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The characteristic thermal energy per microscopic degree-of-freedom scale is $k_{B}T$. Equipartition gives $k_{B}T/2$ per independent quadratic term under its classical assumptions.

Consider two rigid bodies forming an isolated system, with fixed composition and constant whole-body heat capacity at constant volume values $C_1=1\mathrm{kJ}{\mathrm{K}}^{-1}$ and $C_2=3\mathrm{kJ}{\mathrm{K}}^{-1}$. Their initial temperatures are $T_{1,0}=300\mathrm{K}$ and $T_{2,0}=400\mathrm{K}$. Assume each body remains internally uniform, their contact stores negligible energy, no phase transition occurs and no work is done or extracted. These are stipulated model values, not material measurements.

Because $d{U}_i=C_{i}d{T}_i$, write $U_i=a_i+C_i{T}_i$ with arbitrary fixed reference constants $a_i$. The conserved relative internal energy isThis is an energy reference convention, not a claim about actual heat capacities down to zero temperature. At thermal equilibrium the two temperatures equal $T_f$, so the first law of thermodynamics givesThe larger heat capacity pulls the common temperature closer to its initial value: body 1 warms by $75\mathrm{K}$, while body 2 cools by $25\mathrm{K}$.

To predict a timescale, stipulate a constant contact conductance $G=1/R_{\mathrm{th}}=0.1\mathrm{kJ}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, the reciprocal of the contact's thermal resistance. The rate of heat transfer into body 1 is $G(T_2-T_1)$, givingAdding proves conservation of $E$. Subtracting gives exponential decay of the temperature difference, with time constantand thereforeExact equality is approached as $t\to \infty$ in this ideal model. Changing $G$ changes the rate but not $T_f$: equilibrium and the speed of reaching it are separate predictions. Real bodies can have internal temperature gradients, changing heat capacities and contact losses, so the plotted $7.5\mathrm{s}$ is an assumed model timescale.

Two internally equilibrated systems are in thermal equilibrium if bringing them into contact that permits only heat exchange causes no net heat transfer. In ordinary nonrelativistic thermodynamics without appreciable gravitational temperature shifts, their thermodynamic temperatures are equal. Merely preventing heat transfer with insulation does not establish thermal equilibrium: two insulated bodies can have different temperatures. ^[ref]

A thermometer illustrates the definition. Its own temperature settles to the object's temperature after thermal contact, provided its effect on the object is negligible. The existing zeroth law of thermodynamics makes this comparison transitive: two objects separately in thermal equilibrium with the same thermometer are in thermal equilibrium with each other. This lets one calibrated scale compare different materials.

Equal temperatures do not imply equal internal energies or equal heat capacities. Nor does thermal equilibrium by itself imply mechanical equilibrium or chemical equilibrium; these require additional allowed exchanges. There is also a distinction between no net exchange in thermal equilibrium and a constant temperature maintained by equal incoming and outgoing heat. The latter can be a steady state with ongoing entropy production. Microscopic exchanges can continue at thermal equilibrium, but their opposing contributions balance on average.

Historically, OpenStax credits the name “zeroth law” to Ralph Fowler in the 1930s: the already numbered laws needed a logically earlier statement about thermal equilibrium. ^[ref]

Thermal expansion is a change in a body's dimensions as its temperature changes under a specified mechanical constraint. A heated gas under a freely moving loaded piston usually increases its volume at fixed pressure; heating the same gas in a rigid vessel instead raises its pressure. The coefficient of thermal expansion quantifies the first response. Expansion can be negative, so the name does not assert that every substance grows on heating.

For an ideal gas at fixed pressure and particle number, $V=N{k}_{B}T/P$ gives $V_2/V_1=T_2/T_1$. Heating from $300\mathrm{K}$ to $330\mathrm{K}$ therefore increases the volume by exactly $10\%$ within the ideal gas model. For a solid, specify whether the measurement concerns length or volume: for uniform expansion in all three directions, the volumetric coefficient is three times the linear thermal expansion coefficient, because $V\propto L^3$. Direction-dependent expansion requires the separate dimensional responses. ^[ref]

The tensor relating a small unconstrained thermal strain to a temperature change: ${\mathbf{\epsilon }}^{\mathrm{th}}=\mathbf{\alpha }\Delta T$. A crystal orientation rotates this tensor into laboratory coordinates.

A thermal fluctuation is the randomness of an observable in an equilibrium Boltzmann distribution. If the energy contains $-hA$, differentiation of its partition function gives ${\partial }_h⟨A⟩=\beta (⟨A^2⟩-⟨A{⟩}^2)$. Thus the fluctuation variance is related to the response to a conjugate source. In a field theory, the two-point version is the connected correlation function.

The emission coefficient per volume, frequency and solid angle of a Maxwellian plasma has $j_{\nu }\propto n_e{\sum }_i{Z}_i^2{n}_i{T}^{-1/2}{e}^{-h\nu /(k_{B}T)}{\bar{g}}_{\mathrm{ff}}$. Define the angular convention before supplying a numerical coefficient. The high-frequency cutoff measures temperature, while the normalization measures a density-squared emission integral. The radio spectrum is almost flat when optically thin and rises as ${\nu }^2$ when optically thick at fixed projected area.

For a homogeneous material, thermal inertia is $\Gamma =\sqrt{k\rho c_p}$, where $k$ is thermal conductivity, $\rho$ is density and $c_p$ is specific heat capacity. It measures the resistance of the surface temperature to periodic heating and controls the phase lag relevant to the Yarkovsky effect.

For a gas-supported alpha disk with $\kappa \propto {\rho }^{1/3}{T}^{10}$, compare rates at fixed surface density of a disk, radius, $\alpha$ and composition on the thermal timescale of an accretion disk. Assume that the vertical dynamical timescale of a disk is shorter than that thermal time, and that the viscous timescale is longer; the separation is controlled for $\alpha \ll 1$ and small disk aspect ratio. Vertical hydrostatic equilibrium gives $H\propto T^{1/2}$, $\rho \propto T^{-1/2}$, so $\kappa \propto T^{59/6}$. Viscous heating satisfies $Q^{+}\propto T$ and radiative cooling satisfies $Q^{-}\propto T^4/(\kappa \Sigma )\propto T^{-35/6}$. At equilibrium a positive temperature perturbation changes the net heating by $(41/6)(Q_0/T)\delta T>0$. With positive effective heat capacity, it grows rather than relaxes. This is a local thermal statement and uses the fixed-column, rather than fixed-volume-density, comparison.

At fixed surface density, a total-pressure alpha disk supported by radiation pressure has viscous heating $Q^{+}\propto T_c^8$ but radiative cooling $Q^{-}\propto T_c^4$ when electron scattering supplies a nearly constant opacity. Heating therefore wins after a positive temperature perturbation.

thermal insulation reduces unwanted heat exchange. Conduction through supports, convection in gas and thermal radiation require different treatments. vacuum suppresses gas transport but leaves solid conduction and radiation; lowering thermal conductance also increases the temperature rise caused by a fixed internal heat load.

The common boundary between two conducting materials in a thermal interface problem. Perfect thermal contact imposes continuity of temperature and of normal heat flux.

A heat equation problem in adjacent materials with separate thermal conductivities and volumetric heat capacities. Perfect thermal contact imposes continuous temperature and continuous normal heat flux at the interface. Multiplying each material's heat equation by its temperature cancels the interface boundary terms in a weighted energy estimate.

Paramagnetism arising when thermal excitation populates a magnetic manifold above a less magnetic or nonmagnetic ground manifold. Competing thermal population and orientational randomization can produce a maximum in susceptibility at an intermediate temperature.

Thermal motion is microscopic random motion associated with a system's temperature; Brownian motion makes its effects on larger suspended grains visible.

Fluctuations associated with nonzero temperature, including mechanical Brownian motion and electrical noise.

An external electrical load changes the terminal noise. Connecting a purely resistive load $R_L>0$ to the noisy resistor gives the voltage divider ratio $R_L/(R+R_L)$. The source resistor's contribution at the load isIn a rectangular signal bandwidth $B$, the power delivered by that source contribution to the load isbecause $(R-R_L{)}^2\ge 0$. Equality occurs at $R_L=R$: this is matched available noise power. Matching halves the voltage, quarters its power spectral density, and gives $P_{\mathrm{av}}=k_{\mathrm{B}}TB$. Dividing the squared open-circuit voltage by $R$ would instead give four times this available power.

These expressions isolate the source resistor's contribution. A real load resistor also generates thermal electrical noise. At the same temperature its independent contribution produces the equal reverse power flow, so the net thermal power transfer is zero at thermal equilibrium. For example, two equal resistors at the same temperature have total terminal power spectral density $2k_{\mathrm{B}}TR$, equivalent to $4k_{\mathrm{B}}T(R/2)$. One contribution alone is $k_{\mathrm{B}}TR$. The fictitious noiseless series $R$ in the Thévenin equivalent is not an additional physical source: adding its noise again would double count the original resistor.

MIT's laboratory guide mounts resistors on an aluminum box beneath a metal beaker shield, with short twisted leads. Its SR560 amplifier uses nominal gain $1000$ before a Krohn-Hite $1$–$50\mathrm{kHz}$ band-pass filter; an oscilloscope reads root mean square voltage. Calibration sweeps an attenuated sinusoid and measures the actual gain $g(f)$. Students alternate the resistor and a short circuit, subtracting squared readings $V_R^2-V_S^2$, and disconnect the ohmmeter during noise acquisition. Lead and input capacitance enter the gain-squared integral. With settings unchanged, varying electrical resistance or temperature tests the predicted mean square. The guide prescribes repeated readings and warns against averaging noise waveforms to zero. This is a documented teaching protocol, not a supplied dataset of measured outcomes.^[ref]

The subtraction requires care: shorting the input changes the source electrical impedance. An amplifier's input current noise can therefore produce different voltage noise in the two configurations; a shorted baseline alone does not remove that contribution. Gain drift, interference and insufficient observation time also limit the inference. These are limitations of interpreting the measurement, not corrections that can be inferred from the nominal signal bandwidth alone. In an independent analysis one would compare the measured difference withwhere $C$ represents the measured shunt capacitance and $g$ excludes that separately modeled shunt. If calibration already includes the same shunt response, applying the denominator again counts the attenuation twice. The thermal noise worked example and both original figures illustrate ideal models; they claim no measured agreement or measured value of the Boltzmann constant.

Model an ideal resistor of constant electrical resistance $R>0$ at uniform absolute temperature $T$ by its Thévenin equivalent: a zero-mean random voltage source in series with a noiseless $R$. In the classical range its one-sided, open-circuit voltage power spectral density, per hertz, iswhere $k_{\mathrm{B}}$ is the Boltzmann constant. Thus a positive-frequency interval of width $df$ contributes $4k_{\mathrm{B}}TRdf$ to the mean square voltage. This is approximately white noise only where the electrical resistance is effectively constant and $hf\ll k_{\mathrm{B}}T$, with $h$ the Planck constant. The ideal constant spectrum cannot describe arbitrarily high frequencies.^[ref]

For a real voltage, an equivalent two-sided power spectral density is $S_v^{(2)}(f)=2k_{\mathrm{B}}TR$ over signed frequencies. With a real linear time-invariant system of frequency response $H(f)$, the same output mean square isAngle brackets denote an expected value; for this zero-mean noise the mean square is its variance. Do not integrate $4k_{\mathrm{B}}TR$ over both signs of frequency: that counts the same fluctuations twice. All integrals here use frequency $f$ in hertz, not angular frequency $\omega =2\pi f$. The amplitude spectral density is $\sqrt{4k_{\mathrm{B}}TR}$ in $\mathrm{V}/\sqrt{\mathrm{Hz}}$, whereas the root mean square voltage requires a specified measurement signal bandwidth.

For one nondegenerate thermalized particle, excited-state occupation relative to the ground state is set by the excitation gap, not by the absolute energy origin. In an infinite square well of width $L$, the first gap is $3{\pi }^2{\hslash }^2/(2m{L}^2)$. Predominant ground-state occupation requires this gap to be much larger than $k_{B}T$; an explicit tolerated occupation gives a quantitative temperature criterion. State quantum degeneracy adds its statistical weight.