Fill a room with air and every cubic centimetre holds some 2.5\times 10^{19} molecules, each rattling about, colliding billions of times a second. It is tempting to picture them all skittering along at one shared "thermal speed" set by the temperature. They do not. At any instant some molecules are almost stationary, a great many cruise near a few hundred metres per second, and a rare few are screaming along at more than twice that. There is a whole spread of speeds — and the exact shape of that spread is one of the most consequential curves in physics.

That shape controls an astonishing amount of the everyday world. It is why a puddle evaporates at room temperature even though "boiling" is far away: only the fastest molecules in the tail have enough energy to break free, and picking them off cools what remains. It is why chemical reaction rates rocket up when you heat things — reactions need molecules above an activation energy, and warming the gas fattens the high-speed tail. And it is why the Earth has held on to its \text{N}_2 and \text{O}_2 for billions of years while its hydrogen and helium leaked away into space: light molecules move faster, more of them sit in the tail above escape velocity, and over aeons the atmosphere sieves itself. One curve, and it reaches from a drying pavement to the composition of planets.

This page is about that one curve — the Maxwell–Boltzmann speed distribution. We will build it from the Boltzmann distribution on energy, pin down its three characteristic speeds, prove it is properly normalised using the Gamma function, and watch it slide and broaden as you turn up the heat.

From Boltzmann to speeds

The starting point is the Boltzmann factor: in thermal equilibrium at temperature T, the probability of finding a molecule in a state of energy E is weighted by e^{-E/kT}, where k = 1.381\times 10^{-23}\ \text{J/K} is Boltzmann's constant. For a free molecule the only energy is kinetic, E = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m\,(v_x^2 + v_y^2 + v_z^2), so each velocity component carries its own Gaussian weight:

Each of these is a bell curve centred on zero: a molecule is just as likely to drift left as right, so the most probable value of any single component is v_x = 0. Multiply the three together (the components are independent) and the probability of a particular velocity vector \mathbf{v} is

which depends only on the magnitude v = |\mathbf{v}|. So far this is the distribution of velocities. But we usually want the distribution of speeds — how likely a molecule is to have speed near some value v, regardless of direction. That is a different question, and answering it introduces the factor that gives the curve its whole character.

Picture "velocity space", the abstract space whose axes are v_x, v_y, v_z. Every molecule is a point in it, and the exponential above gives the density of points. All the molecules with speed near v lie in a thin spherical shell of radius v and thickness dv. The volume of that shell is its surface area times its thickness, 4\pi v^2\,dv. There are simply more ways to have a large speed than a small one — more directions to point — and that count grows as 4\pi v^2. Multiplying the per-velocity probability by this shell volume gives the speed distribution:

This is the key structural fact, and it is worth saying plainly: f(0) = 0. No molecule sits exactly still, not because rest is forbidden, but because a single point (zero speed) has vanishing "shell volume" — there is no direction to a zero vector. The competition between the rising v^2 and the falling exponential is what makes a peak at all.

Watch it move

The curve below is f(v) for a real gas. Slide the temperature up and the peak marches to the right and the whole hump flattens and spreads — because the area underneath is always exactly 1 (every molecule has some speed), so widening the curve must lower it. Switch the gas between helium, nitrogen and oxygen at fixed temperature and watch the opposite: heavier molecules sit slower and more tightly bunched. Notice too that the curve always starts at the origin — that is the v^2 factor forcing f(0)=0 — and that even when the peak is modest there is a long, thin high-speed tail stretching far to the right.

Everything downstream — evaporation, reaction rates, atmospheric escape — lives in how this curve moves. Warming the gas does not just shift the average; it disproportionately populates the tail, because the tail's height depends exponentially on mv^2/2kT. A modest rise in temperature can multiply the number of molecules above a high threshold speed many-fold, which is exactly why chemistry speeds up so dramatically with heat.

Three speeds, three questions

A curve with a peak and a skewed tail does not have a single "the speed"; it has several natural ones, and they answer subtly different questions. Three matter, and they always come in the same order.

1. The most probable speed v_p — the location of the peak, the single commonest speed. Find it where df/dv = 0. Dropping constants, we differentiate v^2 e^{-mv^2/2kT}:

2. The mean speed \langle v\rangle — the ordinary average, \int_0^\infty v\,f(v)\,dv. The skewed tail drags it above the peak. As we compute below with the Gamma function,

3. The root-mean-square speed v_{\text{rms}} — the speed of the "average kinetic energy" molecule, \sqrt{\langle v^2\rangle}. This one you can get without any integral, straight from equipartition: each of the three translational degrees of freedom carries \tfrac{1}{2}kT of energy, so

All three are \sqrt{kT/m} times a pure number, so they keep a fixed ratio no matter the gas or temperature:

or, dividing through by the smallest,

The order v_p < \langle v\rangle < v_{\text{rms}} is permanent and worth remembering: the peak is the most common speed, the mean is pulled up by the tail, and the rms is highest of all because squaring weights the fast molecules most heavily. Two handy ratios fall straight out: v_{\text{rms}}/v_p = \sqrt{3/2} = \sqrt{1.5} \approx 1.225 and \langle v\rangle / v_p = \sqrt{4/\pi} \approx 1.128.

Why the Gamma function: normalisation and moments

Every claim above — that the curve integrates to 1, that \langle v\rangle = \sqrt{8kT/\pi m}, that \langle v^2\rangle = 3kT/m — is an integral of the form \int_0^\infty v^k e^{-av^2}\,dv, with a = m/2kT. Substituting u = av^2 turns each into the Gamma function \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\,dt. That is the precise reason the Gamma function is a prerequisite here — it is the machine that evaluates all the moments of this distribution:

Normalisation uses k = 2, so we need \Gamma(3/2) = \sqrt{\pi}/2:

Feed that into \int_0^\infty f(v)\,dv with the front constant 4\pi (a/\pi)^{3/2} (since m/2\pi kT = a/\pi):

The constant 4\pi(m/2\pi kT)^{3/2} out front is exactly what makes the total come to one — it is not decoration, it is the normalisation demanded by \Gamma(3/2). The mean speed uses k = 3 and \Gamma(2) = 1:

recovering the mean speed quoted earlier. And the mean square uses k = 4 and \Gamma(5/2) = 3\sqrt{\pi}/4:

so v_{\text{rms}} = \sqrt{3kT/m} — the same answer equipartition gave for free, now confirmed by the integral. Three different questions, one family of Gamma integrals, and the half-integer values \Gamma(3/2) and \Gamma(5/2) doing the real work.

Worked example: nitrogen at room temperature

Take \text{N}_2 at T = 300\ \text{K}. One molecule has mass m = 28\times 1.66\times 10^{-27} = 4.65\times 10^{-26}\ \text{kg}. The recurring quantity is

The three characteristic speeds are then just numerical multiples of that root:

• v_p = \sqrt{2}\,(298) = \sqrt{2kT/m} \approx 422\ \text{m/s};
• \langle v\rangle = \sqrt{8/\pi}\,(298) \approx 1.596\times 298 \approx 476\ \text{m/s};
• v_{\text{rms}} = \sqrt{3}\,(298) \approx 1.732\times 298 \approx 517\ \text{m/s}.

So a typical nitrogen molecule in this room is moving at roughly half a kilometre per second — faster than a rifle bullet — and the three speeds sit in the promised 422 : 476 : 517 ordering, a ratio of about 1 : 1.13 : 1.22.

A ratio question. Suppose you heat this gas from 300\ \text{K} to 1200\ \text{K} — a factor of four. Since every speed scales as \sqrt{T}, all of them grow by \sqrt{4} = 2. The rms speed climbs from 517\ \text{m/s} to about 1034\ \text{m/s}. The curve keeps its shape but stretches out along the speed axis and flattens, its area still exactly one.