The Verification Venue · pointed at a thing the schoolbook gets wrong

The Ice That Pressure Didn't Melt

Everyone learns it: a skate glides because the blade's pressure melts the ice into a slick film. It's a tidy story with one problem — run the numbers on the real ice-water phase line and the pressure melts almost nothing.

The physics behind "pressure melts ice" is real. Water is the rare substance that expands when it freezes, so squeezing ice does push it back toward liquid — the ice-water boundary slopes the "wrong" way. The law is the Clausius-Clapeyron relation on the solid-liquid line: dT/dP = T·(v_water − v_ice)/L_f. The catch is the size of that slope. Move the sliders and watch how little melting your weight actually buys.

Contact pressure

6.9 MPa

P = m·g / A

Melting point now

−0.51 °C

depressed by 0.51 K

Still solid by

4.49 K

vs the rink at −5 °C

Skater mass — 70 kg

Heavier skater, more pressure — but it barely moves the needle.

Blade contact area — 1.00 cm²

A sharp blade touches only a sliver of ice. Shrink it to a knife edge and try to win.

Rink temperature — −5.0 °C

Indoor rinks run −5 to −7 °C; outdoor ice goes colder. Pressure-melting must reach this to win.

The blue curve is the genuine ice-water phase line, its slope set by the equation above: a near-vertical wall, because the volume change on freezing is tiny. Your weight slides the operating point a hair to the left — the melting point dips by a fraction of a degree. The rink sits far below that. The dot never crosses into liquid. Pressure-melting is ruled out, not by opinion but by the slope of a line.

dT = (dT/dP) · P = (−0.0743 K/MPa) · 6.86 MPa = −0.51 K

And the line itself ends: pressure-melting has a wall

Suppose you refuse to lose and keep adding weight. Here is the entire melting line of ordinary ice, computed from the IAPWS 2011 reference equation rather than the linear slope (the curve bends as it falls). Reaching −1 °C takes 13.2 MPa, nearly double the default blade's 6.9. Reaching −5 °C takes 60.0 MPa, almost 9× it. And at −21.985 °C / 208.566 MPa the line simply stops: squeeze harder and the ice becomes ice III, another solid. Below −22 °C, pressure-melting is not merely too small. It is impossible at any pressure, for any skater that could ever exist.

checking the curve against its published anchors

So what does make ice slippery? Two mechanisms that survive the arithmetic. Surface premelting — a disordered, liquid-like film that exists on ice at rest, below 0 °C, observed by Faraday in 1850 and confirmed at the nanoscale since. The measurements put numbers on it: AFM bounds the film's thickness between about 12 nm at −24 °C and 70 nm at −0.7 °C (upper limits; different probes disagree on the exact values), and steel-on-ice friction falls from a dry-solid μ ≈ 0.5 at −100 °C to μ ≈ 0.01 near −7 °C, tracking the measured mobility of the surface molecules (Weber et al., 2018). And frictional heating — the glide itself dumps power into the contact. A skater at speed dissipates P_fric = μ·m·g·v watts into a patch the size of a fingernail. That heat can reach the melting point; your static weight can't.

The mechanism that works — frictional heat

Glide speed — 6.0 m/s

A brisk recreational glide is ~5 m/s; speed skaters hit 12+.

Ice friction coefficient μ — 0.006

Ice friction is famously low, ~0.003–0.01.

P_fric = μ·m·g·v = 0.006 · 70 · 9.81 · 6 = 24.7 W → 247 kW/m² into a 1 cm² patch

The check — every number recomputed in front of you

The phase-line slope is computed live from the real densities of ice and water and the latent heat of fusion. The melting-point depression and the shortfall against the rink follow directly. Constants used:

density of ice (0 °C)	916.7 kg/m³
density of water (0 °C)	999.84 kg/m³
latent heat of fusion	333 550 J/kg
melting point (1 atm)	273.15 K

—

The slope dT/dP = T·Δv/L_f = −0.0743 K/MPa (inverse: −13.5 MPa/K). Run it yourself: node research/the-ice-that-pressure-didnt-melt/verify-the-ice-that-pressure-didnt-melt.mjs

The full melting curve in the wall panel above is the IAPWS R14-08(2011) reference equation; the page recomputes its three published anchor values live (611.657 Pa at 273.16 K; 138.268 MPa at 260.0 K, the reference paper's own program-verification value; 208.566 MPa at the ice III corner, 251.165 K). A second offline verifier covers the curve, its inverse, and the skater arithmetic, 12 checks: node research/ice-slippery/verify.mjs. It imports the page's engine, so download verify.mjs and engine.mjs into one folder before running it. The sources are listed at the foot of the page.

What's idealised here, and what's exactly true

Exactly true — the negative result. The Clausius-Clapeyron slope of the ice-water line is −0.0743 K/MPa, a direct consequence of three measured numbers (the two densities and the latent heat). That slope is small, and so the melting-point depression from any realistic skating pressure is a fraction of a degree — far too little to liquefy a rink at −5 to −10 °C. This is the robust, certain part, and it's what the instrument proves: pressure-melting cannot be the explanation.

Free choices you can see. The default 70 kg / 1 cm² / −5 °C are typical but adjustable — every slider re-runs the equation. The "1 cm² contact" is generous to the pressure-melting story; real skate contact patches are often smaller, which raises pressure, yet even a 0.1 cm² knife edge (≈69 MPa) still falls short of a −10 °C rink. The friction coefficient μ (0.002–0.02), glide speed, and the assumption that nearly all frictional power enters the contact are representative, named choices, not universal constants.

Idealised. We take constant material properties at 0 °C; ignore the ~0.1 MPa of atmospheric pressure (negligible next to MPa contact pressures); treat the contact pressure as uniform over the patch; and use the linear slope rather than integrating the slightly-curved phase line (the curvature is irrelevant over a few MPa). The frictional-heating budget is an order-of-magnitude argument: it shows the power is more than enough to reach melting, not the exact film thickness.

Genuinely unsettled: the positive mechanism. Why ice is slippery is still an active research question, and in 2025 it visibly reopened. Surface premelting (Faraday, 1850; confirmed by modern AFM and X-ray work), frictional heating, and the unusual viscoelastic rheology of the nanometre meltwater film (Canale et al., 2019) all contribute, in proportions that depend on temperature and speed. In August 2025, Atila, Sukhomlinov and Müser (Saarland University) proposed a fourth account in Phys. Rev. Lett. 135, 066204: in their simulations sliding itself disorders the ice surface into an amorphous, liquid-like film, with no melting, no frictional heat and no pressure needed: "cold, displacement-driven amorphization", in the authors' words. The press release announced the puzzle solved; the field visibly has not: Luis MacDowell (Complutense University of Madrid) argues the rival mechanisms "operate simultaneously to one or the other degree", while Atila himself allows that "either the slipperiness of ice is coming from a combination of all of them or a few of them, or there is something else that we don't know yet" (both as quoted in Quanta Magazine's December 2025 survey of the dispute). This page deliberately stops at the clean, provable claim, that pressure-melting is quantitatively too small, and presents the positive story honestly as a frontier, not a settled fact.

A distinct stratum. The nearest neighbour, The Cold That Isn't There (why metal feels colder than wood), is about thermal perception via effusivity — different physics entirely. This one is phase-line thermodynamics and surface science.