Consider the following textbook problem:

An electric car of mass $m = 3500 kg$ has stored $E = 100 kWh$ in its battery. The rolling resistance coefficient is $\mu_R = 0.8$. How far can the car travel at most?

The numerical solution is easy enough to obtain, you simply equate the work done by the rolling friction with the initial energy:

$$ F_R \cdot \Delta x = E \Rightarrow \Delta x = \frac{E}{F_R} = \frac{E}{\mu_R m g} = 13.11\ km.$$

However, upon scrutiny this derivation reveals itself to be flawed: the term $F_R \cdot \Delta x$ merely describes the pseudo-work done by the rolling friction, which a priori is not related in any fixed way to the energy of the car; of course, the pseudo-work done by the resulting force describes the change in center-of-mass kinetic energy per unit of time, however, the resulting force would not only include the rolling friction but also the static friction which is needed to even accelerate the car.

Calculation from first principles

If you start from first principles, you would have the following. The car is a system of particles of mass $m_i$, $i = 1,\ldots,N$, which interact according to some potential $U = U(\overrightarrow{x_1},\ldots,\overrightarrow{x_N})$, and which moreover is acted upon some external forces $\overrightarrow{F_i}$, so that the equations of motion would be

$$ m_i \ddot{\overrightarrow{x_i}} = -\frac{\partial U}{\partial \overrightarrow{x_i}} + \overrightarrow{F_i}.$$

Letting the energy of the system be $$ E = \sum_i \frac{1}{2}m_i \dot{\overrightarrow{x_i}}^2 + U,$$

it then follows from the equations of motion that the change of this energy per unit of time is $$ \frac{dE}{dt} = \sum_i \overrightarrow{F_i} \cdot \dot{\overrightarrow{x_i}},$$ i.e. the work done by the external forces alone.

Moreover, in this special situation, we know that there are two types of external forces acting, the static friction accelerating the car and the rolling friction "slowing down" the car; thus we can write

$$ \overrightarrow{F_i} = \overrightarrow{F_i^s} + \overrightarrow{F_i^r}.$$ These forces are unknown, however, we do know that static friction does no work since it acts upon mass points which are at rest at all times. Therefore we find that

$$\frac{dE}{dt} = \sum_i \overrightarrow{F_i^r}\cdot \dot{\overrightarrow{x_i}}.$$

Up until this point, everything is unproblematic.

The problem

However if we want to continue the calculation, we are troubled by the fact that the forces $\overrightarrow{F_i^r}$ are not known; what is known is only the resulting force

$$ \sum_i \overrightarrow{F_i^r} = - \mu_R m g \frac{\overrightarrow{v_S}}{|\overrightarrow{v_S}|},$$ where $v_s = \dot{\overrightarrow{x_S}}$ is the velocity of the center of mass.

And there is an even a more serious problem. If you think about it, it's not reasonable to assume that the rolling friction should do any work at all; in particular, there's no reason to assume that the work done equals the pseudo-work done by the resulting force, even approximately.

Frictional forces do no work

To see this, consider a block, starting with some initial velocity, sliding on a table. Of course, what happens is that the block slows down and stops eventually due to the frictional force. However that does not allow to conclude that the table does negative energy on the block, as the situation is more or less symmetric; if you don't agree, consider two rotating, equal discs pressed against each other by thermally isolated rollers instead.

What should really happen of course is that the frictional force converts ordered kinetic energy into unordered kinetic energy, i.e. heat, inside the same system, having the effect of both slowing down and heating up the block.

So a more realistic assumption is actually to assume that frictional forces do no work at all.

Dissipational energy to the rescue?

So to summarize we are troubled by two things, first that the work done by the rolling friction is unknown since the forces constituting the resulting frictional force are unknown, and second that that work should actually be zero and in particular is unrelated to the distance travelled by the center of mass.

Ignoring the first problem for the moment, we can remedy the second by including another term in the expression for the energy which makes up for any work done by the frictional forces: $$ E' \equiv \sum_i \frac{1}{2}m_i \dot{\overrightarrow{x_i}}^2 + U - \int dt\ \sum_i \overrightarrow{F_i^r} \cdot \dot{\overrightarrow{x_i}}; $$ this of course implies $\frac{dE'}{dt} = 0$.

This new energy term is problematic since it not only depends on positions or velocities at a given time $t$ but the whole history of the system up until $t$; moreover, it's also not clear what it should represent because the heat would seem to be already included in the kinetic energy term.

However, if we press on, we can make the ad-hoc assumption that the frictional forces are given as $$ \overrightarrow{F_i^r} = -\frac{\partial P}{\partial \overrightarrow{v_i}},$$ in terms of a dissipation function $P = P(\overrightarrow{v_1},\ldots,\overrightarrow{v_N}) = P(\dot{\overrightarrow{x_1}}, \ldots, \dot{\overrightarrow{x_N}}).$ Assuming that $P$ is homogeneous of degree $k$, we can compute the work as $$ \sum_i \overrightarrow{F_i^r} \cdot \overrightarrow{v_i} = \sum_i -\frac{\partial P}{\partial \overrightarrow{v_i}} \cdot \overrightarrow{v_i} = -k P.$$

We would then get that

$$ \frac{dE}{dt} = -k P,$$

which for

$$ P = P(|\overrightarrow{v_S}|) = \mu_R m g |\overrightarrow{v_S}|$$

would imply ($k = 1$) first of all that $\frac{dE}{dt} \leq 0$ and more precisely that $$ dE = -\mu_R m g |dx_S|,$$ which then leads to the same result that we get using the naive calculation if we assume some boundedness of the energy $E$ that excludes the dissipational term, i.e. if we assume some bound on $U$.

My Question

It would seem that the introduction of a "dissipational energy" term resolves the problem posed by the fact that--as introspection suggests--frictional forces shouldn't do work, i.e. should not transfer energy into or out of the system, and that the ansatz for the frictional force resolves the problem that the individual forces making up the rolling friction are not known.

However this "solution" seems rather ad hoc and there are at least three problems, some of which have been already indicated above:

  1. The dissipational energy term seems redundant if heat is interpreted as kinetic energy.
  2. Dissipational energy is a function of the history of the system.
  3. There is no reason to assume that the frictional forces are of the form assumed above.

So my question then would be: What's a proper and realistic treatment of this problem in Newtonian mechanics, and how would it differ from the approach above?


2 Answers 2


The idea of rolling friction is fundamentally a macroscopic, experimental observation, which states that the work is $dw=\mu_R m\vec g\cdot d\vec s$.

As for your questions,

  1. The dissipational energy term seems redundant if heat is interpreted as kinetic energy.

It's not, at least not in the Newtonian approach. The velocity $\vec v$ you write is the velocity of a point "particle" in the "it's small" sense, not particle in the "it's an atom" sense. So, if heat is added to the system, whereas individual atoms might increase their velocity, that does not imply that the variable $\vec v$ does too.

Usually, in dynamics, one thinks of heat simply as waste. But, if you want to include it into your analysis, think of the first law, where heat will alter the internal energy of the system. So, if you redefine the energy of the system to be the kinetic energy $T$, potential energy $V+U$ where $U$ is the internal energy and $V$ is anything else, you therefore have that any heat produced will be accounted for by an increase in internal energy, and that's where that extra term comes from.

  1. Dissipational energy is a function of the history of the system.

Work by a force is path-dependent. Again, think of the first law of thermodynamics, $dU = \delta Q - \delta W$ where the $\delta$ describes a path-dependent differential change. For $\delta W=0$, we have that, $dU = \delta Q$, that is, the internal energy (which is a state function) will be equal to a path-dependent quantity.

  1. There is no reason to assume that the frictional forces are of the form assumed above.

A most unsatisfying answer, but it boils down to: it's experimentally true, so we model the frictional force as such.

  • $\begingroup$ Are you in your comment on the first point impyling that one must never think of the mass points making up a many-particle Newtonian system as elementary (indecomposable) particles and that heat is not to be interpreted as kinetic energy but rather as some internal energy? Of course Newtonian mechanics does not describe physical processes in atoms correctly, but I still find this surprising since I'm used to heat being described simply as irregular, undirected motion of "billiard ball" particles. $\endgroup$ Feb 5 at 9:26
  • $\begingroup$ @BlenderBender yes. even in (classical) thermodynamics problems, we opt to treat velocity as some macroscopic quantity, and any change to temperature is reflected as a change in internal energy. Now, I will finish off by saying this comes from an engineering perspective, which seldom is the most rigorous. But for us, the only question is -- does it work in practice -- and the answer is a yes, it works very well, much to the engineer's relief albeit the physicist's dismay. $\endgroup$
    – user256872
    Feb 5 at 15:56

I've already accepted user256872's answer, but I just wanted to add some remarks to point out the flaws in my initial reasoning.

Conventions about frictional forces

Namely, my troubles seem to mostly stem from a misunderstanding of what is meant by the phrase "the frictional force equals XYZ": this is not a statement about the resulting force that stems from the individual frictional forces that act, but rather a statement about the work done by those frictional forces; to say that "the frictional force equals $\overrightarrow{F}$" is to say that the work done by the frictional forces equals the pseudo-work done by $\overrightarrow{F}$:

$$\sum_i \overrightarrow{F_i} \cdot \overrightarrow{v_i} = \overrightarrow{F} \cdot \overrightarrow{v_S}.$$ Nothing more and nothing less.

Neither the individual frictional forces nor their resulting force is known in general, but this does not matter insofar as the solution to the problem only depends on that work relation to hold; therefore, one may also assume that the individual frictional forces are given as the derivative of some dissipation function, as I've done in my question, without loss of generality.

Finally, in certain circumstances one can conclude that

$$\sum_i \overrightarrow{F_i} = \overrightarrow{F}$$ holds, which for example is the case when $\sum_i \overrightarrow{F_i}$ equals the total resulting force (i.e. $m \dot{\overrightarrow{v_S}} = \sum_i \overrightarrow{F_i}$), the center of mass moves in a straight line (i.e. $\dot{\overrightarrow{v_S}} \parallel \overrightarrow{v_S}$) and the potential and internal kinetic energy are constant (i.e. $\sum_i \overrightarrow{F_i} \cdot \overrightarrow{v_i} = m \dot{\overrightarrow{v_S}} \cdot \overrightarrow{v_S}$).

Regarding the distinction between mechanical energy and heat

Another source of confusion stems from the treatment of heat in Newtonian mechanics. Namely, as @user256872 has pointed out, heat must be treated as a form of energy separate from kinetic energy, if practical calculations are to be possible. However, this directly contradicts the interpretation of heat as (unordered) microscopic motion.

The solution to this puzzle appears to be as follows.

From the viewpoint of microscopic Newtonian physics, a block that slides on a table and eventually comes to rest due to frictional forces will have not changed its kinetic energy: what was "ordered" kinetic energy has simply become "unordered" kinetic energy, i.e. heat.

The problem with that viewpoint is of course that it is completely useless. To make it useful, one must be able to make the distinction merely alluded to, namely to split (microscopic) kinetic energy into "ordered" kinetic energy and heat.

Macroscopic Newtonian mechanics

One possible way to obtain this splitting would be to interpret the given quantitites $\overrightarrow{x_i}$, $\overrightarrow{v_i}$ as microscopic quantities, which are not "observable" themselves since they are subject to fluctuations, and to consider derived macroscopic variables $\overrightarrow{X_i}, \overrightarrow{V_i}$ obtained by averaging:

$$\overrightarrow{X_i}(t) = \left<\overrightarrow{x_i}\right>(t) = \int_{-\infty}^\infty dt'\ \overrightarrow{x_i}(t') \varphi(t'-t).$$

Here $\varphi$ would be some kind of bump function centered around $t = 0$, falling off sharply so that we are taking an average over a short time-interval.

We could then replace microscopic kinetic energy by its average and split it as

$$\left<E_{kin}\right> = \sum_i \frac{1}{2}m_i \left<\overrightarrow{v_i}\right>^2 + \sum_i \frac{1}{2}m_i \left(\left<\overrightarrow{v_i}^2\right> - \left<\overrightarrow{v_i}\right>^2\right) = E_{kin,macro} + E_{heat}.$$

Necessary Non-Determinism

However, it seems that this time-averaging approach is seriously flawed. For example, expected properties of this average do not hold, like averaging being idempotent, unless $\varphi$ is the delta function. What's worse, given this definition, a block all of whose atoms would vibrate with identical frequency, phase and direction would also have heat, which doesn't seem right.

The only way to make this work seems to be if instead of considering time-average we take expectation values (in the sense of stochastics), but this means that we must replace deterministic particle trajectories by stochastic processes, i.e. that we must assume non-determinism from the very beginning.

In short, deterministic Newtonian mechanics seems to be unable to support the phenomenon of heat as a foundational theory. Replacing deterministic by non-deterministic Newtonian mechanics results then in a mathematical apparatus very close to Quantum Mechanics (Feynman-Kac formula).


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