Equipartition theorem — ½k_BT per quadratic degree of freedom

WHAT is a "quadratic degree of freedom"?

WHY is it true? — Derivation from scratch

We use the classical canonical (Boltzmann) distribution: the probability of a microstate of energy $E$ is $\propto e^{-E/k_BT}$ . We want the average energy stored in ONE quadratic term $E_q = \alpha q^2$ .

Step 1 — Write the average as a Boltzmann-weighted integral. $\langle \alpha q^2\rangle = \frac{\displaystyle\int_{-\infty}^{\infty} \alpha q^2\, e^{-\alpha q^2/k_BT}\,dq}{\displaystyle\int_{-\infty}^{\infty} e^{-\alpha q^2/k_BT}\,dq}$

Why this step? The numerator weights each value of $q^2$ by how probable that $q$ is; the denominator normalizes. All other coordinates' factors $e^{-(\dots)/k_BT}$ are identical in numerator and denominator and cancel, so we only deal with the one variable $q$ . This is the crucial simplification.

Step 2 — Substitute to make it dimensionless. Let $\beta = 1/k_BT$ and define $Z(\beta) = \int_{-\infty}^{\infty} e^{-\beta\alpha q^2}\,dq.$ Why this step? Notice the numerator is exactly $-\dfrac{\partial}{\partial\beta}\left(\dfrac{1}{\alpha}\cdot\alpha\, ...\right)$ . More cleanly: $\int \alpha q^2 e^{-\beta\alpha q^2}dq = -\frac{\partial}{\partial\beta}\int e^{-\beta\alpha q^2}dq = -Z'(\beta).$ So $\langle\alpha q^2\rangle = -\dfrac{Z'(\beta)}{Z(\beta)} = -\dfrac{d}{d\beta}\ln Z$ . Why? Differentiating the exponential pulls down $-\alpha q^2$ , manufacturing the factor we need — a standard trick to turn an awkward integral into a derivative.

Step 3 — Evaluate the Gaussian. Using $\int_{-\infty}^\infty e^{-a x^2}dx = \sqrt{\pi/a}$ with $a=\beta\alpha$ : $Z(\beta) = \sqrt{\frac{\pi}{\beta\alpha}} = \sqrt{\frac{\pi}{\alpha}}\,\beta^{-1/2}.$

Step 4 — Take the log-derivative. $\ln Z = \text{const} - \tfrac{1}{2}\ln\beta \implies -\frac{d\ln Z}{d\beta} = \frac{1}{2\beta} = \frac{1}{2}k_BT.$

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

Counting DOF for real gases

Common mistake Steel-manned common errors

Mistake 1: "Each degree of freedom gives $\tfrac12 k_BT$ , so a vibrating bond gives $\tfrac12 k_BT$ ." Why it feels right: "one bond = one motion." The fix: equipartition counts quadratic energy terms, not motions. A vibration stores energy in BOTH kinetic ( $\dot r^2$ ) and potential ( $r^2$ ) forms → two terms → $k_BT$ total.

Mistake 2: "Stiffer/heavier molecules store more thermal energy per mode." Why it feels right: a stiff spring "holds more energy." The fix: the derivation showed $\alpha$ cancels. At fixed $T$ , every quadratic slot holds exactly $\tfrac12k_BT$ , independent of mass or stiffness.

Mistake 3: "It always works." Why it feels right: it's beautifully universal. The fix: it's a classical result. When the energy gap between quantum levels $\gtrsim k_BT$ (e.g. vibration at room temp, rotation of H₂ at low T), that mode is "frozen out" and contributes nothing. Equipartition holds only when $k_BT \gg$ the quantum level spacing.

Mistake 4: "Applies to non-quadratic energy." Why it feels right: you want a universal rule. The fix: the $\tfrac12k_BT$ comes from the Gaussian integral, which needs $E\propto q^2$ . For $E\propto q^4$ you'd get $\tfrac14 k_BT$ instead (try the generalized theorem!).

Recall Feynman: explain to a 12-year-old

Imagine a big room full of bouncing balls, and "temperature" is like how much pocket money each kid gets. Equipartition says: every different way a ball can wiggle gets the SAME amount of pocket money, namely $\tfrac12 k_BT$ . A ball that can move left-right, up-down, and front-back has three ways → three lots of money. A ball connected by a spring also gets money for stretching the spring. It doesn't matter if the spring is strong or weak, or if the ball is heavy or light — the rule shares energy fairly. (The only catch: if a wiggle is too "expensive" to start — like a quantum jump that needs a big chunk at once — that wiggle gets skipped, and equipartition gives it nothing.)

Active-recall checkpoint

Flashcards

What energy does each quadratic degree of freedom carry on average?

\tfrac12 k_BT

What makes a degree of freedom "quadratic"?

Its energy term has the form

\alpha q^2

in a coordinate or momentum.

Why is the result independent of the stiffness/mass constant

\alpha

\alpha

enters

\ln Z

only as an additive constant, which vanishes under

-d/d\beta

; physically a stiffer well narrows

\langle q^2\rangle

just enough to keep

\alpha\langle q^2\rangle=\tfrac12 k_BT

Mean energy of a monatomic ideal gas per atom?

\tfrac32 k_BT

(3 translational DOF), so

C_V=\tfrac32 R

Why does a diatomic gas have

C_V=\tfrac52 R

at room temperature?

3 translation + 2 rotation = 5 quadratic DOF; the 3rd rotation and vibration are frozen out.

Why does a vibrational mode count as 2 quadratic DOF?

It stores both kinetic (

\tfrac12\mu\dot r^2

) and potential (

\tfrac12\kappa r^2

) energy, each quadratic.

Heat capacity of a classical solid (Dulong–Petit)?

C_V=3R

, from 3 KE + 3 PE quadratic DOF per atom =

3k_BT

When does equipartition FAIL?

When the quantum energy-level spacing of a mode is

\gtrsim k_BT

, freezing that mode out (e.g. vibrations at room T).

Starting formula used to derive equipartition?

\langle\alpha q^2\rangle=\dfrac{\int \alpha q^2 e^{-\beta\alpha q^2}dq}{\int e^{-\beta\alpha q^2}dq}=-\dfrac{d\ln Z}{d\beta}

with

Z=\sqrt{\pi/(\beta\alpha)}

Generalized result for

E=\alpha|q|^n

\langle E\rangle=\dfrac{1}{n}k_BT

; equipartition's

\tfrac12

is the

n=2

case.

Why is

\gamma=1.4

for air explained by equipartition?

Diatomic 5 DOF →

C_V=\tfrac52R

C_P=\tfrac72R

\gamma=7/5=1.4

Connections

Boltzmann distribution — supplies the $e^{-E/k_BT}$ weighting the whole proof rests on.
Partition function Z — the $Z(\beta)$ and $-d\ln Z/d\beta$ trick is reused everywhere.
Maxwell–Boltzmann speed distribution — the 3 translational DOF give $\langle\tfrac12 mv^2\rangle=\tfrac32 k_BT$ .
Heat capacity of gases / Dulong–Petit law — direct applications.
Quantum freezing of degrees of freedom — where and why equipartition breaks down.
Ideal gas law PV=NkT — translational equipartition + kinetic theory derives it.
Brownian motion — equipartition fixes the kinetic energy of a suspended particle.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, equipartition theorem ka core idea bahut simple hai: jab system thermal equilibrium me hota hai (temperature $T$ par), to energy har "quadratic slot" me barabar baant jaati hai — exactly $\tfrac12 k_BT$ per slot. "Quadratic slot" matlab koi bhi energy term jo $\alpha q^2$ jaisa dikhe — jaise $\tfrac12 m v_x^2$ (translation), $\tfrac12 I\omega^2$ (rotation), ya $\tfrac12\kappa x^2$ (spring/vibration). Bas square hona chahiye coordinate ya momentum ka.

Sabse mind-blowing baat: yeh $\tfrac12 k_BT$ spring kitni stiff hai ya particle kitna heavy hai uspe depend nahi karta. Derivation me hum Boltzmann factor $e^{-\alpha q^2/k_BT}$ leke Gaussian integral karte hain, aur $\alpha$ sirf ek constant ki tarah $\ln Z$ me aata hai jo $-d/d\beta$ lene par mar jaata hai. Physically: stiff well me particle thoda hi hil paata hai, par utni hi kam jagah me $\alpha\langle q^2\rangle$ wapas $\tfrac12 k_BT$ ho jaata hai. Bilkul fair distribution!

Counting trick yaad rakho: monatomic gas me 3 translation = $\tfrac32 k_BT$ , $C_V=\tfrac32 R$ . Diatomic gas room temperature par 3 translation + 2 rotation = $\tfrac52 k_BT$ , isliye $\gamma=1.4$ — jo air ke liye experiment se match karta hai! Vibration ek special case hai: usme kinetic AUR potential dono quadratic terms hote hain, isliye vo do slots count hota hai ( $k_BT$ , na ki $\tfrac12 k_BT$ ). Solid me har atom ke 3 KE + 3 PE = $3k_BT$ , yeh Dulong–Petit law hai ( $C_V=3R$ ).

Ek important warning: yeh classical result hai. Agar kisi mode ka quantum gap $k_BT$ se bada ho (jaise vibration room temp par), to vo mode "freeze" ho jaata hai aur kuch contribute nahi karta. Isiliye diagram me $C_V$ ek staircase ki tarah badhta hai jaise-jaise $T$ badhta hai — naye modes unfreeze hote jaate hain.

Equipartition theorem — ½k_BT per quadratic degree of freedom

WHAT is a "quadratic degree of freedom"?

WHY is it true? — Derivation from scratch

Counting DOF for real gases

Active-recall checkpoint

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics & Statistical Mechanics (Advanced)

Connections