Regelation: Why Ice Melts Under a Wire and Refreezes Behind It

The anomaly: water's melting line leans backward

For almost every substance, squeezing the solid raises its melting point — press hard enough and a warm solid stays solid. Water breaks this rule. The equation that governs any coexistence line between two phases is the Clausius–Clapeyron relation:

• dP/dT = L / (T · ΔV), where L is the molar (or specific) latent heat of fusion, T is the absolute temperature, and ΔV = V_liquid − V_solid is the volume change on melting.
• For most solids ΔV > 0 (liquid is less dense), so dP/dT > 0 and the melt line slopes to the right — higher pressure, higher melting point.
• For water, ΔV < 0. Ice Iₕ has density ρ_ice ≈ 917 kg/m³ while liquid water at 0 °C is ρ_water ≈ 1000 kg/m³. Melting shrinks the volume, so ΔV is negative and dP/dT < 0: the melt line leans backward.

Invert it to get the quantity regelation cares about: dT/dP = T · ΔV / L. This is the rate at which the melting temperature falls as you crank the pressure — the single number that sets the whole phenomenon.

Putting numbers on the melting-point depression

Everything is measurable. For 1 kg of water freezing/melting at T = 273.15 K:

• Latent heat of fusion: L_f = 334 kJ/kg = 3.34 × 10⁵ J/kg.
• Specific volumes: v_water = 1/1000 = 1.000 × 10⁻³ m³/kg; v_ice = 1/917 = 1.091 × 10⁻³ m³/kg. So Δv = v_water − v_ice = −9.1 × 10⁻⁵ m³/kg.
• Plug in: dT/dP = T · Δv / L_f = (273.15 × (−9.1 × 10⁻⁵)) / (3.34 × 10⁵) ≈ −7.4 × 10⁻⁸ K/Pa.

That is roughly −0.0074 K per bar, or −0.074 K per 10 MPa (≈100 atm). The effect is genuinely small: to depress the melting point by a full degree you need about 13.5 MPa (135 atmospheres). This tiny slope is why the demonstration needs a thin wire and heavy weights to concentrate stress — and why several popular "explanations" that lean on it (skating, glacier sliding) turn out to be exaggerated.

The mechanism: melt below, flow around, refreeze above

Regelation is a two-step, self-sustaining cycle driven by a pressure gradient across a small object embedded in ice:

• Step 1 — pressure melting (high-P face). Under the loaded wire, the contact stress can reach tens of MPa. Locally the melting point drops below the ice's actual temperature (held near 0 °C), so a thin film of meltwater appears. The film thickness is typically ~10 nm to a micron.
• Step 2 — flow. Water is squeezed out of the high-pressure zone under the wire and migrates around its sides to the region above the wire, where the pressure is back to ambient.
• Step 3 — refreezing (low-P face). Above the wire, at ambient pressure, the same water is now below its melting point, so it refreezes, releasing its latent heat.

Here is the crucial energetics: the latent heat L_f = 334 kJ/kg released by refreezing above must be conducted back down through the wire to supply the latent heat absorbed by melting below. The wire is the thermal shortcut. That is why the whole process is rate-limited by heat conduction through the wire, and why the wire's material matters enormously.

Why the wire's material sets the speed

Faraday's original arrangement used a fine wire; later experimenters found the cut rate depends dramatically on thermal conductivity κ. The heat flux the wire can shuttle from the refreezing face to the melting face scales as Q ∝ κ · A · ΔT / d, and the tiny ΔT across the wire is itself set by the melting-point depression, ΔT ≈ |dT/dP| · P.

• A copper wire (κ ≈ 400 W/m·K) cuts through ice far faster than a stainless-steel wire (κ ≈ 15 W/m·K) or a nylon/cotton thread (κ ≈ 0.2 W/m·K) under identical load. Thermally insulating threads may barely move at all — the latent heat cannot get back to the cut.
• Typical laboratory cutting speeds are slow — millimetres per hour to a few cm/hour — precisely because 334 kJ must be pumped through a hair-thin conductor for every kilogram melted.
• Increasing the load raises the pressure melting rate but also drives faster refreezing above; the wire simply descends faster until conduction can't keep up.

This dependence is the fingerprint that distinguishes true regelation from a wire that merely fractures the ice. If the ice cracked, material and conductivity wouldn't matter. Because they do, we know melting and refreezing are really happening.

Where regelation really matters — and where it's a myth

Glacier flow (real). At the base of a glacier, ice slides past bedrock bumps. On the upstream (high-stress) side of a bump, pressure melting lets ice deform and lubricate; meltwater flows to the downstream low-stress side and refreezes. This regelation sliding mechanism, analyzed by J. F. Nye and J. Weertman in the 1950s–60s, dominates for small obstacles (centimetre scale); large obstacles are bypassed instead by plastic creep. The two mechanisms combine to a minimum-resistance "controlling obstacle size" of order 0.1–1 m.

Snowballs and ice fusion (real). Packing a snowball works partly by regelation: pressure at grain contacts melts a little, and on release it refreezes, welding crystals together. Sintering of snow into firn uses the same physics.

• Ice skating (mostly a myth). The classic story — that a skate's pressure melts ice to a slippery film — fails the arithmetic. Even a 70 kg skater on a blade edge of ~10⁻⁴ m² reaches only a few MPa, depressing the melting point by well under 1 K — nowhere near enough to explain gliding at −10 °C. Modern understanding attributes the low friction chiefly to a pre-existing quasi-liquid surface layer (present even without load, studied by Faraday himself) and to frictional heating from the moving blade.

Regime of validity and the limits of ice Iₕ

The −7.4 × 10⁻⁸ K/Pa slope is a near-linear approximation valid only for ice Iₕ, the ordinary hexagonal ice, up to about 209 MPa. At that pressure the melting curve reaches its minimum temperature — the ice Iₕ / ice III / liquid triple point at ≈ −22 °C (251 K) and 209 MPa. Push harder and you enter denser high-pressure ice phases (ice III, V, VI, VII), for which ΔV flips sign and the melting point rises steeply with pressure. So you cannot depress water's freezing point indefinitely: −22 °C is the floor for pressure melting.

• Assumptions in the simple formula: L_f and Δv are treated as constant, and dT/dP is integrated as a straight line. Over the first ~50 MPa this is excellent; near 209 MPa the real curve is noticeably nonlinear.
• Temperature matters: regelation needs the bulk ice near its melting point. In deeply cold ice (say −30 °C), a wire cannot melt anything — the pressure can never drop the melting point below the actual temperature — and it will only fracture or stall.
• The dimensionless comparison is essentially whether |dT/dP| · P_contact exceeds the ice's undercooling (T_melt − T_ice). Only then does the wire descend.

A worked feel: how fast does a wire cut?

Estimate a copper wire of diameter d = 0.5 mm loaded so the contact pressure under it is P ≈ 20 MPa. From dT/dP the melting-point depression is ΔT ≈ 7.4 × 10⁻⁸ × 2 × 10⁷ ≈ 1.5 K — this is the temperature difference the wire's ends see between the melting (bottom) and refreezing (top) faces.

• Heat that must flow per unit length cut: to melt a slab of ice of the wire's width w ≈ d and thickness equal to the descent, the energy is ρ_ice · L_f per unit volume = 917 × 3.34 × 10⁵ ≈ 3.06 × 10⁸ J/m³.
• Heat the copper can deliver: flux q ≈ κ · ΔT / ℓ, with κ_Cu = 400 W/m·K and a conduction path ℓ of order the wire diameter (~5 × 10⁻⁴ m), giving q ≈ 400 × 1.5 / 5 × 10⁻⁴ ≈ 1.2 × 10⁶ W/m².
• Descent speed ≈ q / (ρ_ice · L_f) ≈ 1.2 × 10⁶ / 3.06 × 10⁸ ≈ 4 × 10⁻³ m/s in this crude upper bound — but real cuts are far slower (cm/hour) because most heat leaks sideways into the surrounding ice and the effective ΔT along the wire is much smaller than the full 1.5 K.

The lesson of the estimate is qualitative but robust: swap copper for steel (κ 27× lower) and the wire crawls; swap for nylon and it essentially stops. The latent heat is the currency and the wire's conductivity is the exchange rate.

How much does pressure actually lower ice's melting point? Real numbers along the ice Iₕ–water line.

Pressure	Extra pressure ΔP	ΔT_melt	Melting point	Physical example
0.10 MPa (1 atm)	0	0 K	0.00 °C	Ordinary ice
1 MPa	0.9 MPa	−0.067 K	−0.067 °C	Firm hand squeeze
10 MPa	≈10 MPa	−0.74 K	−0.74 °C	Wire in Faraday's demo
50 MPa	≈50 MPa	−3.7 K	−3.7 °C	Ice-skate blade edge
100 MPa	≈100 MPa	−7.4 K	−7.4 °C	Base of thick glacier
209 MPa	≈209 MPa	−22 K	−22 °C (triple pt)	Limit of ice Iₕ; ice III begins