Equipartition theorem — ½k_BT per quadratic degree of freedom: Visual walkthrough

2.4.14 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Equipartition theorem — ½k_BT per quadratic degree of freedo

Shuru karne se pehle, yahan woh vocabulary hai jo hum earn karenge (koi bhi cheez draw karne se pehle use nahi ki jayegi):

Hamaara ek hi goal hai:

Step 1 — Energy bowl banao

KYA. Hum energy $E_{q} = α q^{2}$ ko coordinate $q$ ke against plot karte hain. Yeh ek parabola hai: ek bowl. Bead iske andar kahin rehta hai, aur position $q$ pe wall ki height batati hai ki wahan rehne ki kitni energy lagti hai.

KYUN. Aage ki saari cheezein is baare mein hain ki bead bowl mein har jagah kitni baar jaata hai. Probability ki baat karne se pehle, hum woh landscape dekhna chahte hain jo bead explore kar raha hai. Bowl stage hai; bead actor hai.

PICTURE. Orange curve $α q^{2}$ hai. $q = 0$ pe (bottom par) cost zero hai — sasta, wahan baithna aasaan. Jaise $q$ kisi bhi direction mein badhta hai, wall tezi se upar uthti hai. Do dashed heights dekho: ek stiffer $α$ (steeper bowl) same $q$ ko kaafi zyada expensive bana deta hai.

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

Step 2 — Boltzmann weight: bead $q$ pe kitni baar jaata hai?

KYA. Nature sabhi jagahon ko equally visit nahi karta. Rule (Boltzmann distribution) yeh hai: coordinate $q$ pe bead milne ki probability proportional hoti hai $P (q) \propto e^{- E_{q} / k_{B} T} = e^{- α q^{2} / k_{B} T} .$

KYUN. Yeh yahi function kyun hai aur koi aur nahi, jaise $1/ E$ ya straight line? Kyunki ek huge thermal bath ke contact mein ek system har state mein $e^{- E / k_{B} T}$ ke proportion mein time spend karta hai — high-energy states exponentially rare hote hain, aur energy aur rarity ke beech "exchange rate" $k_{B} T$ set karta hai. Hum Boltzmann ko yahan re-derive nahi kar rahe; hum ise apna ek imported law maanke use kar rahe hain. Minus sign matter karta hai: zyada energy $\Rightarrow$ chhota exponent $\Rightarrow$ kam likely.

PICTURE. Blue curve visiting probability $e^{- α q^{2} / k_{B} T}$ hai — bowl jaise same axis pe baitha ek bell (Gaussian). Jahan bowl sasta hai ( $q = 0$ ke paas) bell tall hai: frequent visits. Jahan bowl expensive hai (bada $∣ q ∣$ ) bell zero ko hug karti hai: rare visits. Bell, bowl ki ulti chhaya hai.

Step 3 — "Average" ko do integrals ke fraction mein badlo

KYA. Average energy hai: (har jagah ki energy-wahan × kitni-baar-jaate-hain ka sum) divide (total visiting, taki 100% normalize ho). Continuous $q$ ki language mein, "sum" ek integral $\int$ ban jaata hai — infinitely thin slivers ka infinite sum: $⟨ α q^{2} ⟩ = \frac{\int _{- \infty}^{\infty} yahan energy α q ^{2} yahan weight e ^{- α q^{2} / k_{B} T} d q}{\int _{- \infty}^{\infty} yahan weight e ^{- α q^{2} / k_{B} T} d q} .$

KYUN. Average hamesha "weighted total ÷ total weight" hota hai. Top energy add karta hai, har jagah ko uska Boltzmann vote deta hai. Bottom sirf votes add karta hai, taki divide karne se votes proper probability ban jayein. Crucially, ek bade system mein har doosra coordinate same exponential factor contribute karta hai top aur bottom dono mein, toh woh cancel ho jaata hai — hum legitimately sirf ek variable $q$ pe dhyan de sakte hain.

PICTURE. Top panel: numerator integrand $α q^{2} e^{- α q^{2} / k_{B} T}$ — ek do-humped curve (middle mein zero kyunki $q^{2} = 0$ wahan, far out zero kyunki bell usse khatam kar deti hai). Uska shaded area weighted energy hai. Bottom panel: denominator integrand, plain bell, jiska shaded area total weight hai. Answer hai (top area) ÷ (bottom area).

Step 4 — Differentiation trick: $q^{2}$ banao

KYA. Numerator awkward hai extra $q^{2}$ ki wajah se. Hum ek trick se isse hataate hain. Pehle messy exponent ko rename karo: $β = \frac{1}{k _{B} T}$ (Greek "beta"; chhota $β$ = hot, bada $β$ = cold). Bottom integral ko apna naam do: $Z (β) = \int_{- \infty}^{\infty} e^{- β α q^{2}} d q .$ Ab notice karo: agar $Z$ ko $β$ ke respect mein differentiate karo, chain rule exponent se $- α q^{2}$ ka factor neeche kheench laata hai: $\frac{d Z}{d β} = \int_{- \infty}^{\infty} (- α q^{2}) e^{- β α q^{2}} d q .$ Yeh bilkul numerator hai, ek minus sign tak! Isliye $⟨ α q^{2} ⟩ = \frac{- d Z / d β}{Z} = - \frac{d}{d β} ln Z .$

KYUN. Derivative kyun use karte hain? Kyunki "integrand ko $q^{2}$ se multiply karna" aur "pura integral $β$ se differentiate karna" yahan same kaam karte hain — aur doosra far easier hai jab hum $Z$ jaante hain. Derivative ek machine hai jo ek exponential se $q^{2}$ ka factor free mein padh leta hai. Woh aakhri equality $\frac{d}{d β} ln Z = \frac{1}{Z} \frac{d Z}{d β}$ , log-derivative rule, use karti hai.

PICTURE. Hum $Z (β)$ ko ek curve ki tarah dikhate hain, aur kaise ek chhota nudge $Δ β$ (temperature knob thoda cold karna) $Z$ ko shrink karta hai. Us shrink ki slope — red tangent line — height $Z$ se divide karke, precisely hamaari average energy hai. Cooling bell ko narrow karta hai, toh $Z$ (uska area) girta hai; us girne ki rate encode karti hai ki kitni energy us slot mein thi.

Step 5 — Woh ek integral karo jo chahiye (the Gaussian)

KYA. Hume sirf $Z (β) = \int_{- \infty}^{\infty} e^{- β α q^{2}} d q$ evaluate karna hai. Yeh famous Gaussian integral hai. Iska value hai $\int_{- \infty}^{\infty} e^{- a x^{2}} d x = \frac{π}{a}, yahan a = β α,$ $Z (β) = \frac{π}{β α} = \frac{π}{α} β^{- 1/2} .$

KYUN. Ek bell ka area $π / a$ kyun hota hai? Ek taller, wider bell (chhota $a$ ) zyada area enclose karti hai; ek narrow spike (bada $a$ ) kam — aur square-root exactly capture karta hai ki area width ke saath kaise trade karta hai. Exact constant $π$ classic "square it and switch to polar coordinates" proof se aata hai, jise tum yahan faith pe le sakte ho. Ek fact jo hume chahiye: $Z$ scale karta hai $β^{- 1/2}$ ki tarah.

PICTURE. Teen bells draw ki hain hot / medium / cold ke liye ( $a$ chhota / medium / bada). Har ek ke neeche shaded area $Z$ ke barabar hai. Hotter (chhota $a$ ) $\Rightarrow$ fatter bell $\Rightarrow$ bada area. $π / a$ area har ek pe annotate hai.

Step 6 — Log-derivative lo aur dekho $α$ gayab ho jaata hai

KYA. $Z = π / α β^{- 1/2}$ ko $⟨ α q^{2} ⟩ = - d ln Z / d β$ mein plug karo. Pehle log lo — logs product ko sum mein badal dete hain: $ln Z = β mein constant ln \frac{π}{α} - \frac{1}{2} ln β .$ Ab $β$ ke respect mein differentiate karo. Constant term (jahan sirf $α$ rehta hai) ka zero slope hai aur gayab ho jaata hai: $- \frac{d l n Z}{d β} = - (0 - \frac{1}{2 β}) = \frac{1}{2 β} = \frac{1}{2} k_{B} T .$

KYUN. Yahi is pure page ka punchline hai. Differentiation additive constants ko maar deta hai. Kyunki $α$ sirf ek additive constant ke roop mein $ln Z$ mein aaya tha (log ki wajah se product split hua), derivative use poori tarah erase kar deta hai. Stiffness kuch bhi ho, answer wahi $\frac{1}{2} k_{B} T$ hai. Yeh bhi note karo ki end mein temperature mein wapas translate karne ke liye humne $1/ β = k_{B} T$ use kiya.

PICTURE. Hum $ln Z$ ko $β$ ke against plot karte hain: ek straight-ish descending curve jiska slope $- \frac{1}{2} / β$ hai. Horizontal offset (woh $α$ wala part) poori line ke ek vertical shift ke roop mein dikhaya gaya hai — aur ek line ko upar ya neeche shift karne se uski slope kabhi nahi badlti. Do alag $α$ ke liye do lines parallel hain; unki common slope same energy deti hai.

Step 7 — Edge case A: agar slot quadratic NAHI hai toh?

KYA. Maan lo energy $n = 2$ ki jagah kisi power $n$ ke liye $E = α ∣ q ∣^{n}$ hoti. Same $- d ln Z / d β$ machine chalao (Gaussian ek general power integral ban jaata hai, aur $Z \propto β^{- 1/ n}$ ) toh milta hai $⟨ α ∣ q ∣^{n} ⟩ = \frac{1}{n} k_{B} T .$

KYUN. Hume dikhana hoga ki $\frac{1}{2}$ kahan se aaya, warna reader ise galat jagah apply karega. $\frac{1}{2}$ literally $\frac{1}{n}$ hai jab $n = 2$ hai. Ek stiffer-walled quartic bowl ( $n = 4$ ) bead ko zyada confine karta hai, sirf $\frac{1}{4} k_{B} T$ store karta hai. Theorem ka magic number energy well ki shape ka fingerprint hai, koi universal constant nahi.

PICTURE. Teen bowls ek axis pe: $n = 2$ (parabola), $n = 4$ (flat-bottomed, steep walls), $n = 1$ (ek V, sharp point). Har ek ke paas answer $\frac{1}{n} k_{B} T$ hai. $n = 2$ curve highlighted hai — woh equipartition case hai.

Step 8 — Edge case B: classical assumption ( $k_{B} T ≫$ level spacing)

KYA. Haari poori derivation ne states ke upar sum ko smooth $q$ ke upar integral se replace kiya. Yeh tabhi legal hai jab allowed energy levels $k_{B} T$ se kaafi paas packed hon — yaani quantum states ki seedhi ladder smooth ramp jaisi lagti ho. Jab rungs ke beech gap $Δ E$ satisfy karta hai $Δ E ≳ k_{B} T$ , bead pehle excited rung tak pahunch hi nahi sakta: slot frozen out hai aur zero contribute karta hai, $\frac{1}{2} k_{B} T$ nahi.

KYUN. Hume degenerate limit cover karna hoga warna reader room-temperature nitrogen ke liye galat $C_{V} = \frac{7}{2} R$ predict karega. Yahi reason hai ki heat capacities gas garam karne pe steps mein kyun badhte hain — dekho Quantum freezing of degrees of freedom. Step 3 ka integral secretly ek continuum assume karta tha; jab levels coarse hon toh woh assumption fail hoti hai.

PICTURE. Left: closely spaced quantum rungs jisme $k_{B} T$ (ek green bar) gap ke upar towering hai — smooth-integral / equipartition regime, slot "on." Right: widely spaced rungs jisme $k_{B} T$ pehle gap se chhota hai — bead ground rung pe atka hua, slot "off," $0$ contribute karta hai.

Ek-picture summary

KYA. Ek figure jo har step ko chain karta hai: bowl → bell → weighted-energy humps → area $Z$ → slope of $ln Z$ → $\frac{1}{2} k_{B} T$ , $α$ ko enter karte aur phir cross out hote dikhaya gaya hai.

PICTURE. Arrows follow karo left to right. Parabola stage set karta hai; Boltzmann bell batata hai ki har jagah kitni baar visit hoti hai; twin-humped product wahan hai jahan energy actually rehti hai; uska total $Z$ ke through normalize hota hai; $ln Z$ ko cooling knob $β$ se differentiate karne par slope $\frac{1}{2 β}$ milta hai; aur $α$ , jo sirf ek vertical shift ke roop mein aaya tha, cross out kar diya jaata hai.

Recall Pure walkthrough ki Feynman-style retelling

Ek bead ko bowl mein imagine karo. Bowl ki steepness "stiffness" $α$ hai. Bahar ka garam duniya bead ko hilata hai, aur ek golden rule (Boltzmann) kehta hai ki bead har jagah ke paas $e^{- cost / k_{B} T}$ ke proportion mein time spend karta hai — toh woh saste bottom ke aas-paas rehta hai aur expensive walls pe rarely chadta hai. Bead jo average energy carry karta hai woh dhundhne ke liye, hum "energy yahan × time-spent-yahan" add karte hain aur "total time" se divide karte hain — yeh do integrals ka fraction hai. Top mein ek annoying extra $q^{2}$ hai, toh hum ek trick khelate hain: poori cheez ko coldness-knob $β$ se differentiate karna magically exponential se ek $q^{2}$ bahar kheench laata hai. Yeh sab kuch bell curve ke neeche area tak reduce ho jaata hai, ek number jo $β^{- 1/2}$ ki tarah scale karta hai. Uska log lo, uski slope lo, aur baahar aata hai $\frac{1}{2 β} = \frac{1}{2} k_{B} T$ . Khoobsurat baat yeh hai: bowl ki steepness $α$ kabhi bhi sirf log ka ek fixed shift ke roop mein aaya — aur slopes shifts ko ignore karte hain — toh ek stiff bowl aur ek floppy bowl bilkul same energy dete hain. Do footnotes: agar bowl clean parabola nahi hoti (maan lo $q^{4}$ ) toh magic number $\frac{1}{n}$ ho jaata hai; aur agar duniya itni thandi ho ki bead ko sabse neeche rung se utha na sake (quantum freezing), toh slot simply off ho jaata hai aur kuch store nahi karta.