Kinetic theory and molecular velocity distributions

A hole is a selective counter

A gas reservoir is held at 300 K. A tiny hole opens into a vacuum, and a detector counts the molecules that escape. Does the detector sample the same speed distribution as a snapshot of the gas inside?

It does not. A faster molecule can reach the opening from farther away during the same short time interval. Crossing events therefore give extra weight to speed. We will calculate that bias after first distinguishing a velocity component, a speed and a probability density.

The example uses an assigned classical particle mass, rather than a measured fit to a named gas. The reservoir is large enough, and replenished or observed briefly enough, that its temperature and number density can be treated as unchanged. The aperture is thin and much smaller than the mean free path, a condition we check below.

Choose a reading path

For a first reading, follow the reservoir through velocity components, speeds, pressure and the selective opening. The worked examples explain normalization, a probability interval and the escaping count rate. Algebra and logarithms, probability, and momentum conservation are useful preparation; the essential meanings are defined here.

The return path treats drift and a pressure tensor, nonisotropic preparations, collision-conditioned sampling and arrival-time measurements. These extensions explain why the measurement procedure and reference frame must accompany a velocity histogram.

This lesson studies distributions and their mechanical moments. The classical-ideal-gases lesson develops the state-counting and thermodynamic potentials; the Boltzmann-equation lesson develops the actual evolution and transport closure. An equilibrium distribution is not itself an account of how equilibrium was reached.

Name the sample and the density

A velocity is a vector v = (vₓ, vᵧ, v_z). Its components can be positive or negative. Speed v = |v| is a nonnegative magnitude. A resting gas can have mean velocity zero while its mean speed is large: opposite directions cancel as vectors, not as distances travelled.

Use F(v) for a normalized density in three-dimensional velocity space; its argument is the full vector, not only the speed. Then F(v) dvₓ dvᵧ dv_z is the probability of a small velocity cell, and the integral of F over all velocities is one. Multiplying by number density n gives a particle-number density in position and velocity space. Keep those two normalizations separate.

Use fₓ(vₓ) for the signed-component density and g(v) for the speed density. Both have units (m/s)⁻¹ = s/m, but their domains and shapes differ. Probability is an area under a density over an interval; the height at one exact speed is not the probability of that speed. For a continuous distribution, any single exact speed has zero probability.

Obtain the equilibrium components

Assume nonrelativistic particles in classical thermal equilibrium, with a quadratic translational kinetic energy and no velocity-dependent coupling that changes that energy. In the gas rest frame, the equilibrium momentum weight is proportional to exp[−m(vₓ²+vᵧ²+v_z²)/(2k_B T)]. Here k_B = 1.380649 × 10⁻²³ J/K is the Boltzmann constant.

The exponent is a sum of three separate quadratic terms. The weight therefore factors into three Gaussian component weights. Normalizing each with the Gaussian integral gives fₓ(vₓ) = [m/(2πk_B T)]¹ᐟ² exp[−mvₓ²/(2k_B T)]. Its mean is zero and its variance is k_B T/m. The other two components have the same law and are independent under these equilibrium assumptions.

The full density is F(v) = [m/(2πk_B T)]³ᐟ² exp[−mv²/(2k_B T)]. It has units (s/m)³. Rotational symmetry alone would not establish independence or this particular radial law: many non-Gaussian distributions are rotationally symmetric. The factorized equilibrium kinetic-energy weight supplies the additional physical information.

For the assigned mass at 300 K, the component standard deviation is about 298.453 m/s. This width describes the spread of one signed component; it is not the most probable three-dimensional speed.

Count the velocity shell

To find the probability of a speed between v and v + dv, collect every velocity vector whose tip lies in that thin spherical shell. The shell volume in velocity space is 4πv² dv. Because the resting equilibrium density is isotropic, g(v) = 4πv²F(v) for v ≥ 0.

Thus g(v) = 4π[m/(2πk_B T)]³ᐟ² v² exp[−mv²/(2k_B T)]. The integral from zero to infinity is one. The factor v² counts available directions and component combinations at that radius; it is not a new force accelerating slow particles.

At v = 0 the speed density vanishes, although the vector density and each component density are largest at their respective zeros. There are very few velocity-space states in an infinitesimal shell near the origin. The distribution of |vₓ| is instead a folded one-dimensional Gaussian and is not g(v).

The component and speed plots use their own labelled densities. Each plotted curve is part of a normalized distribution on its full infinite domain. The finite figure window does not redefine that normalization.

Three questions, three speeds

The most probable speed asks where g has its maximum. Differentiating ln g for v > 0 gives 2/v − mv/(k_B T) = 0, so v_mp = √(2k_B T/m). This is the mode of the speed density, not an assertion that most particles share one exact speed.

The mean speed weights every speed by its probability: ⟨v⟩ = ∫₀∞ v g(v) dv = √[8k_B T/(πm)]. The root-mean-square speed is v_rms = √⟨v²⟩ = √(3k_B T/m). It weights the high-speed tail more strongly before taking the square root. These are three different questions, with v_mp < ⟨v⟩ < v_rms.

All three scale as √(T/m). Doubling temperature multiplies their values by √2. Reducing particle mass to one quarter at the same temperature doubles them. The probability density becomes broader and lower as its area stays one; a lower peak is not evidence that particles have disappeared.

Use areas and tail probabilities

For a speed interval [v₁, v₂], compute P(v₁ ≤ v ≤ v₂) = ∫ᵥ₁ᵛ² g(v) dv. The two endpoints do not affect the probability because the distribution is continuous. A density times a bin width is only a small-bin approximation, not an exact interval calculation.

With a = k_B T/m and z = v/√(2a), the cumulative probability is G(v) = erf(z) − 2z exp(−z²)/√π. The error function is defined by erf(z) = (2/√π)∫₀ᶻ exp(−s²) ds. A cumulative table or a numerical integral can be used without knowing a closed form.

For the base reservoir, the median speed is about 459.072 m/s and the 90th percentile is 746.215 m/s. About 1.05616% of the molecules have speeds above 1000 m/s. A typical speed is therefore not a hard upper limit.

The plots end at 2000 m/s. The omitted upper-tail probabilities are about 9.6863 × 10⁻¹⁰ for the base gas, 5.2495 × 10⁻⁵ for the warmer gas and 0.0105616 for the quarter-mass gas. The last is visible in an area calculation even though its curve has fallen substantially. Do not renormalize only the displayed part.

Pressure is a momentum flux

Consider a molecule reflecting elastically from a flat stationary wall normal to x. Its x-momentum changes sign, so the momentum-change magnitude is 2m|vₓ|; an equal and opposite impulse is transferred to the wall. The molecules arriving in a short interval occupy a slab whose depth is proportional to their incoming normal speed. Multiplying momentum per event by the arrival rate leads to pₓ = nm⟨vₓ²⟩ for a symmetric resting distribution.

For the equilibrium Gaussian, ⟨vₓ²⟩ = k_B T/m and p = nk_B T. Isotropy makes the same normal pressure apply to every wall orientation. Equivalently p = nm⟨v²⟩/3. The factor one third belongs to the isotropic second moment, not to the mean velocity.

Our base reservoir has p = 4.141947 Pa and mean translational energy per particle 3k_B T/2 ≈ 6.21292 × 10⁻²¹ J. Zero mean velocity does not eliminate either pressure or kinetic energy. Internal rotational or vibrational energy is a separate contribution when a molecular model includes it; the translational pressure calculation does not count it as an extra velocity component.

Count one-sided crossings

Let θ be the angle between a molecule’s velocity and the outward aperture normal. During dt, a velocity class aimed toward a small area A can arrive from a volume A v cosθ dt. Only the outward hemisphere, 0 ≤ θ ≤ π/2, contributes. A molecule nearly parallel to the wall has a small normal reach even if its total speed is large.

For an isotropic resting gas, integrating that geometrical factor over directions and speeds gives the one-sided number flux J = n⟨v⟩/4 = n√[k_B T/(2πm)]. Its units are particles per square metre per second. The factor one quarter comes from the hemisphere and normal projection; arbitrarily assigning one sixth of all particles to a Cartesian direction would not perform this angular average.

If the aperture is thin, collision-free during passage, and weak enough not to disturb the reservoir, the escaping count rate is R = JA. A circular diameter D has A = πD²/4. For the supplied 10 µm opening, A ≈ 7.85398 × 10⁻¹¹ m², J ≈ 1.19065 × 10²³ m⁻² s⁻¹ and R ≈ 9.35138 × 10¹² s⁻¹.

The reservoir initially contains nV = 10¹⁸ particles, so R/(nV) ≈ 9.35138 × 10⁻⁶ s⁻¹. This supports a small initial fractional-loss approximation over a suitably short observation; it does not prove constant conditions for unlimited time. A freely emptying or cooling vessel requires coupled number and energy balances. Our maintained-reservoir model does not describe that evolution.

The aperture biases speed and direction

A crossing counter does not sample molecules uniformly from a spatial snapshot. In an isotropic gas, the extra factor of speed in the arrival volume gives the normalized escaping-event speed density h(v) = v g(v)/⟨v⟩. Written with a = k_B T/m, h(v) = v³ exp[−v²/(2a)]/(2a²). Its integral is one.

The most probable escaping speed is √(3a), the escaping mean speed is 3a/⟨v⟩, and the escaping RMS speed is 2√a. The mean translational energy of an escaping particle is 2k_B T, rather than the reservoir value 3k_B T/2. This is selection by the crossing process, not extra acceleration or a hotter equilibrium reservoir.

The angular probability per unit solid angle is cosθ/π over the outward hemisphere. After integrating over azimuth, the density with respect to θ is 2 sinθ cosθ, with θ measured in radians. Its probability within a cone of half-angle θ₀ is sin²θ₀. Thus half the escaping events lie within 45° of the normal, although the directional density per solid angle is largest on the normal.

These laws refer to counted crossing events. A stationary beam’s local particle density weights slow particles differently because they reside longer in a region. Detector efficiency, angular acceptance, a finite channel and collisions can introduce further selection. State the experiment before identifying a histogram with g or h.

Relative motion controls collisions

Now assign a hard-sphere collision diameter d = 3.70 × 10⁻¹⁰ m. Two identical particles collide if their centres pass within distance d. The collision cross-section is σ = πd², four times the projected disc area of one sphere of radius d/2. The collision diameter d is not the aperture diameter D.

For independently sampled identical Maxwellian velocities at the same temperature, the relative velocity v − v′ has twice the variance of a single velocity in each component. Its mean magnitude is ⟨v_rel⟩ = √2⟨v⟩. Its RMS magnitude is √(6k_B T/m), a different statistic. Substituting the RMS value into a rate that needs the mean would change the answer.

With constant hard-sphere σ, the mean collision frequency per particle is ν̄ = nσ⟨v_rel⟩. The total number of pair-collision events per volume per time is nν̄/2; each event otherwise gets counted once for each participant. The ratio of total molecular distance travelled to participant-collision counts gives ℓ = ⟨v⟩/ν̄ = 1/(√2 nπd²).

For the assigned model, ℓ ≈ 1.64411 mm and ℓ/D ≈ 164.411. The aperture is much smaller than this mean free path, while it is much larger than d. Along with the thin-wall and negligible-disturbance assumptions, these comparisons support the elementary effusion model.

At fixed n and d, this hard-sphere ℓ is independent of temperature, while ν̄ grows as √T. At fixed pressure, n = p/(k_B T) changes with temperature, so ℓ instead grows with T in this same model. Interactions with an energy-dependent scattering cross-section require the corresponding relative-speed average.

Further: separate drift from thermal motion

If the gas moves with mean velocity u, use the peculiar velocity w = v − u to describe thermal motion. A local Maxwellian is Gaussian in w, not generally in the laboratory components around zero. Its covariance is ⟨wᵢwⱼ⟩ = (k_B T/m)δᵢⱼ. Here δᵢⱼ is one when i = j and zero otherwise.

The kinetic pressure tensor is Pᵢⱼ = nm⟨wᵢwⱼ⟩. It describes thermal momentum transport after the mean flow is removed. The laboratory momentum-flux tensor is nm⟨vᵢvⱼ⟩ = Pᵢⱼ + nm uᵢuⱼ. Indices i and j select a momentum component and a direction of transport; both expressions have units Pa.

Adding the assigned uniform drift u = (120, 0, 0) m/s leaves the thermal pressure at 4.141947 Pa. The convective xx contribution is 0.669600 Pa, giving a laboratory xx momentum flux of 4.811547 Pa. It does not raise the thermodynamic temperature merely because another inertial frame measures more kinetic energy.

The zero-drift isotropic aperture formula cannot simply be reused for every moving gas and wall orientation. The correct crossing flux is the integral of the positive wall-relative normal velocity times the actual distribution. Local equilibrium also does not make a spatially varying Maxwellian an exact solution of every transport problem.

Further: an average energy does not fix the distribution

Consider an assigned instantaneous preparation with independent Gaussian components whose variance scales correspond to Tₓ = 600 K and Tᵧ = T_z = 150 K. These are directional kinetic scales, defined by m⟨wᵢ²⟩ = k_B Tᵢ. They are not three thermometers establishing thermal equilibrium.

The average translational energy is k_B(Tₓ+Tᵧ+T_z)/2 = 3k_B(300 K)/2, equal to the base equilibrium value. Yet Pₓₓ is twice the base pressure and Pᵧᵧ and P_zz are each half of it. Its speed law is not the isotropic Maxwell speed law at 300 K.

This preparation makes the logical distinction concrete: a second-moment average does not identify a full distribution. In a freely colliding gas, collisions generally change an anisotropic preparation. We have specified a distribution for a mechanical moment comparison, not a stationary solution or a universal relaxation rate.

Further: a collision-conditioned sample

The mean ν̄ does not mean that every velocity has the same instantaneous collision rate. In the dilute, uncorrelated, homogeneous hard-sphere approximation, a particle currently moving with v has ν(v) = nσ∫ F(v′)|v − v′| d³v′. Even a particle with v = 0 can be struck by moving neighbours.

For a particle maintaining that velocity until its next event in a stationary independent background, a Poisson collision model gives conditional survival exp[−ν(v)t]. Sampling particles at an arbitrary time and averaging over their velocities gives a mixture of these exponentials. It is generally not exp[−ν̄t]. The approximation and the sampling procedure matter.

A collision-event sample weights the velocity distribution by ν(v), whereas a snapshot weights it by F(v). Correspondingly, 1/ν̄ is an aggregate mean interval between participant events in a stationary ensemble. It should not be silently identified with every arbitrary-time residual waiting average. The mean free path above is the ratio of the total distance rate to the participant-event rate.

Detailed collision dynamics, correlations, nonuniform backgrounds and energy-dependent cross-sections require a more complete kinetic description. These limitations are a reason to choose the appropriate Boltzmann-equation or transport model, not to reinterpret every spread in a speed histogram as thermal equilibrium.

Further: arrival time needs a flight model

For a clean change-of-variables example, release an ideal instantaneous effusive pulse into vacuum and treat the tiny aperture as a point source. Imagine a spherical detecting surface of radius L = 0.500 m centred on that point, collecting the outgoing hemisphere with speed-independent per-particle efficiency. Each collision-free radial ray then has the same actual path length L. The source pulse has escaping-event speed density h(v).

For this particular geometry, t = L/v. Conservation of probability gives q(t) dt = h(v)|dv|, so q(t) = h(L/t)L/t² for t > 0. The absolute Jacobian L/t² is essential. q has units s⁻¹, and its integral is one. A flat detector a normal distance L away would instead have t = L/(v cosθ), so a finite angular acceptance would have to be included.

Here h is proportional to v³ exp[−v²/(2a)], giving q proportional to t⁻⁵ exp[−L²/(2at²)]. Its peak is at L/√(5a), not L divided by the peak speed of h. Changing variables changes density heights and their maxima; it preserves interval probabilities.

The mean arrival time is L⟨1/v⟩_escaping = L/⟨v⟩_reservoir, not generally L/⟨v⟩_escaping. A finite source-pulse duration, speed-dependent detection, collisions, gravity when significant, or a different flight geometry requires modifying this model. The plotted time window stops at 5 ms and omits a probability of about 0.00151772 at later times.

Further: check the regime before the formula

The model needs more than a number called temperature. The gas must be sufficiently dilute for the chosen collision approximation, sufficiently nondegenerate for classical statistics, and nonrelativistic for mv²/2 to apply. Its distribution must have the equilibrium or explicitly prepared form used in the calculation.

For an indicative quantum-statistical check, the thermal wavelength is λ_th = h/√(2πmk_B T), using h = 6.62607015 × 10⁻³⁴ J s. For a simple species with no additional unresolved internal degeneracy, nλ_th³ ≪ 1 is a classical dilute-occupation condition. It does not by itself establish negligible interactions or thermal equilibrium.

The separate hard-sphere dilution parameter nd³ is about 5.0653 × 10⁻⁸ here. The thermal-to-rest-energy ratio k_B T/(mc_SI²) is about 9.91083 × 10⁻¹³, where c_SI = 299792458 m/s is the speed of light. These checks concern different approximations and must not be substituted for one another.

Effusion additionally needs an opening and passage length small relative to the appropriate mean free path, with negligible backflow and an adequately maintained reservoir. A long narrow tube, a dense jet, an interacting fluid, a quantum-degenerate gas or a relativistically hot gas belongs to a different regime. The adjacent lessons provide the corresponding models.

Measure the distribution you actually sample

A useful kinetic calculation starts by naming the population: molecules in a volume, events crossing an area, collision participants, or arrivals in a time interval. It then specifies a rest frame, a distribution and the physical weighting of the measurement. Normalization and units help catch a mistaken identification before any large calculation.

Molecular beams, vacuum instruments and gas diagnostics make these distinctions practical. A count rate can test a reservoir-and-aperture model, while a resolved speed or arrival-time histogram tests additional assumptions about selection and propagation. Agreement with one moment, such as pressure, does not prove the whole distribution.

Continue to quantum statistics for population restrictions outside the classical regime, to fluctuations and response for relationships between distributions and measurable variations, and to transport theory and the Boltzmann equation for collisions and nonequilibrium evolution. Keep the present equilibrium, sampling and geometry assumptions with every formula used.

Worked examples

Try it yourself

1. A resting equilibrium gas has ⟨vₓ⟩ = 0. What should you calculate to predict its pressure on an x-normal wall?

   Show the answer and reasoning

   The second moment ⟨vₓ²⟩, giving pₓ = nm⟨vₓ²⟩.

   Opposite velocities cancel in the first moment, but their squared normal components and momentum-transfer contributions are positive.

2. Which density should be integrated to find P(−100 < vₓ < 100 m/s)?

   Show the answer and reasoning

   The signed-component density fₓ over that signed interval.

   g describes the nonnegative magnitude of the full vector. It cannot represent a negative Cartesian component.

3. Explain why the resting speed density is zero at v = 0 although F is largest there.

   Show the answer and reasoning

   The radial shell contributes a factor 4πv².

   A small-radius shell contains very little velocity-space volume. The density per three-dimensional cell and the density per speed interval have different measures.

4. If g is in s/m and a small bin is 2 m/s wide, what are the units of gΔv?

   Show the answer and reasoning

   Dimensionless probability.

   The units cancel. Approximating the bin probability by one height times width also requires that g varies little across that bin.

5. Using the interval table, estimate the fraction above 300 m/s.

   Show the answer and reasoning

   About 0.798736842, or 79.8737%.

   Add the 300–600 and above-600 probabilities, or subtract the below-300 probability from one. Do not read a single curve height.

6. Reduce the particle mass to m/4 at fixed 300 K. Find the new mean speed from the base mean.

   Show the answer and reasoning

   About 952.524 m/s.

   The characteristic scale is proportional to 1/√m, so the factor is two, not four. The full density broadens while retaining unit area.

7. For the base reservoir, put mode, median, mean and RMS speed in increasing order.

   Show the answer and reasoning

   422.076 < 459.072 < 476.262 < 516.936 m/s.

   Each answers a different distribution question. The high-speed tail affects the mean and especially the second moment.

8. If the aperture diameter is halved with all reservoir conditions fixed and the same passage assumptions valid, how does the count rate change?

   Show the answer and reasoning

   It becomes one quarter of the original rate.

   A = πD²/4, whereas J is unchanged. Changing the opening may also change the validity of a real finite-thickness passage model.

9. What fraction of escaping events lie within 30° of the outward normal?

   Show the answer and reasoning

   One quarter.

   The cone probability is sin²θ₀. At θ₀ = 30°, sinθ₀ = 1/2. This is not a uniform distribution in polar angle.

10. Write a normalized speed density for counted escaping events from the supplied isotropic reservoir.

    Show the answer and reasoning

    h(v) = vg(v)/⟨v⟩.

    The extra speed factor comes from the reach of the crossing slab per time. Its normalizer is the reservoir mean speed, so integrating h gives one.

11. What extra balances are needed to predict an unheated finite vessel as it freely effuses for a long time?

    Show the answer and reasoning

    At least coupled particle-number and energy balances, with the evolving reservoir state.

    Escaping particles carry a biased energy sample. The fixed n,T initial or maintained-reservoir count formula alone does not determine the later cooling and depletion.

12. For the assigned identical hard spheres, compare collision cross-section with one sphere’s projected area.

    Show the answer and reasoning

    The collision cross-section is four times the single-particle projected area.

    Collision of centres permits impact parameters up to d, while the physical sphere radius is d/2. The two areas are πd² and πd²/4.

13. Which of 673.536 and 731.057 m/s belongs in ν̄ = nσ⟨v_rel⟩ for the base model?

    Show the answer and reasoning

    673.536 m/s.

    It is the mean relative speed. The larger value is the RMS relative speed; a second moment does not substitute for the first moment in this constant-cross-section rate.

14. Why does the pair-event density rate have a factor one half?

    Show the answer and reasoning

    A collision has two participants.

    Multiplying the per-particle participant rate by particle density counts each identical pair event twice. Divide by two to count events once.

15. Double T at fixed pressure and fixed hard-sphere diameter. What happens to ℓ in this model?

    Show the answer and reasoning

    It doubles.

    The ideal number density halves because n = p/(k_B T), and ℓ is proportional to 1/n. Temperature independence of ℓ required fixed density, not fixed pressure.

16. Can a molecule instantaneously at rest still collide with neighbours? Give the hard-sphere conditional-rate expression at v = 0.

    Show the answer and reasoning

    Yes; ν(0) = nσ⟨v⟩ for the assigned Maxwellian background.

    The relevant speed is the relative speed to a moving partner. Zero laboratory velocity of one particle does not eliminate the partner’s motion.

17. A frame change increases the measured mean squared laboratory speed. Must the gas temperature increase?

    Show the answer and reasoning

    No; remove the mean drift before computing the thermal covariance.

    Thermal energy uses w = v − u. Laboratory momentum and energy include convective terms that depend on frame.

18. Propose a mechanical measurement that distinguishes the two same-energy preparations in the tensor table.

    Show the answer and reasoning

    Compare normal momentum fluxes on differently oriented surfaces, with mean flow removed.

    The isotropic preparation has equal diagonal pressure components. The assigned anisotropic preparation has a larger xx and smaller yy,zz even though its trace agrees.

19. For a flat detector a normal distance L from an uncollimated aperture, what relation replaces t = L/v?

    Show the answer and reasoning

    t = L/(v cosθ) for the straight rays that reach the detector.

    The actual path is longer at nonzero angle. Its angular acceptance and detection weighting must be included; the spherical-detector scalar mapping cannot be reused silently.

20. The arrival plot ends at 5 ms. Should the displayed q(t) be scaled upward until its plotted area is one?

    Show the answer and reasoning

    No.

    About 0.00151772 of the event probability lies later than 5 ms in this model. Retain the full normalized distribution and state the omitted tail.

What to carry forward

• An equilibrium velocity vector, one signed component and its nonnegative speed have different probability densities and measures.
• The speed law contains a radial shell factor; mode, mean, median and RMS speed answer different questions.
• Pressure depends on thermal second moments, so a zero mean velocity can coexist with finite pressure.
• Aperture crossing events weight speed and direction; their distribution differs from a snapshot or a stationary beam-density sample.
• Collision rates involve relative motion and the correct first moment; aggregate intervals, conditional hazards and pair-event counting are distinct.
• Use the stated frame, geometry, sampling law and physical regime before converting a histogram into a temperature, rate or arrival-time prediction.