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The Maxwell-Boltzmann speed distribution

T-020Home CU-104Threads thermo
Statement

kinetic-theory-pressure found the mean-square molecular speed but not how speeds are actually distributed among individual molecules. Treating each velocity component as an independent, Boltzmann-weighted Gaussian and converting to spherical (speed) coordinates gives \(f(v)=4\pi n\left(\tfrac{m}{2\pi kT}\right)^{3/2}v^2e^{-mv^2/2kT}\), from which three distinct characteristic speeds emerge in the fixed ratio \(v_p:\langle v\rangle:v_{\text{rms}}=\sqrt2:\sqrt{8/\pi}:\sqrt3\approx1:1.128:1.225\) — the most probable speed \(v_p\), the mean speed \(\langle v\rangle\), and the root-mean-square speed \(v_{\text{rms}}\) (matching kinetic-theory-pressure's result exactly).

Why it matters

A single average speed hides the fact that at any instant, real gas molecules span a genuine range of speeds, and it is specifically the shape of that distribution's high-speed tail — not the average — that governs several important physical phenomena: evaporation (only the fastest surface liquid molecules escape), atmospheric escape (only the small fraction of light gas molecules exceeding a planet's escape velocity are lost to space over geological time), and, in reaction kinetics, the fraction of molecular collisions energetic enough to overcome an activation barrier (a topic developed fully in a later unit). None of these can be understood from \(v_{\text{rms}}\) or \(\langle v\rangle\) alone; the full distribution is required.

Hypotheses
Each velocity component (\(v_x,v_y,v_z\)) is independently distributed with probability density proportional to the Boltzmann factor \(e^{-KE/kT}\), applied to translational kinetic energy.This is a specific application of the general Boltzmann distribution of statistical mechanics — the probability of a system occupying a microstate of energy \(E\) at thermal equilibrium is proportional to \(e^{-E/kT}\) — here applied to a single molecule's translational kinetic energy \(KE=\tfrac12mv^2\); it is a foundational postulate of statistical mechanics rather than something derived within this result itself. The three velocity components are statistically independent of one another.Since kinetic energy separates additively as \(KE=\tfrac12m(v_x^2+v_y^2+v_z^2)\), the Boltzmann factor \(e^{-KE/kT}\) factors into a product of three exponentials, one per component — mathematically consistent with, and required for, treating \(v_x\), \(v_y\), \(v_z\) as independent random variables each following its own one-dimensional Gaussian distribution.
Proof
1
f(v_x) = \sqrt{\frac{m}{2\pi kT}}\,e^{-mv_x^2/2kT}
The normalised one-dimensional Gaussian (Boltzmann-weighted) distribution for a single velocity component, following directly from Hypotheses; the normalising prefactor ensures \(\int_{-\infty}^{\infty}f(v_x)\,dv_x=1\). A
2
f(v_x,v_y,v_z) = \left(\frac{m}{2\pi kT}\right)^{3/2}e^{-m(v_x^2+v_y^2+v_z^2)/2kT}
By independence (Hypotheses), the joint three-dimensional velocity distribution is the product of three identical one-dimensional Gaussians (Step 1), one per component. A
3
f(v)\,dv = f(v_x,v_y,v_z)\times4\pi v^2\,dv
To find the distribution of speed \(v=\sqrt{v_x^2+v_y^2+v_z^2}\) rather than the full velocity vector, integrate the (isotropic) joint density from Step 2 over all directions at fixed speed — equivalently, multiply by the surface area \(4\pi v^2\) of a thin spherical shell of radius \(v\), thickness \(dv\), in velocity space. This is why \(f(v)\) contains an explicit \(v^2\) factor even though the underlying Boltzmann factor alone is largest at \(v=0\): there is more "room" in velocity space (more directions available) at larger speed. A
4
f(v) = 4\pi n\left(\frac{m}{2\pi kT}\right)^{3/2}v^2e^{-mv^2/2kT}
Substituting Step 2 into Step 3 gives the Maxwell-Boltzmann speed distribution (per molecule, or scaled by \(n\) molecules for a number-density version), the central result of this derivation. A
5
\frac{d}{dv}\left[v^2e^{-mv^2/2kT}\right]=0 \ \Longrightarrow\ v_p=\sqrt{\frac{2kT}{m}}
Setting the derivative of \(f(v)\) to zero and solving locates the distribution's peak (the most probable speed, the single speed shared by the largest fraction of molecules). A
6
\langle v\rangle = \int_0^\infty vf(v)\,dv = \sqrt{\frac{8kT}{\pi m}}, \qquad v_{\text{rms}}=\sqrt{\langle v^2\rangle}=\sqrt{\frac{3kT}{m}}
The mean speed follows from a standard definite integral (evaluated using the Gaussian integral identities, omitted here for brevity); the root-mean-square speed reproduces exactly the value already found independently in kinetic-theory-pressure via the pressure argument — a genuine, non-trivial cross-check that the two derivations (pressure-based and distribution-based) are fully consistent with each other. A
Result
f(v)=4\pi n\left(\frac{m}{2\pi kT}\right)^{3/2}v^2e^{-mv^2/2kT}, \qquad v_p=\sqrt{\frac{2kT}{m}},\ \ \langle v\rangle=\sqrt{\frac{8kT}{\pi m}},\ \ v_{\text{rms}}=\sqrt{\frac{3kT}{m}}

Reading. A single distribution, built from independent Boltzmann-weighted velocity components converted to speed via spherical geometry, yields three genuinely distinct characteristic speeds — not one "average speed," but three different, precisely defined quantities in a fixed, temperature- and mass-independent ratio to one another.

Scope. As with kinetic-theory-pressure, this describes an ideal gas at thermal equilibrium (isotropic, no net flow); it also assumes the gas is dilute enough that intermolecular collisions redistribute energy without correlating individual molecules' velocities in a way that would break the independence assumption.

Corollaries & converses
  • The distribution broadens toward higher speeds as \(T\) increases (all three characteristic speeds scale as \(\sqrt{T}\)) and narrows toward \(v=0\) as \(T\) decreases, consistent with kinetic-theory-pressure's identification of \(T\) with average molecular kinetic energy.
  • At fixed \(T\), a heavier gas's distribution is narrower and shifted to lower speeds than a lighter gas's, exactly as the individual characteristic-speed formulas (each \(\propto1/\sqrt{m}\)) predict — the same mass dependence already established in kinetic-theory-pressure, now shown to apply to the full distribution's shape, not just its mean-square value.
  • Converse: since \(f(v)>0\) for every \(v>0\) (however small the probability), some (vanishingly rare) molecules in principle move at arbitrarily high speed at any positive temperature — it is specifically this small high-speed tail, not the bulk of the distribution near \(v_p\), that determines rates for high-energy-threshold processes such as evaporation, atmospheric escape, and (in later units) reactions with a significant activation energy.
Fails without
  • Drop the Gaussian/Boltzmann-factor form (Hypotheses), assume instead some other ad hoc distribution shape (e.g. a uniform distribution of speeds up to some cutoff): the resulting distribution would not match direct experimental measurements of molecular beam speed distributions (first attempted by Otto Stern in 1920, confirmed with high precision by Miller and Kusch in 1955), which closely follow the specific \(v^2e^{-mv^2/2kT}\) shape derived here — the Maxwell-Boltzmann form is not merely a convenient mathematical choice but a genuinely, independently experimentally verified prediction.
  • Forget the \(v^2\) geometric factor (Step 3), use only the raw Boltzmann factor \(e^{-mv^2/2kT}\) as if it were already the speed distribution: this incorrectly predicts the most probable speed is \(v=0\) (since the bare exponential is maximal there), directly contradicted by every observed and measured molecular speed distribution, which peaks at a nonzero speed \(v_p\) — the \(v^2\) factor, reflecting the growing "room" in velocity space at larger speed, is essential to getting even the qualitative shape right.
Common errors
  • Treating "average speed" as a single, unambiguous quantity, without specifying whether \(v_p\), \(\langle v\rangle\), or \(v_{\text{rms}}\) is meant — the three are numerically distinct (Result), and which one is physically relevant depends on the specific question being asked.
  • Assuming the distribution's peak (\(v_p\)) occurs at \(v=0\) rather than at a positive speed, from forgetting the \(v^2\) factor (Fails without, second bullet).
  • Assuming the fastest few molecules move at some fixed maximum speed; per the Converse, the Maxwell-Boltzmann distribution has no upper cutoff at all — only a rapidly (exponentially in \(v^2\)) decreasing probability at higher speed, never reaching exactly zero at any finite speed.
Discussion

James Clerk Maxwell first derived this distribution in 1860, using essentially the symmetry-and-independence argument presented here, predating by roughly a decade Ludwig Boltzmann's more general statistical-mechanical framework (developed through the 1870s), which later re-derived Maxwell's speed distribution as one specific application of a far more general principle applying to any form of energy, not just translational kinetic energy — hence the combined name "Maxwell-Boltzmann distribution."

Direct experimental confirmation of the full distribution's shape (not merely its average, which pressure measurements alone could already indirectly support via kinetic-theory-pressure) took considerably longer: Otto Stern's pioneering 1920 molecular beam experiment gave qualitative support, but it was not until Miller and Kusch's 1955 experiment, using a precisely rotating slotted-disk velocity selector, that the exact Maxwell-Boltzmann functional form was confirmed with high quantitative precision — a striking, roughly 95-year gap between Maxwell's original theoretical derivation and its most rigorous direct experimental test.

Common misconception: that all molecules in a gas sample move at the same speed (perhaps the \(v_{\text{rms}}\) value used in kinetic-theory-pressure's formulas). \(v_{\text{rms}}\), \(\langle v\rangle\), and \(v_p\) are each single representative numbers extracted from an underlying distribution that genuinely spans a wide, continuous range of individual molecular speeds at any given instant — the representative numbers are useful summaries, not a claim that every molecule shares that exact speed.

Worked examples
1
\text{O}_2\text{ at }298\,\text{K}\ (M=0.032\,\text{kg/mol}): \quad v_p\approx393.5\,\text{m/s}, \quad \langle v\rangle\approx444.0\,\text{m/s}, \quad v_{\text{rms}}\approx482.0\,\text{m/s}
All three characteristic speeds computed from the Result's formulas (using \(R\) and molar mass, the molar-constant pairing per kinetic-theory-pressure's Common errors caution); note \(v_p<\langle v\rangleA
2
\frac{\langle v\rangle}{v_p}=\sqrt{\frac{4}{\pi}}\approx1.128, \qquad \frac{v_{\text{rms}}}{v_p}=\sqrt{\frac{3}{2}}\approx1.225
Both ratios depend only on universal numerical factors (\(4/\pi\) and \(3/2\)), with all dependence on temperature and molecular mass cancelling out identically — the ratios hold for every gas at every temperature, verified directly against the Worked example 1 values: \(444.0/393.5\approx1.128\) and \(482.0/393.5\approx1.225\), exactly matching. A
\text{O}_2,\,298\,\text{K}:\ v_p\approx393.5,\ \langle v\rangle\approx444.0,\ v_{\text{rms}}\approx482.0\,\text{m/s} \quad (\text{ratio }1:1.128:1.225)

Reading. A single set of formulas produces three distinct, precisely related characteristic speeds from nothing more than a gas's molar mass and absolute temperature.

Scope. The universal ratio \(1:1.128:1.225\) holds for any ideal gas at any temperature, a direct consequence of every characteristic speed sharing the identical \(\sqrt{T/M}\) dependence.

Problems
  1. Compute \(v_p\), \(\langle v\rangle\), and \(v_{\text{rms}}\) for carbon dioxide (\(M=0.04401\,\text{kg/mol}\)) at \(350\,\text{K}\).
    Solution\(v_p=\sqrt{\dfrac{2(8.314)(350)}{0.04401}}\approx363.7\,\text{m/s}\). \(\langle v\rangle=v_p\times\sqrt{4/\pi}\approx363.7\times1.128\approx410.3\,\text{m/s}\). \(v_{\text{rms}}=v_p\times\sqrt{3/2}\approx363.7\times1.225\approx445.4\,\text{m/s}\).
  2. At what temperature does nitrogen's most probable speed \(v_p\) reach \(600\,\text{m/s}\)?
    SolutionSolving \(v_p^2=2RT/M\) for \(T\): \(T=\dfrac{v_p^2M}{2R}=\dfrac{(600)^2(0.028014)}{2(8.314)}\approx606\,\text{K}\).
  3. Explain, in terms of the derivation in Step 3, why \(f(0)=0\) exactly, even though the underlying Boltzmann factor \(e^{-mv^2/2kT}\) is maximal (equal to 1) at \(v=0\).
    SolutionThe full speed distribution \(f(v)\) is the product of the Boltzmann factor and the geometric factor \(4\pi v^2\) (Step 3), representing the "amount of velocity-space direction" available at each speed. At exactly \(v=0\), this geometric factor is itself exactly zero (a sphere of zero radius has zero surface area), so despite the Boltzmann factor being at its largest value there, the product \(f(0)=4\pi n\left(\tfrac{m}{2\pi kT}\right)^{3/2}(0)^2e^{0}=0\) exactly — the distribution genuinely vanishes at zero speed, consistent with real molecules always being in some state of motion at any positive temperature.
  4. A sample of argon gas is heated from \(200\,\text{K}\) to \(800\,\text{K}\) (a factor of 4 increase in absolute temperature). By what factor does \(v_{\text{rms}}\) increase?
    SolutionSince \(v_{\text{rms}}\propto\sqrt{T}\), a factor-of-4 increase in \(T\) produces a factor-of-\(\sqrt4=2\) increase in \(v_{\text{rms}}\) — not a factor of 4, since speed scales with the square root of temperature, not temperature directly, exactly as already illustrated in kinetic-theory-pressure's problem set.