The Maxwell–Boltzmann speed distribution

Derivation · the atom

The Maxwell–Boltzmann speed distribution

D-014 maxwell-boltzmann-speeds Home PU-203 Threads matter · chance Depends on D-013 verified

Statement

The distribution of molecular speeds in a classical gas at temperature T is the product of a Boltzmann factor and a geometric factor from the volume of a shell in velocity space.

Why it matters

The distribution is asymmetric with a long high-speed tail, and it is that tail — not the mean — that governs reaction rates, atmospheric escape and evaporation. The most likely velocity is zero, yet the most likely speed is not, and the whole distinction rests on one geometric factor.

Assumptions

Classical, non-degenerate gas.Requires the thermal de Broglie wavelength to be much smaller than the interparticle spacing. Fails for electrons in metals, where Fermi–Dirac statistics apply (PU-302), and for cold atomic gases.

Non-interacting particles.Energy is purely kinetic. Interactions modify the spatial distribution but leave the speed distribution unchanged in equilibrium — a result worth stating because it is surprising.

Isotropic velocity distribution.No bulk flow. In a moving gas the distribution is the same, shifted by the flow velocity.

Derivation

f(v) d³v ∝ e^{−mv²/2k_BT} d³v

Apply D-013 with E = ½mv². The distribution over the velocity vector is a pure Gaussian, peaked at v = 0 — the most likely velocity is zero. A

d³v = 4π v² dv

Convert to speed. All velocity vectors of the same magnitude lie on a spherical shell whose volume grows as v². This purely geometric factor is why the most likely speed is not zero even though the most likely velocity is. B

∫₀^∞ v² e^−av² dv = ¼ √( π / a³ )

Normalise, using the standard Gaussian moment integral with a = m / 2k_BT. Fixing the prefactor so that ∫f(v)dv = 1 turns the proportionality of step 1 into an equality. C

Result

f(v) = 4π ( m / 2πk_BT )^3/2 v² e^{−mv²/2k_BT}

Reading. A competition between two factors: v² rising and the exponential falling. The peak sits where they balance. The long high-speed tail — not the mean — governs reaction rates, atmospheric escape and evaporation.

Units check. [f(v)] must be (m s⁻¹)⁻¹ so that f(v)dv is dimensionless. (m/k_BT)^3/2 has units (s² m⁻²)^3/2 = s³ m⁻³, times v² = m² s⁻², gives s m⁻¹. ✓

The three speeds, and why they differ

Most probable — maximum of f(v): √(2k_BT/m). Ratio 1.000.
Mean — ∫v f(v) dv: √(8k_BT/πm). Ratio 1.128.
RMS — √⟨v²⟩: √(3k_BT/m). Ratio 1.225.
They differ because the distribution is skewed. Only the RMS speed enters the pressure and the energy, since those involve v².

Limiting cases

Low speeds v → 0: f(v) ≈ 4π(m/2πk_BT)^3/2 v² rises as v² — the shell-volume factor dominates and f(0) = 0.
High speeds: the exponential wins and f(v) falls off faster than any power — this is the tail that decides escape and reaction rates.
High temperature or small mass: the whole curve widens and shifts right, since every speed scales as √(k_BT/m).

Breaks when

Quantum degeneracy sets in — replace with Fermi–Dirac or Bose–Einstein statistics.
Speeds approach c — use the relativistic Maxwell–Jüttner distribution, as required in stellar interiors.
The gas is not in equilibrium — a shock front or a rarefied upper atmosphere has a distinctly non-Maxwellian distribution.

Failure modes

Confusing f(v) with f(v). One peaks at zero, the other does not, and the whole distinction is the 4π v² factor.
Using the mean speed where the RMS is needed. The kinetic energy is ½m⟨v²⟩, never ½m⟨v⟩².
Reasoning about escape or reaction rates from the mean speed. Both are tail phenomena and the mean is nearly irrelevant.

Worked number — why Earth kept its atmosphere and lost its hydrogen

At 300 K, nitrogen (28 u) has v_rms ≈ 517 m s⁻¹; hydrogen (2 u) has ≈ 1,930 m s⁻¹. Earth's escape velocity is 11.2 km s⁻¹, so neither mean exceeds it. But escape is decided by the tail: a molecule needs roughly v > 6v_rms for nitrogen versus about 1.6v_rms for hydrogen, and the exponential difference between those two thresholds is enormous. Earth retains nitrogen essentially forever and lost its hydrogen. The same arithmetic explains why the Moon has no atmosphere at all.

Run the check →

Discussion

The result reads as a single competition written in two factors. The v² is a counting term: in three dimensions there are many more distinct velocity vectors with a large speed than with a small one, because the spherical shell in velocity space at radius v has area 4πv². The exponential is an energetic penalty: a molecule of speed v costs energy ½mv², and the Boltzmann factor makes that increasingly improbable. Neither factor alone has a peak away from an endpoint — v² rises without bound, the exponential falls monotonically — but their product turns over at the most probable speed v_mp = √(2k_BT/m), where the fractional gain in phase space just balances the fractional cost in the Boltzmann weight.

The structure is not special to gases; it is what always happens when you take an isotropic Gaussian in a vector and ask for the distribution of its magnitude. The velocity distribution f(v) factorises into three independent one-dimensional Gaussians, one per Cartesian component, each with variance k_BT/m. The speed is the length of that three-component vector, so v/√(k_BT/m) follows a chi distribution with three degrees of freedom. The 4πv² is nothing more than the Jacobian of going from Cartesian to spherical coordinates — a purely geometric object that would appear identically for any isotropic 3D distribution. The physics lives entirely in the Gaussian; the geometry supplies the shell.

This factorisation is also why the speed distribution is startlingly robust. In a real, interacting gas the full partition function separates into a kinetic part and a configurational part, Z = Z_kin Z_conf, because the potential energy depends on positions but not on momenta. Interactions therefore reshape spatial correlations — the pair distribution, the equation of state — while leaving the momentum, and hence the speed, distribution exactly Maxwellian in equilibrium. That is the surprising claim flagged in the assumptions made precise. The same reasoning connects this result to its neighbours by limits: equipartition follows immediately as ½m⟨v²⟩ = ³⁄₂k_BT with each quadratic degree of freedom carrying ½k_BT; restoring quantum statistics recovers Maxwell–Boltzmann only when the occupancy is dilute, so that the ±1 in the Fermi–Dirac and Bose–Einstein denominators is negligible; and letting speeds approach c replaces the Gaussian by the Maxwell–Jüttner form built on the relativistic energy.

The reason the tail, not the mean, does the interesting work is that the Boltzmann factor is exponential in energy while all the transport and loss processes that matter — chemical reaction over an activation barrier, Jeans escape over the gravitational barrier, evaporative cooling — are threshold processes. A modest change in temperature barely moves v_mp (it scales only as √T) but multiplies the population above a fixed high threshold by a large factor, because that population is controlled by e^{−E_thr/k_BT}. This is the microscopic origin of the Arrhenius law and of the extreme mass-selectivity of atmospheric escape.

Common misconceptions. The single most common error is to conflate the velocity distribution f(v), which peaks at v = 0, with the speed distribution f(v), which vanishes at v = 0 and peaks at v_mp — the whole difference is the 4πv² shell factor, and there is no contradiction in "the most likely velocity is zero but the most likely speed is not." A second error is using ⟨v⟩ where v_rms belongs: kinetic energy and pressure involve v², so they are set by ⟨v²⟩, and ⟨v²⟩ ≠ ⟨v⟩² for a spread distribution (indeed v_rms/⟨v⟩ = √(3π/8) ≈ 1.086). A third is estimating escape or reaction rates from the mean speed at all; both are tail phenomena for which the mean is nearly irrelevant.

Worked examples

Example 1 — the three speeds of oxygen at room temperature, and a consistency check. Find v_mp, ⟨v⟩ and v_rms for O₂ (molar mass 32 u) at T = 300 K, and verify that v_rms reproduces the equipartition energy.

m = 32 × (1.66054×10⁻²⁷ kg) = 5.314×10⁻²⁶ kg, k_BT = (1.38065×10⁻²³)(300) = 4.142×10⁻²¹ J

Convert molar mass to molecular mass and evaluate the thermal energy scale.

v_mp = √(2k_BT/m) = √( 2(4.142×10⁻²¹) / 5.314×10⁻²⁶ ) = √(1.559×10⁵) = 394.8 m s⁻¹

Most probable speed — the peak of f(v). Work symbolically, substitute at the end.

⟨v⟩ = √(4/π) · v_mp = 1.1284 × 394.8 = 445.5 m s⁻¹

Mean speed. The ratio √(4/π) is fixed by the distribution, independent of gas or temperature.

v_rms = √(3/2) · v_mp = 1.2247 × 394.8 = 483.6 m s⁻¹

Root-mean-square speed — the only one entering pressure and energy. Order confirmed: v_mp < ⟨v⟩ < v_rms, ratios 1 : 1.128 : 1.225.

½mv_rms² = ½(5.314×10⁻²⁶)(483.6)² = 6.21×10⁻²¹ J vs ³⁄₂k_BT = 1.5(4.142×10⁻²¹) = 6.21×10⁻²¹ J

Consistency check: the mean kinetic energy from v_rms equals the equipartition value, as it must.

v_mp = 395 m s⁻¹, ⟨v⟩ = 446 m s⁻¹, v_rms = 484 m s⁻¹

Answer. For O₂ at 300 K the most probable, mean and rms speeds are 395, 446 and 484 m s⁻¹ respectively. Only v_rms reproduces the equipartition energy ½m⟨v²⟩ = ³⁄₂k_BT = 6.2×10⁻²¹ J.

Example 2 — the tail is exponentially temperature-sensitive. For nitrogen (N₂, 28 u), find the fraction of molecules with speed exceeding V = 1500 m s⁻¹ at 300 K and at 600 K, and compare. This models how a modest heating changes a threshold-limited rate.

F(v > V) = ∫_V^∞ f(v) dv = erfc(x₀) + 2√π x₀ e^−x₀², x₀ ≡ V/v_mp

Integrate the speed distribution above V using x = v/v_mp; the result is a complementary error function plus a tail term.

v_mp(300 K) = √(2k_BT/m) = 422.1 m s⁻¹, x₀ = 1500/422.1 = 3.554

Reduce the threshold to units of the most probable speed at the lower temperature.

F₃₀₀ = erfc(3.554) + 1.1284(3.554)e^−12.63 = 5.0×10⁻⁷ + 1.31×10⁻⁵ = 1.36×10⁻⁵

The threshold sits at 3.55 v_mp, deep in the tail; barely one molecule in 10⁵ qualifies.

v_mp(600 K) = 422.1√2 = 596.9 m s⁻¹, x₀ = 1500/596.9 = 2.513

Doubling T raises v_mp by only √2, moving the threshold to 2.51 v_mp.

F₆₀₀ = erfc(2.513) + 1.1284(2.513)e^−6.315 = 3.8×10⁻⁴ + 5.13×10⁻³ = 5.51×10⁻³

Now roughly one molecule in 180 exceeds V.

F₆₀₀/F₃₀₀ = 5.51×10⁻³ / 1.36×10⁻⁵ ≈ 4.0×10²

Answer. The fraction above 1500 m s⁻¹ rises from 1.4×10⁻⁵ at 300 K to 5.5×10⁻³ at 600 K — a factor of about 400 for a mere doubling of temperature. A 25% shift in v_mp produces a 400-fold change in the tail population: this is why threshold-limited processes obey the near-exponential Arrhenius temperature dependence.

Problems

(Warm-up.) Compute the rms speed of helium atoms (4 u) at 300 K, and state why it is so much larger than the value for O₂ found above.
Solution

m = 4(1.66054×10⁻²⁷) = 6.642×10⁻²⁷ kg. v_rms = √(3k_BT/m) = √( 3(4.142×10⁻²¹)/6.642×10⁻²⁷ ) = √(1.871×10⁶) = 1368 m s⁻¹. It exceeds the O₂ value (484 m s⁻¹) by the factor √(32/4) = √8 = 2.83, because at fixed T all gases share the same mean kinetic energy ³⁄₂k_BT, so v_rms ∝ 1/√m.
(Inverting for temperature.) At what temperature does the most probable speed of O₂ reach 500 m s⁻¹?
Solution

Invert v_mp = √(2k_BT/m): T = mv_mp²/(2k_B) = (5.314×10⁻²⁶)(500)² / (2·1.38065×10⁻²³) = 1.328×10⁻²⁰/2.761×10⁻²³ = 481 K. (A useful check: since v_mp ∝ √T and v_mp(300 K) = 395 m s⁻¹, we expect T = 300(500/395)² = 481 K.)
(Shape of the curve.) By what factor is f(v) smaller at v = 2v_mp than at its peak v = v_mp? Do it without evaluating the normalisation constant.
Solution

Write f(v) ∝ v² e^{−v²/v_mp²}, since mv²/2k_BT = v²/v_mp². In units x = v/v_mp the ratio is

f(2v_mp)f(v_mp) = 2² e⁻⁴1² e⁻¹ = 4e⁻³ = 0.199.

So at twice the most probable speed the distribution has fallen to about 20% of its peak — the exponential has begun to dominate the v² growth.
(Area under a slice.) What fraction of molecules have speeds between v_mp and 1.5v_mp? Use the cumulative form P(<x) = erf(x) − (2/√π)xe^−x² with x = v/v_mp.
Solution

At x = 1: P = erf(1) − 1.1284(1)e⁻¹ = 0.8427 − 0.4151 = 0.3276.

At x = 1.5: P = erf(1.5) − 1.1284(1.5)e^−2.25 = 0.9661 − 0.1784 = 0.7877.

Fraction in the interval = 0.7877 − 0.3276 = 0.460, i.e. about 46% of molecules lie between v_mp and 1.5v_mp — nearly half, reflecting how much of the distribution's weight clusters just above the peak.
(Atmospheric escape.) In Earth's exosphere take atomic hydrogen (1 u) at T = 1000 K. Find v_rms, then estimate the fraction of atoms exceeding the escape speed v_esc = 11.2 km s⁻¹. Comment on why atomic oxygen (16 u) does not escape.
Solution

m = 1.66054×10⁻²⁷ kg, k_BT = 1.381×10⁻²⁰ J. v_rms = √(3k_BT/m) = √(2.494×10⁷) = 4994 m s⁻¹ (≈ 5.0 km s⁻¹), and v_mp = √(2k_BT/m) = 4078 m s⁻¹.

Reduced threshold x₀ = 11200/4078 = 2.746, so x₀² = 7.54. Fraction above escape = erfc(2.746) + (2/√π)(2.746)e^−7.54 = 1.0×10⁻⁴ + 1.64×10⁻³ = 1.7×10⁻³. Roughly one hydrogen atom in 600 in the tail is unbound — enough, integrated over geological time, to strip Earth of hydrogen.

For atomic oxygen the mass is 16× larger, so v_mp is √16 = 4× smaller (1020 m s⁻¹) and x₀ = 11200/1020 = 11.0. The controlling factor e^−x₀² = e⁻¹²¹ ≈ 10⁻⁵³ is utterly negligible: oxygen is retained forever. The escape fraction depends exponentially on m through x₀² = mv_esc²/2k_BT, which is the quantitative root of atmospheric mass-selectivity.