Mean free path, mean speed, RMS speed — derivations: Foundations

We go slowly and in order. Each new symbol is defined in plain words, drawn as a picture, and justified by the question it answers. Nothing is used before it is built.

1. What is a "molecule" and how do we picture it?

Why a ball? Because the only thing that matters for collisions is how close two centres get before they touch. A ball captures that with a single number, $d$ . We do not care about the molecule's real fuzzy shape.

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

$d$ (metres) — the diameter of one molecule. Look at the two touching balls in the figure: the centres are exactly $d$ apart at the moment of contact. This single fact drives the whole mean-free-path derivation.

2. Counting molecules: $N$ , $V$ and number density $n$

Why do we need $n$ and not just $N$ ? Because "crowdedness" is what controls collisions, not the raw count. A million molecules in a stadium is empty; a million in a thimble is a traffic jam. Number density $n$ measures crowding directly.

3. Speed, velocity, and its three components $v_{x}, v_{y}, v_{z}$

Any arrow in 3-D can be split into three sideways pieces: how much it points along $x$ , along $y$ , along $z$ . These pieces are the components $v_{x}, v_{y}, v_{z}$ . Unlike the speed $v$ , each component can be negative — $v_{x} < 0$ just means "moving in the $- x$ direction".

The length of the arrow relates to its pieces by the 3-D Pythagoras rule (note: this involves the plain speed $v$ , not the bold vector):

$v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$

Definition Two facts from isotropy — and they are different

Isotropy means the gas looks the same in every direction — molecules are equally likely to head left as right, up as down. Two separate consequences follow, and the pressure derivation needs both:

Fact 1 — the average component is zero: $\overline{v_{x}} = \overline{v_{y}} = \overline{v_{z}} = 0$ . For every molecule drifting right ( $v_{x} > 0$ ) there's one drifting left ( $v_{x} < 0$ ); the signed values cancel. That is why the gas as a whole doesn't wander off — and why only the squared components (which are always $\geq 0$ and never cancel) can survive into the pressure formula.

Fact 2 — the squared components share equally: $\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}} = \frac{1}{3} \overline{v^{2}}$ . No direction is special, so the three equal squared-averages must add to $\overline{v^{2}}$ .

Together these are exactly what turns the wall-pressure result into the RMS speed in step 5 of the parent derivation.

4. The bar and "mean": what $\overset{v}{ˉ}$ and $\overline{v^{2}}$ mean

Why does the order matter? Squaring before averaging is not the same as averaging then squaring, whenever the speeds are spread out. Fast molecules get blown up more by squaring, so they drag the average of $v^{2}$ upward. This gap is exactly why RMS speed exceeds mean speed — see the next section.

5. The tools that sneak in: integrals, $e^{- x}$ , and the $\propto$ sign

Why an integral here? Molecules don't come in three speeds; they come in a continuous smear of speeds. To average over a smear you can't add finitely — you integrate.

Common mistake Thinking the speed distribution is just

$e^{- m v^{2} /2 k_{B} T}$ Why it feels right: the exponential is the eye-catching part, and it does say "fast is rare". Fix: The true Maxwell–Boltzmann speed distribution is $f (v) d v = 4 π N (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- \frac{m v ^{2}}{2 k _{B} T}} d v .$ Two pieces the bare exponential misses:

a $v^{2}$ prefactor — there are more directions an arrow of length $v$ can point as $v$ grows, so the distribution is $0$ at $v = 0$ , rises, then the exponential pulls it back down. That is what makes it peak at a non-zero speed $v_{p}$ , not at $v = 0$ .
a normalisation constant $4 π N (m /2 π k_{B} T)^{3/2}$ chosen so the fractions add up to all $N$ molecules. The full shape and its origin live in Maxwell-Boltzmann distribution. Here we only borrow the ratio form $v^{2} e^{- a v^{2}}$ , with $a = m / (2 k_{B} T)$ , where the constants cancel.

6. How the mean speed $\overset{v}{ˉ}$ is actually built from the distribution

The parent note states $\overset{v}{ˉ} = 8 k_{B} T / π m$ ; here we show, from zero, why an integral appears and how the number falls out — so the symbol $\overset{v}{ˉ}$ is fully earned.

Because every constant in $f (v)$ is the same in top and bottom, it cancels. Only the shape $v^{2} e^{- a v^{2}}$ survives, with $a = m / (2 k_{B} T)$ :

= \frac{\displaystyle\int_0^\infty v^3 e^{-av^2}\,dv}{\displaystyle\int_0^\infty v^2 e^{-av^2}\,dv}.$$ The two standard results are $\int_0^\infty v^3 e^{-av^2}dv=\dfrac{1}{2a^2}$ and $\int_0^\infty v^2 e^{-av^2}dv=\dfrac{\sqrt\pi}{4a^{3/2}}$. Dividing: $$\bar v = \frac{1/(2a^2)}{\sqrt\pi/(4a^{3/2})} = \frac{2}{\sqrt{\pi a}}.$$ Now put back $a=m/(2k_BT)$: > [!formula] Mean speed (result matches the parent) > $$\boxed{\;\bar v = \frac{2}{\sqrt{\pi a}} = \sqrt{\frac{8k_BT}{\pi m}}=\sqrt{\frac{8RT}{\pi M}}\;}$$ So the phrase "weight by $v^2e^{-av^2}$ and you get the mean speed" is exactly this integral — nothing hand-waved. --- ## 7. The physics constants: $m$, $M$, $T$, $k_B$, $R$, $N_A$, $P$ > [!definition] The core physical symbols > - $m$ — mass of **one** molecule (kg). Tiny, e.g. $\sim 10^{-26}$ kg. > - $M = m\,N_A$ — **molar mass** (kg/mol): mass of one mole of molecules. > - $N_A$ — **Avogadro's number** $\approx 6.022\times10^{23}\ \text{mol}^{-1}$ (units: *per mole*): how many molecules make one mole. > - $T$ — **absolute temperature** (kelvin, K). A measure of average molecular energy. Never in °C here. > - $k_B$ — **Boltzmann constant** $\approx 1.38\times10^{-23}$ J/K: the bridge from temperature to energy *per molecule*. > - $R = k_B N_A \approx 8.314$ J/(mol·K): the same bridge, but *per mole*. > - $P$ — **pressure** (pascal, Pa): force per unit wall area from molecular bombardment. > [!intuition] Why two constants $k_B$ and $R$? > They say the *same thing* at two scales. $k_B T$ is the energy scale of **one molecule**; $R T$ is the energy scale of **one mole**. Which you use just depends on whether you're counting single molecules ($m$, $n$) or moles ($M$). That's why $v_{rms}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$ — top and bottom both get multiplied by $N_A$ (units $\text{mol}^{-1}$), so the ratio is unchanged. > [!mistake] Molar mass in grams > **Why it feels right:** we quote $M$ for nitrogen as "28". > **Fix:** SI needs $M$ in **kg/mol** ($0.028$). Grams give a speed $\sqrt{1000}\approx32\times$ too large. --- ## 8. Collision geometry: cross-section $\sigma$, the moving-target $\sqrt2$, and mean free path $\lambda$ > [!definition] Collision cross-section $\sigma$ > $\sigma = \pi d^2$ — the area of the "danger disc" a moving molecule sweeps. Any molecule whose centre lands inside this disc gets hit. > **Why $d$, not $d/2$?** A collision happens when two *centres* are a distance $d$ apart (each ball has radius $d/2$). So the effective target radius is $d$, and the disc area is $\pi d^2$. > [!intuition] Why a $\sqrt2$ appears — the targets are moving too > The simplest picture pretends *all other molecules stand still* while one flies through them. Then in time $t$ it sweeps a cylinder of volume $\sigma\,\bar v\,t$, hits $n\sigma\bar v\,t$ targets, and the average distance between hits is > $$\lambda_{\text{naive}} = \frac{\bar v\,t}{n\sigma\bar v\,t} = \frac{1}{n\sigma}.$$ > But the targets are **not** standing still — every molecule zooms about. What sets the collision rate is not our molecule's speed against the ground, but its speed **relative to the target**. Averaged over all pairs of independent molecules of equal mean speed, that *relative* speed comes out to $\sqrt2\,\bar v$ (the vector difference of two equal-length random arrows is, on average, $\sqrt2$ times as long). Faster approach → more collisions per second → shorter flights, so we replace $\bar v$ by $\sqrt2\,\bar v$ in the rate: > [!formula] Mean free path (with the moving-target correction) > $$\boxed{\;\lambda = \frac{1}{\sqrt2\,n\sigma} = \frac{1}{\sqrt2\,n\pi d^2}\;}$$ > The $\sqrt2$ in the denominator is *only* there because the other molecules move. Drop it and you overestimate $\lambda$ by a factor $\sqrt2\approx1.41$. > [!definition] Mean free path $\lambda$ > ==Mean free path== $\lambda$ is the average straight-line distance a molecule flies **between two collisions**. Bigger $\lambda$ = emptier gas = longer free flights. These ($\sigma$, the $\sqrt2$, and $\lambda$) are the whole content of Section 4 in the parent; every other symbol above feeds into them. --- ## 9. How it all fits together Read this as a short story, not a wiring diagram. Three streams flow into the topic: > [!intuition] The three streams > **Stream 1 — how fast (one ball).** The velocity **arrow** $\mathbf{v}$ splits into $v_x,v_y,v_z$; isotropy gives $\overline{v_x}=0$ and $\overline{v_x^2}=\tfrac13\overline{v^2}$. Feed this into the bouncing-off-a-wall argument and you get pressure, and from pressure the **RMS speed** $v_{rms}$. Weight the speeds by the Maxwell–Boltzmann shape $v^2e^{-av^2}$ (Section 6) instead and you get the **mean speed** $\bar v$. > > **Stream 2 — how crowded.** The count $N$ in volume $V$ gives density $n=N/V$ — the "how many targets" number. > > **Stream 3 — how big a target.** The diameter $d$ gives the collision disc $\sigma=\pi d^2$. > > **The confluence.** Crowding ($n$), target size ($\sigma$), and how fast a molecule tours the room ($\bar v$) — corrected by the moving-target $\sqrt2$ — together fix the **mean free path** $\lambda$. RMS speed, mean speed, and mean free path — the three headline results — all trace back to these three streams. ```mermaid graph TD V["velocity arrow v"] --> S1["components vx vy vz"] S1 -->|isotropy| WALL["wall bounce pressure"] WALL --> VRMS["v_rms"] V -->|weight by MB shape| VBAR["v_bar mean speed"] N["count N"] --> DEN["density n"] VOL["volume V"] --> DEN DIAM["diameter d"] --> SIG["cross section pi d^2"] DEN --> LAM["mean free path lambda"] SIG --> LAM VBAR --> LAM SQRT2["moving target root 2"] --> LAM ``` If you want the deeper origin of the bell-and-prefactor weighting, that lives in [[Maxwell-Boltzmann distribution]]; the pressure argument is expanded in [[Kinetic theory of pressure]]; the $PV=Nk_BT$ link is [[Ideal gas law]]; the "energy per component" idea is [[Equipartition theorem]]. These same numbers reappear in [[Diffusion and effusion]], [[Viscosity and thermal conductivity of gases]], and [[Speed of sound in gases]]. --- ## Equipment checklist Self-test: can you state each in one sentence *before* revealing? What does the overbar $\overline{(\cdot)}$ mean? ::: Add the quantity for every molecule and divide by $N$ — an average. Difference between $N$ and $n$? ::: $N$ is a plain count; $n=N/V$ is molecules per cubic metre (crowding). How do we write velocity vs speed? ::: Velocity is the bold/arrow vector $\mathbf{v}$ (has direction); speed $v=|\mathbf{v}|$ is its length, always $\ge 0$. Why square before averaging for RMS? ::: Kinetic energy $\propto v^2$; averaging $v^2$ captures typical energy, and squaring weights fast molecules more. What is $\overline{v_x}$ and why? ::: Zero — for every molecule going $+x$ another goes $-x$, so signed components cancel; only squared components survive. Why is $\overline{v_x^2}=\tfrac13\overline{v^2}$? ::: Isotropy — no direction is special, so the three squared components share $\overline{v^2}$ equally. What does the constant $a$ in $e^{-av^2}$ stand for? ::: $a=m/(2k_BT)$, so $av^2$ is kinetic energy over thermal energy. How is $\bar v$ defined as an integral? ::: $\bar v=\int_0^\infty v\,f(v)\,dv \big/ \int_0^\infty f(v)\,dv$ — sum of (speed × how many) over the count. What value does that integral give? ::: $\bar v=2/\sqrt{\pi a}=\sqrt{8k_BT/\pi m}$. Is the speed distribution just $e^{-mv^2/2k_BT}$? ::: No — it has a $v^2$ prefactor and a normalisation constant, which is why it peaks at a non-zero speed. What is $d$ in the collision picture? ::: The molecule's diameter; two molecules collide when their centres are $d$ apart. Why is $\sigma=\pi d^2$ and not $\pi (d/2)^2$? ::: The effective target radius is the full $d$ (sum of two radii), so the disc area uses $d$. Where does the $\sqrt2$ in $\lambda$ come from? ::: The targets also move; the average relative speed is $\sqrt2\,\bar v$, raising the collision rate and shortening $\lambda$. Difference between $k_B$ and $R$? ::: $k_B$ is energy-per-molecule per kelvin; $R=k_B N_A$ is energy-per-mole per kelvin. Units of $M$ in $\sqrt{3RT/M}$? ::: Kilograms per mole (SI), e.g. $0.028$ for $\text{N}_2$. Units of $N_A$? ::: Per mole, $\text{mol}^{-1}$ ($\approx 6.022\times10^{23}\,\text{mol}^{-1}$). What does $\lambda$ physically measure? ::: The average distance a molecule flies between collisions. What does $A\propto B$ tell you? ::: $A$ is a fixed multiple of $B$; double $B$ and $A$ doubles.

1. What is a "molecule" and how do we picture it?

2. Counting molecules: N, V and number density n

3. Speed, velocity, and its three components vx​,vy​,vz​

4. The bar and "mean": what vˉ and v2 mean

5. The tools that sneak in: integrals, e−x, and the ∝ sign

6. How the mean speed vˉ is actually built from the distribution

2. Counting molecules: $N$ , $V$ and number density $n$

3. Speed, velocity, and its three components $v_{x}, v_{y}, v_{z}$

4. The bar and "mean": what $\overset{v}{ˉ}$ and $\overline{v^{2}}$ mean

5. The tools that sneak in: integrals, $e^{- x}$ , and the $\propto$ sign

6. How the mean speed $\overset{v}{ˉ}$ is actually built from the distribution