# Why are there both Static and Kinetic Friction?

When dragging an object, there is a greater start-up force than the force it takes to keep it moving. Why is this? Why are there two different values for friction?

-
I was taught that the most significant factor in most cases is the presence of a thin cushion of air between the two surfaces once they've started to slide. – Harry Johnston Nov 6 '11 at 23:04

This is a deep mystery, nobody knows for sure. It is possibly related to establishing an elastic wave profile along the bottom which is in a dynamic steady-state with the jostling of the contact points, but you need good measurements to make a theory, and good measurements are difficult because of the issue of solid-solid sticking.

The research field which studies questions of this sort is moderately active still, and was very active twenty years ago. It is called depinning. In depinning, there are models which partially explain this effect, although whether these models are relevant for a real solid-on-solid interface remains to be seen.

### Friction phenomenology is counter-intuitve

Sliding friction is approximately constant. This is unusual, because physical quantities tend to go to zero smoothly. For example, the force on a spring is zero at position zero, and it is linearly increasing as a function of displacement. Similarly, the friction force for a slow moving object in a viscous fluid is zero at zero velocity, and is linear in the velocity. This is the general expectation that the friction depend analytically, or at least differentiably, in a small parameter. Solid-on-solid friction violates this expectation.

Another way of saying why a constant friction force is counterintuitive is the belief that there should be a unique steady state motion at each force. If the microscopic contacts are doing some sort of Markov chain, by general principles of probability, it should have a unique steady state distribution at any fixed force. Two different steady states must be separated enough so that they cannot fluctuate into each other.

If this is true, the force must determine the velocity. But this is not true if the force is constant as a function of the velocity--- you can keep an object sliding stably at any velocity with the same force.

So the general expectation is that the friction force is not really completely constant, but slowly varying with the velocity, and that there is a point of statistical non-analyticity, an interesting phase transition, at zero velocity.

### Why constant friction force?

The friction force times the velocity tells you the energy lost to heat per unit time, the work done by friction. This work is the constant friction force times the velocity. This means that friction will remove an equal increment of energy in equal distance.

You can understand this rule heuristically as follows:

The loss in energy is due to an elastic deformation in the solids relaxing. In order to slide to each position, the solid will rearrange its shape slightly to fit into the microscopic best-stick points on the matching surface, and stick to some bump or dirt on the surface.

Each rearrangement of the solid produces sound waves which carry away energy, and transform it into heat. Each sticking and unsticking event produces a given amount of sound, and the sticking and unsticking is a property of the surface only. So that to slide a given distance, you need an average number of sticking/unsticking events, and the friction energy loss is equal to the number of such events times the average energy lost to sound per event.

This model explains why there is a roughly the same amount of work done per unit distance moved. This model of sticking and unsticking in response to forcing is called "depinning" in the condensed matter literature, and it was widely studied in the 1990s.

### Narayan Fisher model

Solid on solid friction is only the most familiar of a host of phenomena which involve sticking and slipping when something is pushed forward. The traditional examples listed in papers on depinning are * Charge density waves * pulling pinned Vortices in superconductors * crack propagation, with starting and stopping

The list of applications for some reason usually excludes solid-on-soild friction, even though this is the most familiar application.

Here is a classic depinning model: consider a position function defined on a two dimensional grid. You can think of the position function as representing the displacement of the surface of contact of the two solids. There is a constant force on every site, and there is an additional elastic force on each site, which is positive when the neighbors are further along and pull the site forward, and negative when the neighbors haven't caught up yet, and hold it back.

The total force is the constant $F$ plus $h(u)+h(d)+h(l)+h(r) - 4 h(c)$ where u,d,l,r are the four neighbors, h is the position function, and c is the center point.

Then there is a sticking force at each point $s(x)$, which is a random number between 0 and 1 (the details of the distribution are not so important, it could be a Gaussian of unit width too, just don't make the sticking distribution a powerlaw).

The law of motion is that wherever the total force is bigger than the sticking force, the site moves forward by one unit.

This model has a second order phase transition at a critical force from a stationary state to a moving state. Past the phase transition, the velocity goes as $(F-F_c)^\alpha$, where F_c is the critical force and $\alpha$ is a critical exponent.

Because the transition is second order, this model does not explain static friction. Nor does it produce a constant friction with velocity, the force goes up with velocity. But it does produce a critical force which is nonzero, and determined by the sticking force distribution in steady state.

### Fisher's Overshoots

The model above fails to account for static friction, and its analog in other depinning models. To fix this, Fisher introduced the concept of a stress overshoot. The idea here is that right after a site depins and hops, it has made sound-waves all around it which jostle the site for a while, and these sound waves momentarity increase the force randomly in various directions. This means that the mean elastic force doesn't have to overcome the pinning force by itself, it can get some momentary help from the transient soundwave profile.

The model for this is that right after a site hops, on the next time step, there is a parameter M called the overshoot parameter, which weakens the on-site pinning force, and that of the nearest neighbors. The total force only has to exceed the pinning force minus M in order to get the site to hop again.

This model takes into account the local stress profile, and Fisher's physical intuition led him to propose that the transition in this model will be first order, not second order, because there should be hysteresis.

The reason is simple: while the object is moving, it feels the overshoots from the transient stresses, and when it stops it doesn't, because the sound modes die away. So it should take more force to get something to move than to keep in moving once it starts. This means that you have to increase the force to more than the amount required to maintain the motion in order to start the motion, and this is a hysteresis loop.

The hysteresis loop is usually a property of first order transitions.

### Phony hysteresis

Fisher's physical intuition is correct in natural models, but the original conclusion that the transition is first order is not correct. To explain this, the statistical theory of the steady state motion is a renormalization group fixed point, and the M perturbation is much like a mass term--- it adds a little bit of inertia to a site. A site which hopped tends to hop again.

Mass terms in viscous systems are generically irrelevant perturbations. They do not affect the statistical description at long wavelengths in any way. But these perturbations do produce hysteresis in some models, so there seems to be a contradiction with the general principles of renormalization group theory.

The resolution of this mini-paradox is a little subtle. It is the subject of this paper http://arxiv.org/abs/cond-mat/0301495 (nonfree version is slightly different in terminology, but not in content: http://prl.aps.org/abstract/PRL/v92/i25/e255502, I coauthored this with Jennifer Schwarz). I will run through the argument in the paper, which shows that

• The transition is hysteretic--- you need more force to get the thing started than to keep it moving.
• The transition is still second order, and in the same universality class as the Fisher Narayan model.

These two conclusions are not contradicting.

The demonstration is easiest by considering a series of models, each of which is progressively less obvious, but each of which have the same behavior.

• Model 1: Global overshoot model

This model is not physical at all, it is just used for the sake of mathematical argument. consider the discrete Narayan Fisher hopping model on a lattice (the stick-force lattice model described above without overshoots), and add to it an overshoot force which is global and nonadditive--- whenever at least one site somewhere has hopped, all sites feel an overshoot force M.

When the system is moving, this means that the pinning force distribution is just shifted over by a constant M, and the steady state is exactly the same as the Narayan Fisher model with the shifted distribution. When the system is stopped, you need to make the force M units bigger than normal to get one site to hop, but the moment it does, the force on every site goes down by M, so you can quickly drop the external force by $M-\epsilon$ and keep it moving.

So there is exactly M units of hysteresis in this model, but because of the stupid global nature of the force, it is manifestly obvious that this model is completely equivalent to the Narayan Fisher model, and it has the same second order phase transition.

I lobbied to call this behavior "phony hysteresis", because it is so fake. The hysteresis loop is not an indication of a first order phase transition, it's just a stupid thing overlayed on top of a fundamentally second order transition. PRL's referees didn't like phony hysteresis, but this is I what I will call it below.

• Model 2: local nonadditive overshoot model

The local nonadditive overshoot model says that when a site hops, it feels M units of overshoot force, and its neighbors do too, on the next time step. This force is nonadditive, meaning that it's the same M if one neighbor hops or if all four neighbors hop.

This model is superficially different from the global overshoot model, but they are actually the same in steady state. When you have a steady state motion, the only sites that hop on time t are those whose local environment has changed, and these are the sites whose neighbors have hopped. These sites feel the overshoot force.

So when the steady state is set up, the hopping sites all feel the overshoot. Further, the sites which do not hop felt the overshoot at the last time they or something around them moved, so that their stability at that time means that they don't move even if you put the overshoot on them too.

So you might as well apply a nonadditive M overshoot to all the sites. It makes no difference in steady state. This proves the equivalence of the nonadditive model to the global model.

You conclude that the nonadditive model has exactly M units of phony hysteresis, just like the global model. But the nonadditive model is local, so phony hysteresis is actually happens in realistic things. The transition is still second order, and completely rigorously equivalent to Narayan Fisher

• Model 3: Additive stress overshoot

Now you get to Fisher's original overshoot model, where the overshoot force is additive, and equal to M times the number of neighbors that hopped on the last timestep (where neighbor includes yourself if you hopped then too--- this is crucial). In this case, you can separate out the model into two parts:

• M units of nonadditive stress overshoot, which is felt by every site whose neighborhood changed the last timestep.
• an additional stress overshoot equal to M times the number of hopping neighbors minus 1.

You analyze the two contributions separately. The second overshoot can be simulated directly, and it has a second order phase transition in the same universality class as the Narayan Fisher model. So this behaves as expected from renormalization theory, it has no effect on the continuum limit.

The first contribution just tacks on exactly M units of phony hysteresis on top of the no-hysteresis second model. So the additive local overshoot model, the realistic model, has M units of phony hysteresis disguising a completely fine second order transition point.

This is a realistic model with static friction greater than dynamic friction. It is not a correct model, because forces in a solid are nonlocal.

### Complications

Fishers original model had M units of overshoot stress on the neighbors, but no overshoot stress on the site that just hopped. This is because you don't think that a site which just hopped would need to hop again. While this is true, the double-hops are rare, the lack of the self-overshoot ruins the proof of the phony hysteresis theorem, so you don't have phony hysteresis, and the system eventually settles down to a nonhysteretic steady state.

But it takes forever because the system is so close to the system with true phony hysteresis. This led to much confusion.

### Is static friction phony hysteresis?

The phony hysteresis means that the transition going down in force is second order, but you see a hysteresis going up, just because you didn't set up the overshoots. The overshoots are models of sound waves in the bottom of the material, which locally help push the material, to overcome the sticking points.

These soundwaves always fall off in amplitude slower than elastic deformations. So the overshoots are always important in a larger range than the elasticity. This is exactly the domain where you get phony hysteresis, of magnitude equal to the minimum of the overshoot stress on the moving filament of active sites. The filament is narrow, and the soundwaves are broad, so this minimum should be nonzero.

Because of this, I do believe that the static friction is the phony hysteresis. This means that the transition to zero velocity when going down in force is the true statistical stick-slip dynamics of depinning, while you need an extra force to get the thing moving because the soundwaves have not been set up yet.

This is not a complete answer, because the actual magnitude of the hysteresis loop has not been estimated, and there is no real demonstration that the minimum overshoot on the active filament is nonzero. Further, this predicts that the force for sliding friction should go up with a nonzero critical exponent as a function of the velocity, although when the exponent small as $1.6$ you can model the force as effectively constant, especially since the range of variations is defined by the enormous scale of the speed of sound in the material.

-
Is "maximum friction" an accepted expression for static friction? – Georg Nov 6 '11 at 22:06
@Georg: It may not be standard, but its clear enough, I think. – Ron Maimon Nov 7 '11 at 4:01

I have always been under the impression that since there are two vectors involved one the force of friction and the other your force. Since these two forces if equal will cancel out there arises the need to overcome that original force of friction. After that has been done can you decrease your force to that of friction and still continue to move. The reason you can continue to move is because the object now has kinetic energy and will not settle into the tiny "cracks" of the surface it is on. That us how I have been taught.

-
This is a common explanation, but it is not at all obvious that it is correct. Is crack-settling the mechanism of friction, or is it Van-Der-Waals sticking? Is the time scale for crack settling really large compared to the sticking and unsticking time for contacts? You want a quantitative model that tells you what happens at the surface. My answer is too glib. There was work done on similar problems by many people (including myself--- shame on me), so that a little bit is known. – Ron Maimon Dec 8 '11 at 10:26