Maxwell-Boltzmann distribution — full derivation: Visual walkthrough

Step 0 — What is a "velocity" and what are we counting?

WHAT. A molecule at one instant has a velocity: an arrow pointing where it's going, whose length is its speed. In 3D that arrow has three shadows on three walls — its components $v_{x}, v_{y}, v_{z}$ (how fast it moves left-right, forward-back, up-down).

WHY. Because the speed — the length of the arrow — hides three independent numbers inside it. To understand the speed we first understand each shadow, then reassemble.

PICTURE. The blue arrow is one molecule's velocity. Its three coloured shadows are the components. The length of the arrow is $v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ (Pythagoras in 3D).

Figure — Maxwell-Boltzmann distribution — full derivation

Step 1 — Each component follows a bell curve, and why it must

WHAT. We claim the probability of a molecule having sideways speed near $v_{x}$ is a Gaussian (a symmetric bell centred at zero):

g (v_{x}) = A e^{- b v_{x}^{2}} .

WHY this shape and not another? Two physical facts pin it down:

Independence — nothing links the left-right speed to the up-down speed. So the chance of all three is the product $g (v_{x}) g (v_{y}) g (v_{z})$ .
Isotropy — space has no favourite direction, so the joint chance can depend only on the arrow's length, i.e. on $v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ .

A product on the left that depends only on a sum of squares on the right — that is the exponential's signature. Taking $ln$ turns the product into a sum: $ln g (v_{x}) + ln g (v_{y}) + ln g (v_{z})$ must be a function of $v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ . The only way is $ln g (v_{x}) = a - b v_{x}^{2}$ each. Undo the $ln$ and you get the Gaussian.

PICTURE. The bell peaks at $v_{x} = 0$ (most molecules drift only gently sideways) and dies for large $∣ v_{x} ∣$ . It is symmetric: a molecule is equally likely to go left or right.

Step 2 — Nail the constants: normalization + temperature

WHAT. Two unknowns remain: $A$ and $b$ . We fix them with two facts.

WHY (fact 1: total = 1). All the probability must add to one. The area under a Gaussian is a classic Gaussian integral:

\int_{- \infty}^{\infty} A e^{- b v_{x}^{2}} d v_{x} = A \frac{π}{b} = 1 \Rightarrow A = \frac{b}{π} .

WHY (fact 2: temperature). We need $b$ to know how hot the gas is. The equipartition theorem says each sideways motion carries average energy $\frac{1}{2} k_{B} T$ :

⟨ \frac{1}{2} m v_{x}^{2} ⟩ = \frac{1}{2} k_{B} T \Rightarrow ⟨ v_{x}^{2} ⟩ = \frac{k _{B} T}{m} .

For a Gaussian, the spread (variance) is $⟨ v_{x}^{2} ⟩ = \frac{1}{2 b}$ . Setting them equal:

\frac{1}{2 b} = \frac{k _{B} T}{m} \Rightarrow b = \frac{m}{2 k _{B} T} .

PICTURE. Two temperatures. The hot bell is shorter and wider (bigger spread, smaller $b$ ); the cold bell is tall and narrow. Both enclose exactly the same area = 1.

Step 3 — Stack three bells: the cloud in velocity space

WHAT. The full velocity distribution is the product of three identical bells:

F (v) = (\frac{m}{2 π k _{B} T})^{3/2} e^{- m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2}) /2 k_{B} T} .

WHY. Independence again — the joint chance is the product. The three squared terms recombine into $v^{2}$ , so $F$ depends only on distance from the origin. That means the probability cloud is spherically symmetric: densest at the centre, fading outward equally in all directions.

PICTURE. A fuzzy ball of dots in velocity space. Dense near the centre (slow molecules common), thinning outward (fast molecules rare). No direction is favoured — a perfect fuzzy sphere.

Step 4 — The trick that makes the $v^{2}$ appear: counting the shell

WHAT. We want speed, not the arrow. Many different arrows share the same length $v$ — they all sit on a sphere of radius $v$ . To get the chance of speed near $v$ we must add up the cloud density over that entire sphere.

WHY. The density $F (v)$ is the same everywhere on the sphere (spherical symmetry). So the total is just density × surface area of the shell:

f (v) d v = density F (v) \times shell volume 4 π v^{2} d v .

The surface area of a sphere of radius $v$ is $4 π v^{2}$ . This is where the $v^{2}$ is born — it's pure geometry, "how much room exists at speed $v$ ."

PICTURE. A thin spherical shell of radius $v$ , thickness $d v$ . Its area $4 π v^{2}$ grows as the shell gets bigger. At $v = 0$ the shell shrinks to a point — zero area, zero room.

Step 5 — The tug-of-war: two curves fight, one wins

WHAT. Collect everything into the Maxwell-Boltzmann speed distribution:

f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- m v^{2} /2 k_{B} T} .

WHY the hill shape? Two competing factors:

$v^{2}$ — the geometric factor, rising from 0, wanting large $v$ .
$e^{- m v^{2} /2 k_{B} T}$ — the energy factor (from the Boltzmann factor), crushing large $v$ .

Multiply a rising parabola by a falling bell and you get a curve that starts at 0, climbs, peaks, then dies — a lopsided hill.

PICTURE. Three curves overlaid: the rising $v^{2}$ (green), the falling exponential (red), and their product $f (v)$ (blue). Watch how the product is dragged to zero at both ends but bulges in the middle.

Step 6 — Edge cases: the two ends of the curve

WHAT. Check the boundaries the reader will hit.

Case $v = 0$ (the still molecule). $f (0) = 4 π (\dots)^{3/2} \cdot 0^{2} \cdot 1 = 0$ . Even though the exponential is maximal ( $e^{0} = 1$ ), the $v^{2}$ shell has zero area. No molecule is exactly frozen — not because it's energetically forbidden, but because there's no room at $v = 0$ .

Case $v \to \infty$ (the superfast molecule). Here $v^{2}$ grows without bound but $e^{- m v^{2} /2 k_{B} T}$ dies far faster (exponential always beats any power). So $f (v) \to 0$ . No molecule is infinitely fast.

PICTURE. Zoom on both tails: at the left the parabola pins the curve to zero; at the right the exponential drags it back down. The curve kisses zero at both ends.

Step 7 — Where the peak sits (the balance point)

WHAT. The peak $v_{p}$ is where the rising $v^{2}$ and falling $e^{- α v^{2}}$ (with $α = m /2 k_{B} T$ ) exactly balance. Set the derivative to zero:

\frac{d}{d v} (v^{2} e^{- α v^{2}}) = 2 v e^{- α v^{2}} - 2 α v^{3} e^{- α v^{2}} = 0 \Rightarrow 1 = α v^{2} \Rightarrow v_{p} = \frac{2 k _{B} T}{m} .

WHY the derivative? The peak is a flat spot — where the curve stops rising and starts falling, its slope is zero. The derivative is the slope. It splits into "geometry pushing right" ( $2 v$ ) minus "energy pulling left" ( $2 α v^{3}$ ); they cancel exactly at $v_{p}$ .

PICTURE. The two forces as arrows at three points: below $v_{p}$ geometry wins (curve rises), at $v_{p}$ they tie (flat top), above $v_{p}$ energy wins (curve falls).

The one-picture summary

Everything on one canvas: the fuzzy velocity cloud (Step 3) → its spherical shell (Step 4) → the $v^{2}$ vs exponential tug-of-war (Step 5) → the final hill with $v_{p} < ⟨ v ⟩ < v_{r m s}$ marked (Steps 6–7).

Recall Feynman retelling — the whole walkthrough in plain words

Picture every molecule as an arrow in a 3D graph. Because nothing links left-right to up-down, and no direction is special, the cloud of arrow-tips is a fuzzy ball — dense in the middle, thin at the edges, and each of the three shadows follows the same simple bell curve. Making the bell finite forces it to decay; matching it to temperature fixes exactly how fast it decays (the $m /2 k_{B} T$ inside).

Now I don't care which way an arrow points, only its length — its speed. All arrows of the same length live on a sphere, and bigger spheres have more surface, growing like $v^{2}$ . So the chance of a speed is (how dense the cloud is there) × (how much room the sphere gives). Density falls off exponentially; room grows like $v^{2}$ . The two fight: room wins at small speeds (so zero speed is impossible — no room), energy wins at big speeds (so infinite speed is impossible). They tie at the peak $v_{p} = 2 k_{B} T / m$ . Because the right tail is long, the mean and rms speeds sit a little past the peak. Heat the gas and the whole hill slides right and spreads out. That hill is the Maxwell-Boltzmann distribution.

Recall Self-test

Why is $f (0) = 0$ despite $e^{0} = 1$ ? ::: The $v^{2}$ shell-area factor is zero at $v = 0$ — no room, even though the energy cost is lowest. What single external fact did we import? ::: The Boltzmann factor: high energy is exponentially unlikely, giving the $e^{- m v^{2} /2 k_{B} T}$ . Where does the $4 π v^{2}$ come from? ::: The surface area of the sphere of radius $v$ in velocity space (Step 4). Why must $b > 0$ ? ::: Otherwise the Gaussian grows without bound and can't normalize.

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