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The Boltzmann distribution assigns probabilities to individual microstates in a canonical ensemble, a statistical ensemble for thermal contact. The system exchanges energy with a thermal reservoir at fixed positive absolute temperature $T$, while its particle number $N$ and external parameters, such as volume, stay fixed. The thermal reservoir is large enough that these exchanges barely change its temperature; the coupling permits thermal equilibrium but contributes negligible interaction energy. All the microstates considered have the same $N$. No chemical potential term is needed in their weights; fixed $N$ does not imply that the chemical potential is zero.

For a finite discrete set of microstates with energies $E_i$, writeHere $k_B$ is the Boltzmann constant, $\beta$ is the inverse temperature, $e^{-\beta E_i}$ is a Boltzmann factor, and $Z$ is the partition function. The Boltzmann factor is a weight; dividing by the partition function makes it a probability. ^[ref]

The useful surprise is that a fixed temperature still permits thermal fluctuations in the system's energy. The Boltzmann distribution predicts their probabilities and expected values without tracking every microscopic exchange. Its exponential function also makes rare excited microstates sensitive to temperature: an energy gap much larger than $k_{B}T$ gives strong Boltzmann suppression, while warming increases their relative populations. A population ratio can therefore reveal a gap or serve as a thermometer, provided thermal equilibrium and the relevant state degeneracies are known.

Historically, Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) presents the canonical ensemble in Chapter IV. His equations 90–92 give an exponential function of energy and its normalization, in classical phase space. This classical formulation exhibits the same weighting principle. ^{[ref] [ref]}

Group microstates by their energy $E$. The state degeneracy $g(E)$ counts how many microstates have that energy; their aggregate probability isHere $P(E)$ is the probability of the macrostate retaining only energy, while $p_i$ is the probability of one individual microstate. For two such macrostates,Counting more microstates can compensate for a smaller Boltzmann factor. Omitting the state degeneracy confuses a sum over microstates with a sum over distinct energies. ^[ref]

Consider two distinguishable, noninteracting two-level systems, each with energies $0$ and $\Delta >0$. Their four microstates $(0,0),(0,1),(1,0),(1,1)$ have total energies $0,\Delta ,\Delta ,2\Delta$. Thus the three energy macrostates have state degeneracies $1,2,1$. This is a stipulated finite model, not a measured material.

Set $q=e^{-\beta \Delta }$. The four individual Boltzmann factors are $1,q,q,q^2$, so the partition function and aggregate probabilities areEach of the two middle microstates has probability $q/Z$, while their shared energy macrostate has probability $2q/Z$.

At the temperature satisfying $\Delta =k_{B}T\ln 2$, the Boltzmann factor is $q=1/2$ and the partition function is $Z=9/4$. ThereforeThe middle energy is as likely as the ground state energy because its state degeneracy is twice as large. Yet each individual middle microstate is only half as likely as the ground state microstate. The diagram places the four energies beside their individual probabilities, then shows the sums within each macrostate.

The expected value of energy follows directly by weighting the three possible energies:At $q=1/2$ this is $2\Delta /3$. An individual energy measurement still returns $0$, $\Delta$ or $2\Delta$; the expected value describes the ensemble average, rather than a fourth possible measured energy.

Keep the gap $\Delta$ fixed and vary positive temperature. With $x=k_{B}T/\Delta$, the Boltzmann factor is $q=e^{-1/x}$. In the four-microstate example, as $T\to 0^{+}$, $q\to 0$: the unique ground state has probability tending to one, and the other microstates have probabilities tending to zero. If several microstates instead shared the lowest energy, their state degeneracy would split that limiting population equally.

As $T\to \infty$ in this fixed finite model, all four individual Boltzmann factors tend to one, so all four individual microstates have probability tending to $1/4$. The three aggregate probabilities instead approachThe equal-population limit applies to this finite set of microstates. An infinite collection of microstates cannot all have the same positive normalized probability, so the statement cannot be extended to an unbounded energy spectrum. Real systems may also acquire additional accessible microstates as temperature rises.

The plotted crossing occurs at $x=1/\ln 2$, where $P(0)=P(\Delta )=4/9$. For larger positive temperatures, the middle macrostate is more probable than the lowest-energy macrostate, although the lowest-energy individual microstate remains more probable than either individual middle microstate. The expected value of energy rises from zero toward $\Delta$ in this finite model.

A dimensionless atomic collision measure used to parameterize electron-impact transition coefficients. In the Maxwellian approximation used here, $q_{ji}=8.6\times 10^{-12}{\Omega }_{ji}/(g_j\sqrt{T_e})$ in ${\mathrm{m}}^3{\mathrm{s}}^{-1}$, and detailed balance fixes the excitation coefficient. Its value and temperature dependence encode atomic scattering physics beyond the elementary Boltzmann factor.

A reversed bond costs $2J$, so its Boltzmann factor $w=e^{-2J/(k_{B}T)}$ is a natural thermal coordinate near the zero-temperature chain fixed point. Exact length-two decimation gives $w^{\prime }\simeq 2w$, and the One-dimensional Ising correlation function gives $\xi /a=-1/\log \tanh K\sim 1/(2w)$. This is a length exponent one with respect to $w$. The correlation length instead diverges exponentially in inverse physical temperature; claiming a conventional finite power $T^{-1}$ would confuse scaling coordinates. Choosing $e^{-K}$ gives multiplier $\sqrt{2}$ and power two in that alternative coordinate, the same physical length.

The Einstein model of a solid represents the vibrations of $N$ chemical atoms by $3N$ independent one-dimensional quantum harmonic oscillators, all with the same positive angular frequency $\omega$. Equivalently, each chemical atom has three independent oscillator directions. The common angular frequency is a simplifying model assumption, not a claim about every vibration of a real solid. Hold the volume, $N$ and $\omega$ fixed. Define the characteristic temperature ${\theta }_E=\hslash \omega /k_B$ and the dimensionless ratio $x={\theta }_E/T$. The vibrational internal energy isThe first term is the zero-point energy $3N\hslash \omega /2$. It is constant at fixed $\omega$, so it contributes no heat capacity at constant volume. The temperature-dependent term givesThis is the vibrational heat capacity at constant volume of the model, with no expansion or other contributions included. ^[ref]

For a worked comparison, choose the model parameter ${\theta }_E=300\mathrm{K}$. The normalized heat capacity at constant volume depends only on $T/{\theta }_E$:

At $T={\theta }_E$, one mole of chemical atoms therefore has vibrational heat capacity at constant volume $3R(0.92067)\simeq 22.965\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, using the unrounded model value. Even at twice the characteristic temperature, the model remains about $2.06\%$ below the Dulong-Petit law limit. These are calculations for the ideal Einstein solid, not material measurements.

A short derivation uses the thermal partition function of a quantum harmonic oscillator. The energy eigenvalues $E_n=\hslash \omega (n+1/2)$ and their Boltzmann factors form a geometric series, giving the one-oscillator canonical partition function $z=e^{-x/2}/(1-e^{-x})$. The mean occupation is $1/(e^x-1)$, hence the displayed $U$ after summing over $3N$ independent quantum harmonic oscillators. Taking its derivative with $dx/dT=-x/T$ gives the displayed heat capacity at constant volume. ^[ref]

The two limits explain the shape. At high temperature, a Taylor series in small $x$ givesAt low temperature, $1-e^{-x}\to 1$, soThe zero-point energy remains even when the thermal response is tiny: freeze-out means that changing temperature scarcely changes the occupations, not that the chemical atoms have zero energy. This is the useful lesson of the Einstein model of a solid: the classical counting rule reappears when many levels are accessible, while discrete excitation spacings explain its failure upon cooling.

The exponential low-temperature decay is also the model's limitation. Real three-dimensional crystalline solids have acoustic phonon branches with arbitrarily small angular frequencies in the bulk limit. The Debye model includes these collective normal modes and predicts a low-temperature vibrational contribution $C_V\propto T^3$, which the single-frequency Einstein solid cannot reproduce. This qualification concerns the vibrational contribution; additional excitations can change a material's total heat capacity. ^[ref]

Einstein introduced the quantum treatment of solid vibrations in his 1907 paper Die Plancksche Theorie der Strahlung und die Theorie der spezifischen Wärme. Its historical importance is that quantum excitation energies explained why the classical Dulong-Petit law could fail; the modern quantum harmonic oscillator formulation above includes the zero-point energy explicitly. ^[ref]

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.

Take a cycle graph with four labeled graph vertices and four distinct edges $\{1,2\},\{2,3\},\{3,4\},\{4,1\}$. Set $h=0$, $J>0$, and impose periodic boundary conditions by writing ${\sigma }_5={\sigma }_1$. The square drawing represents a one-dimensional ring: it has no diagonal interactions and is not a two-dimensional lattice.

Call an edge between opposite Ising spins a domain wall. If $d$ is their number, the four bond products sum to $4-2d$, so the energy isEach domain wall costs $2J$ relative to an aligned bond. Returning around the ring must return to the starting Ising spin, so $d$ is even: equivalently, ${\prod }_{i=1}^4({\sigma }_i{\sigma }_{i+1})=1$. Choosing an even set of $d$ wall edges and then choosing ${\sigma }_1$ determines the entire pattern. The state degeneracy is therefore $g_d=2\left(\genfrac{}{}{0ex}{}{4}{d}\right)$, using a binomial coefficient.

The twelve middle microstates consist of eight patterns with one exceptional Ising spin and four patterns with two adjacent $+1$ Ising spins and two adjacent $-1$ Ising spins. The two alternating patterns have four domain walls. All $2+12+2=16$ microstates are shown below.

Writing $K=\beta J$, the partition function and the probability of each energy class areFor an original exact example, choose $K=(\ln 2)/4$. Then $e^{4K}=2$, $Z=17$, and $P(d=0)=4/17$, $P(d=2)=12/17$, $P(d=4)=1/17$. Each aligned microstate has probability $2/17$, each middle microstate $1/17$, and each alternating microstate $1/34$: multiply by state degeneracy before comparing whole classes. As $K\to 0$, all sixteen microstates become equally likely, but the class probabilities approach $1/8,3/4,1/8$. As $K\to \infty$, the two all-aligned microstates each have limiting probability $1/2$.

A hard-sphere gas is a classical model of identical impenetrable spheres, each of diameter $\sigma$ and radius $\sigma /2$. Its pair potential is $u(r)=+\infty$ for centre separation $r<\sigma$ and $u(r)=0$ for $r\ge \sigma$; the convention at exact contact has no effect on the integral. It isolates the effect of particle size without an attractive potential energy.

The upper-left panel is a cross-section of a three-dimensional forbidden region: the second centre cannot enter a sphere of radius $\sigma$ around the first centre. The particles themselves have radius $\sigma /2$. Consequently, with particle volume $v_{\mathrm{s}}$ and pair excluded volume $v_{\mathrm{ex}}$,The last equality follows directly from the Mayer f-function: $f=-1$ throughout the forbidden volume and zero outside it. Confusing particle radius with contact separation changes the answer by a factor of eight; omitting the unordered-pair factor changes it by a further factor of two.

Define the packing fraction $\eta =n{v}_{\mathrm{s}}$, where $n$ is number density. Then the virial expansion beginsFor example, at the toy value $\eta =0.01$, the leading correction predicts $Z\approx 1.04$; the omitted terms still matter if accuracy beyond this approximation is needed. The classical second virial coefficient is independent of temperature because the geometry of forbidden configurations does not change when the gas is warmed. The pressure itself still scales with temperature at fixed number density.

This model makes the positive correction particularly clear: eliminating overlapping configurations changes the dependence of the available configuration space on volume, even though the allowed configurations all have zero interaction energy. The figure's other panels add a toy attractive shell and show how its Boltzmann factor competes with that geometry. The hard-sphere gas is a useful reference model, while a low-order virial expansion remains a dilute-gas approximation rather than a dense sphere packing formula.

Let $u_s=⟨ϵ{⟩}_s$ and $v_s={\mathrm{Var}}_s(ϵ)$ for the conditional distributions of parahydrogen and orthohydrogen, with $s=\mathrm{p},\mathrm{o}$. Differentiating the normalized Boltzmann factors gives $d{u}_s/dt=v_s/t^2$. Thus the rotational heat capacity at constant volume per molecule of a composition frozen at $x$ isAt conversion equilibrium, the law of total variance adds a nonnegative contribution from the changing species populations:The last term is the contribution from conversion. Compare the first two terms with a frozen sample at the same instantaneous composition; a frozen $3:1$ sample generally has different weights too. This formula follows directly from the variance of a mixture: subtract the overall mean inside each species, expand the square, and the cross terms vanish because each conditional deviation has zero mean. It also proves nonnegative rotational heat capacity in either model.

For an original worked example at $t=1/2$, summing the allowed ladders gives $q_{\mathrm{p}}\approx 1.00003072$, $q_{\mathrm{o}}\approx 0.05494692$, and $x_{\mathrm{o}}^{\mathrm{eq}}\approx 0.14150982$. The calculated rotational heat capacities are $c_{\mathrm{eq}}/k_{\mathrm{B}}\approx 1.94719633$ and $c_{\mathrm{fr}}/k_{\mathrm{B}}\approx 0.00110733$ for frozen normal hydrogen. The contrast is large because changing species accesses the $J=0$ to $J=1$ separation, whereas fixed species must use the larger $J=0$ to $2$ or $J=1$ to $3$ separations. These are model numbers at a chosen dimensionless temperature, not measured hydrogen data.

Both rotational heat capacities approach zero as $t\to 0$, although frozen normal hydrogen retains nonzero rotational internal energy. Both approach $k_{\mathrm{B}}$ as $t\to \infty$ within the rotor model. The high-temperature limit is a formal limit with all other excitations suppressed; actual molecular hydrogen eventually needs molecular vibration, nonrigidity, and dissociation. Multiplying the per-molecule result by the Avogadro constant gives rotational molar heat capacity in units of the molar gas constant $R$. A dilute classical ideal gas additionally contributes $3R/2$ from translation to its heat capacity at constant volume.

Inverse temperature is $\beta =1/(k_{B}T)$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature. A canonical ensemble weights an energy $E$ by its Boltzmann factor $e^{-\beta E}$. Calling couplings dimensionless means absorbing this factor $\beta$ into their energy coefficients. The alternative convention $k_B=1$ makes inverse temperature $1/T$; physical and dimensionless couplings must not be mixed.

In a thermal Euclidean worldline path integral with $\hslash =1$, $\beta$ is also the circumference of the imaginary-time circle.

A macrostate groups microstates sharing prescribed coarse data. The grouping depends on what is retained: here retain only total energy. For the two distinguishable two-level systems, the macrostates $E=0,\Delta ,2\Delta$ have state degeneracies $1,2,1$. Although two units are not macroscopic in size, they make this loss of information visible: knowing $E=\Delta$ does not tell us which unit is excited.

At inverse temperature $\beta \Delta =\ln 3$, each unit's excited-state probability is $p=1/4$. Their independence gives microstate probabilities $9/16,3/16,3/16,1/16$ in the order above. Adding within each macrostate givesThe total expected value of energy is therefore $\Delta (6/16)+2\Delta (1/16)=\Delta /2$, matching two units each with expected value of energy $\Delta /4$. Equal state degeneracies would give the wrong answer: macrostate probabilities combine counting with the Boltzmann factor. At very high positive temperature, all four microstates become equally likely, but these three macrostate probabilities approach $1/4,1/2,1/4$.

The Maxwell-Boltzmann distribution describes the random velocities and speeds of particles in a classical ideal gas at thermal equilibrium. Individual particles keep changing velocity through collisions, while the equilibrium probability distribution stays fixed. For one species of particle mass $m$ at positive temperature $T$, the characteristic scale iswhere $k_B$ is the Boltzmann constant. A signed velocity component follows a normal distribution; the nonnegative magnitude follows the Maxwell speed distribution. Their peaks differ because a speed measurement groups together all directions of motion.

Use the gas rest frame, three spatial dimensions, and nonrelativistic particles, with $k_{B}T\ll m{c}^2$ where $c$ is the speed of light. Assume a dilute gas whose interactions between encounters can be neglected in the ideal gas approximation and whose translational motion is classical. A sufficient condition for negligible quantum exchange effects is $n{\lambda }^3\ll 1$, where $n$ is the number density and $\lambda$ the thermal de Broglie wavelength. Diluteness alone does not guarantee thermal equilibrium; the gas must have equilibrated. These are the assumptions behind the classical Maxwell-Boltzmann velocity distribution. ^{[ref] [ref]}

The connection to Maxwell–Boltzmann statistics is the Boltzmann factor for translational kinetic energy, $E_{\mathrm{kin}}=m|\mathbf{v}{|}^2/2$. Normalizing $e^{-E_{\mathrm{kin}}/(k_{B}T)}$ over velocity gives a probability density; an occupation factor $e^{(\mu -E)/(k_{B}T)}$ instead specifies a mean occupation, with chemical potential $\mu$. Their normalizations serve different purposes. ^[ref]

This connects microscopic motion to measurements: a gas thermometer uses the ideal gas law, collisions and the mean free path govern transport, and measurements of spectral lines use thermal Doppler broadening to infer temperature. Each probes a different aspect of the same thermal motion.

In 1860, James Clerk Maxwell presented the component and speed laws in Part I, Proposition IV of Illustrations of the Dynamical Theory of Gases, pp. 22–23. His argument combined independence of the three velocity components with isotropy, and he explicitly compared the component law with the normal distribution of observational errors. Independence is an assumption of that argument; perpendicular coordinate axes alone do not establish it. ^[ref]

The Maxwell speed distribution is the probability density function for the magnitude $v=|\mathbf{v}|$. Unlike a signed velocity component, speed cannot be negative:For $0\le a<b$, the probability of a speed in $[a,b]$ is ${\int }_a^{b}g(v)dv$, and ${\int }_0^{\infty }g(v)dv=1$. The density has units of inverse speed, so its height alone is not a probability. ^[ref]

All velocities with magnitudes from $v$ to $v+dv$ occupy a thin spherical shell of volume $4\pi v^{2}dv$ in velocity space. The isotropy of the Maxwell-Boltzmann velocity distribution makes its density constant over each shell, giving $g(v)=4\pi v^{2}f(v{\mathbf{e}}_x)$ for any fixed unit direction ${\mathbf{e}}_x$. A larger radius offers more directions and more volume. The increasing shell factor initially wins over the decreasing Boltzmann factor, so $g(0)=0$ and the most probable speed is positive. This is the geometric reason that the peaks of the vector and speed densities disagree.

With the dimensionless quantity $u=v/s$, every such gas has the same rescaled speed density:For comparison, $w=v_x/s$ has the standard normal distribution density $\varphi (w)=e^{-w^2/2}/\sqrt{2\pi }$ on the whole real line. The Maxwell speed distribution is neither the positive half of this curve nor the distribution of $|v_x|$: it combines three components.

The first two moments of the Maxwell speed distribution give the mean and root mean square:Consequently the mean translational kinetic energy is $m⟨v^2⟩/2=3k_{B}T/2$, as required by the equipartition theorem. ^[ref] The mean speed weights fast particles more than the density peak does, and the root mean square weights them more strongly still.

Changing temperature or particle mass only changes $s\propto \sqrt{T/m}$. At fixed mass, four times the temperature doubles every characteristic speed; at fixed temperature, four times the mass halves them. A stretched density becomes lower so that its area stays one. For two species at the same temperature, equal mean translational kinetic energy therefore means different typical speeds.

The Mayer f-function measures how a pair potential changes a pair's Boltzmann factor relative to the noninteracting value one:It vanishes where $u=0$, lies between $-1$ and $0$ for repulsion, and is positive for attraction. Even an infinite hard-core potential energy produces the finite value $f=-1$. This is why expanding in Mayer f-functions handles a hard-sphere gas more naturally than a Taylor series in $u$. ^[ref]

For pairwise-additive potential energy $U={\sum }_{i<j}u(r_{ij})$, the exact identityturns the configurational part of the canonical partition function into a sum of products associated with pairs. Drawing one edge for each chosen factor $f(r_{ij})$ gives the graphs used in the cluster expansion. The second virial coefficient uses just one pair's relative-position integral, $B_2=-\frac{1}{2}\int f{d}^3\mathbf{r}$.

For the square-well toy pair potential in second virial coefficient, $f=-1$ in the core, $f=e^{1/T^{\ast }}-1$ in the attractive shell, and $f=0$ beyond it. Its positive plateau need not be small: low number density, rather than small $f$ at every separation, controls the dilute-gas approximation. The integral weights each plateau by spherical-shell volume; equal-looking unweighted areas on a plot of $f$ versus $r$ need not cancel in $B_2$.

Choose a vibrational Raman shift magnitude ${\tilde{\nu }}_v=1000\mathrm{c}{\mathrm{m}}^{-1}$, excitation vacuum wavelength ${\lambda }_0=532\mathrm{nm}$ and temperature $T=300\mathrm{K}$. These are original illustrative inputs, not a measured 1928 optical spectrum or a claim about a particular liquid. They giveTake reciprocal lengths in consistent units before converting to nanometres. At $300\mathrm{K}$, $r\simeq 0.00826337$ and $\overline{v}\simeq 0.00833222$. The predicted line-integrated radiant intensity ratio is $0.0126522$, whereas the ideal photon count-rate ratio is $0.0113740$. Thus the radiometric ratio is about $1.27$ percent, rather than the $0.826$ percent obtained by using the Boltzmann factor alone.

For a finite-range isotropic Heisenberg model, the Mermin-Wagner theorem excludes spontaneous magnetic order in one dimension at every nonzero temperature. Long-wavelength rotations cost very little energy: a smooth orientational fluctuation has $\delta F=({\rho }_s/2)\int dx({\partial }_x\theta {)}^2$, so its thermal variance contains $(k_{\mathrm{B}}T/{\rho }_s)\int dq/q^2$, which diverges in the infrared. In a Heisenberg ferromagnet, the same obstruction appears as a divergent thermal occupation of quadratic long-wave spin waves. This explains why a finite magnetization cannot be supported, although correlations can be strong over a long finite distance. The rigorous statement requires short-range interactions and the stated continuous symmetry; interchain coupling and anisotropy change the problem. Symmetry forcing the mean to zero in a finite chain alone would not prove absence of thermodynamic order. ^[ref]

For the Ising model, use dimensionless $s_i=2S_i^z=\pm 1$, $h=g{\mu }_{\mathrm{B}}B$, $K=\beta J/2$ and $b=\beta h/2$. Its pair Boltzmann factor isMultiplying N such factors around the periodic ring counts each bond once and each field term twice with weight one half. Summing the closed spin-index sequence therefore gives the transfer matrix for the one-dimensional Ising modelThe two eigenvalues areAt finite temperature every entry of P is positive, so the Perron–Frobenius theorem gives ${\lambda }_{+}>|{\lambda }_{-}|$. ConsequentlyThe exact finite-N correction to the extensive expression is $-k_{\mathrm{B}}T\ln [1+({\lambda }_{-}/{\lambda }_{+}{)}^N]$. No assumption that ${\lambda }_{-}$ is positive is needed.

Differentiate at fixed temperature. For $D=\sqrt{{\sinh }^{2}b+e^{-2\beta J}}$, ${\partial }_b\ln (\cosh b+D)=\sinh b/D$. The Ising-chain magnetization with spin-half normalization gives total magnetic momentThis is the thermodynamic-limit result; the moment per site is M/N. For every finite $T>0$, ${\lim }_{B\to 0}M/N=0$, even with the thermodynamic limit taken first. The zero-field magnetic susceptibility per site is $(g{\mu }_{\mathrm{B}}{)}^2{e}^{\beta J}/(4k_{\mathrm{B}}T)$, large but finite.

The Ising-chain domain-wall energy convention explains the result for $J>0$: changing a parallel bond into an antiparallel bond costs J in this spin-half normalization. Domain walls, occurring in pairs on a ring, acquire a nonzero density at any nonzero temperature. Indeed the zero-field correlation length is $\xi =-a/\ln [\tanh (\beta J/2)]\sim (a/2)e^{\beta J}$ at low temperature. The Ising symmetry is discrete, so this is a separate one-dimensional domain-wall mechanism rather than an application of the continuous-symmetry theorem. At zero temperature a ferromagnetic chain can select a fully polarized state; the $B\to 0$ and $T\to 0$ limits do not commute.

Let $\beta =1/(k_{B}T)$ and use the two spin values $+1,-1$ as row and column labels. Define the spin-chain transfer matrixIt is a real symmetric matrix because $E_0(s,t)=E_0(t,s)$; every entry is strictly positive for finite bond energies at $T>0$. The Boltzmann factor of a spin configuration factorizes into the product $W_{{\sigma }_1{\sigma }_2}\cdots W_{{\sigma }_L{\sigma }_1}$. Summing the intermediate spin labels is exactly matrix multiplication, and the final closure imposes periodic boundary conditions. ThusWriting $W=\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$, its eigenvalues areA positive-entry matrix need not have two positive eigenvalues; ${\lambda }_{-}$ can be negative. The Perron–Frobenius theorem ensures ${\lambda }_{+}>|{\lambda }_{-}|$ here. Therefore the Helmholtz free energy is $F_L=-k_{B}T\ln Z_L$, and its value per spin in the thermodynamic limit is $-k_{B}T\ln {\lambda }_{+}$.

For an open chain there are $L-1$ bonds, with both endpoint Ising spins independently summed. Retain the same spin-chain transfer matrix, but replace the trace by endpoint vectors. With $1=(1,1{)}^{\mathsf{T}}$,This follows directly by expanding the product of $L-1$ Boltzmann factors. If ${\mathbf{v}}_{\pm }$ are orthonormal eigenvectors of the symmetric matrix $W$, the equivalent spectral decomposition isFixed endpoint Ising spins replace the corresponding $1$ by a basis vector; extra endpoint energies replace its entries by the corresponding Boltzmann factors. The leading bulk Helmholtz free energy per spin agrees with the periodic result, because the boundary changes only subextensive terms. In particular, $L=1$ gives $Z_1^{\mathrm{open}}=1^{\mathsf{T}}1=2$, as required for one free Ising spin.