Why Ice Skates Glide

Equivalent forms

P = \frac{mg}{A}

T_{\text{glide}} \uparrow \text{ from frictional heating } \dot Q = \mu N v

A famous case where the 'obvious' Clausius-Clapeyron explanation fails quantitatively — a lesson in checking the numbers, not just the sign.

Symbol	Name	SI unit	Dimension	Range
$P$	Contact pressureoutput Pressure of blade on ice	Pa	M L⁻¹ T⁻²	100000 – 100000000
$m$	Skater mass Mass pressing on the blade	kg	M	30 – 120
$A$	Contact area Blade-ice contact area	m²	L²	0.00005 – 0.002
$v$	Glide speed Speed of the skate over the ice	m/s	L T⁻¹	0 – 12

Unit systems

SI:: P in Pa, T in K
natural:: reduced pressure
CGS:: P in $dyn/cm^{2}$

Where it holds

Pressure-melting via Clausius-Clapeyron is rigorous but tiny; the dominant glide mechanisms (frictional heating, intrinsic premelting layer) are still active research.

Discovery

James Thomson / Michael Faraday (premelting) · 1850

James Thomson predicted pressure-melting in 1850 and brother Lord Kelvin confirmed it; but Faraday's 1850 premelting idea and modern friction studies show pressure is far too weak to explain skating.

Try this

What makes ice so slippery — and almost nothing else?

The textbook answer (pressure melts the ice) is mostly wrong: the math gives a temperature drop of hundredths of a degree. The real culprits are frictional heating and a liquid-like premelting layer.

Research status: active

Common misconceptions

'Pressure from the blade melts the ice' is quantitatively false at normal temperatures; glide comes mainly from frictional melting plus an intrinsic surface premelting layer.

Derivation

Clausius-Clapeyron for the solid-liquid line:

dP/dT = L_f/(T \Delta V

_melt).

Ice is unusual because

\Delta V

_melt < 0 (water is denser than ice), so dT_m/dP < 0.

Plugging

T = 273\,\mathrm{K}

,

L_f = 334\,\mathrm{kJ/kg}

,

\Delta V = -9\times 10^{-5} m^{3}/kg

gives

dT_m/dP \approx -7.4\times 10^{-3} K/atm

— about 0.1 K even at several MPa.

Limiting cases

Realistic skating pressure

(\sim few MPa)

⟶ Melting-point drop

\sim 0.1\,\mathrm{K}

— far too small to melt ice well below

0^{\circ}C

High speed⟶ Frictional heating

\mu Nv

dominates; thicker meltwater film, more slip

Very cold ice (<

-30^{\circ}C)

⟶ Premelting layer thins and ice becomes noticeably less slippery

Clausius-Clapeyron Equation→Why Ice Skates Glide→First Law of Thermodynamics→Fourier's Law of Heat Conduction→