2.4.14 · D5Thermodynamics & Statistical Mechanics (Advanced)
Question bank — Equipartition theorem — ½k_BT per quadratic degree of freedom
Toolbox — the symbols and the one identity these traps rely on
Before the traps, here is every piece of notation used on this page, defined in plain words and anchored to a picture. Do not skip this — several "traps" are traps precisely because people misread these symbols.
The right panel above is the other picture you need: heat capacity climbs in steps as temperature rises and each mode "unfreezes". Keep it in mind for the edge-case questions.
True or false — justify
A heavier gas molecule stores more thermal energy per translational mode than a lighter one at the same temperature.
False — the mass constant cancels in ; every translational quadratic slot holds exactly regardless of mass. The heavy molecule just moves slower to store the same energy.
A stiffer spring (larger ) in a vibrating bond raises the average vibrational energy.
False — stiffness sets the width of the Gaussian bell in , not its energy content. A stiffer well confines more, so stays pinned at .
Equipartition assigns to every coordinate of a molecule.
False — it assigns to every quadratic energy term, not every coordinate. One vibration coordinate carries two terms (KE in and PE in ), hence .
Equipartition is a purely classical result.
True — it comes from a continuous Gaussian integral over ; once energy levels are discrete and spaced , the integral is wrong and the mode is frozen out (see Quantum freezing of degrees of freedom).
The theorem requires the potential to be exactly quadratic.
True for the factor specifically — a term gives instead, so only yields one-half.
A monatomic ideal gas and a diatomic gas at the same temperature have the same average energy per molecule.
False — both share per quadratic DOF, but they have different DOF counts (3 vs 5 at room temp), so the diatomic molecule holds more total energy.
Adding a magnetic field that makes a charged particle circle does not add a new quadratic energy term for the circular motion's speed.
True — the magnetic force does no work and the kinetic energy is still in the same three velocity components; no new quadratic term appears, so equipartition and hence are unchanged.
Two identical vibrating bonds each contribute when active, so total is .
True — each bond independently supplies one KE and one PE quadratic term, giving each; independent quadratic terms simply add.
Spot the error
"A diatomic molecule has 6 degrees of freedom (3 translation + 3 rotation), so ."
The rotation about the bond axis has a negligible moment of inertia, so its quantum spacing is huge and it is frozen out — only 2 rotational DOF count, giving 5 total and .
"Vibration adds one DOF, so a hot diatomic gas has 6 DOF and ."
Vibration adds two quadratic terms (kinetic using the reduced mass , and potential ), so it is DOF and , not .
"Because cancels, energy is independent of temperature too."
Only the stiffness/mass constant cancels; survives explicitly as (since ), so the energy is linear in temperature — that linearity is what gives a constant heat capacity.
" means ."
The theorem constrains the mean of , not ; in fact by symmetry of the Gaussian bell, while .
"Since equipartition is universal, it predicts for a solid at 5 K."
At low the phonon (harmonic) modes freeze out and , violating the classical of Dulong–Petit law; equipartition holds only when level spacing.
"The partition-function trick works for any energy function."
The identity gives the total mean energy generally, but the clean -per-term result needs each term to be quadratic so the Gaussian integral applies.
"Real air is mostly N₂ and O₂, so its should be ."
is the monatomic value; diatomics at room temperature have 5 DOF giving , which is what is measured for air.
Why questions
Why does the constant vanish from the final energy but not from the width of the position distribution?
In , is only inside the additive constant and dies under ; but still shows the bell narrowing with larger .
Why can we treat one quadratic term in isolation instead of the whole many-particle energy?
The Boltzmann factor factorizes over independent terms, so every other coordinate's factor is identical in numerator and denominator and cancels — leaving a single-variable integral (see Boltzmann distribution).
Why does temperature act like a universal "price" of one quadratic slot?
Because the derivation gives the same for any quadratic term regardless of what physical quantity it describes, so alone fixes the energy per slot across all systems in equilibrium.
Why does a vibrational mode "cost more" to activate than a rotational one in the same molecule?
The vibrational energy-level spacing is much larger than the rotational spacing, so much higher is needed for to exceed it and unfreeze it (see Quantum freezing of degrees of freedom).
Why does the Gaussian integral, not some other integral, appear in the derivation?
Because the energy term is , so the Boltzmann weight is — a Gaussian bell; a non-quadratic term would give a different integral and a different fraction of .
Why does the heat capacity, not the energy itself, jump in "steps" as rises through the freezing thresholds?
As each mode unfreezes it adds a linear-in- chunk of energy, so its slope turns on abruptly, producing the plateaus you saw in the right panel of the figure.
Why does equipartition connect the Maxwell–Boltzmann speed distribution to a single number ?
The Maxwell–Boltzmann distribution is exactly the Gaussian bell in each velocity component, and integrating against it gives per component — three of them sum to .
Edge cases
At , what does equipartition predict for the energy per mode, and is it right?
It predicts ; classically that's fine, but quantum mechanics keeps a zero-point energy , so equipartition (a classical high- result) does not describe the ground state.
For an energy term instead of , what is the mean energy?
Using generalized equipartition with gives , half the quadratic value.
What happens to equipartition for a single particle in a box with only translation (a genuine ideal gas), letting the box shrink to atomic size?
When the box is small enough that the quantum energy gaps approach , translation itself freezes and equipartition fails — this is the quantum-gas regime.
Does the third rotation of a diatomic molecule ever contribute?
In principle only at absurdly high where exceeds its tiny-moment-of-inertia level spacing, but the molecule dissociates first, so effectively it never contributes.
What does equipartition say about a free particle undergoing Brownian motion suspended in a fluid at temperature ?
Its three translational velocity components still each carry , so — the same as a gas atom, which is why a Brownian particle "reports" the fluid temperature.
For a two-dimensional gas (particles confined to a plane), what is the energy per atom and ?
Only 2 translational quadratic terms exist, so per atom and per mole.
Is the ideal gas law [[Ideal gas law PV=NkT|]] itself a consequence of equipartition?
Not directly — follows from the kinetic pressure argument, but equipartition supplies the matching , so the two are consistent descriptions of the same translational energy.
Recall One-line summary to carry away
Answer ::: Equipartition gives per quadratic energy term — universal in the classical, high-temperature limit, blind to mass and stiffness, but silenced by quantum freezing whenever a mode's level spacing beats .