# Modelling friction as a conservative force

Friction is usually considered as a non-conservative force, but by considering the microscopic movement of particles which produce the friction, it seems we can model friction as conservative force field (the lost energy transform in kinetic energy of microparticles, what we call "heat").

For example, consider the uni-dimensional movement of a macroscopic solid block sliding on the floor, let call its position as $$X(t)$$. On the other hand, the floor is modeled as a huge collection of $$N$$ material particles, like a thermal reservoir/bath, interacting with each other and with the block. As these particles move very little from their equilibrium position $$x_i = 0$$, we can approximate the potential energy associated with their interaction by a Taylor series up to quadratic terms:

$$V(x_i) = \frac{1}{2}\sum_{i, j=1}^N k_{ij}x_ix_j$$

The interaction of the block and the floor is assumed to be represented by a potential $$V_f(x_i,X,\dot{X})$$ ("potential associated with friction"). So, the Lagrangian of this system would be:

$$L(X,x_i,\dot{X},\dot{x}_i) = \frac{M}{2}\dot{X}^2 + \frac{m_i}{2}\sum_{i=1}^N \dot{x}_i^2 - V_f(x_i,X,\dot{X}) - \frac{1}{2} \sum_{i, j=1}^N k_{ij}x_ix_j .$$

Then for almost every solution with initial conditions $$\dot{x}_i = 0$$ the energy flows to the floor and we have $$\dot{X}\to 0$$.

My question are:

1. What forms could the potential $$V_f(x_i,X,\dot{X})$$ take?

2. How could we show that most solutions (with the above-mentioned initial conditions) end up with the block stopped on the floor?

My guess for the question 1 is that $$V_f(x_i,X,\dot{X}) = -(\alpha\ \text{sgn}(\dot{X})X) \sum_{i=1} \text{rect}(X-x_i)$$ or some modification will work, because this guarantees that each particle only produces influence during the time there is contact between the block and itself.

• Given ‘heat’ you can’t recover all the work that went into generating it from thermodynamics. Therefore it is not conservative. Jun 11 at 22:59
• @JonCuster, We certainly cannot recover it, but not because the energy is not conserved, but because it is extremely unlikely from the equations that the system evolves in the direction that there is a net flow of degrees of freedom from the ground to the block that ended up stopping. Jun 11 at 23:03
• Thermodynamics holds. It is not ‘unlikely’ - once it is ‘heat’ you can’t be 100% efficient turning it back into work. Period. Jun 12 at 2:30
• Jun 22 at 17:54
• To do what you want you will need to model both the floor and the box as networks of springs and masses and have some interaction between particles in the box and particles in the floor. You will probably need to do something else to me sure both the box and floor have some sort of expected "equilibrium" shape (rectangular). If the floor and box are flat at the interface I think it's possible you won't see any dissipation (forces pulling the box backwards to stop it will balance forces pulling the box forward). You probably need jagged edges on both. Not sure how to implement that. Jun 22 at 22:17

Adding friction to a Lagrangian is not commonly taught. It is non-trivial, but not too difficult in the end. The key element is producing a "dissipation function" $$D$$ that we can use to augment the Euler Lagrange equations. This approach is well described in this tutorial, but I will summarize the key points here.

In this approach the Euler Lagrange equations are modified as follows: $$\frac{d}{dt}\frac{\partial L}{\partial \dot q}=\frac{\partial L}{\partial q}-\frac{\partial D}{\partial \dot q}$$ where $$D$$ is the "dissipation function" and represents the rate at which power is leaving the system due to dissipation. It is related to the generalized frictional force by $$Q=-\frac{\partial D}{\partial \dot q}$$

As you mentioned, the dissipation function is a sort of a potential, but with the potential being related to the generalized velocities instead of the generalized coordinates. That makes it non-conservative.

For a frictional force which is proportional to the n-th power of the velocity the dissipation function can be written $$D=\frac{1}{n+1}c v^{n+1}$$ where $$c$$ is usually determined by comparison with the frictional force law which gives $$Q$$.

For example, with a typical normal friction force we have $$F=-\mu m g$$ which is proportional to the 0-th power of the velocity and therefore yields $$D=\mu m g \dot x$$ Or for a typical drag friction force we have $$F=\frac{1}{2}\rho C_D A v^2$$ which is proportional to the 2-nd power of the velocity and therefore yields $$D=\frac{1}{6}\rho C_D A \dot x^3$$

• Note, however, that the dissipation function is not part of the Lagrangian as the OP has in mind (i.e., you don't substitute $L \to L + D$ to add friction.) It's effectively added to the Euler-Lagrange equation post hoc as a generalized force instead. Jun 22 at 18:09
• Yes, that is what I was trying to imply regarding augmenting the Euler Lagrange equations instead of augmenting the Lagrangian
– Dale
Jun 22 at 19:54

For systems with a Hamiltonian formulation, the Poincaré recurrence theorem (PRT) would indicate that most trajectories will eventually evolve back to a state arbitrarily close to their initial conditions. Under such circumstances if $$X = X_0 \neq 0$$ initially then the system will eventually evolve back to some state where $$X \approx X_0$$. In such a case, to say that $$X \to 0$$ in the strict mathematical sense of $$\lim_{t \to \infty} X = 0$$ can't be true; the best you can do is to say that $$X \approx 0$$ for most of the time for "most" trajectories.

If you want to avoid the PRT, your system will have to either (a) not have a Hamiltonian formulation, or (b) have trajectories that are unbounded in phase space. Option (b) is probably not what you want on physical grounds, so your best hope for avoiding the consequences of the PRT rests on whether a Hamiltonian formulation of the system exists. This, in turn, will depend on the form of your "potential" $$V_f(x_i, X, \dot{X})$$; you can probably avoid it for a "non-smooth" version of $$V_f$$, but I would worry that it can't be avoided for a "nice" $$V_f$$ function.

• Option b) isn't as unphysical. Sufficiently singular potential function (e.g. Newtonian gravitational energy between mass points) makes the available phase space volume infinite. But even without this, the recurrence times in PRT may be so large that Poincare recurrences pose no real contradiction to observations. Jun 22 at 22:31

It's not possible to do what you are attempting. For friction to be a conservative force, then the net work done by moving along a path where the start and endpoints are the same is zero which would mean the net heat generated would be zero. If this were true, then an object could be cooled to its original temperature before motion by simply reversing its path back to its origin: $$F_{friction}(d_{forward} + d_{back}) = 0 = \Delta Q$$

The reason is that friction is a destructive force. A wooden block moving along sandpaper is losing mass left as wood dust. This is the part of the friction process that is irreversible. So if you could model friction as a conservative force, you would have to add equations to cover the loss in mass and heat released from the breaking of bonds.

• In fact, the second law of thermodynamics is a statistic law, if we wait a huge amoung of time we can observe sporadic violations, including seeing how heat flows from a colder body to a warmer one. So I conjectured that by considering a sufficiently high N we could obtain that only for ultra-specific initial conditions we would see such violations of the second law. Jun 22 at 18:23
• Friction is a destructive force. In reality, objects lose mass during friction (wood dust from sanding a wooden block) and the destruction of surface bonds is how heat is generated. This is the part that is irreversible and must be addressed if friction is treated as a conservative force Jun 22 at 18:33
• @Davius You'll be waiting a long time. In Schroeder's Thermal Physics text, he argues that the probability that n particles of an ideal gas occupy the left-half of a volume is $2^{-n}$. For $n=100$, that's about $7 \times 10^{-31}$.. "you would have to check a trillion times per second for the age of the universe before finding an arrangement even once. For $n=10^{23}$, the probability is a very small number." WolframAlpha gives $2^{-10^{23}}\approx 10^{-10^{22.48}}$. Jun 23 at 1:09