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Which speed distribution does an arrival counter measure? Suppose an equilibrium, nonrelativistic ideal gas escapes by effusion through a thin opening into vacuum. The opening is small compared with the mean free path; the reservoir remains at temperature $T$. Let $f(v)$ denote its normalized Maxwell speed distribution, with mass $m$ and $a=m/(2k_{B}T)$:At a given direction toward the opening, the number crossing per unit time is proportional to the normal component $v\cos \theta$. Integrating over a fixed accepted cone contributes an angular factor independent of $v$. Thus the normalized probability density function for speeds among arriving particles isIts integral is one because ${\int }_0^{\infty }{v}^3{e}^{-a{v}^2}dv=1/(2a^2)$. The reservoir mode is $v=\sqrt{2k_{B}T/m}$; the arrival mode is $\sqrt{3k_{B}T/m}$. Faster particles get more opportunities to cross the opening. This is a selection effect, not heating after emission.

Specify the observable before applying the formula. An instantaneous count in a fixed segment of the beam weights the arrival flux by the time spent there, proportional to $1/v$; in this ideal model its normalized speed law returns to $f(v)$. A detector's speed-dependent efficiency, source collisions or a velocity selector changes the arriving law again. Counting arrivals and counting particles currently present are different measurements.

A gas thermometer infers temperature from the equation of state of a confined gas. The constant-volume version measures pressure; the constant-pressure version measures volume. For an ideal gas,Here $N$ is particle number, $n$ is amount of substance, $k_B$ is the Boltzmann constant, and $R$ is the molar gas constant. The reference and test states must be internally equilibrated, and $T_0$ must be known on the absolute thermodynamic temperature scale. A pressure ratio cancels the unknown gas inventory and bulb size in the ideal constant-volume limit. This cancellation is the method's appeal: a mechanical measurement transfers an absolute temperature scale between different materials. The Maxwell speed distribution and ideal gas pressure from wall collisions explain the microscopic origin of the response. ^[ref]

For an original ideal-model calculation, choose $T_0=300\mathrm{K}$, $P_0=120\mathrm{kPa}$ and $P=160\mathrm{kPa}$. Then $T=400\mathrm{K}$. These are stipulated values, not observations. Both pressures must be absolute, referred to vacuum. If the surrounding atmospheric pressure were $100\mathrm{kPa}$, the corresponding gauge pressures would be $20$ and $60\mathrm{kPa}$; their ratio is three, which would incorrectly predict $900\mathrm{K}$. A pressure difference from the atmosphere cannot be substituted into the ideal gas law.

A motivating calibration experiment. Schooley's account of the NBS/NIST work describes a helium-filled metal bulb, a capillary, a capacitance diaphragm gauge, and a mercury pressure standard. Circulating bath liquid passed the bulb assembly and platinum resistance thermometers. The resistance probes checked the surrounding temperature; gas pressure supplied the independent thermodynamic comparison. In an early series, workers filled the cleaned bulb with purified helium at the highest test temperature, checked pressure over three hours, then measured at lower temperatures and near a reference at $0{}^{\circ }\mathrm{C}$. Six filling pressures allowed extrapolation toward zero pressure. ^[ref]

To turn this principle into a readable measurement protocol, use the following controls. These steps are a proposed teaching interpretation of the cited apparatus, rather than additional reported NBS observations:

The corresponding NBS apparatus used a protective case with an outer helium layer held close to the bulb pressure, reducing pressure deformation while aiding thermal contact. Thermal expansion was measured on bulb material. Gas in the capillary remained partly outside the hot region, requiring a measured tube temperature profile and geometry. Pure working gas also mattered: contaminants that sorb on surfaces can change the freely circulating inventory. These are physical effects to measure or bound, rather than scatter that disappears by averaging. ^[ref]

For illustration, neglect connecting-tube gas and describe a uniform bulb with a dilute virial expansion in particle number density ${\rho }_N=N/V$:Here $Z=PV/(N{k}_{B}T)$ and $B_2$ is the second virial coefficient in the particle-density convention; a molar-density convention rescales the coefficient. The second relation follows exactly from the definition of $Z$ for the same $N$, and shows the signs of the corrections. If the bulb expands by $0.1\%$ and $Z=1$ at both states, the toy pressure ratio above gives $400.4\mathrm{K}$, rather than $400\mathrm{K}$. Improved expansion and helium nonideality models really changed the temperatures inferred from the 1989 NIST data: Harvey's 2023 reanalysis found expansion updates of roughly $1$–$3\mathrm{mK}$ and nonideality updates of roughly $-0.1$ to $-0.5\mathrm{mK}$. Those are reported corrections to historical results, not the much larger illustrative correction just calculated. ^[ref]

With equal corrected bulb volumes, negligible or corrected connecting-tube gas, fixed reference and test temperatures, and a dilute classical single gas phase,Repeat the reference/test comparison for several sealed fills, reducing $N$ between comparisons but keeping it fixed within each pair. The extrapolation varies density at fixed endpoint temperatures; it does not cool one fill toward zero temperature. Real gases have different finite-density corrections but share the classical ideal gas intercept under these assumptions. The limit does not remove leaks, bulb expansion, gauge offsets, or the gas thermometer dead-space correction. At very low pressures, thermal transpiration can make remote sensor and bulb pressures differ through a nonisothermal capillary; this must be corrected too. The NBS work measured this pressure correction using a separate comparison of narrow and wider tubes. ^[ref]

At $x=1$, the toy models give $T_{\mathrm{app}}=407.92\mathrm{K}$ and $388.12\mathrm{K}$, respectively, although their imposed true test temperature is the same. Their zero-density intercept is the reusable quantity. Measuring several low-density points permits a fit and a check for curvature; asserting a straight line from one finite-density point supplies neither. A practical fit must balance nonideality against increasing relative pressure error and the capillary corrections, and keep its extrapolation uncertainty.

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]

In the gas rest frame, the Maxwell-Boltzmann velocity distribution is the probability density function per unit volume in velocity space:Thus $f(\mathbf{v})d^3\mathbf{v}$ gives the probability of a small three-dimensional velocity region. For a gas with bulk mean velocity $\mathbf{V}$, replace $\mathbf{v}$ by $\mathbf{v}-\mathbf{V}$; the Maxwell speed distribution then describes the magnitude relative to that bulk motion.

For any one Cartesian velocity component $v_x$, integrating out the other two givesThe factorization $f(\mathbf{v})=q(v_x)q(v_y)q(v_z)$ establishes independence: all three components are centered normal random variables, with variance $s^2$. Both the component density $q$ and the vector density $f$ have their largest values at zero. This concerns densities for small equal-sized regions; any single exact velocity still has probability zero. ^[ref]

The standard deviation $s$ measures thermal motion even though the mean velocity is zero. Positive and negative components cancel in the mean, but their squares add. This is why thermal Doppler broadening measures a nonzero line width in a gas with no bulk motion.

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.

A mode is a value at which a probability mass function or specified probability density function attains its largest value. There may be more than one. For the normal distribution the unique mode equals the mean; the Maxwell speed distribution has mode $\sqrt{2k_{B}T/m}$ and a larger mean speed. For a continuous probability distribution, a mode maximizes density rather than the probability of an exact value, which is zero. Density modes can change under a nonlinear change of variables, as the distinction between Maxwell-Boltzmann velocity distribution and Maxwell speed distribution illustrates.

The most probable speed is the mode, or density maximum, of the Maxwell speed distribution:For small equal-width speed bins, the bin near $v_{\mathrm{mp}}$ contains the largest expected fraction of particles. This does not mean that any exact speed has positive probability, or that every particle has this speed. The three common summaries obeyso $v_{\mathrm{mp}}<⟨v⟩<v_{\mathrm{rms}}$. The first describes a density peak, the second the mean speed, and the third the root mean square associated with kinetic energy. ^[ref]

For a concrete scale $s=300\mathrm{m}{\mathrm{s}}^{-1}$, these are approximately $424$, $479$, and $520\mathrm{m}{\mathrm{s}}^{-1}$. The most probable velocity vector and each most probable component remain zero in the gas rest frame. A zero vector labels one small region of velocity space, whereas a positive speed labels an entire spherical shell.

A short check uses the derivative of the logarithm of the Maxwell speed distribution for $v>0$:It is positive below $\sqrt{2}s$ and negative above it, locating the unique peak. The factor $v^2$ from the spherical shell supplies the term $2/v$; omitting it would incorrectly move the speed peak to zero.

In the canonical ensemble, a one-particle velocity state has kinetic energy $E(\mathbf{v})=m|\mathbf{v}{|}^2/2$ and hence probability density proportional to $e^{-\beta E}$, where $\beta =(k_{B}T{)}^{-1}$. The three Cartesian Gaussian integrals normalize the density to the Maxwell-Boltzmann velocity distributionIntegrating over a spherical shell of radius $v$ and area $4\pi v^2$ gives the Maxwell speed distributionSinceits unique positive maximum is the most probable speed in the Maxwell distribution

The Maxwell speed distribution is proportional to $v^2{e}^{-m{v}^2/(2k_{B}T)}$. Setting its logarithmic derivative to zero gives the most probable speed:Use molecular oxygen, with $M\simeq 0.032\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, and $T=300\mathrm{K}$.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

Let the gas have a small bulk velocity $u_x(y)$, an approximately isotropic local thermal velocity distribution, and a mean free path $\lambda$ short compared with the flow-gradient scale. A molecule transports the mean streamwise momentum acquired near its previous collision. To first order, its nonequilibrium momentum relative to the destination layer is approximately $-m{\tau }_c{v}_{y}d{u}_x/dy$. The resulting transverse momentum flux isThus the dynamic viscosity is $\eta \simeq \rho {\tau }_c⟨v_y^2⟩$. With isotropy, a representative collision time ${\tau }_c\sim \lambda /⟨v⟩$, and a chosen thermal speed average, this has the formThe dimensionless $K$ depends on how speeds and collision times are averaged; a simple one-speed estimate gives $K=1/3$. Assumptions include dilute binary collisions, a locally equilibrated distribution, weak shear, and negligible boundary slip. The kinetic theory identity $p=\rho ⟨v^2⟩/3$ then givesFor the Maxwell speed distribution, $⟨v^2⟩/⟨v{⟩}^2=3\pi /8$, soThis explains both the units of inverse time and the relation to molecular collision frequency, up to the averaging convention.

In the cylinder apparatus, distinguish the vector mean molecular velocity, $⟨\mathbf{v}⟩=\mathbf{u}$, from the mean thermal speed $⟨|\mathbf{v}-\mathbf{u}|⟩$. For a slow nearly isothermal Couette flow, the thermal speed is approximately position-independent, whereas the bulk mean is azimuthal and changes from the local rotating-wall speed to zero at the fixed wall. In the side gap, with $y$ measured radially from the inner wall,Across either end gap, at cylindrical radius $s$, the corresponding profile is $u_{\phi }(s,z)\simeq \omega s(1-z/h)$, with $z$ measured away from the moving end face. The no-slip boundary condition and small $h/R,h/H$ underlie these linear profiles.

The side-wall shear stress has magnitude $\eta \omega R/h$, producing resisting torque $2\pi \eta \omega R^{3}H/h$. The original diagram also has a gap at both circular end faces. At either face, the stress is $\eta \omega s/h$ and its ring contribution isAdding the side and both ends gives the end-face torque in narrow-gap cylindrical viscometry:The motor supplies the opposite of this viscous resisting torque. Corner flow and the small axle area are neglected. Retaining only the cylindrical wall would omit the $\pi R^4/h$ term shown by the actual closed-ended geometry.

For hard-sphere Ar atoms of collision diameter $d_{\mathrm{Ar}}$, $\lambda =1/(\sqrt{2}n\pi d_{\mathrm{Ar}}^2)$ and $\rho =n{m}_{\mathrm{Ar}}$. ThereforeUsing $m_{\mathrm{Ar}}=40m_u$, $T=273.15\mathrm{K}$ and the simple estimate $K=1/3$ givesThis is a kinetic estimate with a model-dependent numerical prefactor, rather than a precise atomic size measurement. Since the density cancels and the hard-sphere collision diameter is taken constant, the prediction is $\eta \propto \sqrt{T}$. The collision cross-section of a real atom can vary with speed, so its exact temperature dependence need not follow this power law.

For isotropic three-dimensional diffusion with coefficient $D$, the vector mean displacement is zero, the radial RMS is $\sqrt{6Dt}$ and the mean radial magnitude is $4\sqrt{Dt/\pi }$. The radial distribution is the Maxwell speed distribution arising from three independent Gaussian coordinate increments. State the statistic before calling a quantity an average distance; it is not the cumulative path length.

A speed distribution is the probability distribution of the magnitude of velocity. When it has a probability density function $f$, the fraction of particles with speeds between $v$ and $v+dv$ is $f(v)dv$, and ${\int }_0^{\infty }f(v)dv=1$. The Maxwell speed distribution describes an equilibrium ideal gas; the effusive beam speed distribution weights that population by its rate of arrival.