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The three-dimensional Debye model normalizes its phonon density of states to $3N$ modes: $g(\omega )=9N{\omega }^2/{\omega }_D^3$ below the Debye frequency ${\omega }_D$. The thermal energy is $U={\int }_0^{{\omega }_D}g(\omega )\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]d\omega$. Differentiate at fixed volume and substitute $x=\hslash \omega /(k_{B}T)$ to obtain the displayed heat capacity at constant volume.

At $T\ll {\Theta }_D$, integration by parts gives ${\int }_0^{\infty }{x}^4{e}^x/(e^x-1{)}^{2}dx=4{\int }_0^{\infty }{x}^3/(e^x-1)dx=4{\pi }^4/15$. Therefore $C_V\sim (12{\pi }^4/5)N{k}_B(T/{\Theta }_D{)}^3$. At $T\gg {\Theta }_D$, each mode approaches the equipartition theorem value $k_B$ of heat capacity, so $C_V\to 3N{k}_B$, the Dulong-Petit law. Fixed harmonic zero-point energy does not contribute to either limit.

In the classical high-temperature limit, each independent harmonic oscillator contributes $k_B$ to the heat capacity at constant volume, where $k_B$ is the Boltzmann constant. A three-dimensional crystalline solid with $N$ chemical atoms has $3N$ vibrational normal modes in the bulk harmonic approximation. Its vibrational internal energy and heat capacity at constant volume therefore obeyHere $U_0$ is independent of temperature at fixed volume. Thus the Dulong-Petit law predicts $C_V\simeq 3N{k}_B$ for a real solid in this regime. Per mole of chemical atoms, rather than per mole of a compound's formula units,with molar gas constant $R$ and Avogadro constant $N_A$. A formula unit containing $r$ chemical atoms consequently has a classical vibrational prediction of $3rR$ per mole of formula units.

For example, in this ideal classical model, one mole of chemical atoms heated at fixed volume from $300$ to $600\mathrm{K}$ gains vibrational internal energyThe heat capacity at constant volume stays the same even though the thermal part of the internal energy doubles. These temperatures illustrate the stipulated classical model; they do not assert that every real solid is classical at $300\mathrm{K}$.

The proof outline is the equipartition theorem: one classical harmonic oscillator has a quadratic kinetic energy and a quadratic potential energy, each with mean $k_{B}T/2$. Their sum is $k_{B}T$, whose derivative with respect to temperature is $k_B$. Summing over $3N$ normal modes gives $3N{k}_B$. The classical result does not require equal angular frequencies: each harmonic normal mode contributes the same amount once it is classical. Overall rigid motions give only a finite-size correction to the bulk count. This independence from the individual angular frequencies makes the Dulong-Petit law useful as a quick estimate of a solid's heating requirement.

Experiments on freely expanding solids often measure the heat capacity at constant pressure $C_P$, whereas the harmonic prediction is for the heat capacity at constant volume $C_V$. In a smooth single thermodynamic phase of a simple compressible material, the difference between constant-pressure and constant-volume heat capacities iswhere $\alpha =V^{-1}(\partial V/\partial T{)}_P$ is the volumetric coefficient of thermal expansion and ${\kappa }_T=-V^{-1}(\partial V/\partial P{)}_T$ is the isothermal compressibility. Using measured $C_P$ as an approximation to $C_V$ requires this correction to be small compared with $C_V$. ^[ref]

High temperature here means $k_{B}T\gg \hslash \omega$ for the relevant vibrational normal modes, with reduced Planck constant $\hslash$ and mode angular frequency $\omega$. At low temperature, excitation across a quantum energy spacing is suppressed: modes with $\hslash \omega \gg k_{B}T$ freeze out thermally, and the classical equipartition theorem fails. The Einstein model of a solid makes this loss of vibrational heat capacity explicit. The Dulong-Petit law is therefore an ideal high-temperature vibrational limit, not a statement that all solids have the same heat capacity at every temperature. Other contributions and departures from the harmonic approximation can matter even at high temperature. ^[ref]

Historically, Petit and Dulong's 1819 experiments compared the specific heat capacities of several elemental solids and found an approximately constant product with their relative atomic weights. Their original empirical regularity came before the classical harmonic explanation; it suggested that counting chemical atoms reveals a simpler pattern than comparing equal masses. ^[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.

For a circular domain of radius $R_0$ with line tension $\gamma$, the stiffness of a real polar sine or cosine amplitude with integer angular harmonic $m$ is $K_m=\pi \gamma (m^2-1)/R_0$, when the enclosed area is fixed. The equipartition theorem gives $⟨u_m^2⟩=k_{B}T{R}_0/[\pi \gamma (m^2-1)]$ for $m\ge 2$. A positive-frequency complex Fourier mode coefficient has half this variance. The uniform mode is forbidden and $m=1$ is translation.

For a tension-dominated cylindrical lipid vesicle of fixed length $L$ and volume, a real axisymmetric sine amplitude $u_q$ has stiffness $K_q=\pi \sigma L[(q{R}_0{)}^2-1]/R_0$. Fixed volume requires a mean-radius shift $-u_q^2/(4R_0)$. Stable modes have $⟨u_q^2⟩=k_{B}T/K_q$ by the equipartition theorem. Modes with $q{R}_0<1$ have the Rayleigh–Plateau instability, rather than a negative equilibrium variance. Bending modifies very short wavelengths and near-marginal modes.

Heat capacity measures the heat supplied per small temperature increase along a specified process:The subscript $X$ records the constraint, such as fixed volume or fixed pressure. The units are $\mathrm{J}{\mathrm{K}}^{-1}$. The symbol $\delta Q$ emphasizes that heat is a transfer along a path, rather than a state function with a differential $dQ$. Consequently, an unspecified heat capacity does not determine the required heat.

For a closed system of fixed composition undergoing a quasistatic process, assume that only pressure-volume work occurs, with no friction and no change in bulk kinetic energy or potential energy. The first law of thermodynamics and $H=U+pV$, where $U$ is internal energy and $H$ is enthalpy, then giveThese partial derivatives define heat capacity at constant volume and heat capacity at constant pressure, with the amount and composition also held fixed. Integrating along the selected constraint gives $Q={\int }_{T_1}^{T_2}{C}_X(T)dT$ within a single thermodynamic phase; $Q=C_X\Delta T$ requires constant $C_X$. ^[ref]

For a computed example, take a monatomic gas in the classical ideal gas regime, with amount of substance $n=1\mathrm{mol}$ and stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The equipartition theorem gives $U=3nRT/2$, and the ideal gas law gives $pV=nRT$. ThusFor the same temperature increase $\Delta T=10\mathrm{K}$, the two processes requireHere work done by the gas is positive. At fixed pressure, the extra heat supplies expansion work; the internal energy increase is the same because internal energy of an ideal gas depends only on temperature. Their final volume values differ. ^[ref]

The practical value of heat capacity is that a measured heat input and temperature response can determine it. Once its constraint and temperature dependence are known, it predicts heating requirements and supplies entropy changes through entropy from heat capacity. ^[ref]

A freely orientable classical magnetic dipole moment of fixed magnitude $m$ in a uniform magnetic field $B$ has angular canonical partition function $Z=4\pi \sinh x/x$, where $x=mB/(k_{B}T)$. Its internal energy is $U=-mB\coth x+k_{B}T$, and its fixed-field heat capacity is $C_B=k_B[1-x^2/{\sinh }^{2}x]$. At low temperature, two locally Cartesian tangent coordinates on the orientation sphere each contribute $k_{B}T/2$ by the equipartition theorem, giving $C_B\to k_B$. A discrete quantum spin in a nonzero magnetic field instead has a spectral gap and its heat capacity vanishes exponentially at sufficiently low temperature.

The heat capacity ratio compares heat capacity at constant pressure $C_p$ with heat capacity at constant volume $C_V$ at the same equilibrium state and fixed composition:Here $C_p,C_V$ refer to the whole sample, in $\mathrm{J}{\mathrm{K}}^{-1}$; molar heat capacities $C_{p,\mathrm{m}}=C_p/n$ and $C_{V,\mathrm{m}}=C_V/n$ are per mole; specific heat capacities $c_p=C_p/m$ and $c_V=C_V/m$ are per unit mass. The amount of substance $n$ or mass $m$ cancels only when both capacities use the same normalization. Thus heat capacity ratio is dimensionless and independent of sample size. For a fixed-composition ideal gas with positive finite $C_V$, Mayer's relation giveswhere $R$ is the molar gas constant. More ways to store internal energy make a given temperature increase require more heat, bringing heat capacity ratio closer to unity. ^[ref]

The equipartition theorem predicts $C_{V,\mathrm{m}}=fR/2$ when $f$ independent quadratic terms in molecular energy are thermally active and their number stays fixed; consequently $\gamma =1+2/f$. A classical monatomic gas with no appreciable electronic excitation has $f=3$, giving $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. For an ideal gas of diatomic molecules with three translational and two rotational terms active, but its vibrational normal mode effectively frozen, $f=5$ gives $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. A fully classical harmonic vibration adds both kinetic energy and potential energy, giving $f=7$ and $\gamma =9/7$. These are model regimes, not universal constants of every gas: rotational freezing, electronic excitation, and changing composition invalidate the simple count.^[ref]

The mechanical significance of heat capacity ratio follows from the first law of thermodynamics. For a fixed amount of ideal gas in a reversible adiabatic process, $C_V(T)dT=-pdV$. Combining this with the ideal gas law gives the local relationThe partial derivative holds at fixed entropy and composition even when heat capacity ratio varies with temperature. The finite-path formulas $p{V}^{\gamma }=\mathrm{constant}$ and $T{V}^{\gamma -1}=\mathrm{constant}$ require the additional approximation of constant heat capacity ratio. A rapid compression also needs internal equilibration to approximate this path; rapid motion alone does not ensure reversibility. The same local response gives isentropic compressibility $1/(\gamma p)$ and, for small sound disturbances in this ideal gas regime, speed of sound $a=\sqrt{\gamma p/\rho }$, where $\rho$ is mass density. Thus thermal storage controls mechanical stiffness, sound propagation and the measurements in Rüchardt experiment and Clément-Desormes experiment.

A strongly stretched inextensible worm-like chain has two small transverse tangent components. Its quadratic energy is $\frac{1}{2}\int [A|{\partial }_s{\mathbf{t}}_{\perp }{|}^2+f|{\mathbf{t}}_{\perp }{|}^2]ds-fL$. For a long bulk chain, the equipartition theorem assigns each Fourier mode covariance $k_{B}T/(A{q}^2+f)$. Integrating over $q$ and summing both components gives $⟨|{\mathbf{t}}_{\perp }{|}^2⟩=k_{B}T/\sqrt{Af}$, hence the extension deficit is half that value. With persistence length $L_p=A/(k_{B}T)$, the deficit is $\sqrt{k_{B}T/(4L_{p}f)}$. It requires small slopes and negligible stretching of the backbone.

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

Internal energy $U$ is the energy stored in the microscopic degrees of freedom of a thermodynamic system: random molecular kinetic energy, internal molecular excitations, and interaction potential energy. Its value requires a specified system and an energy reference. In thermodynamic equilibrium, internal energy is a state function; it records the present state, without remembering whether a previous change came from stirring, compression or heat transfer. This makes internal energy the common accounting quantity connecting mechanical experiments to thermal ones.^[ref]

Heat and work describe energy crossing the system boundary during a process. They are not separately stored contents of the system. For a closed system whose bulk kinetic energy and external potential energy have unchanged endpoint values, the first law of thermodynamics givesHere $Q_{\mathrm{in}}$ is positive for incoming heat, and $W_{\mathrm{on}}=-W_{\mathrm{out}}$ is positive for net work done on the system. Two routes between the same equilibrium states have the same $\Delta U$, although their heat and work can differ. The infinitesimal notation $dU=\delta Q_{\mathrm{in}}+\delta W_{\mathrm{on}}$ distinguishes a state function change from path-dependent transfers.^[ref]

For a calculated example, take one mole of a classical monatomic gas in the ideal gas regime, at fixed volume, from $300$ to $310\mathrm{K}$. With stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, its internal energy change is $3nR\Delta T/2=124.71\mathrm{J}$. A purely thermal route supplies $Q_{\mathrm{in}}=124.71\mathrm{J}$ with $W_{\mathrm{on}}=0$. An ideal insulated stirring route supplies $W_{\mathrm{on}}=124.71\mathrm{J}$ with $Q_{\mathrm{in}}=0$, measured between equilibrium endpoints after the motion dies away. The same stored increment has two different transfer histories; neither history defines a separate kind of stored energy.

The familiar internal energy of an ideal gas is a useful special case. Neglect intermolecular interactions, hold the amount of substance $n$ and composition fixed, and suppose that $f$ independent quadratic terms per particle obey the classical equipartition theorem. Choosing the constant reference contribution as zero givesHere $T$ is absolute temperature, $R$ is the molar gas constant, and $C_V$ is the whole sample's heat capacity at constant volume. A monatomic gas has $f=3$ translational terms. A rigid diatomic molecule has $f=5$ when its two rotational terms are active and vibration is frozen; activating one harmonic oscillator vibration adds two terms, one kinetic energy and one potential energy, giving $f=7$. The counting is of quadratic contributions, so a vibrational coordinate is not just one contribution.^[ref]

This constant-$f$ formula applies only over a regime with those same active terms, classical translational statistics, nonrelativistic motion and unchanged composition. Quantum freezing, quantum degeneracy, chemical reactions or appreciable intermolecular interactions invalidate that simple counting. More generally, at fixed amount and composition an ideal gas can still have $U=U(T)$ with varying heat capacity at constant volume, so use $\Delta U={\int }_{T_1}^{T_2}{C}_V(T)dT$. A nonzero constant reference adds $U_{\ast }$ to $fnRT/2$ without altering $\Delta U$.

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.

This coefficient multiplies the squared mean-curvature term $\frac{\kappa }{2}\int (2H-C_0{)}^{2}dA$ in the Helfrich membrane energy. It has units of energy, unlike a beam's flexural rigidity. For small height modes of a nearly flat membrane of area $A$, with Fourier normalization $h_q=A^{-1}\int h(\mathbf{x})e^{-i\mathbf{q}\cdot \mathbf{x}}{d}^{2}x$, quadratic bending and tension energies give $⟨|h_q{|}^2⟩=k_{B}T/[A(\kappa q^4+\sigma q^2)]$ by the equipartition theorem. Thermal-flicker measurements can therefore separate short-wavelength bending from longer-wavelength tension effects.

For small-slope height fluctuations of a nearly planar membrane, $A$ is projected area, $\sigma$ is membrane tension, $\kappa$ is membrane bending modulus, and $q$ is a wavenumber. With a Fourier convention in which the mode amplitude $h_{\mathbf{q}}$ has units of length, the quadratic Helfrich membrane energy and the equipartition theorem give the displayed thermal variance. Bending dominates for $q\gg \sqrt{\sigma /\kappa }$, giving a $q^{-4}$ spectrum. Tension dominates at smaller $q$, giving $q^{-2}$. The corresponding membrane elastic length is $\sqrt{\kappa /\sigma }$ when length is measured as $1/q$; a wavelength convention multiplies it by $2\pi$.

A monatomic gas consists of individual atoms or ions rather than molecules. In the classical dilute ideal-gas regime, its three translational degrees of freedom give $U=3N{k}_{B}T/2$ by equipartition theorem. Thus $C_V=3N{k}_B/2$, while $C_P-C_V=N{k}_B$ follows by differentiating the ideal-gas enthalpy $H=U+PV=5N{k}_{B}T/2$. Consequently the adiabatic index is $\gamma =C_P/C_V=5/3$ and the specific internal energy is $e=3P/(2\rho )$. The same translational result applies to a fully ionized mixture when the particle numbers are fixed. This excludes additional excitation, ionization and radiation contributions to the heat capacity.

Write $s_V^2$ for the detector's position-related voltage variance, after characterizing independent detector noise and the acquisition response. The equipartition theorem estimate is $k=k_{B}T/({\beta }^2{s}_V^2)$. For small independent input standard uncertainties, first-order propagation givesIn this expression $u_{s_V^2}$ is the uncertainty of the variance itself. If the assumed independent relative uncertainties are $1\%$, $2\%$, and $3\%$, respectively, the resulting relative standard uncertainty is $\sqrt{26}\%\simeq 5.10\%$. This invented budget excludes model bias. Shared calibration inputs introduce covariance terms: a force estimate $F=k\beta \overline{V}$ using this same thermal estimate has $F=k_{B}T\overline{V}/(\beta s_V^2)$, so treating $k$ and $\beta$ as independent would count their sensitivity incorrectly.

Correlated Brownian motion also makes sample count a poor proxy for independent information. For a long continuous stationary Gaussian record of duration $\mathcal{T}$, estimating ${\sigma }^2$ from ${\mathcal{T}}^{-1}\int x(t{)}^{2}dt$ gives, to leading order,The Gaussian distribution fourth-moment identity gives $\mathrm{Cov}(x(t{)}^2,x(0{)}^2)=2C_x(t{)}^2$; integrating this covariance over the observation interval gives the result. Subtracting a fitted mean changes higher-order terms when $\mathcal{T}\gg \tau$. The ideal $10\mathrm{s}$ example gives about $0.97\%$ relative uncertainty in the variance, even though $200000$ samples were recorded. This ideal statistical estimate does not include drift, detector noise or incorrect drag.

Add the random thermal force to the deterministic optical trap relaxation time model. In one stationary coordinate of a harmonic optical trap, the overdamped Langevin equation isHere $\xi$ is Gaussian white noise understood through its integrals, $\delta$ is the Dirac delta function, and $\zeta$ is the linear friction coefficient. The covariance coefficient has units ${\mathrm{N}}^2\mathrm{s}$; multiplying by $\delta (t-t^{\prime })$, of units ${\mathrm{s}}^{-1}$, gives squared force. This fluctuation-dissipation relation for a Langevin particle makes the stationary variance ${\sigma }^2=k_{B}T/k$, consistent with the equipartition theorem. The solution is an Ornstein-Uhlenbeck process with autocovariance functionIndeed, at positive lag the noise has zero correlation with the initial position, so $\zeta C_x^{\prime }=-k{C}_x$; its initial value is ${\sigma }^2$. This supplies both a decay time and an amplitude, making a stronger calibration test than a position histogram alone.

Specify the Fourier transform convention before using a power spectral density. For frequency $f$ in hertz, define $S_x^{(2)}(f)={\int }_{-\infty }^{\infty }{C}_x(t)e^{-2\pi ift}dt$ and the one-sided density $S_x^{(1)}(f)=2S_x^{(2)}(f)$ for $f>0$. Integration of the exponential givesThe position power spectral density has units ${\mathrm{m}}^2/Hz$; its integral is a variance, not a standard deviation. At $f=f_c$ it is half its zero-frequency plateau. Angular frequency conventions require a corresponding $2\pi$ conversion in both axis and integration measure. If $x=\beta (V-V_0)$ with position sensitivity $\beta$ in $m/V$, then $S_V^{(1)}=S_x^{(1)}/{\beta }^2$ in ${\mathrm{V}}^2/Hz$ before independent detector noise is added.

For an original invented calibration example, choose $a=0.500\mu \mathrm{m}$, dynamic viscosity $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, $T=300\mathrm{K}$, $k=0.0200pN/nm=2.00\times 10^{-5}N/m$, and $\beta =100nm/V$. The unbounded no-slip Stokes drag law givesThe corresponding voltage standard deviation is $0.14391\mathrm{V}$, and the position power spectral density plateau is $0.390369\mathrm{n}{\mathrm{m}}^2/Hz$, or $3.90369\times 10^{-5}{\mathrm{V}}^2/Hz$ in the detector. Recovering both $k=k_{B}T/{\sigma }^2$ and $k=2\pi \zeta f_c$ is an internal model check; actual agreement also tests the assumed temperature, drag and detector response.

The generator uses the exact stationary update $x_{j+1}=\rho x_j+\sigma \sqrt{1-{\rho }^2}{Z}_j$, with $\rho =e^{-\Delta t/\tau }$ and independent standard normal random variables $Z_j$. For a demeaned block of $N$ samples and its unnormalized discrete Fourier transform $X_j$, the two-sided periodogram is $\Delta t|X_j{|}^2/N$; doubling positive bins except the zero and Nyquist endpoints gives a one-sided density. Summing those bins times $\Delta f=1/(N\Delta t)$ returns the block mean square by discrete Parseval identity. Independent detector noise, finite exposure and aliasing would alter a measured spectrum and must enter a fit before interpreting $f_c$.

In the classical thermal model, a coordinate in a harmonic potential obeys $\gamma \dot{x}+kx=\xi$, where $\gamma$ is the drag coefficient and $⟨\xi (t)\xi (t^{\prime })⟩=2\gamma k_{B}T\delta (t-t^{\prime })$. Its stationary one-sided spectral density per hertz is the displayed Lorentzian line shape. The integral over positive frequencies equals $k_{B}T/k$, consistent with the equipartition theorem. On a log-log plot the spectrum is flat for $f\ll f_c$ and has slope $-2$ for $f\gg f_c$. The crossover is determined by the restoring stiffness and drag; an added detector response multiplies the mechanical spectrum by the squared magnitude of its transfer function.

For a classical monatomic homogeneous fluid with pairwise central pair potential $v(r)$, the mean interaction energy is one half the sum over neighbours of a tagged particle. Using the radial distribution function gives the displayed formula; the factor one half prevents counting each pair twice. The kinetic term follows from the equipartition theorem. Its use requires the relevant integral to exist and a thermodynamic state specified by number density and temperature. Quantum kinetic contributions, many-body forces and internal molecular modes require modifications.

For a high-temperature linear rigid rotor, $q_{\mathrm{rot}}=T/{\theta }_{\mathrm{rot}}$ up to temperature-independent symmetry/spin factors. Therefore $q^{\prime }/q=1/T$ and $q^{\prime \prime }=0$, givingThis agrees with the equipartition theorem for two quadratic rotational coordinates. It is the rotational contribution; adding free three-dimensional translation gives $C_{V,m}=5R/2$ when vibrations remain unexcited. A constant rotational symmetry number does not affect this derivative.