) e^{- β α q^{2}} d q .$ Isliye numerator $- Z^{'} (β)$ ke equal hai, aur $⟨ α q^{2} ⟩ = - \frac{Z ^{'} ( β )}{Z ( β )} = - \frac{d l n Z}{d β} .$ Yeh trick kyun: yeh ek awkward $q^{2}$ -weighted integral ko ek simple Gaussian ke plain derivative mein convert kar deta hai. (Yahi log-derivative move Partition function Z ke saath use hota hai.)

Step 3 — Gaussian evaluate karo. $a = β α$ rakho: $Z (β) = \frac{π}{β α} = \frac{π}{α} β^{- 1/2} .$

Step 4 — log-derivative lo. $ln Z = constant in β \frac{1}{2} ln \frac{π}{α} - \frac{1}{2} ln β ⟹ - \frac{d l n Z}{d β} = \frac{1}{2 β} = \frac{1}{2} k_{B} T .$ Stiffness $α$ sirf additive constant $\frac{1}{2} ln (π / α)$ mein tha, jo $d / d β$ ke under vanish ho jaata hai. Isliye $⟨ α q^{2} ⟩ = \frac{1}{2} k_{B} T,$ $α$ se independent. $■$

L5.2 — $E = α ∣ q ∣^{n}$ tak generalise karo

$E = α ∣ q ∣^{n}$ ke liye argument repeat karo aur dikhao ki $⟨ E ⟩ = \frac{1}{n} k_{B} T$ . (Hint: $β$ -dependence extract karne ke liye $u = β^{1/ n} q$ substitute karo.)

Recall Solution

$Z (β) = \int_{- \infty}^{\infty} e^{- β α ∣ q ∣^{n}} d q .$ Substitute karo $u = β^{1/ n} q$ , toh $q = β^{- 1/ n} u$ aur $d q = β^{- 1/ n} d u$ : $Z (β) = β^{- 1/ n} constant in β \int_{- \infty}^{\infty} e^{- α ∣ u ∣^{n}} d u = C β^{- 1/ n} .$ Phir $ln Z = ln C - \frac{1}{n} ln β$ , aur $⟨ E ⟩ = - \frac{d l n Z}{d β} = \frac{1}{n β} = \frac{1}{n} k_{B} T .$ $n = 2$ rakhne par familiar $\frac{1}{2} k_{B} T$ wapas milta hai; $n = 4$ deta hai $\frac{1}{4} k_{B} T$ (L3.2 se match karta hai). $■$

Related deep dives: Heat capacity of gases, Brownian motion.