Canonical ensemble ($p_s=Z^{-1}{e}^{-\beta E_s}$)

The canonical ensemble is the probability distribution for a system in thermal equilibrium with a heat bath at fixed temperature $T$, with fixed particle number and external parameters. It allows energy exchange, so individual energy measurements fluctuate. With Boltzmann constant $k_B$ and inverse temperature $\beta =1/(k_{B}T)$,The heat bath is idealized as large enough that exchanges barely change its temperature; coupling must permit equilibration while contributing negligible interaction energy. A finite heat bath approximates this description only when its temperature change is negligible over the relevant exchanges. Strong coupling or a small heat bath can require a different description.

For one two-level system at positive temperature, the Boltzmann weights are $1,e^{-\beta \Delta }$, henceAt $\beta \Delta =\ln 3$, $Z_1=4/3$ and $p=1/4$. This is a probabilistic thermal mixed state. For quantum levels, its density matrix is $\rho =(1-p)|0⟩⟨0|+p|1⟩⟨1|$. The pure state $\sqrt{1-p}|0⟩+\sqrt{p}|1⟩$ has the same energy measurement probabilities, but also has quantum coherence between the energy eigenstates. Matching those probabilities does not make the states equivalent.

The canonical ensemble assigns the density matrix $e^{-\beta H}/Z$, with $Z=\mathrm{Tr}{e}^{-\beta H}$. Its expectation value of an observable is the displayed normalized trace. In imaginary time, cyclicity of this trace gives periodic bosonic correlation functions; it does not make the imaginary-time Heisenberg operator itself periodic.

For a dilute classical ideal gas whose atoms carry independent nondegenerate internal levels $0,ϵ$, the canonical partition function factorizes into translation and $(1+e^{-\beta ϵ}{)}^N$. The upper-level probability is $p=(1+e^{\beta ϵ}{)}^{-1}$. Taking two derivatives of its logarithm gives the title sum of independent variances. The internal contribution freezes exponentially at low temperature and tends to $N{ϵ}^2/4$ at high temperature. The translational classical result cannot be continued through quantum degeneracy without changing the model.

For $Z={\sum }_i{e}^{-\beta E_i}$ and $U=⟨E⟩$, the entropy is $S=k_B\ln Z+U/T$. It follows directly by substituting canonical probabilities into $S=-k_B{\sum }_i{p}_i\ln p_i$.

Differentiating the canonical partition function at fixed external parameters gives $⟨E⟩=-{\partial }_{\beta }\ln Z$ and ${\partial }_{\beta }^2\ln Z=⟨E^2⟩-⟨E{⟩}^2$. Consequently the heat capacity isThe identity assumes an energy spectrum independent of temperature and the appropriate fixed-volume canonical ensemble. It relates a thermal response to equilibrium variance.

A quantum Gibbs state is the density operator of a canonical ensemble with inverse temperature $\beta =1/(k_{B}T)$. It exists when the canonical partition function $Z=\mathrm{Tr}(e^{-\beta H})$ is finite. The spectral theorem assigns probabilities $e^{-\beta E_j}/Z$ to an orthonormal eigenbasis of the Hamiltonian. An additive constant in the Hamiltonian cancels between numerator and denominator. For $H=-\mu B{\sigma }_z$, the exponential of a Pauli vector, continued to an imaginary argument, gives ${\rho }_{\beta }=[I+\tanh (\beta \mu B){\sigma }_z]/2$.

Two independently available parts of a thermodynamic system are noninteracting in the canonical description when their state space is a Cartesian product and their energy is additive, $E(a,b)=E_A(a)+E_B(b)$. Both conditions are needed for the canonical partition function to factor:An additive expression on a constrained set of allowed state pairs need not give this factorization. Relabeling the states to make their allowed labels independent can instead introduce an interaction into the energy. For two two-state labels, an additive energy necessarily satisfies the rectangular identity $E_{11}+E_{22}=E_{12}+E_{21}$.

For N independent units with nondegenerate levels 0 and $\Delta >0$, the canonical partition function is $(1+e^{-\beta \Delta }{)}^N$. Its energy is $N\Delta /(1+e^{\beta \Delta })$ and heat capacity is $N{k}_B{x}^2{e}^x/(1+e^x{)}^2$, with $x=\Delta /(k_{B}T)$. Both are activated at low temperature. At high temperature the energy saturates at $N\Delta /2$ and heat capacity decays as $T^{-2}$. The equal-degeneracy Schottky peak is at $x\simeq 2.39936$ and height $0.43923N{k}_B$. A common energy shift leaves heat capacity unchanged; unequal level degeneracies require a different partition function.

For independently labelled particles with nondegenerate levels, choosing which $N_{+}$ particles occupy the upper level gives a binomial multiplicity. With $x=(N_{+}-N_{-})/N$, its large-number entropy is $N{k}_B[\ln 2-(1+x)\ln (1+x)/2-(1-x)\ln (1-x)/2]$. This counting assumes distinguishable or localized degrees of freedom, not indistinguishable particles sharing one pair of quantum orbitals.

A model in which each excitation changes volume by a fixed increment $v$. If the all-ground configuration has volume $V_0$, then $V=V_0+v⟨N_{\mathrm{ex}}⟩$. Energies and statistical weights must be specified under the relevant mechanical constraint; volume endpoints alone do not define arbitrary intermediate configurations.

For a specified optical transition, the on-resonance saturation parameter is the light intensity divided by the saturation intensity. The independent-beam Doppler cooling force model assumes weak total excitation and defines $s$ separately for each beam.

The saturation intensity is the characteristic light intensity used to normalize the on-resonance saturation parameter of a specified optical transition. Its value depends on the transition and the stated illumination convention.

The partition function $Z(\beta )={\sum }_s{e}^{-\beta E_s}$ is the sum of the Boltzmann weights over individual microstates. Dividing by it makes their probabilities sum to one. When summing distinct energies instead, include the state degeneracy: $Z={\sum }_{E}g(E)e^{-\beta E}$. For two independent distinguishable two-level systems,The middle coefficient counts two microstates, exactly as in the preceding figure. For $N$ such units, $Z_N=Z_1^N$; this factorization requires additive energy and independence, and generally fails with interactions.

The usefulness goes beyond normalization. For fixed energies, derivatives of the natural logarithm of the partition function yield measurable quantities:Here $U$ is internal energy, the second expression is the energy variance, and $S$ is Gibbs entropy. For one two-level system at fixed $\Delta$, its heat capacity $C=\partial U/\partial T$ and entropy areAt $\beta \Delta =\ln 3$, these give $U/\Delta =0.25$, $C/k_B\simeq 0.2263$ and $S/k_B\simeq 0.5623$. For independent units, all three quantities add.

The figure shows why heat capacity peaks. At low temperature, the excited energy is rarely occupied; at high temperature, the probabilities are almost equal and further warming barely changes internal energy. The largest response lies between these limits, at $k_{B}T/\Delta \simeq 0.4168$, where $C/k_B\simeq 0.4392$. Meanwhile entropy rises from zero toward $k_B\ln 2$, reflecting two equally likely alternatives. This high-temperature limit belongs to the stipulated two-level system; extra accessible energy eigenstates would change it.

Counting permutations of identical classical particles as different states produces a spurious entropy of mixing when otherwise identical gases are combined. The dilute ideal gas partition function $Z_N=Z_1^N/N!$ removes that label multiplicity and gives an extensive entropy in the thermodynamic limit. For different species, genuinely distinct composition information still produces a entropy of mixing. quantum statistics supplies the underlying identical-particle counting.

The single-molecule state sum $q={\sum }_s{e}^{-\beta E_s}$ for specified energy references and external constraints. In a dilute classical ideal gas, indistinguishability gives $Q_N=q^N/N!$; translation supplies the factor proportional to volume.

If energies are measured upward from a species reference ${\epsilon }_0$, its absolute single-particle sum is $q_{\mathrm{abs}}=e^{-\beta {\epsilon }_0}q$. The ideal-gas chemical potential is then ${\epsilon }_0-k_{B}T\ln (q/N)$. A zero-point factor can be included either in $q$ or in the reference energy, but not both.

For a free molecule in a uniform ideal gas, $q_i$ is proportional to volume. The ratio $f_i=q_i/V$ is independent of the container volume at fixed temperature and has units of inverse volume.

The electronic partition function includes electronic excitation energies and degeneracies $g_i$. If only an exactly degenerate ground manifold of size $g$ is populated, $q_{\mathrm{el}}=g$ and its molar entropy is $R\log g$. Resolved splittings require a state sum rather than replacing all components by one degeneracy.

A vibrational partition function sums over molecular vibrational energy levels. A single quantum harmonic oscillator, with energies measured relative to its ground level and vibrational temperature ${\theta }_v=h\nu /k$, has $q_{\mathrm{vib}}=(1-e^{-{\theta }_v/T}{)}^{-1}$. Retaining its absolute zero-point energy multiplies this by $e^{-{\theta }_v/(2T)}$. That factor shifts the internal energy but contributes no entropy.

The harmonic-oscillator entropy contribution for one vibrational coordinate, with $x={\theta }_{\mathrm{vib}}/T$. Contributions from independent normal modes add, with one term for each component of a degenerate mode. A temperature-independent zero-point energy does not change this entropy.

The rotational partition function sums Boltzmann factors over rotational energy levels, including their degeneracies and the appropriate indistinguishability convention. For a linear rigid rotor at high temperature, $q_{\mathrm{rot}}=8{\pi }^{2}IkT/(\sigma h^2)$, with moment of inertia $I$ and rotational symmetry number $\sigma$.

The entropy contribution from populated molecular rotational states. A nonlinear rigid rotor has $q_{\mathrm{rot}}\propto T^{3/2}$ at high temperature, giving $S_{\mathrm{rot},\mathrm{m}}=R(\ln q_{\mathrm{rot}}+3/2)$. Its rotational symmetry number removes redundant orientations.

For a free molecule of mass $m$ in a large three-dimensional volume $V$, replacing the translational state sum by an integral gives the displayed partition function. For identical classical molecules the many-molecule canonical partition function is $q_{\mathrm{trans}}^N/N!$, not simply $q_{\mathrm{trans}}^N$. Differentiating its Helmholtz free energy at fixed volume gives the ideal-gas translational entropy.

The trace of imaginary-time evolution is the canonical partition function of a quantum system. In a coordinate basis it is the integral of the diagonal Euclidean path integral kernel. Inserting a symmetry operator instead gives a weighted thermal trace.

For two equal-frequency quantum harmonic oscillators, the ordinary thermal trace is $r/(1-r{)}^2$ and the trace with the parity operator is $r/(1+r{)}^2$, where $r=e^{-\beta \omega }$. Total excitation $N$ has degeneracy $N+1$ and parity $(-1{)}^N$. Therefore the traces expand as ${\sum }_{n\ge 1}(\pm 1{)}^{n-1}n{r}^n$. The alternating weights come from parity insertion in a bosonic state space, not fermionic statistics.

Canonical differentiation gives

Canonical energy variance equals ${\partial }_{\beta }^2\log Z=k_B{T}^2{C}_V$.

At fixed particle number and volume, a classical monatomic ideal gas has $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$. The canonical energy fluctuation identity $\mathrm{Var}(E)=k_B{T}^2{C}_V$ therefore gives ${\sigma }_U/U=\sqrt{2/(3N)}$. Absolute fluctuations grow as $\sqrt{N}$, while relative fluctuations decrease as $N^{-1/2}$. A particle reservoir changes the fluctuation problem because energy and particle-number fluctuations are then correlated.

Forput $a=\kappa R^2/2$. The normalized classical one-particle partition function isIts mean total and potential energies are

If $a\gg k_{B}T$, the wall is thermally inaccessible and $⟨E⟩\sim 2k_{B}T$, as four quadratic terms predict. If $a\ll k_{B}T$, the spatial distribution is nearly uniform in the disk andso only the two kinetic quadratic terms contribute at leading order.

For $N$ indistinguishable noninteracting classical particles, $Z_N=z_1^N/N!$, where $z_1$ is the normalized one-particle phase-space integral.

For a structureless particle with Hamiltonian $p^2/(2m)+ar$, $a>0$, the classical phase space integral gives the displayed one-particle partition function. Its spatial integral is $4\pi {\int }_0^{\infty }{r}^2{e}^{-ar/(k_{B}T)}dr=8\pi (k_{B}T/a{)}^3$, while the momentum integral gives ${\lambda }_T^{-3}$. For $N$ indistinguishable dilute particles, $Z_N=Z_1^N/N!$, $F\simeq -N{k}_{B}T[\ln (Z_1/N)+1]$, $U=(9/2)N{k}_{B}T$ and $S\simeq N{k}_B[\ln (Z_1/N)+11/2]$. These thermodynamic formulas use leading Stirling accuracy; the classical phase-space integral additionally requires temperatures above the discrete trap level scale.

For a dilute classical trapped gas, $\mu \simeq k_{B}T\ln (N/Z_1)$ with energy zero at the potential minimum. Thus $z=e^{\mu /(k_{B}T)}=N/Z_1$. Quantum occupation effects begin when $z$ is no longer small, giving the displayed order-of-magnitude particle threshold. Equivalently, the central density $n(0)=N/[8\pi (k_{B}T/a{)}^3]$ satisfies $n(0){\lambda }_T^3\sim 1$. This is a degeneracy crossover criterion rather than an exact Bose condensation threshold. In the continuum approximation, spinless bosons have excited-state saturation $N_c=\zeta (9/2)Z_1$, from the trap density of states proportional to $E^{7/2}$. A quantum ground energy should be subtracted consistently when discreteness matters.

At fixed atom number, the entropy of the classical ideal gas in a spherical linear trap depends on temperature and strength through $Z_1\propto T^{9/2}{a}^{-3}$. Reversible adiabatic change therefore holds this product constant, giving the displayed relation. Reducing trap strength lowers the temperature while increasing cloud size. Its peak classical phase-space occupation $N/Z_1$ remains constant, so this cooling alone does not make the gas more quantum degenerate. Heat exchange, particle loss or irreversible entropy production would change this relation.

For the potential $U=r^{2n}/V^{2n/3}$, rescaling $r=V^{1/3}(kT{)}^{1/(2n)}u$ in the classical one-particle partition function gives the formula above. The parameter-conjugate force is $P=NkT/V$, the internal energy is $(3/2+3/(2n))NkT$, and $C_P-C_V=Nk$. The kinetic term contributes $3NkT/2$, while the potential term contributes $3NkT/(2n)$. For indistinguishable particles the $N!$ factor changes free energy and entropy but not these force and heat-capacity formulas.

The three-dimensional translational momentum integral is $1/{\lambda }^3$ per unit volume, where

Equating thermal de Broglie wavelength $h/\sqrt{2\pi m{k}_{B}T}$ to mean particle spacing $n^{-1/3}$ gives the displayed quantum-degeneracy scale. It is an onset estimate rather than a universal phase-transition temperature. Different typical-momentum conventions change its order-one factor; density, mass, spin degeneracy and interactions determine the detailed crossover or condensation criterion.

The translational concentration scale for a nonrelativistic particle in three dimensions, per internal state. It is the inverse cube of the thermal de Broglie wavelength. A classical component with number density $n$ and external one-particle energy $U$ has chemical potential $\mu =k_{B}T\ln (n/n_Q)+U$. An additional internal degeneracy must be included explicitly; it is not hidden in this definition. The dilute classical condition is $n/n_Q\ll 1$ for each occupied internal component.

For energy $pc+U(x)$,If $U(r)=r^{2n}/V^{2n/3}$, thenConsequently $pV=N{k}_{B}T$, while the mean energy and variance are $NA{k}_{B}T$ and $NA(k_{B}T{)}^2$ with $A=3+3/(2n)$.

The radial density is proportional to $r^2{e}^{-r^{2n}/(V^{2n/3}{k}_{B}T)}$, whose mode is

In the classical dilute regime, a monatomic ideal gas has entropy

For $N$ distinguishable noninteracting spins, let a fraction $\alpha$ occupy the level $+ϵ$ and the remainder occupy $-ϵ$. The Stirling formula givesThe relation $1/T=\partial S/\partial E$ yieldsPositive temperature corresponds to $0\le \alpha <1/2$; a population inversion $1/2<\alpha \le 1$ has negative temperature.

For $N$ independent spins with $s_z\in \{-1,0,1\}$ and one-spin Hamiltonian $-\mu B{s}_z$,Writing $x=\beta \mu B$, its heat capacity is

For $N$ noninteracting units with energies $-ϵ,0,ϵ$, put $x=\beta ϵ$. Thenand

For one three-level unit with energies $-\mu B,0,\mu B$, the high-temperature heat capacity is $C\simeq 2{\mu }^2{B}^2/(3k_B{T}^2)$. A measured coefficient $C=a/T^2$ therefore gives the displayed magnetic moment. If the heat capacity instead represents $N$ independent units, replace $a$ by $a/N$; the normalization must be established before extracting a single-unit moment.

Negative absolute temperature can occur only for a system with an energy spectrum bounded above. Since $\beta =\partial S/\partial E<0$, higher-energy states are more populated than lower-energy states.

Instantaneously reversing a Hamiltonian $H(B)$ satisfying $H(-B)=-H(B)$ leaves populations unchanged but rewrites them as a canonical distribution with ${\beta }^{\prime }=-\beta$. When the partition function is unchanged, the free energy changes sign and the heat capacity is unchanged.

A negative-temperature system is hotter than every positive-temperature system. If heat $\delta Q$ passes from it to a positive-temperature body, the total entropy change is

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

Rough elastic collisions can exchange rotational and translational kinetic energy. For three-dimensional rigid balls with all three rotations thermalized, these six quadratic terms contribute $3k_{B}T$ per particle. A uniform gravitational height distribution adds $k_{B}T$, giving total mechanical mean energy $4k_{B}T$.

If each initially stationary ball is released at height $h$ and the subsequent isolated dilute assembly reaches effective classical equilibrium, energy conservation gives $4k_{B}T=mgh$. The height-tail probability above the release height is then $e^{-4}$. The effective temperature refers to mechanical motion, not the material's internal temperature.

For a classical canonical coordinate with confining potential energy $U=A|x{|}^s$, $A>0$ and $s>0$, integration by parts in the Boltzmann distribution gives $⟨x{U}^{\prime }(x)⟩=k_{B}T$. Homogeneity then gives $⟨U⟩=k_{B}T/s$. The usual quadratic contribution $k_{B}T/2$ is the special case $s=2$.

In the dilute Maxwell--Boltzmann regime, a species has number density proportional to $g(m{k}_{B}T{)}^{3/2}{e}^{(\mu -m{c}^2)/(k_{B}T)}$.

At inverse temperature $\beta =1/(k_{B}T)$, the Boltzmann factor $e^{-\beta E}$ is the relative canonical ensemble weight of a microstate with energy $E$.