# How to show directly that ##\vec{F}=-\mu\vec{v}## increases entropy?

• greypilgrim
In summary, the process of friction involving a force whose direction somehow depends on the direction of the velocity, such as ##\vec{F}=-\mu\cdot\vec{v}##, increases entropy.
greypilgrim
Hi.

Processes involving a friction force whose direction somehow depends on the direction of the velocity, such as ##\vec{F}=-\mu\cdot\vec{v}##, aren't symmetric with respect to time reversal. If you play it backwards, this force would be accelerating.

On the other hand, friction dissipates heat, and that increases entropy. So this is in agreement with thermodynamics.

I wonder: Is it really necessary to assume that friction dissipates heat to see that entropy is increased in such a process? Or is there a more direct or fundamental way to derive that the occurrence of a force like ##\vec{F}=-\mu\cdot\vec{v}## increases entropy, without getting into technicalities?

Are you asking about the origin of friction, or about the relationship of friction to energy and entropy?

By the way, a force does not change either entropy or energy. You need force times distance first. And in that light, I don't see any difference between friction force and any other kind of force. Energy is conserved. When you do work, the energy must come from somewhere and go somewhere.

It is true that friction is not time reversible, but neither is it fundamental to physics. Friction is a proxy for the average of many mechanisms that happen on the molecular level, especially the electromagnetic force.

greypilgrim said:
If you play it backwards, this force would be accelerating.

And objects fly up instead of fall down.

And objects fly up instead of fall down.
Which they also do given enough initial upwards velocity, so this is a time reversal symmetric process. In contrast, we never observe an object on a table accelerating due to friction.

anorlunda said:
By the way, a force does not change either entropy or energy. You need force times distance first.
Sure. I was talking about "processes".

anorlunda said:
And in that light, I don't see any difference between friction force and any other kind of force. Energy is conserved. When you do work, the energy must come from somewhere and go somewhere.
Yes. And it's all clear if this energy "goes somewhere" as heat, because then it definitely increases entropy. But I wonder if it's necessary to assume that the energy ##\vec{F}\cdot d\vec{s}=-\mu\vec{v}\cdot d\vec{s}## is converted to heat to show that it increases entropy or if this connection is more fundamental, for which the time reversal asymmetry might be an indication.

Maybe I should ask the converse: Can there be a force of the Form ##\vec{F}=-\mu\vec{v}## in a process where the entropy does not increase?

I'm trying to put this more clearly. In a process involving a friction force like ##\vec{F}=-\mu\cdot\vec{v}##, there are two notions of the direction of time:
1. It "doesn't look right" if you videotape it and play it backwards. This is due to the fact that the friction force is proportional to the velocity and now acts in the other direction. Gravity for example always acts in the same direction, regardless if the videotape is played backwards or not.
2. Entropy increases. This is due to the heat dissipated by friction.
It's not straightforward how the second notion follows from the first. Rather, it needs the additional assumption that the work done by the friction force gets converted into heat.

My question is if there is a more fundamental relation between those two notions that doesn't require this explicit assumption.

greypilgrim said:
I'm trying to put this more clearly. In a process involving a friction force like ##\vec{F}=-\mu\cdot\vec{v}##, there are two notions of the direction of time:
1. It "doesn't look right" if you videotape it and play it backwards. This is due to the fact that the friction force is proportional to the velocity and now acts in the other direction. Gravity for example always acts in the same direction, regardless if the videotape is played backwards or not.
2. Entropy increases. This is due to the heat dissipated by friction.
It's not straightforward how the second notion follows from the first. Rather, it needs the additional assumption that the work done by the friction force gets converted into heat.

My question is if there is a more fundamental relation between those two notions that doesn't require this explicit assumption.
For a fundamental development on how viscous friction translates into entropy generation (this is only one of the mechanisms responsible for entropy generation), see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11, problem 11.D.1.

As reckoned from the frame of reference of the body, the ground is moving in the same direction as the frictional force, so ##F=\mu v##, and the rate at which the force is doing work on the interface is ##W=Fv=\mu v^2##. So it doesn't matter what direction the velocity has. If we apply the first law of thermodynamics to the (massless) interface, we have Q=-W (where -Q represents rate of heat flow from from the interface to the body), so heat is leaving the interface and entering the body. This heat flow represents entropy transfer from the interface to the body. If the portion of the body adjacent to the interface heats up substantially due to the frictional heating, there will also be entropy generation within the body as a result of temperature gradients within the body near the interface. So the entropy change of the body will be greater than just the total heat flow divided by the average boundary temperature. This is how the Clausius inequality comes into play: ##\Delta S>\frac{|Q|}{T_B}##

Chestermiller said:
For a fundamental development on how viscous friction translates into entropy generation (this is only one of the mechanisms responsible for entropy generation), see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11, problem 11.D.1.
This is kind of exactly the opposite of what I want: I don't want to introduce any further concepts like heat or the properties of fluids. Given shall just be ##\vec{F}=-\mu\vec{v}##, no further assumptions about what causes this force or into what form of energy the work done by it gets converted.

Chestermiller said:
As reckoned from the frame of reference of the body, the ground is moving in the same direction as the frictional force, so ##F=\mu v##, and the rate at which the force is doing work on the interface is ##W=Fv=\mu v^2##. So it doesn't matter what direction the velocity has.
I think one needs to be careful here, as the frame of reference of the body might very well not be inertial.
But I also don't see the point; if we play the videotape backwards, the ground is moving in the other direction and the force points into the other direction as well, so the force is still time-asymmetric. We just don't have the minus in the formula.

Chestermiller said:
If we apply the first law of thermodynamics to the (massless) interface, we have Q=-W (where -Q represents rate of heat flow from from the interface to the body), so heat is leaving the interface and entering the body.
Maybe the problem is that I want to apply the thermodynamic concept of entropy, but without assuming basic thermodynamic quantities such as heat, which might just be nonsense.
Or maybe a more statistical approach might help, like Boltzmann or Gibbs entropy. So the question would be why the work done by the force ##\vec{F}=-\mu\vec{v}## leads to an increase of microstates.

greypilgrim said:
I think one needs to be careful here, as the frame of reference of the body might very well not be inertial.
On the other hand, suppose it is inertial (body moving at constant velocity by an additional external force being applied to the body, equal to the frictional force).
But I also don't see the point; if we play the videotape backwards, the ground is moving in the other direction and the force points into the other direction as well, so the force is still time-asymmetric. We just don't have the minus in the formula.
The point is that the work the ground does on the interface is positive definite, irrespective of the direction of the force.

Maybe the problem is that I want to apply the thermodynamic concept of entropy, but without assuming basic thermodynamic quantities such as heat, which might just be nonsense.
You can say that again.
Or maybe a more statistical approach might help, like Boltzmann or Gibbs entropy. So the question would be why the work done by the force ##\vec{F}=-\mu\vec{v}## leads to an increase of microstates.
Good luck with that.

The problem is that with that simplified point of view you cannot even define entropy, because you only look at the equation of motion for the average velocity of your particle. To get a useful definition of entropy, you'd need some random element in the discussion and then define entropy with the corresponding probability distribution.

One way is to think in terms of a Langevin equation, leading to a Fokker-Planck equation for the probability distribution. This is a simplification of the more general Boltzmann equation for a heavy ("Brownian") particle moving in a heat bath consisting of very many light particles (like the pollen immersed in water looked at under a microscope by Brown leading to the name-giving phenomenon of "Brownian motion").

It's an interesting problem to investigate the Fokker-Planck equation and calculate the entropy and checking, whether it's increasing. Maybe I'll do that soon, if I find the time (seems to be a nice topic for an Insights article).

hutchphd

## 1. How can I experimentally demonstrate that ##\vec{F}=-\mu\vec{v}## increases entropy?

To show this relationship directly, you would need to set up an experiment in which you can measure both the force, ##\vec{F}##, and the velocity, ##\vec{v}##, of a system. This could be done using various methods such as tracking the movement of particles or using force sensors. By varying the force and velocity and measuring the resulting change in entropy, you can demonstrate the direct relationship between the two variables.

## 2. Can you provide an example of a system where ##\vec{F}=-\mu\vec{v}## increases entropy?

One example is a gas in a container with a piston. If the gas is compressed by applying a force on the piston, its velocity will increase, and the entropy of the system will also increase. This is because the gas molecules will have more possible positions and velocities, leading to an increase in disorder and entropy.

## 3. How does the equation ##\vec{F}=-\mu\vec{v}## relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. The equation ##\vec{F}=-\mu\vec{v}## is a direct representation of this law, as it shows that an increase in force and velocity will lead to an increase in entropy.

## 4. Is the relationship between ##\vec{F}## and ##\vec{v}## always linear in terms of increasing entropy?

No, the relationship between ##\vec{F}## and ##\vec{v}## is not always linear. In some cases, the increase in force may not result in a proportional increase in velocity, leading to a non-linear relationship. However, the overall trend will still be an increase in entropy.

## 5. How does the value of ##\mu## affect the relationship between ##\vec{F}## and ##\vec{v}## in terms of increasing entropy?

The value of ##\mu##, which represents the coefficient of friction, will affect the strength of the relationship between ##\vec{F}## and ##\vec{v}##. A higher value of ##\mu## will result in a stronger negative relationship, meaning that an increase in force will lead to a larger decrease in velocity and a larger increase in entropy.

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