# Ice skating, how does it really work?

Some textbooks I came across, and a homework assignment I had to do several years ago, suggested that the reason we can skate on ice is the peculiar $$p(T)$$-curve of the ice-water boundary. The reasoning is that due to the high pressure the skates put on the ice, it will melt at temperaturs below $$273 K$$ and thus provide a thin film of liquid on which we can skate. It was then mentioned as fun fact that you could ice-skate on a planet with lakes of frozen dioxide because that gas has the $$p(T)$$-curve the other way round.

My calculations at that time told me that this was, pardon my french, bollocks. The pressure wasn't nearly high enough to lower the melting point to even something like $$-0.5$$ degrees Celsius.

I suppose it is some other mechanism, probably related to the crystal structure of ice, but I'd really appreciate if someone more knowledgeable could tell something about it.

• I'm pretty sure this was a problem on the final exam for my undergrad thermodynamics class ;-) so at least I think your calculation is reasonable. I don't know/remember what the real reason is though. Commented Dec 7, 2010 at 5:31
• Well, this analysis completely ignores that when you skate you are not standing but you are actually moving. There should be some friction between the skates and ice and this should provide enough heat to melt the ice and create a thin water film. At least this is my intuition (perhaps completely wrong). Commented Dec 7, 2010 at 11:44
• The binding energy near a surface is different than the binding energy in bulk, and it is possible that you melt a thin surface layer without melting the bulk. Commented Aug 21, 2011 at 14:14
• A new publication on the issue: phys.org/news/2018-05-slipperiness-ice.html Commented Sep 6, 2018 at 12:06
• Even intuitively it wouldn’t work by the pressure theory. You need some grip/non-slippery-ness to propel yourself and to be able to take a corner. If higher pressure makes ice more slippery, sharpening the blade (which increases local pressure) would make skating more difficult - which is of course complete nonsense.
– W_vH
Commented Jun 28 at 11:54

Yup, this is true that the pressure is too small, but the true explanation is not justified yet. Nevertheless the common sense is that there is a lubricating film of water or at least anomalous ice. For an overview, see: http://lptms.u-psud.fr/membres/trizac/Ens/L3FIP/Ice.pdf

• The pressure is too small for bulk melting, but surface melting is different, and this is the relevant issue. Compressing a surface of water-ice will melt some surface, but compressing the surface of other materials will solidify any surface liquid, because liquid ice has a smaller volume. The explanation is fundamentally correct, the bulk melting is irrelevant. Commented Sep 23, 2011 at 17:10
• @RonMaimon: If there are any irregularities in the ice, or the blade, wouldn't the pressure at those points be nearly infinite unless or until the H2O underneath them gave way? I would think at least some of the weight of the skater would be borne by liquid water unless the compressed water liquified, reshaped itself to a lower-pressure configuration, and refroze. Would skis be effective at -35 on a polished frozen sheet of ice? Commented Nov 29, 2014 at 16:14
• @supercat: The pressure won't be infinite because Young modulus of ice is not infinite - it is not infinitely rigid; it's compressible to a degree and will yield (compress) under pressure a little. It may also locally break destructively and get displaced as dust/shards without phase change.
– SF.
Commented May 17, 2018 at 22:59
• @SF.: I would regard compression, breaking, and melting as forms of "giving way". My point is that even if there wouldn't be enough pressure for ice to melt if weight were uniformly applied on a skate, some areas under a skate will usually be under much higher pressure than others. Commented May 17, 2018 at 23:11

This question has been hotly contested for ages.

Calderon & Mohazzabi $$^{[1]}$$ give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their paper "Premelting, Pressure Melting, and Regelation of ice revisited"

They offer both theoretical and experimental evidence that neither pressure melting nor friction melting alone explain the phenomenon and conclude from atomic force microscopy, among other evidence, that there is a pre-melting Quasi-​Liquid surface layer with special properties - This was originally proposed by Faraday and Thompson back in the 1850s - that, in conjuxtion with some pressure melting, make ice skating possible.

Actually they point to other research that shows ice is not the only solid behaving differently on the surface when near its melting point. The main reasons we notice ice is because it is one of the few substances we encounter that is near its melting point when we do encounter it and its abundance.

Ski-ing is also helped along by friction melting once movement starts.The weighted wire cutting through the ice, however, is pressure melting and regelation.

Another paper that summarises previous research well is that by Dash et. al. $$^{[2]}$$

Both papers referenced also give a good set of references for further reading.

References

1. Calderon, C. and Mohazzabi, P. (2018) "Premelting, Pressure Melting, and Regelation of Ice Revisited." Journal of Applied Mathematics and Physics, 6, 2181-2191. https://doi.org/10.4236/jamp.2018.611183
1. Drake, J.G., Fu, H. and Wettlaufer, J.S. (1995) "The Premelting of Ice and Its Environmental Consequences." Reports on Progress in Physics,58, 115. es. Reports on Progress in Physics,58, 115. https://doi.org/10.1088/0034-4885/58/1/003
• Fantastic. Great to see that there's still new material being published on these topics. Thank you for sharing the paper. Commented Jul 4, 2020 at 19:23
• No problem. It's been bugging scientists and engineers since the 19th century, don't know if this is the final word, but I haven't seen anything new on the topic over the last year or so.
– Gwyn
Commented Jul 4, 2020 at 20:31

It was shown that surface water molecules vibrate more strongly than those in the bulk, having less neighbor molecules to interact with. Apparently, this creates a nanometric film of quasi-liquid water that reduces friction.

• this is only true up to certain temperature below which you need not have a layer of water. Commented Nov 10, 2012 at 18:41

I remember reading in a book (on surface physics) during my grad study on this topic. There was a diagram on friction of a steel "skate" on solid argon at and below argon melting temperature. The diagram was qualitatively identical to the same experiment for ice. Friction dropped to low values when temperature aproached melting point. Argon melts regularly, for that reason pressure melting is not possible. I regret that I did not memorize the title and author of that book :=( Georg

Another fact against 2pressure melting": how does skiing work? The pressure under a ski is very low.

• That doesn't argue against pressure melting at all. Why would you expect skiing and skating to exploit the same mechanism? Why would you expect snow and solid ice to have the same properties? Commented Feb 12, 2014 at 14:41

Well, having a solid block of ice. Atatch weights to a string on both ends and hang it over the ice. The string will go trough the ice over a period of time, without actually cutting the entire block. How does this happen? possibly the pressure melting minescule ammounts of ice underneath the string and the water refreezing above the string.

The assertion that the skate does not exert enough pressure to melt ice is wrong. Imagine that the skate is lowered vertically until it touches a perfectly flat surface of ice. The initial contact area (before the blade starts to sink into the ice) would be incalculably small and the initial pressure incalculably large because of curvatures. A typical freestyle blade’s “rocker” has a radius of 6 feet; its “hollow” of 7/16 to 10/16 inch. The blade is typically 0.15 inch thick, so its two edges have “bite” angles of 7 to 10 degrees. The rate at which an edge could melt ice and sink in would be limited by heat conduction. In a dynamic situation, with the skater gliding along at a good speed, viscous dissipation in the thin layer of lubricating water would generate some of the heat. If the skater’s trajectory is curved but the rocker’s curvature multiplied by sin(tilt) is poorly matched to the curvature of the trajectory, then there will be additional friction and sound effects as the edge chews up the ice.

Your calculations are right, I think. I'm going to do the math here too and the pressure required is great. Your model, however, is wrong if you are imagining a piston compressing an ice cube. This is not a representation of what happens. Some points on the surface that touches the ice, in very small areas, provide the necessary pressure and the water formed is simply expelled from that place, giving way to more melting.

This problem, coincidentally, is literally a textbook case. Tester and Modell presents it as the problem 15-1:

The Clausius-Clapeyron equation describes the relationship between the pressure and temperature at the phase boundary between two phases of matter. Here's a step-by-step derivation of the Clausius-Clapeyron equation:

### Step-by-Step Derivation

1. Consider a Phase Equilibrium: Let's consider a system in equilibrium between two phases: phase 1 (solid) and phase 2 (liquid). At equilibrium, the chemical potentials of the two phases are equal: $$\mu_1(T, P) = \mu_2(T, P)$$ where $$\mu_1$$ and $$\mu_2$$ are the chemical potentials of the solid and liquid phases, respectively.

2. Differential Change at Equilibrium: For small changes in temperature $$dT$$ and pressure $$dP$$, the differentials of the chemical potentials are: $$d\mu_1 = \left( \frac{\partial \mu_1}{\partial T} \right)_P dT + \left( \frac{\partial \mu_1}{\partial P} \right)_T dP$$ $$d\mu_2 = \left( \frac{\partial \mu_2}{\partial T} \right)_P dT + \left( \frac{\partial \mu_2}{\partial P} \right)_T dP$$ Since the chemical potentials must remain equal at equilibrium, their differentials must also be equal: $$d\mu_1 = d\mu_2$$

3. Using Thermodynamic Identities: The partial derivatives of the chemical potential with respect to temperature and pressure can be related to entropy and volume: $$\left( \frac{\partial \mu}{\partial T} \right)_P = -S_m$$ $$\left( \frac{\partial \mu}{\partial P} \right)_T = V_m$$ where $$S_m$$ is the molar entropy and $$V_m$$ is the molar volume.

4. Applying the Identities: Substitute these thermodynamic identities into the differentials: $$-S_{m1} dT + V_{m1} dP = -S_{m2} dT + V_{m2} dP$$

5. Rearranging the Equation: Collect terms involving $$dT$$ on one side and terms involving $$dP$$ on the other side: $$(V_{m2} - V_{m1}) dP = (S_{m2} - S_{m1}) dT$$

6. Solving for $$\frac{dP}{dT}$$: Finally, divide both sides by $$dT$$ and $$(V_{m2} - V_{m1})$$: $$\frac{dP}{dT} = \frac{S_{m2} - S_{m1}}{V_{m2} - V_{m1}}$$

7. Relating Entropy Change to Latent Heat: The change in molar entropy $$S_{m2} - S_{m1}$$ can be related to the latent heat $$L$$ of the phase transition: $$S_{m2} - S_{m1} = \frac{L}{T}$$ where $$L$$ is the latent heat per mole and $$T$$ is the absolute temperature.

8. Substituting the Entropy Change: Substitute this relation into the previous equation: $$\frac{dP}{dT} = \frac{L/T}{V_{m2} - V_{m1}}$$

9. Final Form of the Clausius-Clapeyron Equation: $$\frac{dP}{dT} = \frac{L}{T \Delta V_m}$$ where $$\Delta V_m = V_{m2} - V_{m1}$$ is the change in molar volume during the phase transition.

The Clausius-Clapeyron equation is derived by considering the equilibrium condition between two phases, using the thermodynamic identities for the chemical potentials, and relating the changes in entropy to the latent heat of the transition. The final form of the equation is:

$$\frac{dP}{dT} = \frac{L}{T \Delta V_m}$$

This equation describes how the pressure of a phase boundary changes with temperature, taking into account the latent heat and the change in volume between the two phases.

To solve Problem 15.1 from the Tester/Modell book, we need to understand the phenomenon of regelation, which involves a wire cutting through ice under the influence of attached weights without leaving a trace. This problem is often explained by the pressure-melting point relationship of ice.

### Explanation and Analysis

Regelation refers to the process where ice melts under pressure and refreezes when the pressure is reduced. This phenomenon can be explained using the principles of thermodynamics, particularly the Clausius-Clapeyron relation, which describes the phase boundary between ice and water.

### Clausius-Clapeyron Relation

The Clausius-Clapeyron equation is given by:

$$\frac{dP}{dT} = \frac{L}{T \Delta V}$$

Where:

• $$\frac{dP}{dT}$$ is the slope of the phase boundary in the pressure-temperature ( $$P$$- $$T$$) diagram.
• $$L$$ is the latent heat of fusion.
• $$T$$ is the absolute temperature.
• $$\Delta V$$ is the change in volume during the phase transition.

For water and ice:

• The volume of water decreases when it freezes (ice has a larger volume than water at 0°C).
• Therefore, applying pressure lowers the melting point of ice.

### Explanation of the Phenomenon

1. Pressure Application:

• When a wire with weights is placed over the ice, it exerts a high pressure on the ice directly beneath the wire.
• According to the Clausius-Clapeyron relation, this high pressure lowers the melting point of the ice.
2. Melting and Cutting:

• The ice under the wire melts because the pressure has lowered its melting point below the current temperature.
• The wire sinks through the melted water, continuously melting the ice beneath it as the pressure is maintained.
3. Refreezing:

• Once the wire has passed through, the pressure is removed, and the water refreezes back to ice because the pressure is no longer lowering the melting point.
• This explains why the wire can cut through the ice without leaving a trace.

### Comparison to Ice Skating

Similar reasoning applies to the formation of a thin water layer beneath ice skates:

• The pressure exerted by the thin blade of the skate lowers the melting point of the ice.
• A thin layer of water forms under the skate, reducing friction and allowing smooth gliding.

### Conclusion

The explanation provided in many physics texts, which attributes regelation to the high pressure exerted by the wire lowering the ice's freezing point, is consistent with thermodynamic principles. This reasoning also applies to the phenomenon observed in ice skating.

The wire cuts through the ice due to the high pressure it exerts, which lowers the melting point of the ice directly beneath it. The ice melts under the wire, allowing the wire to pass through. Once the wire has passed, the pressure is released, and the water refreezes, leaving no trace of the wire's path.

This phenomenon is well-explained by the Clausius-Clapeyron relation and is an excellent demonstration of the interplay between pressure, temperature, and phase changes in thermodynamics.

To calculate the weight necessary on each side of the nylon line to cut through a cube of ice, we need to determine the pressure required to lower the melting point of ice below the ambient temperature. Here are the steps to approach this problem:

1. Determine the Required Pressure to Lower the Melting Point: Use the Clausius-Clapeyron equation to find the relationship between pressure and melting point.

2. Calculate the Force Required: Convert the pressure into the force exerted by the nylon line.

3. Convert Force into Weight: Calculate the mass (weight) required to exert this force.

### Step 1: Determine the Required Pressure

First, we'll use the Clausius-Clapeyron equation to find the pressure required to lower the melting point of ice to below 0°C.

The Clausius-Clapeyron equation is: $$\frac{dP}{dT} = \frac{L}{T \Delta V}$$

Where:

• $$L$$ is the latent heat of fusion of ice, $$334 \, \text{kJ/kg} = 334,000 \, \text{J/kg}$$
• $$T$$ is the melting temperature of ice in Kelvin, $$273 \, \text{K}$$
• $$\Delta V$$ is the change in volume per unit mass when ice melts, $$\Delta V = V_{\text{liquid}} - V_{\text{solid}}$$

The density of ice is $$\rho_{\text{ice}} = 0.917 \, \text{g/cm}^3 = 917 \, \text{kg/m}^3$$, and the density of water is $$\rho_{\text{water}} = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3$$.

$$\Delta V = \frac{1}{\rho_{\text{ice}}} - \frac{1}{\rho_{\text{water}}} = \frac{1}{917} - \frac{1}{1000} \, \text{m}^3/\text{kg}$$

$$\Delta V = 0.0010904 - 0.001 = 0.0000904 \, \text{m}^3/\text{kg}$$

Now, plug these values into the Clausius-Clapeyron equation:

$$\frac{dP}{dT} = \frac{334,000 \, \text{J/kg}}{273 \, \text{K} \times 0.0000904 \, \text{m}^3/\text{kg}}$$

$$\frac{dP}{dT} = \frac{334,000}{24.6912}$$

$$\frac{dP}{dT} \approx 13,529 \, \text{Pa/K}$$

To lower the melting point by 1°C (1 K), the pressure required is approximately 13,529 Pa.

### Step 2: Calculate the Force Required

Assuming we need to lower the melting point by approximately 1°C (enough to initiate melting under the line), the pressure required is:

$$P = 13,529 \, \text{Pa}$$

The area of contact of the nylon line with the ice depends on the diameter of the line. Let's assume a standard nylon fishing line with a diameter of 0.5 mm (0.0005 m).

The contact area $$A$$ of the line with the ice is the width times the length of the side of the cube:

$$A = \text{diameter} \times \text{length}$$

$$A = 0.0005 \, \text{m} \times 0.3 \, \text{m} = 0.00015 \, \text{m}^2$$

The force $$F$$ required is given by:

$$F = P \times A$$

$$F = 13,529 \, \text{Pa} \times 0.00015 \, \text{m}^2$$

$$F \approx 2.029 \, \text{N}$$

### Step 3: Convert Force into Weight

To convert this force into the equivalent weight, we use the relation:

$$F = m g$$

Where $$g \approx 9.81 \, \text{m/s}^2$$.

$$m = \frac{F}{g} = \frac{2.029 \, \text{N}}{9.81 \, \text{m/s}^2}$$

$$m \approx 0.207 \, \text{kg}$$

Therefore, the weight necessary on each side of the line to cut through a 30 cm cube of ice using a 0.5 mm diameter nylon line is approximately 0.207 kg.

Regelation-Regelation is the phenomenon of melting under pressure and freezing again when the pressure is reduced. Many sources state that regelation can be demonstrated by looping a fine wire around a block of ice, with a heavy weight attached to it.

Ice skaters shoes :

Whole weight of skater is concentrated at this small portion of area,thus ice under shoes melts quickly [due to Regelation] converting ice to water (notice that due to high pressure ice converts to water without increase of temperature , generally ice melts at 0℃). Hence due to replacing of some amount of ice with water, friction of surface decrease and skater moves easily.

Why to use term regelation? Since due to pressure (or) strain , small amount of ice is coverted to water,whole ice does not break(,melt),making skating possible.

Also: people tried to add on wiki,edit summary

• The part on ice skating was added to the wikipedia entry as a misconception. Unless you can prove otherwise I think it is unwise to cite that wikipedia entry as a source.
– JMac
Commented Mar 10, 2017 at 14:13
• It was removed 10 years ago and still not added back. It was also cited with a source in the misconceptions (I don't currently have access to the source). Unless you can provide some good proof of this, you are going against the other things said and the purpose of this thread. The point being that the burden is on you to prove this pressure is enough for the stated effects. Many sources don't believe it is a good enough explanation.
– JMac
Commented Mar 10, 2017 at 14:21
• @JMac c'mon , regelation is called when when we apply pressure on ice and that turns into water. But when that pressure is removed,water again coverts into ice. Don't you this this is what happens while skating? Commented Mar 10, 2017 at 14:35
• The pressures from skates don't seem to be nearly high enough to temporarily melt ice at temperatures even like -1° C. The issue isn't that regulation isn't a thing. The issue is that quantitative analysis of the situation shows that the effect isn't great enough to even cause localized melting. You would need pressures you wont achieve by skating on ice, so additional factors are required to describe the phenomenon. Your answer provides nothing beyond what the OP already described, and then described his issue with it. Unless you can mathematically prove otherwise, this doesn't answer it
– JMac
Commented Mar 10, 2017 at 15:05