checking…
The Newtonian Drift · a bench of heat

Where Pressure Comes From

The same hard discs that relax to the Maxwell–Boltzmann bell next door also drum on the walls. Each bounce flips one velocity component and hands the wall a kick of 2m|v⊥| — nothing more. Add up those kicks per unit wall, per unit time, and out falls a pressure. This bench measures that drumbeat live and shows it settling onto P·A = N·kT — the exact law the Carnot engine one wing over simply assumes. No law is plugged in; it is counted off the walls. Squeeze the box and watch the pressure climb to hold the line.

The box

discs N
wall hits
temperature ⟨½mv²⟩
kT
box area A = W·H

The walls flash on each hit, brighter for a harder kick (∝ 2|v⊥|). Colour is speed — slow to fast. Drag the right wall to squeeze the box into a rectangle.

The wall drumbeat  →  pressure, vs. the law N·kT/A

The amber line is N·kT/A — the ideal-gas law for the temperature the gas reads on itself. The teal trace is the running average of the wall drumbeat, P = Σ 2|v⊥| / (perimeter · time), scrolling and settling onto the line. Nothing is fitted: P is counted off the walls.

Head-to-head · P from collisions vs P from the law · & the dimensionless cross with Carnot
P_wall · counted off the walls
Z_sim = —
P_law = N·kT/A · the law
Z_carnot = —
measuring…

Controls — drag any of these and the law re-measures live

Try “½-count the kick”: a wall reflection flips the velocity and delivers 2m|v⊥|, not m|v⊥|. Count only one mirror and P falls to exactly half — the trace drops far below the amber line and the head-to-head flips red. That factor of two is the physics.

One bounce, one kick: 2m|v⊥|

A disc hits a wall and bounces: its perpendicular velocity flips sign, v⊥ → −v⊥. The wall absorbs the change in momentum, Δp = |(−v⊥) − v⊥| = 2|v⊥| (with mass m=1). That is the whole atom of pressure — exact, no statistics. Sum these kicks over every wall hit, divide by the perimeter you spread them over and the time you waited, and you have a force per unit length: a 2-D pressure, P = Σ2|v⊥| / (perimeter·t).

Why it lands on N·kT/A

The virial makes it exact in the idealization. Average the wall flux over a thermal gas and the bookkeeping collapses to P·A = N·kT — the same ⟨½v²⟩=kT equipartition the M–B bell next door is fitted to. The self-test proves it with zero sim noise: on a synthetic thermal velocity set the virial identity P·A/N = kT holds to machine precision. The live run then has only finite-time scatter to shed — which it does, the residual shrinking toward ~1%.

Tiny discs ≈ a point gas

The discs render as fine dust on purpose. The law P·A=N·kT is exact only in the point limit; fatter discs would show the van der Waals excess — the wall is shoved a little harder when discs take up room (excluded area). At the M–B page's 6% packing the residual is a large, time-stable +15%; here at ~0.2% packing it falls to ≲1%, genuinely dominated by finite-time scatter. The residual is biased slightly positive and never hits zero — it is convergent, never perfect.

The honest part: what “agrees with Carnot” really means

It is one law under one dictionary — particle count N ↔ moles n, Boltzmann k_B ↔ gas constant R, with N·k_B = n·R (R = N_A·k_B). Carnot ships a 3-D gas with the real R = 8.314 baked into its pressure(n,T,V); this bench is the 2-D ideal law with k_B ≡ 1 (so kT carries energy and A is an area). A literal pressure(N,kT,A) would read off by exactly that factor of R — so the cross is not a lucky meeting of two formulas. It is the single dimensionless law PV/(NkT) = 1, asserted once on each side of the dictionary: Z_carnot = 1 exactly (it is nRT/V over nRT/V), and Z_sim → 1 off the wall count. Same law, twice.

Self-test detail