Mean free path, mean speed, RMS speed — derivations: Visual walkthrough

Before we start, the words we will use constantly — each pinned to a picture below:

A molecule = a tiny hard ball. Its width across is called its diameter, written $d$ .
Number density $n$ = how many molecules sit in each cubic metre of space, i.e. $n = N / V$ (count $N$ of them, divide by the volume $V$ they live in).
Mean speed $\overset{v}{ˉ}$ = the everyday-average speed of a molecule (derived in the parent note as $\overset{v}{ˉ} = 8 k_{B} T / π m$ ). We treat it as a single known number — a speed in metres per second — and nothing more.
The average of a quantity — written with a bar on top, like $\overline{v^{2}}$ — just means: measure it for every molecule, add them all up, divide by how many there are. So $\overline{v^{2}}$ is "the average of the squared speeds". We will lean on this notation from Step 6 on.

Step 1 — One ball, its width, and what "touching" means

WHAT. We first decide when two molecules count as colliding.

WHY. The whole derivation is about counting crashes. We cannot count crashes until we agree what a crash is. A ball has size, so "touching" is not "centres at the same point" — it is "centres a certain distance apart".

PICTURE. Two balls each of diameter $d$ (so each has radius $d /2$ ). They just touch when their centres are exactly $d$ apart — because $d /2 + d /2 = d$ .

Figure — Mean free path, mean speed, RMS speed — derivations

This is the single most-fumbled point, so we frame it now: the effective "target radius" is $d$ .

Step 2 — Shrink one ball to a point, grow the target to radius $d$

WHAT. We replace the real picture (two fat balls) with an easier one (a point molecule chasing fat targets).

WHY. Two moving fat balls are hard to track. But a collision only depends on the distance between centres. So we can keep the geometry identical while making the moving molecule a dot and giving the target molecule the combined radius $d$ . Same collisions, simpler bookkeeping.

PICTURE. The moving molecule becomes a dot. Every other molecule becomes a disc of radius $d$ . A crash happens exactly when the dot's path enters one of those discs.

Step 3 — The dot drags a tube: the collision cylinder

WHAT. As the dot flies in a straight line, it sweeps out a long thin tube. Any target whose centre lies inside the tube gets hit.

WHY. We want to count how many targets the dot meets. That count equals how many target-centres fall inside the swept region. A straight-flying dot with reach $d$ sweeps a cylinder of radius $d$ .

PICTURE. Let $t$ be a stretch of time we watch the molecule for (measured in seconds). In that time the dot travels a distance = (its speed) × (that time) = $\overset{v}{ˉ} t$ . The tube therefore has:

radius $d$ (the reach from Step 1),
length $\overset{v}{ˉ} t$ (speed × time).

Why $π d^{2}$ and not $π (d /2)^{2}$ ? Because the tube radius is the reach $d$ from Step 1, not the physical radius $d /2$ . Area of a circle is $π r^{2}$ with $r = d$ .

Step 4 — Count the targets inside the tube

WHAT. Count how many molecules sit inside the swept tube.

WHY. Each target inside the tube is one collision. Counting them gives the number of crashes in the time $t$ .

PICTURE. Fill the surrounding gas with $n$ molecules per cubic metre. The tube's volume is (face area)×(length). Multiply by $n$ to get the head-count inside.

$Volume of tube = face area σ \times length \overset{v}{ˉ} t = π d^{2} \overset{v}{ˉ} t$

$Number of collisions = targets per m^{3} n \times tube volume π d^{2} \overset{v}{ˉ} t = nπ d^{2} \overset{v}{ˉ} t$

Step 5 — Distance ÷ crashes = mean free path (first draft)

WHAT. Divide the total distance flown by the number of crashes to get the average distance between crashes. We give that average distance a name: the mean free path, written $λ$ (the Greek letter "lambda").

WHY. "Mean free path" literally means the average straight run between two collisions. If you fly a total length $\overset{v}{ˉ} t$ and hit things $nπ d^{2} \overset{v}{ˉ} t$ times, then per hit you covered:

PICTURE. A zig-zag path chopped into segments by collisions; $λ$ is the average segment length.

$λ = \frac{total distance}{number of collisions} = \frac{v ˉ t}{nπ d ^{2} v ˉ t}$

Now watch the cancellation — this is the satisfying part. The $\overset{v}{ˉ}$ cancels (top and bottom), and the $t$ cancels (top and bottom):

$λ = \frac{v ˉ t}{nπ d ^{2} v ˉ t} = \frac{1}{nπ d ^{2}}$

This is the first draft — good, but it hides a simplification we fix next.

Step 6 — The hidden simplification: the targets are moving too

WHAT. In Step 3 we treated the targets as frozen. They are not — they are zooming around just like our dot. We correct for this.

WHY. The collision rate depends on how fast the dot approaches the target relative to that target, not on the dot's ground speed. Two moving molecules close in on each other faster (on average) than one moving toward a still one.

PICTURE. Draw the dot's velocity $v_{1}$ and a target's velocity $v_{2}$ as arrows. What matters is the relative velocity $v_{rel} = v_{1} - v_{2}$ (the arrow from tip of $v_{2}$ to tip of $v_{1}$ ).

Recall the bar means "average over all molecules" (defined at the top). The clean quantity to average is the squared relative speed, because squaring turns the vector subtraction into a dot product we can expand:

$\overline{v_{rel}^{2}} = \overline{(v_{1} - v_{2})^{2}} = ball 1 \overline{v_{1}^{2}} + ball 2 \overline{v_{2}^{2}} - = 0 (random directions) 2 \overline{v_{1} \cdot v_{2}} = 2 \overline{v^{2}}$

The middle term $\overline{v_{1} \cdot v_{2}}$ averages to zero: the two molecules pick their directions independently, so their arrows point every-which-way and the dot product is as often positive as negative. Both balls have the same speed statistics, so $\overline{v_{1}^{2}} = \overline{v_{2}^{2}} = \overline{v^{2}}$ , giving $\overline{v_{rel}^{2}} = 2 \overline{v^{2}}$ .

Intuition From

$\overline{v_{rel}^{2}} = 2 \overline{v^{2}}$ to the honest $2$ A quick warning: the root of the average square, $\overline{v_{rel}^{2}} = 2 \overline{v^{2}}$ , is the RMS relative speed — not exactly the mean relative speed $\overline{∣ v_{rel} ∣}$ we actually need for the collision rate. The two are close but not identical. The precise calculation is worth stating: if you feed two independent Maxwell–Boltzmann velocities into $∣ v_{1} - v_{2} ∣$ and take the mean, the difference of two Maxwell distributions is itself a Maxwell distribution with twice the variance. Its mean speed is exactly $2$ times the single-molecule mean speed. So the honest, distribution-averaged result agrees with the crude estimate: $\overline{∣ v_{rel} ∣} = 2 \overset{v}{ˉ} .$ That is why $2$ (and not, say, $8/ π$ -flavoured mess) is the exact factor. See Maxwell-Boltzmann distribution for the "difference of two Maxwells" fact.

Step 7 — Put the $2$ where it belongs

WHAT. In the collision rate (crashes per time), the speed that counts is the mean relative speed $2 \overset{v}{ˉ}$ , not $\overset{v}{ˉ}$ . But the distance the dot actually travels through space is still $\overset{v}{ˉ} t$ (its own ground speed).

WHY. Distance flown = own speed × time. Collisions suffered = relative speed × time × (targets per volume × area). Different speeds go in different places.

PICTURE. Same tube as Step 5, but the crash-counter now ticks $2$ times faster because targets approach.

$λ = \frac{distance dot flies}{crashes} = \frac{v ˉ t}{nπ d ^{2} ( 2 v ˉ ) t} = \frac{v ˉ}{2 nπ d ^{2} v ˉ}$

Cancel $\overset{v}{ˉ}$ once more:

Step 8 — Rewrite in lab quantities (pressure and temperature)

WHAT. Swap the microscopic $n$ (which you can't measure directly) for pressure and temperature (which you can, with a gauge and a thermometer).

WHY. $n$ is invisible; the pressure and temperature of a gas are on the wall of every lab. To make the swap we need three more physical symbols, introduced here in plain words:

Pressure $P$ — the push per unit area the gas exerts on its container walls (measured in pascals, Pa).
Absolute temperature $T$ — the hotness of the gas measured from absolute zero (in kelvin, K).
Boltzmann constant $k_{B}$ — a fixed conversion number ( $1.38 \times 1 0^{- 23}$ joules per kelvin) that turns "one molecule's worth of temperature" into energy. It is the bridge between the microscopic and the everyday.

The Ideal gas law ties these together as $P V = N k_{B} T$ . Dividing both sides by $V k_{B} T$ gives exactly the number density we need:

$n = \frac{N}{V} = \frac{P}{k _{B} T}$

Substitute this into the boxed result:

$λ = \frac{1}{2 nπ d ^{2}} = \frac{1}{2 π d ^{2}} \cdot \frac{1}{n} = \frac{1}{2 π d ^{2}} \cdot \frac{k _{B} T}{P}$

Edge & degenerate cases (what happens at the extremes)

The one-picture summary

Everything on this page in one frame: a dot flies through a field of radius- $d$ target discs, dragging a cylinder of cross-section $π d^{2}$ ; the targets rush in too (the $2$ ); distance ÷ crashes gives $λ$ .

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Imagine you're a marble rolling blind across a floor scattered with other marbles. First question: when do you "hit" one? Not when your centres meet — when they get within one marble-width of each other, because both marbles are fat. So we play a trick: pretend you are a tiny dot and every other marble is puffed up to double size. Now, as you roll, you drag an invisible tube behind you the width of that puffed target. Any marble whose centre sits inside your tube is a crash. Count the crashes: it's how crowded the floor is, times the tube's face size ( $π d^{2}$ ), times how far you rolled. Divide the distance you rolled by the number of crashes and — magic — your speed and time both cancel, leaving $λ = 1/ (nπ d^{2})$ . One last honesty check: the other marbles aren't sitting still, they're rolling at you too, so you meet them a bit more often — about $2$ times as often. Slip that $2$ into the bottom and you're done: $λ = 1/ (2 nπ d^{2})$ . Bigger marbles or a more crowded floor ⇒ shorter runs between bumps. Simple.

Recall Quick self-test

Why is the tube radius $d$ and not $d /2$ ? ::: Collision happens at centre-separation $d$ (sum of two radii), so the reach is a full diameter.
Why do $\overset{v}{ˉ}$ and $t$ cancel in $λ$ ? ::: Fly faster/longer and you cover more distance and hit more targets in the same proportion — the ratio is unchanged.
Where does $2$ come from? ::: Targets move too; averaging $∣ v_{1} - v_{2} ∣$ over two Maxwell distributions gives $2 \overset{v}{ˉ}$ .
What happens to $λ$ as $P \to 0$ ? ::: $n \to 0$ , so $λ \to \infty$ — free flight, no collisions.

Flashcards

Mean free path in terms of n and d

λ = \frac{1}{2 nπ d ^{2}}

Mean free path in terms of P and T

λ = \frac{k _{B} T}{2 π d ^{2} P}

Why collision cross-section is

π d^{2}

not

π (d /2)^{2}

Collision occurs at centre separation

d

, so effective target radius is

d

Origin of the

2

factor

Mean relative speed of two molecules is

2 \overset{v}{ˉ}

(difference of two Maxwell distributions)

Dependence of

λ

on pressure at fixed T

λ \propto 1/ P