Visual walkthrough — Kinetic theory — pressure derivation, temperature as mean KE

Before we start, three words in plain English (we will use them, so we define them first):

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Step 1 — Put one molecule in a box

WHAT. We build a cube of side length $L$. Its volume is $V=L^3$. We drop in one molecule of mass $m$ moving with velocity that we split into three arrows: $v_x$ (sideways), $v_y$ (up), $v_z$ (in/out). We stare only at the right wall, the one facing the $x$-arrow.

WHY. A whole gas is chaos. But every molecule obeys the same simple rule, so if we fully understand one, we can add up the rest by multiplication. Start small.

PICTURE. The red arrow is the full velocity. The cyan arrow is just its $x$-part, $v_x$ — that is the only part that pushes the right wall. The up/down and in/out parts slide along the wall and never hit it.

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Step 2 — One bounce: how much momentum changes

WHAT. The molecule reaches the right wall and bounces back elastically (speed unchanged, only direction reversed). Its sideways velocity flips from $+v_x$ to $-v_x$. The momentum change is $\Delta p_x=\underset{\text{after: heading left}}{\underbrace{m{v}_x}}-\underset{\text{before: heading right}}{\underbrace{(-m{v}_x)}}=2m{v}_x$

Read the equation left to right: $m{v}_x$ is the momentum after the bounce (now negative, i.e. pointing left, but we wrote the magnitude with the sign built into the subtraction); $(-m{v}_x)$ is what it had before; subtracting a negative doubles it. The wall received this whole $2m{v}_x$ as a kick.

WHY. Pressure is ultimately momentum delivered to the wall. So the very first thing we must know is: how much momentum does one hit deliver? That is $\Delta p_x$. Because the collision is elastic (see Elastic Collisions and Momentum), no speed is lost — the full amount reverses, giving the maximum possible $2m{v}_x$, not something smaller.

PICTURE. Before: cyan arrow points right, length $m{v}_x$. After: cyan arrow points left, same length. The amber arrow shows the difference — a right-pointing kick of size $2m{v}_x$ that the molecule gave the wall.

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Step 3 — How often does the same molecule come back?

WHAT. After bouncing, the molecule must fly across to the far wall (distance $L$) and return (another $L$) before it can hit the right wall again. That round trip is $2L$, travelled at sideways speed $v_x$: $\Delta t=\frac{2L}{v_x}$

Term by term: $2L$ is the round-trip distance, $v_x$ is the sideways speed, and distance÷speed = time. A bigger $v_x$ makes $\Delta t$ smaller — faster molecules pound the wall more often.

WHY. One kick isn't pressure; a steady rain of kicks is. To turn kicks into a steady force we need to know how much time passes between kicks. That gap is $\Delta t$.

PICTURE. Follow the dashed path: right wall → left wall → back. The label marks the total length $2L$; the clock shows the elapsed time $\Delta t=2L/v_x$.

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Step 4 — One molecule's steady average force

WHAT. Force is momentum delivered per unit time. One molecule delivers $\Delta p_x=2m{v}_x$ every $\Delta t=2L/v_x$ seconds, so its average force on the wall is $F_1=\frac{\Delta p_x}{\Delta t}=\frac{2m{v}_x}{2L/v_x}=\frac{m{v}_x^2}{L}$

Watch the algebra: the two $2$'s cancel; dividing by $2L/v_x$ means multiplying by $v_x/2L$; one $v_x$ from the top times the $v_x$ from the flip gives $v_x^2$. Note the square — this is the first appearance of $v_x^2$, and it is why $\overline{v^2}$ (not average speed) will run the whole show.

WHY. We use Newton's second law in its momentum form $F=\Delta p/\Delta t$ (rate of momentum transfer) rather than $F=ma$, because a collision has no smooth acceleration — just discrete kicks. Rate-of-momentum is exactly the tool for "many tiny impulses averaged into a steady push."

PICTURE. The staircase graph: each vertical jump is one kick of $2m{v}_x$; the steps are spaced $\Delta t$ apart. The straight amber line is their average slope — a smooth steady force $F_1=m{v}_x^2/L$.

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Step 5 — Add up all $N$ molecules

WHAT. A real gas has a huge number $N$ of molecules, each with its own sideways speed $v_{x1},v_{x2},\dots$. Forces add, so $F=\frac{m}{L}(v_{x1}^2+v_{x2}^2+\cdots +v_{xN}^2)=\frac{m}{L}N\overline{v_x^2}$

The bracket is a sum of $N$ squared speeds. We define $\overline{v_x^2}$ as that sum divided by $N$ — the average of $v_x^2$. So the sum equals $N\overline{v_x^2}$, and we pull it out cleanly. This is the standard "sum = count × average" move.

WHY. We cannot track a trillion molecules individually. But we don't have to: only their average squared sideways speed matters for the total force. Averaging is what lets one clean number stand in for an ocean of molecules.

PICTURE. A cloud of molecules, each a small cyan dot with its own little $v_x^2$ bar. The amber bar on the right is their average $\overline{v_x^2}$ — one number summarising the swarm.

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Step 6 — Isotropy: trade $\overline{v_x^2}$ for $\frac{1}{3}\overline{v^2}$

WHAT. The gas is random — no direction is special. So on average the sideways, up/down, and in/out jiggles are equal: $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$ The full speed obeys Pythagoras in 3D, $v^2=v_x^2+v_y^2+v_z^2$. Averaging both sides:

\quad\Rightarrow\quad \boxed{\;\overline{v_x^2} = \tfrac13\,\overline{v^2}\;}$$ Term by term: the left $\overline{v^2}$ is the average of the *whole* speed squared; on the right the three equal pieces collapse to $3\overline{v_x^2}$; divide by 3 to isolate the sideways piece. **WHY.** We *want* $\overline{v^2}$, not $\overline{v_x^2}$, in our final answer — because $\overline{v^2}$ is what connects to **energy** ($\tfrac12 m\overline{v^2}$) and hence temperature. This step is the bridge from "sideways" to "total." The famous factor $\tfrac13$ is nothing mysterious: it is literally **the 3 spatial dimensions**. **PICTURE.** A 3D box with the red velocity arrow and its three cyan shadow-arrows on the axes. The three shadows are, on average, the same length — that equality *is* isotropy. ![[deepdives/dd-physics-1.7.09-d2-s06.png]] > [!intuition] Where the $\tfrac13$ really comes from > In a flat 2D world (only $x,y$) you'd have $\overline{v^2}=2\overline{v_x^2}$, giving $\tfrac12$. In our 3D world it's $\tfrac13$. The number *counts the directions a molecule can move*. It is geometry, never a fudge factor. --- ## Step 7 — Divide by area: pressure at last **WHAT.** Pressure is force per area, and the wall's area is $L^2$. Substitute $\overline{v_x^2}=\tfrac13\overline{v^2}$ from Step 6: $$P = \frac{F}{L^2} = \frac{\dfrac{m}{L}\,N\,\overline{v_x^2}}{L^2} = \frac{m\,N\,\tfrac13\overline{v^2}}{L^3} = \frac{1}{3}\,\frac{N m\,\overline{v^2}}{V}$$ Term by term: numerator $NmL^{-1}\overline{v_x^2}$ is the total force from Step 5; dividing by $L^2$ (the area) spreads it over the wall; the three $L$'s downstairs combine into $L^3 = V$, the volume. The $\tfrac13$ walked in from isotropy. **WHY.** Force alone can't be compared between a fist-sized box and a room; pressure — force per area — is the fair, size-independent quantity we actually measure with a gauge (see [[Pressure as Force per Area]]). This division is the final translation into a lab-measurable number. **PICTURE.** The finished cube: total force $F$ smeared evenly across the $L^2$ wall as a field of small amber arrows, with the boxed result printed beside it. ![[deepdives/dd-physics-1.7.09-d2-s07.png]] > [!formula] The result, three equivalent faces > $$\boxed{\,P = \frac{1}{3}\frac{N m\,\overline{v^2}}{V}\,} > \qquad P = \tfrac13\rho\,\overline{v^2} > \qquad P = \tfrac13\rho\,v_{rms}^2$$ > with **density** $\rho = Nm/V$ and **rms speed** $v_{rms}=\sqrt{\overline{v^2}}$. > This is exactly what pairs with $PV=Nk_BT$ to give temperature — see [[Ideal Gas Law PV=nRT]] and [[Boltzmann Constant and Equipartition]]. --- ## Step 8 — Edge and degenerate cases (never leave a gap) **WHAT.** We check the corners so no reader hits an unshown scenario. - **A molecule with $v_x = 0$** (moving purely up/down/in/out): it never reaches the right wall, so it contributes $F_1 = mv_x^2/L = 0$. Correct — it should push the *side* walls, not this one. No division-by-zero problem because $F_1$ uses $v_x^2$, not $1/v_x$. - **All molecules frozen, $\overline{v^2}=0$** (absolute zero, ideal limit): then $P=0$. A gas with zero jiggle exerts zero pressure — the formula behaves. - **Angled hits.** A molecule striking at a slant still only reverses its $v_x$; the $v_y,v_z$ parts glide along the wall unchanged, delivering no perpendicular momentum. So our whole derivation, built on $v_x$ alone, already covers every angle. - **Very fast molecules** dominate: because force scales as $v_x^2$, the fast ones contribute disproportionately — which is *why* the correct average is $\overline{v^2}$ (rms), heavier-weighting fast molecules, not the plain average speed. **WHY.** A formula you trust only in the "nice" case is a formula you don't trust. Testing $v_x=0$, $\overline{v^2}=0$, and slanted hits shows the derivation was complete, not lucky. **PICTURE.** Three mini-panels: (left) a molecule sliding along the wall with $v_x=0$ giving no hit; (middle) a slant hit with only its cyan $v_x$ reversing; (right) the $P$-vs-$\overline{v^2}$ line passing cleanly through the origin. ![[deepdives/dd-physics-1.7.09-d2-s08.png]] --- ## The one-picture summary **WHAT.** Everything above, compressed into a single flow: bounce → kick $2mv_x$ → repeat every $2L/v_x$ → force $mv_x^2/L$ → sum to $\tfrac{m}{L}N\overline{v_x^2}$ → isotropy $\tfrac13\overline{v^2}$ → divide by area → $P=\tfrac13\frac{Nm\overline{v^2}}{V}$. ![[deepdives/dd-physics-1.7.09-d2-s09.png]] ```mermaid flowchart LR A["one bounce reverses v_x"] --> B["kick = 2 m v_x"] B --> C["repeats every 2L over v_x"] C --> D["force = m v_x squared over L"] D --> E["sum over N gives N times mean v_x squared"] E --> F["isotropy gives one third mean v squared"] F --> G["divide by area L squared"] G --> H["P = one third N m mean v squared over V"] ``` > [!recall]- Feynman retelling — say it in plain words > Imagine one tiny ball trapped in a box, flying side to side. Every time it smacks the right wall it bounces straight back, and in doing so it shoves the wall — the shove is *twice* its sideways punch, because it went from heading in to heading out. Then it has to travel all the way across and back before it can shove again, and how long that takes depends on how fast it's going sideways. Punch-per-shove divided by time-between-shoves is a steady push from that one ball. Now fill the box with trillions of balls; their pushes add up, and since they all move randomly, the sideways jiggle is exactly one-third of the total jiggle. Spread that total push over the wall's area and you've got pressure. The punchline: **pressure is just molecular jiggle, counted, averaged, and spread over the wall** — and the mysterious "one-third" is nothing but the three directions space offers. > [!recall]- Rebuild the chain (cover the answers) > What one quantity flips in a wall bounce, and what is $\Delta p_x$? ::: only $v_x$ flips; $\Delta p_x = 2mv_x$ > Time between hits on the same wall? ::: $\Delta t = 2L/v_x$ > Force from one molecule? ::: $F_1 = mv_x^2/L$ > Force from all $N$? ::: $F = \tfrac{m}{L}N\overline{v_x^2}$ > Isotropy gives which substitution? ::: $\overline{v_x^2}=\tfrac13\overline{v^2}$ > Final pressure formula? ::: $P=\tfrac13\,Nm\overline{v^2}/V$ > [!mnemonic] Remember the skeleton > **"Two kicks, round trip, square it, sum it, third it, spread it."** > $2mv_x \to 2L/v_x \to v_x^2 \to N\overline{v_x^2} \to \tfrac13\overline{v^2} \to \div\,\text{area}$. --- Where this leads next: pair $P=\tfrac13\frac{Nm\overline{v^2}}{V}$ with the gas law to get temperature-as-energy — then explore [[Maxwell-Boltzmann Speed Distribution]], [[Internal Energy of Ideal Gas]], and [[Degrees of Freedom and Molar Heat Capacity]].