Equipartition theorem — ½k_BT per quadratic degree of freedom: Question bank

2.4.14 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Equipartition theorem — ½k_BT per quadratic degree of freedo

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

Upar ka right panel woh doosri picture hai jo tumhe chahiye: heat capacity steps mein badhti hai jaise temperature badhti hai aur har mode "unfreeze" hoti hai. Ise edge-case questions ke liye yaad rakho.

True ya False — justify karo

Ek bhaari gas molecule, same temperature par, ek halki molecule se har translational mode mein zyada thermal energy store karti hai.

False — mass constant

- d ln Z / d β

mein cancel ho jaata hai; har translational quadratic slot exactly

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

hi rakhta hai, mass se koi farak nahi padta. Bhaari molecule sirf dhimi chaal se same energy store karti hai.

Vibrating bond mein ek stiffer spring (bada

κ

) average vibrational energy badhata hai.

False — stiffness

q

ki Gaussian bell ki width set karti hai, uski energy content nahi. Stiffer well

q

ko zyada confine karti hai, isliye

κ ⟨ q^{2} ⟩

exactly

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

par tika rehta hai.

Equipartition molecule ke har coordinate ko

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

assign karta hai.

False — yeh har quadratic energy term ko

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

assign karta hai, na ki har coordinate ko. Ek vibration coordinate

r

mein do terms hote hain (KE

\overset{r}{˙}

mein aur PE

r

mein), isliye total

k_{B} T

hota hai.

Equipartition purely classical result hai.

True — yeh

q

par ek continuous Gaussian integral se aata hai; jab energy levels discrete ho jaate hain aur

≳ k_{B} T

spaced hote hain, tab integral galat hai aur mode freeze out ho jaata hai (dekho Quantum freezing of degrees of freedom).

Theorem

⟨ α q^{2} ⟩ = \frac{1}{2} k_{B} T

ke liye potential ka exactly quadratic hona zaroori hai.

True specifically

\frac{1}{2}

factor ke liye — ek term

α ∣ q ∣^{n}

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

ki jagah

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

deta hai, toh sirf

n = 2

one-half deta hai.

Same temperature par ek monatomic ideal gas aur ek diatomic gas ka average energy per molecule same hota hai.

False — dono

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

per quadratic DOF share karte hain, lekin unke DOF counts alag hote hain (room temp par 3 vs 5), toh diatomic molecule zyada total energy rakhta hai.

Ek magnetic field jo charged particle ko circle mein ghuma deti hai, circular motion ki speed ke liye koi naya quadratic energy term add nahi karti.

True — magnetic force koi kaam nahi karti aur kinetic energy abhi bhi teen velocity components mein

\frac{1}{2} m v^{2}

hai; koi naya quadratic term nahi aata, isliye equipartition aur

C_{V}

unchanged rehte hain.

Do identical vibrating bonds, active hone par har ek

k_{B} T

contribute karte hain, toh total

2 k_{B} T

hai.

True — har bond independently ek KE aur ek PE quadratic term supply karta hai, jo

k_{B} T

each deta hai; independent quadratic terms simply add ho jaate hain.

Error pakdo

"Ek diatomic molecule mein 6 degrees of freedom hain (3 translation + 3 rotation), toh

C_{V} = 3 R

Bond axis ke around rotation ka moment of inertia negligible hai, toh uski quantum spacing bahut badi hai aur woh freeze out ho jaati hai — sirf 2 rotational DOF count hote hain, jo total 5 dete hain aur

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

"Vibration ek DOF add karta hai, toh ek hot diatomic gas mein 6 DOF aur

C_{V} = 3 R

hota hai."

Vibration do quadratic terms add karta hai (kinetic

\frac{1}{2} μ \overset{r}{˙}^{2}

reduced mass

μ

use karke, aur potential

\frac{1}{2} κ r^{2}

), toh yeh

5 + 2 = 7

DOF aur

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

hota hai,

3 R

nahi.

"Kyunki

α

cancel ho jaata hai, energy bhi temperature se independent hai."

Sirf stiffness/mass constant cancel hota hai;

T

explicitly

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

ke roop mein bachta hai (kyunki

\frac{1}{2 β} = \frac{1}{2} k_{B} T

), toh energy temperature mein linear hai — yahi linearity constant heat capacity deti hai.

⟨ α q^{2} ⟩ = \frac{1}{2} k_{B} T

ka matlab hai

⟨ q ⟩ = k_{B} T / (2 α)

Theorem

q^{2}

ke mean ko constrain karta hai,

q

ko nahi; actually

⟨ q ⟩ = 0

Gaussian bell ki symmetry se, jabki

⟨ q^{2} ⟩ = k_{B} T / (2 α)

"Kyunki equipartition universal hai, yeh 5 K par ek solid ka

C_{V}

predict karta hai."

Kam

T

par phonon (harmonic) modes freeze out ho jaate hain aur

C_{V} \to 0

, jo Dulong–Petit law ke classical

3 R

ko violate karta hai; equipartition sirf tabhi hold karta hai jab

k_{B} T ≫

level spacing.

"Partition-function trick

⟨ α q^{2} ⟩ = - d ln Z / d β

kisi bhi energy function ke liye kaam karti hai."

- d ln Z / d β

identity generally total mean energy deti hai, lekin clean

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

-per-term result ke liye har term ka quadratic hona zaroori hai taaki Gaussian integral

Z = π / (β α)

apply ho sake.

"Real air mostly N₂ aur O₂ hai, toh iska

γ

\frac{5}{3}

hona chahiye."

\frac{5}{3}

monatomic value hai; diatomics at room temperature ke 5 DOF hote hain jo

γ = \frac{7}{5} = 1.4

dete hain, aur yahi air ke liye measure kiya jaata hai.

Why questions

α

final energy se kyun gayab ho jaata hai, lekin position distribution ki width se nahi?

ln Z = const - \frac{1}{2} ln β

mein,

α

sirf additive constant ke andar hai aur

- d / d β

ke under mar jaata hai; lekin

⟨ q^{2} ⟩ = k_{B} T / (2 α)

abhi bhi dikhata hai ki bell bade

α

ke saath narrow hoti hai.

Hum poore many-particle energy ki jagah ek quadratic term ko akele treat kyun kar sakte hain?

Boltzmann factor

e^{- β E}

independent terms par factorize hota hai, toh har doosre coordinate ka factor numerator aur denominator mein identical hota hai aur cancel ho jaata hai — ek single-variable integral bachta hai (dekho Boltzmann distribution).

Temperature ek quadratic slot ki universal "price" ki tarah kyun act karta hai?

Kyunki derivation se har quadratic term ke liye same

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

milta hai, chahe woh kisi bhi physical quantity ko describe kare, toh sirf

T

hi har system mein equilibrium par har slot ki energy fix karta hai.

Ek vibrational mode ko activate karna usi molecule ki rotational mode se zyada "costly" kyun hai?

Vibrational energy-level spacing

ℏ ω

rotational spacing se bahut zyada hai, toh vibrational mode ko unfreeze karne ke liye bahut zyada

T

chahiye jab

k_{B} T

usse exceed kare (dekho Quantum freezing of degrees of freedom).

Derivation mein koi aur integral ki jagah Gaussian integral kyun aata hai?

Kyunki energy term

α q^{2}

hai, toh Boltzmann weight

e^{- β α q^{2}}

hai — ek Gaussian bell; ek non-quadratic term alag integral aur

k_{B} T

ka alag fraction deta.

Heat capacity mein, energy khud mein nahi, "steps" kyun aate hain jab

T

freezing thresholds se upar badhta hai?

Jab har mode unfreeze hoti hai, woh

T

mein linear energy ka ek chunk add karti hai, toh iska slope

C_{V} = \partial U / \partial T

abruptly on ho jaata hai, jo figure ke right panel mein dikhe plateaus produce karta hai.

Equipartition Maxwell–Boltzmann speed distribution ko ek single number

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

se kyun connect karta hai?

Maxwell–Boltzmann distribution exactly har velocity component mein Gaussian bell hai, aur

\frac{1}{2} m v^{2}

ko uske against integrate karne par

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

per component milta hai — teenon sum hokar

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

dete hain.

Edge cases

T = 0

par, equipartition har mode ke liye energy ke baare mein kya predict karta hai, aur kya yeh sahi hai?

Yeh

\frac{1}{2} k_{B} T \to 0

predict karta hai; classically yeh theek hai, lekin quantum mechanics ek zero-point energy

\frac{1}{2} ℏ ω

rakhti hai, toh equipartition (ek classical high-

T

result) ground state describe nahi karta.

α q^{2}

ki jagah

E = α q^{4}

energy term ke liye, mean energy kya hai?

Generalized equipartition

⟨ α ∣ q ∣^{n} ⟩ = \frac{1}{n} k_{B} T

mein

n = 4

rakhne par

\frac{1}{4} k_{B} T

milta hai, jo quadratic value ka aadha hai.

Ek single particle ke liye ek box mein sirf translation ke saath (genuine ideal gas), box ko atomic size tak shrink karne par equipartition ka kya hoga?

Jab box itna chhota ho jaata hai ki quantum energy gaps

\sim h^{2} / (m L^{2})

k_{B} T

ke paas aa jaate hain, tab translation khud freeze ho jaata hai aur equipartition fail ho jaata hai — yeh quantum-gas regime hai.

Kya diatomic molecule ka teesra rotation kabhi contribute karta hai?

Principle mein sirf absurdly high

T

par jahan

k_{B} T

uski tiny-moment-of-inertia level spacing se exceed kare, lekin pehle hi molecule dissociate ho jaata hai, toh effectively yeh kabhi contribute nahi karta.

Temperature

T

par fluid mein suspended Brownian motion karne wale ek free particle ke baare mein equipartition kya kehta hai?

Uske teen translational velocity components mein abhi bhi

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

hai, toh

⟨ \frac{1}{2} m v^{2} ⟩ = \frac{3}{2} k_{B} T

— bilkul ek gas atom jaisa, isliye ek Brownian particle fluid temperature "report" karta hai.

Ek two-dimensional gas (particles ek plane mein confined) ke liye, energy per atom aur

C_{V}

kya hai?

Sirf 2 translational quadratic terms exist karte hain, toh

⟨ E ⟩ = k_{B} T

per atom aur

C_{V} = R

per mole.

Kya ideal gas law [[Ideal gas law PV=NkT|

P V = N k T

]] khud equipartition ka consequence hai?

Directly nahi —

P V = N k T

kinetic pressure argument se follow karta hai, lekin equipartition matching

⟨ \frac{1}{2} m v^{2} ⟩ = \frac{3}{2} k_{B} T

supply karta hai, toh dono same translational energy ke consistent descriptions hain.

Recall Ek-line summary jo saath le jaao

Answer ::: Equipartition $\frac{1}{2} k_{B} T$ deta hai har quadratic energy term ko — classical, high-temperature limit mein universal, mass aur stiffness se andha, lekin quantum freezing se chup ho jaata hai jab bhi kisi mode ki level spacing $k_{B} T$ ko beat kar de.