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## 1  Kinetic Energy

A lot of people think there is a unique, well-defined notion of “the” kinetic energy ... but in fact there is a range of different concepts all of which are sometimes called “the" KE:

• At one extreme there is what we might call the microscopic or ultramicroscopic viewpoint, where we take into account the motion of every microscopic particle in the system:

 KE[microscopic] = ∑½ pi2 / mi              (1)

where pi is the momentum of the ith particle, mi is the mass of the ith particle, and the sum runs over all particles.

• At the opposite extreme there is what we might call the macroscopic or holoscopic viewpoint, where we look only at the common-mode motion of the whole object, using only the total momentum P := ∑pi and the total mass M := ∑mi, so that

 KE[holoscopic] = ½ P2 / M              (2)

• There are innumerably many gray areas between these extremes. We might call these mesoscopic viewpoints. Among these is the view that divorces mechanics from thermodynamics, as exemplified by a ordinary spinning flywheel: in mechanics class, if somebody asks you to calculate “the" KE of the flywheel you probably aren’t expected to include the KE of the ultramicroscopic thermal agitation, just the rotational KE. For a flywheel,
• The holoscopic KE is zero, because the center of mass isn’t moving;
• The mesoscopic KE is the usual rotational KE; and
• The ultramicroscopic KE includes the above plus the thermal agitation. In this viewpoint, the thermal KE is the largest contribution unless the flywheel is unusually cold and/or rotating unusually fast.

On the other side of the same coin: If somebody asks you to calculate the thermal KE, you aren’t expected to include the organized rotational KE. So you would have to evaluate KE[microscopic] and subtract off KE[mesoscopic].

Note that we don’t consider the flywheel to be mesoscopic; rather, we partition an ordinary flywheel into mesoscopic cells when we evaluate KE[mesoscopic].

• For a discussion of what “kinetic energy” means in special relativity, see reference 1 and reference 2.

A more-general approach would be to specify a length-scale “λ" specifying the resolution, i.e. how closely we are going to look at things. Then we partition the object into cells of size λ, and define

 KE[λ] := ∑½ pk2 / mk              (3)

where the sum runs over all cells.

• When λ is infinitesimal, we recover the ultramicroscopic KE.
• When λ is larger than the size of the object, we get a single cell and we recover the holoscopic KE.
• Intermediate values of λ are useful, too.
• You can have a hierarchy of lengthscales. The cells of size λ1 can be subdivided into subcells of size λ2 and so on.

Remarkably, the value of KE[λ] is not very sensitive to the choice of λ, over a wide range, as we now discuss.

Consider a flywheel in the form of a solid cube with edge-length L = 1 meter. Choose λ = 1 cm; that is, partition the object in to a million cubelets each 1 cm on a side.

We can make a scaling argument. The moment of inertia scales like r2 m. Since the cube and cubelets all have similar shape (similar in the strict sense of Euclidean geometry), we don’t need to worry about dimensionless factors in front of the scaling formula.

The moment of inertia of each cubelet scales like λ5 ... three factors of λ for the mass and two factors for the r2 in r2 dm. The number of cubelets scales like 1/λ3, so when we sum over cubelets we find that the total KE[є] tied up strictly inside the cubelets scales like λ2 (not including the center-of-mass motion of the cubelet). That is,

 KE[є] − KE[λ] ≈ (λ/L)2 KE[є]              (4)

where є is some length-scale small compared to λ but large enough to wash out any ultramicroscopic motions (e.g. thermal agitation).

In our numerical example, λ = 1 cm, so

 KE[1cm] ≈ 99.99% KE[є]              (5)

for any є that is small compared to 1 cm but still large compared to atoms.

## 2  Work

In mechanics, the definition of work is ambiguous, but only mildly so. (By way of contrast, in thermodynamics, the ambiguities are more numerous and much more serious, as discussed in reference 3. Thermodynamics is beyond the scope of this document.)

In mechanics, all notions of “work” have something to do with force times distance.

The conventional definition of work done on an object is:

work =
 Γ
F · ds              (6)

where the integral runs along some path Γ, namely the path taken by the point of application of the force, and where the displacement ds is a step along the path.

The laws of physics require us to know the direction and magnitude of the force, and also the point of application of the force.

• For a structureless pointlike particle, the point of application is obvious.
• For an extended object, the point of application must be specified.

Beware: If you want to calculate the work, it is generally not safe to multiply the “average” force by the “average” displacement. Sometimes you can get away with that, but sometimes you can’t. For example, consider the wheel shown in figure 1. The hands pull on the strings (shown in blue) which in turn pull on the wheel, causing it to spin faster. The average force on the wheel is zero and the average displacement is zero, but the work being done on the wheel is definitely nonzero. Figure 1: Work Done on an Extended Object
• When in doubt, break the object into a large number of small cells, and apply equation 6 to each cell separately:

work =  ∑ i
 Γ
Fi · dsi              (7)

where the sum runs over all cells in the object. The idea here is that as each cell becomes smaller, the energy associated with internal motion within the cell becomes small ... not just small, but disproportionately small, so that even after summing over all cells the internal motions make a negligible contribution to the total energy. For example, in figure 1, the center of mass of the wheel as a whole is not moving, and the rotational kinetic energy of the wheel is considered “internal” to the wheel. In contrast, if we break the wheel into small cells, the center of mass of each cell is moving, and these center-of-mass motions carry the lion’s share of the kinetic energy. (Each cell is also rotating, but these “internal” rotational energies are disproportionately small, and don’t add up to much, in accordance with the scaling argument that leads to equation 4.)

• Sometimes you can get away with using the average force and the average displacement, if you are careful. This is important, because the concept of work wouldn’t be practical if every time you wanted to use it, you had to evaluate everything cell by cell, using atomic-sized or subatomic-sized cells. It is easy to identify two simple cases:
• As we shall see in section 4.2, if a given system or subsystemis rigid and nonrotating, so that all parts move at the same velocity. then it doesn’t matter where you apply the force. It is OK to use total force and average motion.
• As we shall see in section 4.3, if the force is evenly distributed across all the cells, in proportion to the mass of each cell, then the internal motions of the cell don’t matter. The system doesn’t even need to be rigid or nonrotating. You can get away with using total force and center-of-mass velocity.
• There are undoubtedly other sufficient conditions.

We now introduce another definition of work. This definition is somewhat more sophisticated.

Rather than talking about the work done on the object, we talk about the work done on the boundary of the object, and more specifically on various parts of the boundary.

For example: suppose I am pushing a car up a ramp at a steady rate. I am pushing forward on the car, while other forces (notably a component of the gravitational force) are pushing backwards. It certainly feels to me like I am doing work. Indeed I am doing work, in the sense that energy is crossing the boundary from me into the car, via the part of the car I am pushing on. This energy flows through the car without accumulating in the car. That is, as quickly as the energy flows into the car (from me) it flows out again (into the gravitational field). If we consider the car as a whole, the work is zero in this situation, but if we divide the boundary into parts, there can be nonzero work on this-or-that part.

## 3  The Work / Kinetic Energy Theorems

For each lengthscale λ, we can establish a work[λ]/KE[λ] theorem. Specifically, for each cell, we define work using equation 6, and the total work[λ] is just the sum over cells in the obvious way. As we shall see in section 7, the theorem states

 Δ KE[λ] = work[λ]              (8)

We speak of the work/KE theorems, plural, because there is a separate theorem for each lengthscale λ.

When λ is large, work[λ] is sometimes called the pseudowork. See e.g. reference 4.

Returning to the example of pushing a car up a ramp: The KE of the car is not changing, which is consistent with the pseudowork/KE theorem, because the total work done on the car is zero ... even though my local contribution to the work is nonzero.

## 4  Common Mode and Differential Mode

### 4.1  Quick Version

Another bit of terminology that may be helpful. For any object (or cell or subcell):

• Define “common mode" motion to refer to the motion of the center of mass. The common-mode momentum is P = ∑pi. The common-mode KE is

 KEcm = ½ P2 / M.              (9)

• Define “differential mode" motion to refer to the motion relative to an observer comoving with the center of mass. In the lab frame, the center of mass has velocity

 V = P/M              (10)

The differential-mode momentum of the ith particle is

 p′i = pi − V mi              (11)

The differential-mode KE of the particle is ½ pi2 / mi and the differential-mode KE of the object is

 KEdm := ∑½ p′i2 / mi              (12)

This is useful because

 KE = KEcm + KEdm              (13)

which in turn helps you understand the work[λ]/KE[λ] theorem.

### 4.2  Variations; Uniform Velocity

Starting from equation 7, we can re-express the total work as:

work =
 ∑ i
w~i dt
w~i := F~i · vi
(14)

where vi is the velocity of the point of application of the ith force, i.e. the force applied to the ith cell.

Here we have written a tilde over the F~i, to remind us that force is extensive, so that the force on an average cell is that cell’s share of the total force. Ditto for w~i, which represents the cell’s share of power. This stands in contrast to the velocity, which gets no tilde because it is intensive.

We can define the average velocity as:

v =
(1/N
 ∑ i
vi
(15)

Similarly we can define the average share of the force:

F~ =
(1/N
 ∑ i
F~i
(16)

We now define the variations:

 δvi := vi − ⟨v⟩ δF~i := F~i − ⟨F~⟩
(17)

Hence

 vi = δvi + ⟨v⟩ F~i = δF~i + ⟨F~⟩
(18)

and of course

 ∑ δvi = 0 ∑ δF~i = 0
(19)

Plugging equation 18 into equation 14 we find:

w~i = F~i · vi
= (⟨F~⟩ + δF~i)· (⟨v⟩ + δvi)
= F~⟩·⟨v⟩ + ⟨F~⟩·δvi + δF~i·⟨v⟩ + δF~i·δvi
 d(work) dt
= F·⟨v⟩ +  δF~i·δvi
(20)

where F is the total force (not the average share of force). Note that the two terms that were linear in the variations dropped out, because of the sum rule, equation 19.

The last term in equation 20 can be considered an inner product twice over. The explicit dot product is an ordinary three-dimensional real-space inner product ... but the sum over i can also be considered an inner product, in an abstract N-dimensional space. This term goes to zero if the variations in force are perpendicular (in real space) to the variations in velocity, in which case every term in the sum over i is separately zero ... but it also goes to zero if these terms are individually nonzero but add up to zero, which is to say that the terms are uncorrelated.

In any case, whenever the sum over i goes to zero, it means we can think about the system in macroscopic terms. That is, we can calculate the work using the total force F and the average velocity ⟨v⟩.

Perhaps the simplest situation in which the sum goes to zero is the situation where all the cells in the system are moving with the same velocity, so that δvi = 0 for all i.

### 4.3  Variations; Uniform Acceleration

We can learn something new from equation 14 if we divide the force by mass, and multiply the velocity by mass:

 w~i := F~i · vi = (F~i / m~i) · (m~i vi) = ai · p~i
(21)

where ai is the acceleration of the ith particle, m~i is its share of the mass, and p~i is its share of the momentum.

The rest of the calculation runs closely parallel to section 4.2. The only difference is that we are using differently-weighted averages. We obtain:

 d(work) dt
= a⟩ ·p δai·δp~i
(22)

where p is the total momentum of the system.

Perhaps the simplest situation in which this sum over i vanishes is when all cells have the same acceleration, such as might result from a uniform gravitational field. In such a case we can multiply and divide by the total mass to obtain:

 d(work) dt
= F·vcm
(23)

where F is the total force and vcm is the center-of-mass velocity.

## 5  Remarks

• Everything in this document is restricted to nonrelativistic mechanics only. I tried generalizing it in the obvious way ... without success.
• Nothing we can ever do is sufficient to salvage the ghastly “thermodynamic" formula δ E = W + Q, in which W stands for the so-called work. As explained in reference 3, that formula is a Bad Idea and no amount of tinkering with the definition of W will fix it.
• This is tangentially related to the physics of friction. The energy that is dissipated via friction is hiding in the sum over i in equation 22: there are billions of tiny variations in the force, slightly correlated with tiny variations in the motion.

## 6  Momentum

Let’s take a slight detour to talk about momentum.

Energy is important. Momentum is important. Each obeys a local conservation law. The two concepts are intimately related, but they are not the same.

For instance, consider a box containing 13 particles moving to the left, plus 13 particles moving to the right, all with comparable speeds.

 Each of the 26 particles has some momentum, but the momentum of the left-moving particles is opposite to the momentum of the right-moving particles, so the system as a whole has little if any overall momentum. Each of the 26 particles has some kinetic energy, and all of them make a positive contribution to the total energy of the system.

 If we apply a force to the system, it changes the momentum. If all we care about is the momentum of the system, it doesn’t matter where we apply the force; any force applied anywhere in the system has the same effect on the overall momentum. If we care about the energy, it matters a great deal where we apply the force. A leftward force applied to a leftward-moving particle increases the system energy; the same force applied to a rightward-moving particle decreases the system energy.

Specifically, the change in total momentum is:

P :=
 ∑ i
pi
=
 ∑ i
Fi dt
= Ftot dt
(24)

where the index i runs over all particles in the system, pi is the momentum of the ith particle, Fi is the force applied to that particle, and where Ftot := ∑Fi refers to the total force on the system.

Specifically, the change in microscopic kinetic energy is

KE =
 ∑ i
½ pi2 / m
=
 ∑ i
(Fi · dxi)
(25)

where xi is the position of the ith particle, and mi is its mass. This is called the work / kinetic-energy theorem. The notion of work is discussed in section 7; see also reference 3.

 To summarize: Momentum is denoted p and is related to force times time. Kinetic energy is p2/2m and is related to force times distance.

## 7  Momentum Squared, Kinetic Energy, and Work

As discussed in section 1, we divide the system into cells. Let’s look at the square of the momentum of one of the cells, and see how it changes when we apply a force:

 d (p2) = 2 p · dp = 2 m v · dp = 2 m v · F dt = 2 m F · v dt = 2 m F · dx
(26)

or, simply,

 d (p2) = 2 m F · dx
(27)

where m is the total mass of this cell, v := p/m is the velocity of its center of mass, x is the distance traveled by the center of mass, and F is the total force (i.e. net force) applied to the cell.

We rearrange it so it has dimensions of energy:

 d (p2 / 2m) = F · dx
(28)

And then sum over all cells

 ∑ k
(pk2 / 2mk)
=
 ∑ k
Fk · dxk
(29)

where the sum runs over all cells.

This proves the work/KE theorem, in particular the work[λ]/KE[λ] theorem, equation 8.

Equation 24 is useful if you know a certain force is applied for a certain time; equation 27 is useful if you know a certain force is applied while the cell moves a certain distance.

You might be tempted to take the limit of ultramicroscopic cells (λ → 0). In theory, this would make the theorem very powerful, in the sense that you would have accounted for all the kinetic energy. In practice, the drawback is that such a theorem would be very hard to apply, because it would require knowing every detail of the motion and every detail of what force is applied at which point. For example, when calculating the KE of a flywheel, do you include the KE of electrons whizzing around inside individual atoms? You could, but it would be unconventional and almost certainly not worth the effort.

At the other extreme (large λ) the theorem is easy to apply, but you have to remind yourself (and remind all other stakeholders) that you are calculating the common-mode kinetic energy. You shouldn’t call it “the” kinetic energy unless it is super-clear from context that that’s what you mean.

Remarks:

• You can write the kinetic energy of any system as the common-mode kinetic energy plus the differential-mode kinetic energy. That is, the kinetic energy observed by Joe (in the lab frame) is equal to the kinetic energy observed by Moe (comoving with the center of mass) plus ½ M V2 where V is the Moe-Joe relative velocity. This simple expression utterly depends on a special property of the center-of-mass frame. If you have two observers neither of which is comoving with the center of mass, the energy transformation equations are more complicated.
• The holoscopic KE (i.e. common-mode KE) is a lower bound on the mesoscopic KE. In turn the mesoscopic KE is a lower bound on the microscopic KE. In general, the closer you look, the more KE you will find.
• Given cells at a certain lengthscale, if you know the kinetic energy of each cell separately, you can calculate the kinetic energy at a coarser lengthscale by combining cells. But if all you know is the total KE at one lengthscale, you can’t calculate anything but a bound on the total KE at another lengthscale.
• The change in KE on one length scale is not a bound on the change in KE on any other lengthscale.
• You can have a large energy transfer without a large momentum transfer, or vice versa, as can be seen reference 5.

Here’s another interesting contrast:

KE[holoscopic] =
 1 M

 ∑ i
(Fi)  ·
 ∑ j
(dxj mj)
(30)

KE[microscopic] =
 ∑ i
(Fi · dxi)
(31)

 All the forces are summed (and all the displacements are summed) before multiplying. The sum over displacements is a weighted sum, weighted by mass. Each microscopic force is multiplied by the corresponding microscopic displacement before summing.

In general, it makes a huuuuge difference whether the multiplication occurs before or after the summation. It also makes a huuuuge difference whether or not a weighted sum is used.

## 8  References

1.
John Denker,
“Welcome to Spacetime”
www.av8n.com/physics/spacetime-welcome.htm

2.
John Denker,
“Spacetime Kinetic Energy – An Exercise in Numerical Methods” www.av8n.com/physics/spacetime-kinetic-energy.htm

3.
John Denker,
“Work” (chapter in Modern Thermodynamics)
./thermo-laws.htm#sec-def-work

4.
B. A. Sherwood and W. H. Bernard,
“Work and heat transfer in the presence of sliding friction”
Am. J. Phys 52(11) (1984).
http://www4.ncsu.edu/~basherwo/docs/Friction1984.pdf

5.
John Denker,
“What Makes the Car Go”
./car-go.htm
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