2.3 The Maxwell–Boltzmann distribution

The pressure and equipartition results used the averages $\langle v_x^2\rangle$ and $\langle\varepsilon\rangle$ without ever needing the full distribution of velocities. That distribution — how many molecules move at each speed — is fixed not by mechanics but by statistical equilibrium, and it is the Maxwell–Boltzmann distribution.

Fixing the form

In equilibrium the joint distribution of the three velocity components $(v_x, v_y, v_z)$ must satisfy three conditions: it is factorisable across components (the components are independent), it is isotropic (it depends only on the speed $|\mathbf v|$ ), and it is consistent with $\langle\varepsilon\rangle = \tfrac32 k_B T$ from 2.2. Only one functional form meets all three.

▶ Why the component distribution is Gaussian Derivation

Independence means the joint density factorises, $F(v_x, v_y, v_z) = f(v_x)\,f(v_y)\,f(v_z)$ . Isotropy means it depends only on $v^2 = v_x^2 + v_y^2 + v_z^2$ . A function of a sum that is also a product of functions of the separate terms must be an exponential: $\ln F = \ln f(v_x) + \ln f(v_y) + \ln f(v_z)$ depends only on $v_x^2 + v_y^2 + v_z^2$ , which forces $\ln f(v_x) = a + b\,v_x^2$ . The distribution must be normalisable, so $b<0$ ; writing $b = -m/(2k_B T)$ and fixing $a$ by normalisation,

f(v_x) \;=\; \left(\frac{m}{2\pi k_B T}\right)^{1/2} e^{-m v_x^2/(2k_B T)}.

The choice $b = -m/(2k_B T)$ is exactly what makes $\langle\tfrac12 m v_x^2\rangle = \tfrac12 k_B T$ , the equipartition condition. ✓

To get the distribution of speeds rather than components, multiply the component density by the volume of the spherical shell of radius $v$ in velocity space, $4\pi v^2\,dv$ :

\boxed{\;f(v) \;=\; 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2\, e^{-m v^2/(2k_B T)}.\;}

The $v^2$ from the shell pushes the peak away from zero; the exponential cuts off the high-speed tail. The distribution is therefore skewed, with a long tail toward high speeds.

v_p (most probable)410 m/s

⟨v⟩ (mean)463 m/s

v_rms502 m/s

c (speed of sound)343 m/s

Gas: Temperature: 293 K (20 °C)

The most-probable, mean, and RMS speeds always stand in the ratio √2 : √(8/π) : √3 ≈ 1.41 : 1.60 : 1.73. The speed of sound is the smallest of the four: it equals √(γ/3) ≈ 0.68 of v_rms for diatomic gas. Drop the temperature and the whole distribution contracts toward zero like √T.

Slide the temperature: the distribution shifts and broadens as $\sqrt{T}$ . Switch gases: at the same temperature, lighter molecules (H₂, He) reach far higher thermal speeds, since $\langle\varepsilon\rangle$ is fixed but $m$ is smaller.

Three characteristic speeds

The skew means there is no single “the” speed; three are useful, each the right one for a different question:

Most-probable speed $v_p = \sqrt{2 k_B T/m}$ — the peak, the typical speed of a randomly chosen molecule.
Mean speed $\langle v\rangle = \sqrt{8 k_B T/(\pi m)}$ — the average, the right speed for collision-rate arguments (rate $\propto \langle v\rangle$ ).
Root-mean-square speed $v_\text{rms} = \sqrt{3 k_B T/m}$ — the right speed for pressure and energy, which depend on $\langle v^2\rangle$ .

They sit in the fixed ratio $\sqrt2 : \sqrt{8/\pi} : \sqrt3 \approx 1.41 : 1.60 : 1.73$ , independent of gas or temperature — a direct fingerprint of the distribution’s shape.