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:
| 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.
| KE[holoscopic] = ½ P2 / M (2) |
Note that we don’t require the flywheel to be mesoscopic; rather, we partition an ordinary flywheel into mesoscopic cells when we evaluate KE[mesoscopic].
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.
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 m. The number of cubelets scales like 1/λ3, so when we sum over cubelets we find that the total KE[є] tied up inside the cubelets scales like λ2. 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) |
In mechanics, the definition of work is ambiguous. All versions have something to do with force times distance.
| work = | ∫ |
| F · ds (6) |
where ds is a step along the path. In particular, if the object in question is a cell of size λ, then F is the total force on the cell, and ds measures the displacement of the center of mass of the cell.
It must be emphasized that in this version, F is the total force on the cell.
In particular, 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 on the car, and according to the local definition that’s true. Energy is crossing the boundary from me to the car.
In contrast, the overall work is zero.
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] (7) |
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 1.
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.
Another bit of terminology that may be helpful. For any object (or cell or subcell):
| KEcm = ½ P2 / M. (8) |
| V = P/M (9) |
The differential-mode momentum of the ith particle is
| p’i = pi − V mi (10) |
The differential-mode KE of the particle is ½ p’i2 / mi and the differential-mode KE of the object is
| KEdm := ∑½ p’i2 / mi (11) |
This is useful because
| KE = KEcm + KEdm (12) |
which in turn helps you understand the work[λ]/KE[λ] theorem.
All of the above is restricted to nonrelativistic mechanics only. I tried generalizing it in the obvious way ... without success.
Also, none of the above is sufficient to salvage the ghastly “thermodynamic" formula δ E = W + Q, in which W stands for the so-called work. That formula is a Bad Idea and no amount of tinkering with the definition of W will fix it.
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:
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
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 2. |
| 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. |
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:
| (15) |
or, simply,
| (16) |
where m is the total mass of this cell, v := p/m is the velocity of its center of mass, x is the distance travelled 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:
| (17) |
And then sum over all cells
| (18) |
where the sum runs over all cells.
This proves the work/KE theorem, in particular the work[λ]/KE[λ] theorem, equation 7.
Equation 13 is useful if you know a certain force is applied for a certain time; equation 16 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 enery. You shouldn’t call it “the” kinetic energy unless it is super-clear from context that that’s what you mean.
Remarks:
Here’s another interesting contrast:
| 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.