Contents
Section 3.1 - Momentum In One Dimension
Section 3.2 - Force In One Dimension
Section 3.3 - Resonance
Section 3.4 - Motion In Three Dimensions

Forces transfer momentum to the girl.

Chapter 3. Conservation of Momentum

I think, therefore I am.\par{}I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery. -- René Descartes

3.1 Momentum In One Dimension

3.1.1 Mechanical momentum

In the martial arts movie Crouching Tiger, Hidden Dragon, those who had received mystical enlightenment are able to violate the laws of physics. Some of the violations are obvious, such as their ability to fly, but others are a little more subtle. The rebellious young heroine/antiheroine Jen Yu gets into an argument while sitting at a table in a restaurant. A young tough, Iron Arm Lu, comes running toward her at full speed, and she puts up one arm and effortlessly makes him bounce back, without even getting out of her seat or bracing herself against anything. She does all this between bites.

Although kinetic energy doesn't depend on the direction of motion, we've already seen on page 88 how conservation of energy combined with Galilean relativity allows us to make some predictions about the direction of motion. One of the examples was a demonstration that it isn't possible for a hockey puck to spontaneously reverse its direction of motion. In the scene from the movie, however, the woman's assailant isn't just gliding through space. He's interacting with her, so the previous argument doesn't apply here, and we need to generalize it to more than one object. We consider the case of a physical system composed of pointlike material particles, in which every particle interacts with every other particle through an energy U(r) that depends only on the distance r between them. This still allows for a fairly general mechanical system, by which I mean roughly a system made of matter, not light. The characters in the movie are made of protons, neutrons, and electrons, so they would constitute such a system if the interactions among all these particles were of the form U(r).¹ We might even be able to get away with thinking of each person as one big particle, if it's a good approximation to say that every part of each person's whole body moves in the same direction at the same speed.

The basic insight can be extracted from the special case where there are only two particles interacting, and they only move in one dimension, as in the example shown in figure b. Conservation of energy says

For simplicity, let's assume that the interactions start after the time we're calling initial, and end before the instant we choose as final. This is true in figure b, for example. Then U_i=U_f, and we can subtract the interaction energies from both sides, giving.

As in the one-particle argument on page 88, the trick is to require conservation of energy not just in one particular frame of reference, but in every frame of reference. In a frame of reference moving at velocity u relative to the first one, the velocities all have u added onto them:²

When we square a quantity like (v_1i+u)², we get the same v_1i² that occurred in the original frame of reference, plus two u-dependent terms, 2v_1iu+u². Subtracting the original conservation of energy equation from the version in the new frame of reference, we have

This is a statement that when you add up mv for the whole system, that total remains constant over time. In other words, this is a conservation law. The quantity mv is called momentum, notated p for obscure historical reasons. Its units are kg⋅m/s.

Unlike kinetic energy, momentum depends on the direction of motion, since the velocity is not squared. In one dimension, motion in the same direction as the positive x axis is represented with positive values of v and p. Motion in the opposite direction has negative v and p.

This example points to an intuitive interpretation of conservation of momentum, which is that interactions are always mutual. That is, Jen Yu can't change Lu's momentum without having her own momentum changed as well.

3.1.2 Nonmechanical momentum

So far, it sounds as though conservation of momentum can be proved mathematically, unlike conservation of mass and energy, which are entirely based on observations. The proof, however, was only for a mechanical system, with interactions of the form U(r). Conservation of momentum can be extended to other systems as well, but this generalization is based on experiments, not mathematical proof. Light is the most important example of momentum that doesn't equal mv --- light doesn't have mass at all, but it does have momentum. For example, a flashlight left on for an hour would absorb about 10^-5 kg⋅m/s of momentum as it recoiled.

The reason for bringing this up is not so that you can plug numbers into formulas in these exotic situations. The point is that the conservation laws have proven so sturdy exactly because they can easily be amended to fit new circumstances. The momentum of light will be a natural consequence of the discussion of the theory of relativity in chapter 7.

3.1.3 Momentum compared to kinetic energy

Momentum and kinetic energy are both measures of the quantity of motion, and a sideshow in the Newton-Leibniz controversy over who invented calculus was an argument over whether mv (i.e., momentum) or mv² (i.e., kinetic energy without the 1/2 in front) was the “true” measure of motion. The modern student can certainly be excused for wondering why we need both quantities, when their complementary nature was not evident to the greatest minds of the 1700s. The following table highlights their differences.

Kinetic energy... & Momentum...
\hline\hline doesn't depend on direction. & depends on direction.
\hline

is always positive, and cannot cancel out. & cancels with momentum in the opposite direction.
\hline

can be traded for forms of energy that do not involve motion. Kinetic energy is not a conserved quantity by itself. & is always conserved in a closed system.
\hline

is quadrupled if the velocity is doubled. & is doubled if the velocity is doubled.
\hline

Here are some examples that show the different behaviors of the two quantities.

Discussion Questions

3.1.4 Collisions in one dimension

Physicists employ the term “collision” in a broader sense than in ordinary usage, applying it to any situation where objects interact for a certain period of time. A bat hitting a baseball, a cosmic ray damaging DNA, and a gun and a bullet going their separate ways are all examples of collisions in this sense. Physical contact is not even required. A comet swinging past the sun on a hyperbolic orbit is considered to undergo a collision, even though it never touches the sun. All that matters is that the comet and the sun interacted gravitationally with each other.

The reason for broadening the term “collision” in this way is that all of these situations can be attacked mathematically using the same conservation laws in similar ways. In our first example, conservation of momentum is all that is required.

The above example was simple because both cars had the same velocity afterward. In many one-dimensional collisions, however, the two objects do not stick. If we wish to predict the result of such a collision, conservation of momentum does not suffice, because both velocities after the collision are unknown, so we have one equation in two unknowns.

Conservation of energy can provide a second equation, but its application is not as straightforward, because kinetic energy is only the particular form of energy that has to do with motion. In many collisions, part of the kinetic energy that was present before the collision is used to create heat or sound, or to break the objects or permanently bend them. Cars, in fact, are carefully designed to crumple in a collision. Crumpling the car uses up energy, and that's good because the goal is to get rid of all that kinetic energy in a relatively safe and controlled way. At the opposite extreme, a superball is “super” because it emerges from a collision with almost all its original kinetic energy, having only stored it briefly as interatomic electrical energy while it was being squashed by the impact.

Collisions of the superball type, in which almost no kinetic energy is converted to other forms of energy, can thus be analyzed more thoroughly, because they have K_f=K_i, as opposed to the less useful inequality K_f<K_i for a case like a tennis ball bouncing on grass.

Often, as in example 11, the details of the algebra are the least interesting part of the problem, and considerable physical insight can be gained simply by counting the number of unknowns and comparing to the number of equations. Suppose a beginner at pool notices a case where her cue ball hits an initially stationary ball and stops dead. “Wow, what a good trick,” she thinks. “I bet I could never do that again in a million years.” But she tries again, and finds that she can't help doing it even if she doesn't want to. Luckily she has just learned about collisions in her physics course. Once she has written down the equations for conservation of energy and no loss of kinetic energy, she really doesn't have to complete the algebra. She knows that she has two equations in two unknowns, so there must be a well-defined solution. Once she has seen the result of one such collision, she knows that the same thing must happen every time. The same thing would happen with colliding marbles or croquet balls. It doesn't matter if the masses or velocities are different, because that just multiplies both equations by some constant factor.

The discovery of the neutron

This was the type of reasoning employed by James Chadwick in his 1932 discovery of the neutron. At the time, the atom was imagined to be made out of two types of fundamental particles, protons and electrons. The protons were far more massive, and clustered together in the atom's core, or nucleus. Electrical attraction caused the electrons to orbit the nucleus in circles, in much the same way that gravity kept the planets from cruising out of the solar system. Experiments showed, for example, that twice as much energy was required to strip the last electron off of a helium atom as was needed to remove the single electron from a hydrogen atom, and this was explained by saying that helium had two protons to hydrogen's one. The trouble was that according to this model, helium would have two electrons and two protons, giving it precisely twice the mass of a hydrogen atom with one of each. In fact, helium has about four times the mass of hydrogen.

Chadwick suspected that the helium nucleus possessed two additional particles of a new type, which did not participate in electrical interactions at all, i.e., were electrically neutral. If these particles had very nearly the same mass as protons, then the four-to-one mass ratio of helium and hydrogen could be explained. In 1930, a new type of radiation was discovered that seemed to fit this description. It was electrically neutral, and seemed to be coming from the nuclei of light elements that had been exposed to other types of radiation. At this time, however, reports of new types of particles were a dime a dozen, and most of them turned out to be either clusters made of previously known particles or else previously known particles with higher energies. Many physicists believed that the “new” particle that had attracted Chadwick's interest was really a previously known particle called a gamma ray, which was electrically neutral. Since gamma rays have no mass, Chadwick decided to try to determine the new particle's mass and see if it was nonzero and approximately equal to the mass of a proton.

g / Chadwick's subatomic pool table. A disk of the naturally occurring metal polonium provides a source of radiation capable of kicking neutrons out of the beryllium nuclei. The type of radiation emitted by the polonium is easily absorbed by a few mm of air, so the air has to be pumped out of the left-hand chamber. The neutrons, Chadwick's mystery particles, penetrate matter far more readily, and fly out through the wall and into the chamber on the right, which is filled with nitrogen or hydrogen gas. When a neutron collides with a nitrogen or hydrogen nucleus, it kicks it out of its atom at high speed, and this recoiling nucleus then rips apart thousands of other atoms of the gas. The result is an electrical pulse that can be detected in the wire on the right. Physicists had already calibrated this type of apparatus so that they could translate the strength of the electrical pulse into the velocity of the recoiling nucleus. The whole apparatus shown in the figure would fit in the palm of your hand, in dramatic contrast to today's giant particle accelerators.

Unfortunately a subatomic particle is not something you can just put on a scale and weigh. Chadwick came up with an ingenious solution. The masses of the nuclei of the various chemical elements were already known, and techniques had already been developed for measuring the speed of a rapidly moving nucleus. He therefore set out to bombard samples of selected elements with the mysterious new particles. When a direct, head-on collision occurred between a mystery particle and the nucleus of one of the target atoms, the nucleus would be knocked out of the atom, and he would measure its velocity.

Suppose, for instance, that we bombard a sample of hydrogen atoms with the mystery particles. Since the participants in the collision are fundamental particles, there is no way for kinetic energy to be converted into heat or any other form of energy, and Chadwick thus had two equations in three unknowns:

equation #1: conservation of momentum

equation #2: no loss of kinetic energy

unknown #1: mass of the mystery particle

unknown #2: initial velocity of the mystery particle

unknown #3: final velocity of the mystery particle

The number of unknowns is greater than the number of equations, so there is no unique solution. But by creating collisions with nuclei of another element, nitrogen, he gained two more equations at the expense of only one more unknown:

equation #3: conservation of momentum in the new collision

equation #4: no loss of kinetic energy in the new collision

unknown #4: final velocity of the mystery particle in the new collision

He was thus able to solve for all the unknowns, including the mass of the mystery particle, which was indeed within 1% of the mass of a proton. He named the new particle the neutron, since it is electrically neutral.

Discussion Questions

3.1.5 The center of mass

Figures h and j show two examples where a motion that appears complicated actually has a very simple feature. In both cases, there is a particular point, called the center of mass, whose motion is surprisingly simple. The highjumper flexes his body as he passes over the bar, so his motion is intrinsically very complicated, and yet his center of mass's motion is a simple parabola, just like the parabolic arc of a pointlike particle. The wrench's center of mass travels in a straight line as seen from above, which is what we'd expect for a pointlike particle flying through the air.

j / In this multiple-flash photograph, we see the wrench from above as it flies through the air, rotating as it goes. Its center of mass, marked with the black cross, travels along a straight line, unlike the other points on the wrench, which execute loops. (PSSC Physics)

The highjumper and the wrench are both complicated systems, each consisting of zillions of subatomic particles. To understand what's going on, let's instead look at a nice simple system, two pool balls colliding. We assume the balls are a closed system (i.e., their interaction with the felt surface is not important) and that their rotation is unimportant, so that we'll be able to treat each one as a single particle. By symmetry, the only place their center of mass can be is half-way in between, at an x coordinate equal to the average of the two balls' positions, x_cm=(x₁+x₂)/2.

Figure i makes it appear that the center of mass, marked with an ×, moves with constant velocity to the right, regardless of the collision, and we can easily prove this using conservation of momentum:

Since momentum is conserved, the last expression is constant, which proves that v_cm is constant.

Rearranging this a little, we have p_total=m_totalv_cm. In other words, the total momentum of the system is the same as if all its mass was concentrated at the center of mass point.

Sigma notation

When there is a large, potentially unknown number of particles, we can write sums like the ones occurring above using symbols like “+…,” but that gets awkward. It's more convenient to use the Greek uppercase sigma, Σ, to indicate addition. For example, the sum 1²+2²+3²+4²=30 could be written as

read “the sum from j=1 to n of j².” The variable j is a dummy variable, just like the dx in an integral that tells you you're integrating with respect to x, but has no significance outside the integral. The j below the sigma tells you what variable is changing from one term of the sum to the next, but j has no significance outside the sum.

As an example, let's generalize the proof of p_total=m_totalv_cm to the case of an arbitrary number n of identical particles moving in one dimension, rather than just two particles. The center of mass is at

where x₁ is the mass of the first particle, and so on. The velocity of the center of mass is

What about a system containing objects with unequal masses, or containing more than two objects? The reasoning above can be generalized to a weighted average:

3.1.6 The center of mass frame of reference

A particularly useful frame of reference in many cases is the frame that moves along with the center of mass, called the center of mass (c.m.) frame. In this frame, the total momentum is zero. The following examples show how the center of mass frame can be a powerful tool for simplifying our understanding of collisions.

Discussion Questions

3.2 Force In One Dimension

3.2.1 Momentum transfer

For every conserved quantity, we can define an associated rate of flow. An open system can have mass transferred in or out of it, and we can measure the rate of mass flow, dm/dt in units of kg/s. Energy can flow in or out, and the rate of energy transfer is the power, P=dE/dt, measured in watts.³ The rate of momentum transfer is called force,

The units of force are kg⋅m/s², which can be abbreviated as newtons, 1 N=kg⋅m/s². Newtons are unfortunately not as familiar as watts. A newton is about how much force you'd use to pet a dog. The most powerful rocket engine ever built, the first stage of the Saturn V that sent astronauts to the moon, had a thrust of about 30 million newtons. In one dimension, positive and negative signs indicate the direction of the force --- a positive force is one that pushes or pulls in the direction of the positive x axis.

This is also the principle of airbags in cars. The time required for the airbag to decelerate your head is fairly long: the time it takes your face to travel 20 or 30 cm. Without an airbag, your face would have hit the dashboard, and the time interval would have been the much shorter time taken by your skull to move a couple of centimeters while your face compressed. Note that either way, the same amount of momentum is transferred: the entire momentum of your head.

Force is defined as a derivative, and the derivative of a sum is the sum of the derivatives. Therefore force is additive: when more than one force acts on an object, you add the forces to find out what happens. An important special case is that forces can cancel. Consider your body sitting in a chair as you read this book. Let the positive x axis be upward. The chair's upward force on you is represented with a positive number, which cancels out with the earth's downward gravitational force, which is negative. The total rate of momentum transfer into your body is zero, and your body doesn't change its momentum.

Discussion Question

3.2.2 Newton's laws

Although momentum is the third conserved quantity we've encountered, historically it was the first to be discovered. Isaac Newton formulated a complete treatment of mechanical systems in terms of force and momentum. Newton's theory was based on three laws of motion, which we now think of as consequences of conservation of mass, energy, and momentum.

Newton’s laws in one dimension:

Newton’s first law: If there is no force acting on an object, it stays in the same state of motion.

Newton’s second law: The sum of all the forces acting on an object determines the rate at which its momentum changes,Ftotalderpdert.

Newton’s third law: Forces occur in opposite pairs. If object A interacts with object B, then A’s force on B and B’s force on A are related byFABFBA.

{}The second law is the definition of force, which we've already encountered.⁴ The first law is a special case of the second law --- if dp/dt is zero, then p=mv is a constant, and since mass is conserved, constant p implies constant v. The third law is a restatement of conservation of momentum: for two objects interacting, we have constant total momentum, so =F_BA+F_AB.

Using the third law correctly

If you've already accepted Galilean relativity in your heart, then there is nothing really difficult about the first and second laws. The third law, however, is more of a conceptual challenge. The first hurdle is that it is counterintuitive. Is it really true that if a fighter jet collides with a mosquito, the mosquito's force on the jet is just as strong as the jet's force on the mosquito? Yes, it is true, but it is hard to believe at first. That amount of force simply has more of an effect on the mosquito, because it has less mass.

A more humane and practical experiment is shown in figure c. A large magnet and a small magnet are weighed separately, and then one magnet is hung from the pan of the top balance so that it is directly above the other magnet. There is an attraction between the two magnets, causing the reading on the top scale to increase and the reading on the bottom scale to decrease. The large magnet is more “powerful” in the sense that it can pick up a heavier paperclip from the same distance, so many people have a strong expectation that one scale's reading will change by a far different amount than the other. Instead, we find that the two changes are equal in magnitude but opposite in direction, so the upward force of the top magnet on the bottom magnet is of the same magnitude as the downward force of the bottom magnet on the top magnet.

To students, it often sounds as though Newton's third law implies nothing could ever change its motion, since the two equal and opposite forces would always cancel. As illustrated in figure d, the fallacy arises from assuming that we can add things that it doesn't make sense to add. It only makes sense to add up forces that are acting on the same object, whereas two forces related to each other by Newton's third law are always acting on two different objects. If two objects are interacting via a force and no other forces are involved, then both objects will accelerate --- in opposite directions, as shown in figure e!

Here are some suggestions for avoiding misapplication of Newton's third law:

1. It always relates exactly two forces, not more.
2. The two forces involve exactly two objects, in the pattern A on B, B on A.
3. The two forces are always of the same type, e.g., friction and friction, or gravity and gravity.

Directions of forces

We've already seen that momentum, unlike energy, has a direction in space. Since force is defined in terms of momentum, force also has a direction in space. For motion in one dimension, we have to pick a coordinate system, and given that choice, forces and momenta will be positive or negative. We've already used signs to represent directions of forces in Newton's third law, F_AB=-F_BA.

There is, however, a complication with force that we were able to avoid with momentum. If an object is moving on a line, we're guaranteed that its momentum is in one of two directions: the two directions along the line. But even an object that stays on a line may still be subject to forces that act perpendicular to the line. For example, suppose a coin is sliding to the right across a table, g, and let's choose a positive x axis that points to the right. The coin's motion is along a horizontal line, and its momentum is positive and decreasing. Because the momentum is decreasing, its time derivative d p/d t is negative. This derivative equals the horizontal force of friction F₁, and its negative sign tells us that this force on the coin is to the left.

But there are also vertical forces on the coin. The Earth exerts a downward gravitational force F₂ on it, and the table makes an upward force F₃ that prevents the coin from sinking into the wood. In fact, without these vertical forces the horizontal frictional force wouldn't exist: surfaces don't exert friction against one another unless they are being pressed together.

To avoid mathematical complication, we want to postpone the full three-dimensional treatment of force and momentum until section 3.4. For now, we'll limit ourselves to examples like the coin, in which the motion is confined to a line, and any forces perpendicular to the line cancel each other out.

Discussion Questions

3.2.3 What force is not

Violin teachers have to endure their beginning students' screeching. A frown appears on the woodwind teacher's face as she watches her student take a breath with an expansion of his ribcage but none in his belly. What makes physics teachers cringe is their students' verbal statements about forces. Below I have listed several dicta about what force is not.

Force is not a property of one object.

A great many of students' incorrect descriptions of forces could be cured by keeping in mind that a force is an interaction of two objects, not a property of one object.

Force is not a measure of an object's motion.

If force is not a property of a single object, then it cannot be used as a measure of the object's motion.

Force is not energy.

Force is not stored or used up.

Because energy can be stored and used up, people think force also can be stored or used up.

Forces need not be exerted by living things or machines.

Transforming energy from one form into another usually requires some kind of living or mechanical mechanism. The concept is not applicable to forces, which are an interaction between objects, not a thing to be transferred or transformed.

A force is the direct cause of a change in motion.

I can click a remote control to make my garage door change from being at rest to being in motion. My finger's force on the button, however, was not the force that acted on the door. When we speak of a force on an object in physics, we are talking about a force that acts directly. Similarly, when you pull a reluctant dog along by its leash, the leash and the dog are making forces on each other, not your hand and the dog. The dog is not even touching your hand.

Discussion Questions

3.2.4 Forces between solids

Conservation laws are more fundamental than Newton's laws, and they apply where Newton's laws don't, e.g., to light and to the internal structure of atoms. However, there are certain problems that are much easier to solve using Newton's laws. As a trivial example, if you drop a rock, it could conserve momentum and energy by levitating, or by falling in the usual manner.⁵ With Newton's laws, however, we can reason that a=F/m, so the rock must respond to the gravitational force by accelerating.

Less trivially, suppose a person is hanging onto a rope, and we want to know if she will slip. Unlike the case of the levitating rock, here the no-motion solution could be perfectly reasonable if her grip is strong enough. We know that her hand's interaction with the rope is fundamentally an electrical interaction between the atoms in the surface of her palm and the nearby atoms in the surface of the rope. For practical problem-solving, however, this is a case where we're better off forgetting the fundamental classification of interactions at the atomic level and working with a more practical, everyday classification of forces. In this practical scheme, we have three types of forces that can occur between solid objects in contact:

A normal force,Fn,	is perpendicular to the surface of contact, and prevents objects from passing through each other by becoming as strong as necessary (up to the point where the objects break). “Normal” means perpendicular.
Static friction,Fs,	is parallel to the surface of contact, and prevents the surfaces from starting to slip by becoming as strong as necessary, up to a maximum value ofFsmax. “Static” means not moving, i.e., not slipping.
Kinetic friction,Fk,	is parallel to the surface of contact, and tends to slow down any slippage once it starts. “Kinetic” means moving, i.e., slipping.

If you put a coin on this page, which is horizontal, gravity pulls down on the coin, but the atoms in the paper and the coin repel each other electrically, and the atoms are compressed until the repulsion becomes strong enough to stop the downward motion of the coin. We describe this complicated and invisible atomic process by saying that the paper makes an upward normal force on the coin, and the coin makes a downward normal force on the paper. (The two normal forces are related by Newton's third law. In fact, Newton's third law only relates forces that are of the same type.)

If you now tilt the book a little, static friction keeps the coin from slipping. The picture at the microscopic level is even more complicated than the previous description of the normal force. One model is to think of the tiny bumps and depressions in the coin as settling into the similar irregularities in the paper. This model predicts that rougher surfaces should have more friction, which is sometimes true but not always. Two very smooth, clean glass surfaces or very well finished machined metal surfaces may actually stick better than rougher surfaces would, the probable explanation being that there is some kind of chemical bonding going on, and the smoother surfaces allow more atoms to be in contact.

Finally, as you tilt the book more and more, there comes a point where static friction reaches its maximum value. The surfaces become unstuck, and the coin begins to slide over the paper. Kinetic friction slows down this slipping motion significantly. In terms of energy, kinetic friction is converting mechanical energy into heat, just like when you rub your hands together to keep warm. One model of kinetic friction is that the tiny irregularities in the two surfaces bump against each other, causing vibrations whose energy rapidly converts to heat and sound --- you can hear this sound if you rub your fingers together near your ear.

For dry surfaces, experiments show that the following equations usually work fairly well:

and

where μ_s, the coefficient of static friction, and μ_k, the coefficient of kinetic friction, are constants that depend on the properties of the two surfaces, such as what they're made of and how rough they are.

j / 1. The cliff's normal force on the climber's feet. 2. The track's static frictional force on the wheel of the accelerating dragster. 3. The ball's normal force on the bat.

3.2.5 Fluid friction

Try to drive a nail into a waterfall and you will be confronted with the main difference between solid friction and fluid friction. Fluid friction is purely kinetic; there is no static fluid friction. The nail in the waterfall may tend to get dragged along by the water flowing past it, but it does not stick in the water. The same is true for gases such as air: recall that we are using the word “fluid” to include both gases and liquids.

Unlike kinetic friction between solids, fluid friction increases rapidly with velocity. It also depends on the shape of the object, which is why a fighter jet is more streamlined than a Model T. For objects of the same shape but different sizes, fluid friction typically scales up with the cross-sectional area of the object, which is one of the main reasons that an SUV gets worse mileage on the freeway than a compact car.

Discussion Question

3.2.6 Analysis of forces

Newton's first and second laws deal with the total of all the forces exerted on a specific object, so it is very important to be able to figure out what forces there are. Once you have focused your attention on one object and listed the forces on it, it is also helpful to describe all the corresponding forces that must exist according to Newton's third law. We refer to this as “analyzing the forces” in which the object participates.

Often you may be unsure whether you have missed one of the forces. Here are three strategies for checking your list:

The following is another example in which we can profit by checking against our physical intuition for what should be happening.

I believe that constructing the type of table described in this section is the best method for beginning students. Most textbooks, however, prescribe a pictorial way of showing all the forces acting on an object. Such a picture is called a free-body diagram. It should not be a big problem if a future physics professor expects you to be able to draw such diagrams, because the conceptual reasoning is the same. You simply draw a picture of the object, with arrows representing the forces that are acting on it. Arrows representing contact forces are drawn from the point of contact, noncontact forces from the center of mass. Free-body diagrams do not show the equal and opposite forces exerted by the object itself.

Discussion Questions

3.2.7 Transmission of forces by low-mass objects

You're walking your dog. The dog wants to go faster than you do, and the leash is taut. Does Newton's third law guarantee that your force on your end of the leash is equal and opposite to the dog's force on its end? If they're not exactly equal, is there any reason why they should be approximately equal?

If there was no leash between you, and you were in direct contact with the dog, then Newton's third law would apply, but Newton's third law cannot relate your force on the leash to the dog's force on the leash, because that would involve three separate objects. Newton's third law only says that your force on the leash is equal and opposite to the leash's force on you,

and that the dog's force on the leash is equal and opposite to its force on the dog

Still, we have a strong intuitive expectation that whatever force we make on our end of the leash is transmitted to the dog, and vice-versa. We can analyze the situation by concentrating on the forces that act on the leash, F_dL and F_yL. According to Newton's second law, these relate to the leash's mass and acceleration:

The leash is far less massive then any of the other objects involved, and if m_L is very small, then apparently the total force on the leash is also very small, F_dL + F_yL≈ 0, and therefore

Thus even though Newton's third law does not apply directly to these two forces, we can approximate the low-mass leash as if it was not intervening between you and the dog. It's at least approximately as if you and the dog were acting directly on each other, in which case Newton's third law would have applied.

In general, low-mass objects can be treated approximately as if they simply transmitted forces from one object to another. This can be true for strings, ropes, and cords, and also for rigid objects such as rods and sticks.

m / If we imagine dividing a taut rope up into small segments, then any segment has forces pulling outward on it at each end. If the rope is of negligible mass, then all the forces equal +T or -T, where T, the tension, is a single number.

If you look at a piece of string under a magnifying glass as you pull on the ends more and more strongly, you will see the fibers straightening and becoming taut. Different parts of the string are apparently exerting forces on each other. For instance, if we think of the two halves of the string as two objects, then each half is exerting a force on the other half. If we imagine the string as consisting of many small parts, then each segment is transmitting a force to the next segment, and if the string has very little mass, then all the forces are equal in magnitude. We refer to the magnitude of the forces as the tension in the string, T. Although the tension is measured in units of Newtons, it is not itself a force. There are many forces within the string, some in one direction and some in the other direction, and their magnitudes are only approximately equal. The concept of tension only makes sense as a general, approximate statement of how big all the forces are.

If a rope goes over a pulley or around some other object, then the tension throughout the rope is approximately equal so long as there is not too much friction. A rod or stick can be treated in much the same way as a string, but it is possible to have either compression or tension.

Since tension is not a type of force, the force exerted by a rope on some other object must be of some definite type such as static friction, kinetic friction, or a normal force. If you hold your dog's leash with your hand through the loop, then the force exerted by the leash on your hand is a normal force: it is the force that keeps the leash from occupying the same space as your hand. If you grasp a plain end of a rope, then the force between the rope and your hand is a frictional force.

A more complex example of transmission of forces is the way a car accelerates. Many people would describe the car's engine as making the force that accelerates the car, but the engine is part of the car, so that's impossible: objects can't make forces on themselves. What really happens is that the engine's force is transmitted through the transmission to the axles, then through the tires to the road. By Newton's third law, there will thus be a forward force from the road on the tires, which accelerates the car.

Discussion Question

3.2.8 Work

Energy transferred to a particle

To change the kinetic energy, K=(1/2)mv², of a particle moving in one dimension, we must change its velocity. That will entail a change in its momentum, p=mv, as well, and since force is the rate of transfer of momentum, we conclude that the only way to change a particle's kinetic energy is to apply a force.⁶ A force in the same direction as the motion speeds it up, increasing the kinetic energy, while a force in the opposite direction slows it down.

Consider an infinitesimal time interval during which the particle moves an infinitesimal distance dx, and its kinetic energy changes by dK. In one dimension, we represent the direction of the force and the direction of the motion with positive and negative signs for F and dx, so the relationship among the signs can be summarized as follows: \begin{center}

F0	derx0	derK0
F0	derx0	derK0
F0	derx0	derK0
F0	derx0	derK0

\end{center} This looks exactly like the rule for determining the sign of a product, and we can easily show using the chain rule that this is indeed a multiplicative relationship:

In general, when energy is transferred by a force,⁷ we use the term work to refer to the amount of energy transferred. This is different from the way the word is used in ordinary speech. If you stand for a long time holding a bag of cement, you get tired, and everyone will agree that you've worked hard, but you haven't changed the energy of the cement, so according to the definition of the physics term, you haven't done any work on the bag. There has been an energy transformation inside your body, of chemical energy into heat, but this just means that one part of your body did positive work (lost energy) while another part did a corresponding amount of negative work (gained energy).

Work in general

I derived the expression Fdx for one particular type of kinetic-energy transfer, the work done in accelerating a particle, and then defined work as a more general term. Is the equation correct for other types of work as well? For example, if a force lifts a mass m against the resistance of gravity at constant velocity, the increase in the mass's gravitational energy is d(mgy)=mgdy=Fdy, so again the equation works, but this still doesn't prove that the equation is always correct as a way of calculating energy transfers.

Imagine a black box⁸, containing a gasoline-powered engine, which is designed to reel in a steel cable, exerting a certain force F. For simplicity, we imagine that this force is always constant, so we can talk about Δx rather than an infinitesimal dx. If this black box is used to accelerate a particle (or any mass without internal structure), and no other forces act on the particle, then the original derivation applies, and the work done by the box is W=FΔx. Since F is constant, the box will run out of gas after reeling in a certain amount of cable Δx. The chemical energy inside the box has decreased by -W, while the mass being accelerated has gained W worth of kinetic energy.⁹

Now what if we use the black box to pull a plow? The energy increase in the outside world is of a different type than before; it takes the forms of (1) the gravitational energy of the dirt that has been lifted out to the sides of the furrow, (2) frictional heating of the dirt and the plowshare, and (3) the energy needed to break up the dirt clods (a form of electrical energy involving the attractions among the atoms in the clod). The box, however, only communicates with the outside world via the hole through which its cable passes. The amount of chemical energy lost by the gasoline can therefore only depend on F and Δx, so it is the same -W as when the box was being used to accelerate a mass, and thus by conservation of energy, the work done on the outside world is again W.

This is starting to sound like a proof that the force-times-distance method is always correct, but there was one subtle assumption involved, which was that the force was exerted at one point (the end of the cable, in the black box example). Real life often isn't like that. For example, a cyclist exerts forces on both pedals at once. Serious cyclists use toe-clips, and the conventional wisdom is that one should use equal amounts of force on the upstroke and downstroke, to make full use of both sets of muscles. This is a two-dimensional example, since the pedals go in circles. We're only discussing one-dimensional motion right now, so let's just pretend that the upstroke and downstroke are both executed in straight lines. Since the forces are in opposite directions, one is positive and one is negative. The cyclist's total force on the crank set is zero, but the work done isn't zero. We have to add the work done by each stroke, W=F₁Δx₁+F₂Δx₂. (I'm pretending that both forces are constant, so we don't have to do integrals.) Both terms are positive; one is a positive number multiplied by a positive number, while the other is a negative times a negative.

This might not seem like a big deal --- just remember not to use the total force --- but there are many situations where the total force is all we can measure. The ultimate example is heat conduction. Heat conduction is not supposed to be counted as a form of work, since it occurs without a force. But at the atomic level, there are forces, and work is done by one atom on another. When you hold a hot potato in your hand, the transfer of heat energy through your skin takes place with a total force that's extremely close to zero. At the atomic level, atoms in your skin are interacting electrically with atoms in the potato, but the attractions and repulsions add up to zero total force. It's just like the cyclist's feet acting on the pedals, but with zillions of forces involved instead of two. There is no practical way to measure all the individual forces, and therefore we can't calculate the total energy transferred.

To summarize, is a correct way of calculating work, where F_j is the individual force acting on particle j, which moves a distance dx_j. However, this is only useful if you can identify all the individual forces and determine the distance moved at each point of contact. For convenience, I'll refer to this as the work theorem. (It doesn't have a standard name.)

There is, however, something useful we can do with the total force. We can use it to calculate the part of the work done on an object that consists of a change in the kinetic energy it has due to the motion of its center of mass. The proof is essentially the same as the proof on 158, except that now we don't assume the force is acting on a single particle, so we have to be a little more delicate. Let the object consist of n particles. Its total kinetic energy is , but this is what we've already realized can't be calculated using the total force. The kinetic energy it has due to motion of its center of mass is

Figure q shows some examples of the distinction between K_cm and K. Differentiating K_cm, we have

I'll call this the kinetic energy theorem --- like the work theorem, it has no standard name.

Discussion Questions

r / Discussion question B.

3.2.9 Simple machines

Conservation of energy provided the necessary tools for analyzing some mechanical systems, such as the seesaw on page 85 and the pulley arrangements of the homework problems on page 119, but we could only analyze those machines by computing the total energy of the system. That approach wouldn't work for systems like the biceps/forearm machine on page 85, or the one in figure s, where the energy content of the person's body is impossible to compute directly. Even though the seesaw and the biceps/forearm system were clearly just two different forms of the lever, we had no way to treat them both on the same footing. We can now successfully attack such problems using the work and kinetic energy theorems.

3.2.10 Force related to interaction energy

In section 2.3, we saw that there were two equivalent ways of looking at gravity, the gravitational field and the gravitational energy. They were related by the equation dU=mgdr, so if we knew the field, we could find the energy by integration, , and if we knew the energy, we could find the field by differentiation, g=(1/m)dU/dr.

The same approach can be applied to other interactions, for example a mass on a spring. The main difference is that only in gravitational interactions does the strength of the interaction depend on the mass of the object, so in general, it doesn't make sense to separate out the factor of m as in the equation dU=mgdr. Since F=mg is the gravitational force, we can rewrite the equation in the more suggestive form dU=Fdr. This form no longer refers to gravity specifically, and can be applied much more generally. The only remaining detail is that I've been fairly cavalier about positive and negative signs up until now. That wasn't such a big problem for gravitational interactions, since gravity is always attractive, but it requires more careful treatment for nongravitational forces, where we don't necessarily know the direction of the force in advance, and we need to use positive and negative signs carefully for the direction of the force.

In general, suppose that forces are acting on a particle --- we can think of them as coming from other objects that are “off stage” --- and that the interaction between the particle and the off-stage objects can be characterized by an interaction energy, U, which depends only on the particle's position, x. Using the kinetic energy theorem, we have dK=Fdx. (It's not necessary to write K_cm, since a particle can't have any other kind of kinetic energy.) Conservation of energy tells us dK+dU=0, so the relationship between force and interaction energy is dU=-Fdx, or

3.3 Resonance

Resonance is a phenomenon in which an oscillator responds most strongly to a driving force that matches its own natural frequency of vibration. For example, suppose a child is on a playground swing with a natural frequency of 1 Hz. That is, if you pull the child away from equilibrium, release her, and then stop doing anything for a while, she'll oscillate at 1 Hz. If there was no friction, as we assumed in section 2.5, then the sum of her gravitational and kinetic energy would remain constant, and the amplitude would be exactly the same from one oscillation to the next. However, friction is going to convert these forms of energy into heat, so her oscillations would gradually die out. To keep this from happening, you might give her a push once per cycle, i.e., the frequency of your pushes would be 1 Hz, which is the same as the swing's natural frequency. As long as you stay in rhythm, the swing responds quite well. If you start the swing from rest, and then give pushes at 1 Hz, the swing's amplitude rapidly builds up, as in figure a, until after a while it reaches a steady state in which friction removes just as much energy as you put in over the course of one cycle.

What will happen if you try pushing at 2 Hz? Your first push puts in some momentum, p, but your second push happens after only half a cycle, when the swing is coming right back at you, with momentum -p! The momentum transfer from the second push is exactly enough to stop the swing. The result is a very weak, and not very sinusoidal, motion, b.

\formatlikesubsubsection{Making the math easy}
This is a simple and physically transparent example of resonance: the swing responds most strongly if you match its natural rhythm. However, it has some characteristics that are mathematically ugly and possibly unrealistic. The quick, hard pushes are known as impulse forces, c, and they lead to an x-t graph that has nondifferentiable kinks. Impulsive forces like this are not only badly behaved mathematically, they are usually undesirable in practical terms. In a car engine, for example, the engineers work very hard to make the force on the pistons change smoothly, to avoid excessive vibration. Throughout the rest of this section, we'll assume a driving force that is sinusoidal, d, i.e., one whose F-t graph is either a sine function or a function that differs from a sine wave in phase, such as a cosine. The force is positive for half of each cycle and negative for the other half, i.e., there is both pushing and pulling. Sinusoidal functions have many nice mathematical characteristics (we can differentiate and integrate them, and the sum of sinusoidal functions that have the same frequency is a sinusoidal function), and they are also used in many practical situations. For instance, my garage door zapper sends out a sinusoidal radio wave, and the receiver is tuned to resonance with it.

A second mathematical issue that I glossed over in the swing example was how friction behaves. In section 3.2.4, about forces between solids, the empirical equation for kinetic friction was independent of velocity. Fluid friction, on the other hand, is velocity-dependent. For a child on a swing, fluid friction is the most important form of friction, and is approximately proportional to v². In still other situations, e.g., with a low-density gas or friction between solid surfaces that have been lubricated with a fluid such as oil, we may find that the frictional force has some other dependence on velocity, perhaps being proportional to v, or having some other complicated velocity dependence that can't even be expressed with a simple equation. It would be extremely complicated to have to treat all of these different possibilities in complete generality, so for the rest of this section, we'll assume friction proportional to velocity

simply because the resulting equations happen to be the easiest to solve. Even when the friction doesn't behave in exactly this way, many of our results may still be at least qualitatively correct.

3.3.1 Damped, free motion

Numerical treatment

An oscillator that has friction is referred to as damped. Let's use numerical techniques to find the motion of a damped oscillator that is released away from equilibrium, but experiences no driving force after that. We can expect that the motion will consist of oscillations that gradually die out. In section 2.5, we simulated the undamped case using our tried and true Python function based on conservation of energy. Now, however, that approach becomes a little awkward, because it involves splitting up the path to be traveled into n tiny segments, but in the presence of damping, each swing is a little shorter than the last one, and we don't know in advance exactly how far the oscillation will get before turning around. An easier technique here is to use force rather than energy. Newton's second law, a=F/m, gives a=(-kx-bv)/m, where we've made use of the result of example 38 for the force exerted by the spring. This becomes a little prettier if we rewrite it in the form

which gives symmetric treatment to three terms involving x and its first and second derivatives, v and a. Now instead of calculating the time Δt=Δx/v required to move a predetermined distance Δx, we pick Δt and determine the distance traveled in that time, Δx=vΔt. Also, we can no longer update v based on conservation of energy, since we don't have any easy way to keep track of how much mechanical energy has been changed into heat energy. Instead, we recalculate the velocity using Δv=aΔt.

import math
k=39.4784 # chosen to give a period of 1 second
m=1.
b=0.211 # chosen to make the results simple
x=1.
v=0.
t=0.
dt=.01
n=1000
for j in range(n):
  x=x+v*dt
  a=(-k*x-b*v)/m
  if (v>0) and (v+a*dt<0) :
    print("turnaround at t=",t,", x=",x)
  v=v+a*dt
  t=t+dt

turnaround at t= 0.99 , x= 0.899919262445
turnaround at t= 1.99 , x= 0.809844934046
turnaround at t= 2.99 , x= 0.728777519477
turnaround at t= 3.99 , x= 0.655817260033
turnaround at t= 4.99 , x= 0.590154191135
turnaround at t= 5.99 , x= 0.531059189965
turnaround at t= 6.99 , x= 0.477875914756
turnaround at t= 7.99 , x= 0.430013546991
turnaround at t= 8.99 , x= 0.386940256644
turnaround at t= 9.99 , x= 0.348177318484

The spring constant, k=4π=39.4784 N/m, is designed so that if the undamped equation was still true, the frequency would be 1 Hz. We start by noting that the addition of a small amount of damping doesn't seem to have changed the period at all, or at least not to within the accuracy of the calculation.¹⁰ You can check for yourself, however, that a large value of b, say 5 N⋅s/m, does change the period significantly.

We release the mass from x=1 m, and after one cycle, it only comes back to about x=0.9 m. I chose b=0.211 N⋅s/m by fiddling around until I got this result, since a decrease of exactly 10% is easy to discuss. Notice how the amplitude after two cycles is about 0.81 m, i.e., 1 m times 0.9²: the amplitude has again dropped by exactly 10%. This pattern continues for as long as the simulation runs, e.g., for the last two cycles, we have 0.34818/0.38694=0.89982, or almost exactly 0.9 again. It might have seemed capricious when I chose to use the unrealistic equation F=-bv, but this is the payoff. Only with -bv friction do we get this kind of mathematically simple exponential decay.

Analytic treatment

Taking advantage of this unexpectedly simple result, let's find an analytic solution for the motion. The numerical output suggests that we assume a solution of the form

where the unknown constants ω_f and c will presumably be related to m, b, and k. The constant c indicates how quickly the oscillations die out. The constant ω_f is, as before, defined as 2π times the frequency, with the subscript f to indicate a free (undriven) solution. All our equations will come out much simpler if we use ωs everywhere instead of s from now on, and, as physicists often do, I'll generally use the word “frequency” to refer to ω when the context makes it clear what I'm talking about. The phase angle δ has no real physical significance, since we can define t=0 to be any moment in time we like.

The factor A for the initial amplitude can also be omitted without loss of generality, since the equation we're trying to solve, ma+bv+kx = 0 is linear. That is, v and a are the first and second derivatives of x, and the derivative of Ax is simply A times the derivative of x. Thus, if x(t) is a solution of the equation, then multiplying it by a constant gives an equally valid solution. This is another place where we see that a damping force proportional to v is the easiest to handle mathematically. For a damping force proportional to v², for example, we would have had to solve the equation ma+bv²+kx = 0, which is nonlinear.

For the purpose of determining ω_f and c, the most general form we need to consider is therefore x = e^-ctsin ω_f t , whose first and second derivatives are and a = e^-ct≤ft(c² sin ω_f t. Plugging these into the equation ma+bv+kx = 0 and setting the sine and cosine parts equal to zero gives, after some tedious algebra,

Intuitively, we expect friction to “slow down” the motion, as when we ride a bike into a big patch of mud. “Slow down,” however, could have more than one meaning here. It could mean that the oscillator would take more time to complete each cycle, or it could mean that as time went on, the oscillations would die out, thus giving smaller velocities.

Our mathematical results show that both of these things happen. The first equation says that c, which indicates how quickly the oscillations damp out, is directly related to b, the strength of the damping.

The second equation, for the frequency, can be compared with the result from page 115 of for the undamped system. Let's refer to this now as ω_o, to distinguish it from the actual frequency ω_f of the free oscillations when damping is present. The result for ω_f will be less than ω_o, due to the presence of the b²/4m² term. This tells us that the addition of friction to the system does increase the time required for each cycle. However, it is very common for the b²/4m² term is negligible, so that ω_f≈ω_o.

Figure g shows an example. The damping here is quite strong: after only one cycle of oscillation, the amplitude has already been reduced by a factor of 2, corresponding to a factor of 4 in energy. However, the frequency of the damped oscillator is only about 1% lower than that of the undamped one; after five periods, the accumulated lag is just barely visible in the offsetting of the arrows. We can see that extremely strong damping --- even stronger than this --- would have been necessary in order to make ω_f≈ω_o a poor approximation.

3.3.2 The quality factor

It's usually impractical to measure b directly and determine c from the equation c=b/2m. For a child on a swing, measuring b would require putting the child in a wind tunnel! It's usually much easier to characterize the amount of damping by observing the actual damped oscillations and seeing how many cycles it takes for the mechanical energy to decrease by a certain factor. The unitless quality factor, Q, is defined as Q=ω_o/2c, and in the limit of weak damping, where ω≈ω_o, this can be interpreted as the number of cycles required for the mechanical energy to fall off by a factor of e^2π=535.49… Using this new quantity, we can rewrite the equation for the frequency of damped oscillations in the slightly more elegant form .

The decay of a musical sound is part of what gives it its character, and a good musical instrument should have the right Q, but the Q that is considered desirable is different for different instruments. A guitar is meant to keep on sounding for a long time after a string has been plucked, and might have a Q of 1000 or 10000. One of the reasons why a cheap synthesizer sounds so bad is that the sound suddenly cuts off after a key is released.

3.3.3 Driven motion

The driven case is extremely important in science, technology, and engineering. We have an external driving force F=F_m sin ω t, where the constant F_m indicates the maximum strength of the force in either direction. The equation of motion is now

After the driving force has been applied for a while, we expect that the amplitude of the oscillations will approach some constant value. This motion is known as the steady state, and it's the most interesting thing to find out; as we'll see later, the most general type of motion is only a minor variation on the steady-state motion. For the steady-state motion, we're going to look for a solution of the form

In contrast to the undriven case, here it's not possible to sweep A and δ under the rug. The amplitude of the steady-state motion, A, is actually the most interesting thing to know about the steady-state motion, and it's not true that we still have a solution no matter how we fiddle with A; if we have a solution for a certain value of A, then multiplying A by some constant would break the equality between the two sides of the equation of motion. It's also no longer true that we can get rid of δ simply be redefining when we start the clock; here δ represents a difference in time between the start of one cycle of the driving force and the start of the corresponding cycle of the motion.

The velocity and acceleration are v=ωAsin(ω t+δ) and a=-ω²Acos(ω t+δ), and if we plug these into the equation of motion, \eqref{eqn:resonancemotion}, and simplify a little, we find

The sum of any two sinusoidal functions with the same frequency is also a sinusoidal, so the whole left side adds up to a sinusoidal. By fiddling with A and δ we can make the amplitudes and phases of the two sides of the equation match up.

Steady state, no damping

A and δ are easy to find in the case where there is no damping at all. There are now no cosines in equation \eqref{eqn:steadystate} above, only sines, so if we wish we can set δ to zero, and we find A=F_m/(k-mω²)=F_m/m(ω_o²-ω²). This, however, makes A negative for ω>ω_o. The variable δ was designed to represent this kind of phase relationship, so we prefer to keep A positive and set δ=π for ω>ω_o. Our results are then A = \frac{F_m}{m\left|\omega^2-\omega_\zu{o}^2\right|} \ \intertext{and} \delta = \left\{\begin{array}{ll} 0, & \omega<\omega_\zu{o}\ \pi, & \omega>\omega_\zu{o}\end{array}\right. .

The most important feature of the result is that there is a resonance: the amplitude becomes greater and greater, and approaches infinity, as ω approaches the resonant frequency ω_o. This is the physical behavior we anticipated on page 168 in the example of pushing a child on a swing. If the driving frequency matches the frequency of the free vibrations, then the driving force will always be in the right direction to add energy to the swing. At a driving frequency very different from the resonant frequency, we might get lucky and push at the right time during one cycle, but our next push would come at some random point in the next cycle, possibly having the effect of slowing the swing down rather than speeding it up.

The interpretation of the infinite amplitude at ω=ω_o is that there really isn't any steady state if we drive the system exactly at resonance --- the amplitude will just keep on increasing indefinitely. In real life, the amplitude can't be infinite both because there is always some damping and because there will always be some difference, however small, between ω and ω_o. Even though the infinity is unphysical, it has entered into the popular consciousness, starting with the eccentric Serbian-American inventor and physicist Nikola Tesla. Around 1912, the tabloid newspaper The World Today credulously reported a story which Tesla probably fabricated --- or wildly exaggerated --- for the sake of publicity. Supposedly he created a steam-powered device “no larger than an alarm clock,” containing a piston that could be made to vibrate at a tunable and precisely controlled frequency. “He put his little vibrator in his coat-pocket and went out to hunt a half-erected steel building. Down in the Wall Street district, he found one --- ten stories of steel framework without a brick or a stone laid around it. He clamped the vibrator to one of the beams, and fussed with the adjustment [presumably hunting for the building's resonant frequency] until he got it. Tesla said finally the structure began to creak and weave and the steel-workers came to the ground panic-stricken, believing that there had been an earthquake. Police were called out. Tesla put the vibrator in his pocket and went away. Ten minutes more and he could have laid the building in the street. And, with the same vibrator he could have dropped the Brooklyn Bridge into the East River in less than an hour.”

The phase angle δ also exhibits surprising behavior. As the frequency is tuned upward past resonance, the phase abruptly shifts so that the phase of the response is opposite to that of the driving force. There is a simple interpretation for this. The system's mechanical energy can only change due to work done by the driving force, since there is no damping to convert mechanical energy to heat. In the steady state, then, the power transmitted by the driving force over a full cycle of motion must average out to zero. In general, the work theorem dE=Fdx can always be divided by dt on both sides to give the useful relation P=Fv. If Fv is to average out to zero, then F and v must be out of phase by ±π/2, and since v is ahead of x by a phase angle of π/2, the phase angle between x and F must be zero or π.

Given that these are the two possible phases, why is there a difference in behavior between ω<ω_o and ω>ω_o? At the low-frequency limit, consider ω=0, i.e., a constant force. A constant force will simply displace the oscillator to one side, reaching an equilibrium that is offset from the usual one. The force and the response are in phase, e.g., if the force is to the right, the equilibrium will be offset to the right. This is the situation depicted in the amplitude graph of figure h at ω=0. The response, which is not zero, is simply this static displacement of the oscillator to one side.

At high frequencies, on the other hand, imagine shaking the poor child on the swing back and forth with a force that oscillates at 10 Hz. This is so fast that there is essentially no time for the force F=-kx from gravity and the chain to act from one cycle to the next. The problem becomes equivalent to the oscillation of a free object. If the driving force varies like sin(ω t), with δ=0, then the acceleration is also proportional to the sine. Integrating, we find that the velocity goes like minus a cosine, and a second integration gives a position that varies as minus the sine --- opposite in phase to the driving force. Intuitively, this mathematical result corresponds to the fact that at the moment when the object has reached its maximum displacement to the right, that is the time when the greatest force is being applied to the left, in order to turn it around and bring it back toward the center.

i / Example 43: a viola without a mute (left), and with a mute (right). The mute doesn't touch the strings themselves.

Steady state, with damping

The extension of the analysis to the damped case involves some lengthy algebra, which I've outlined on page 844 in appendix 2. The results are shown in figure j. It's not surprising that the steady state response is weaker when there is more damping, since the steady state is reached when the power extracted by damping matches the power input by the driving force. The maximum amplitude, at the peak of the resonance curve, is approximately proportional to Q.

j / Dependence of the amplitude and phase angle on the driving frequency. The undamped case is Q=∞, and the other curves represent Q=1, 3, and 10. F_m, m, and ω_o are all set to 1.

What is surprising is that the amplitude is strongly affected by damping close to resonance, but only weakly affected far from it. In other words, the shape of the resonance curve is broader with more damping, and even if we were to scale up a high-damping curve so that its maximum was the same as that of a low-damping curve, it would still have a different shape. The standard way of describing the shape numerically is to give the quantity Δω, called the full width at half-maximum, or FWHM, which is defined in figure k. Note that the y axis is energy, which is proportional to the square of the amplitude. Our previous observations amount to a statement that Δω is greater when the damping is stronger, i.e., when the Q is lower. It's not hard to show from the equations on page 844 that for large Q, the FWHM is given approximately by

Another thing we notice in figure j is that for small values of Q the frequency ω_res of the maximum A is less than ω_o.¹¹ At even lower values of Q, like Q=1, the A-ω curve doesn't even have a maximum near ω>0.

Physical reason for the relationship between Q and the FWHM

What is the reason for this surprising relationship between the damping and the width of the resonance? Fundamentally, it has to do with the fact that friction causes a system to lose its “memory” of its previous state. If the Pioneer 10 space probe, coasting through the frictionless vacuum of interplanetary space, is detected by aliens a million years from now, they will be able to trace its trajectory backwards and infer that it came from our solar system. On the other hand, imagine that I shove a book along a tabletop, it comes to rest, and then someone else walks into the room. There will be no clue as to which direction the book was moving before it stopped --- friction has erased its memory of its motion. Now consider the playground swing driven at twice its natural frequency, figure m, where the undamped case is repeated from figure b on page 168. In the undamped case, the first push starts the swing moving with momentum p, but when the second push comes, if there is no friction at all, it now has a momentum of exactly -p, and the momentum transfer from the second push is exactly enough to stop it dead. With moderate damping, however, the momentum on the rebound is not quite -p, and the second push's effect isn't quite as disastrous. With very strong damping, the swing comes essentially to rest long before the second push. It has lost all its memory, and the second push puts energy into the system rather than taking it out. Although the detailed mathematical results with this kind of impulsive driving force are different,¹² the general results are the same as for sinusoidal driving: the less damping there is, the greater the penalty you pay for driving the system off of resonance.

Transients

What about the motion before the steady state is achieved? When we computed the undriven motion numerically on page 169, the program had to initialize the position and velocity. By changing these two variables, we could have gotten any of an infinite number of simulations.¹³ The same is true when we have an equation of motion with a driving term, ma+bv+kx = F_m sin ωt (p. , equation \eqref{eqn:resonancemotion}). The steady-state solutions, however, have no adjustable parameters at all --- A and δ are uniquely determined by the parameters of the driving force and the oscillator itself. If the oscillator isn't initially in the steady state, then it will not have the steady-state motion at first. What kind of motion will it have?

The answer comes from realizing that if we start with the solution to the driven equation of motion, and then add to it any solution to the free equation of motion, the result,

is also a solution of the driven equation. Here, as before, ω_f is the frequency of the free oscillations (ω_f≈ω_o for small Q), ω is the frequency of the driving force, A and δ are related as usual to the parameters of the driving force, and A' and δ' can have any values at all. Given the initial position and velocity, we can always choose A' and δ' to reproduce them, but this is not something one often has to do in real life. What's more important is to realize that the second term dies out exponentially over time, decaying at the same rate at which a free vibration would. For this reason, the A' term is called a transient. A high-Q oscillator's transients take a long time to die out, while a low-Q oscillator always settles down to its steady state very quickly.

Overdamped motion

The treatment of free, damped motion on page 171 skipped over a subtle point: in the equation , Q<1/2 results in an answer that is the square root of a negative number. For example, suppose we had k=0, which corresponds to a neutral equilibrium. A physical example would be a mass sitting in a tub of syrup. If we set it in motion, it won't oscillate --- it will simply slow to a stop. This system has Q=0. The equation of motion in this case is ma+bv=0, or, more suggestively,

One can easily verify that this has the solution , and integrating, we find . In other words, the reason ω_f comes out to be mathematical nonsense¹⁴ is that we were incorrect in assuming a solution that oscillated at a frequency ω_f. The actual motion is not oscillatory at all.

In general, systems with Q<1/2, called overdamped systems, do not display oscillatory motion. Most cars' shock absorbers are designed with Q≈1/2, since it's undesirable for the car to undulate up and down for a while after you go over a bump. (Shocks with extremely low values of Q are not good either, because such a system takes a very long time to come back to equilibrium.) It's not particularly important for our purposes, but for completeness I'll note, as you can easily verify, that the general solution to the equation of motion for 0<Q<1/2 is of the form x=Ae^-ct+Be^-dt, while Q=1/2, called the critically dampedcase, gives x=(A+Bt)e^-ct.

3.4 Motion In Three Dimensions

3.4.1 The Cartesian perspective

When my friends and I were bored in high school, we used to play a paper-and-pencil game which, although we never knew it, was Very Educational --- in fact, it pretty much embodies the entire world-view of classical physics. To play the game, you draw a racetrack on graph paper, and try to get your car around the track before anyone else. The default is for your car to continue at constant speed in a straight line, so if it moved three squares to the right and one square up on your last turn, it will do the same this turn. You can also control the car's motion by changing its Δ x and Δ y by up to one unit. If it moved three squares to the right last turn, you can have it move anywhere from two to four squares to the right this turn.

b / French mathematician René Descartes invented analytic geometry; Cartesian (xyz) coordinates are named after him. He did work in philosophy, and was particularly interested in the mind-body problem. He was a skeptic and an antiaristotelian, and, probably for fear of religious persecution, spent his adult life in the Netherlands, where he fathered a daughter with a Protestant peasant whom he could not marry. He kept his daughter's existence secret from his enemies in France to avoid giving them ammunition, but he was crushed when she died of scarlatina at age 5. A pious Catholic, he was widely expected to be sainted. His body was buried in Sweden but then reburied several times in France, and along the way everything but a few fingerbones was stolen by peasants who expected the body parts to become holy relics.

The fundamental way of dealing with the direction of an object's motion in physics is to use conservation of momentum, since momentum depends on direction. Up until now, we've only done momentum in one dimension. How does this relate to the racetrack game? In the game, the motion of a car from one turn to the next is represented by its Δ x and Δ y. In one dimension, we would only need Δ x, which could be related to the velocity, Δ x/Δ t, and the momentum, mΔ x/Δ t. In two dimensions, the rules of the game amount to a statement that if there is no momentum transfer, then both mΔ x/Δ t and mΔ y/Δ t stay the same. In other words, there are two flavors of momentum, and they are separately conserved. All of this so far has been done with an artificial division of time into “turns,” but we can fix that by redefining everything in terms of derivatives, and for motion in three dimensions rather than two, we augment x and y with z:

We call these the x, y, and z components of the velocity and the momentum.

There is both experimental and theoretical evidence that the x, y, and z momentum components are separately conserved, and that a momentum transfer (force) along one axis has no effect on the momentum components along the other two axes. On page 88, for example, I argued that it was impossible for an air hockey puck to make a 180-degree turn spontaneously, because then in the frame moving along with the puck, it would have begun moving after starting from rest. Now that we're working in two dimensions, we might wonder whether the puck could spontaneously make a 90-degree turn, but exactly the same line of reasoning shows that this would be impossible as well, which proves that the puck can't trade x-momentum for y-momentum. A more general proof of separate conservation will be given on page 209, after some of the appropriate mathematical techniques have been introduced.

c / Bullets are dropped and shot at the same time.

As an example of the experimental evidence for separate conservation of the momentum components, figure c shows correct and incorrect predictions of what happens if you shoot a rifle and arrange for a second bullet to be dropped from the same height at exactly the same moment when the first one left the barrel. Nearly everyone expects that the dropped bullet will reach the dirt first, and Aristotle would have agreed, since he believed that the bullet had to lose its horizontal motion before it could start moving vertically. In reality, we find that the vertical momentum transfer between the earth and the bullet is completely unrelated to the horizontal momentum. The bullet ends up with p_y<0, while the planet picks up an upward momentum p_y>0, and the total momentum in the y direction remains zero. Both bullets hit the ground at the same time. This is much simpler than the Aristotelian version!

Discussion Questions

f / Discussion question A.

3.4.2 Rotational invariance

The Cartesian approach requires that we choose x, y, and z axes. How do we choose them correctly? The answer is that it had better not matter which directions the axes point (provided they're perpendicular to each other), or where we put the origin, because if it did matter, it would mean that space was asymmetric. If there was a certain point in the universe that was the right place to put the origin, where would it be? The top of Mount Olympus? The United Nations headquarters? We find that experiments come out the same no matter where we do them, and regardless of which way the laboratory is oriented, which indicates that no location in space or direction in space is special in any way.¹⁵

This is closely related to the idea of Galilean relativity stated on page 60, from which we already know that the absolute motion of a frame of reference is irrelevant and undetectable. Observers using frames of reference that are in motion relative to each other will not even agree on the permanent identity of a particular point in space, so it's not possible for the laws of physics to depend on where you are in space. For instance, if gravitational energies were proportional to m₁m₂ in one location but to (m₁m₂)^1.00001 in another, then it would be possible to determine when you were in a state of absolute motion, because the behavior of gravitational interactions would change as you moved from one region to the other.

Because of this close relationship, we restate the principle of Galilean relativity in a more general form. This extended principle of Galilean relativity states that the laws of physics are no different in one time and place than in another, and that they also don't depend on your orientation or your motion, provided that your motion is in a straight line and at constant speed.

The irrelevance of time and place could have been stated in chapter 1, but since this section is the first one in which we're dealing with three-dimensional physics in full generality, the irrelevance of orientation is what we really care about right now. This property of the laws of physics is called rotational invariance. The word “invariance” means a lack of change, i.e., the laws of physics don't change when we reorient our frame of reference.

3.4.3 Vectors

Remember the title of this book? It would have been possible to obtain the result of example 53 by applying the Pythagorean theorem to d x, d y, and d z, and then dividing by d t, but the rotational invariance approach is simpler, and is useful in a much broader context. Even with a quantity you presently know nothing about, say the magnetic field, you can infer that if the components of the magnetic field are B_x, B_y, and B_z, then the physically useful way to talk about the strength of the magnetic field is to define it as . Nature knows your brain cells are precious, and doesn't want you to have to waste them by memorizing mathematical rules that are different for magnetic fields than for velocities.

When mathematicians see that the same set of techniques is useful in many different contexts, that's when they start making definitions that allow them to stop reinventing the wheel. The ancient Greeks, for example, had no general concept of fractions. They couldn't say that a circle's radius divided by its diameter was equal to the number 1/2. They had to say that the radius and the diameter were in the ratio of one to two. With this limited number concept, they couldn't have said that water was dripping out of a tank at a rate of 3/4 of a barrel per day; instead, they would have had to say that over four days, three barrels worth of water would be lost. Once enough of these situations came up, some clever mathematician finally realized that it would make sense to define something called a fraction, and that one could think of these fraction thingies as numbers that lay in the gaps between the traditionally recognized numbers like zero, one, and two. Later generations of mathematicians introduced further subversive generalizations of the number concepts, inventing mathematical creatures like negative numbers, and the square root of two, which can't be expressed as a fraction.

In this spirit, we define a vector as any quantity that has both an amount and a direction in space. In contradistinction, a scalar has an amount, but no direction. Time and temperature are scalars. Velocity, acceleration, momentum, and force are vectors. In one dimension, there are only two possible directions, and we can use positive and negative numbers to indicate the two directions. In more than one dimension, there are infinitely many possible directions, so we can't use the two symbols + and - to indicate the direction of a vector. Instead, we can specify the three components of the vector, each of which can be either negative or positive. We represent vector quantities in handwriting by writing an arrow above them, so for example the momentum vector looks like this, , but the arrow looks ugly in print, so in books vectors are usually shown in bold-face type: p. A straightforward way of thinking about vectors is that a vector equation really represents three different equations. For instance, conservation of momentum could be written in terms of the three components,

or as a single vector equation,¹⁶

The following table summarizes some vector operations.

operation	definition
mathbfvector	sqrtvectorx2vectory2vectorz2
mathbfvectormathbfvector	Add component by component.
mathbfvectormathbfvector	Subtract component by component.
mathbfvector cdot scalar	Multiply each component by the scalar.
mathbfvector scalar	Divide each component by the scalar.

{}The first of these is called the magnitude of the vector; in one dimension, where a vector only has one component, it amounts to taking the absolute value, hence the similar notation.

Geometric representation of vectors

A vector in two dimensions can be easily visualized by drawing an arrow whose length represents its magnitude and whose direction represents its direction. The x component of a vector can then be visualized, j, as the length of the shadow it would cast in a beam of light projected onto the x axis, and similarly for the y component. Shadows with arrowheads pointing back against the direction of the positive axis correspond to negative components.

In this type of diagram, the negative of a vector is the vector with the same magnitude but in the opposite direction. Multiplying a vector by a scalar is represented by lengthening the arrow by that factor, and similarly for division.

All the vector techniques can be applied to any kind of vector, but the graphical representation of vectors as arrows is particularly natural for vectors that represent lengths and distances. We define a vector called r whose components are the coordinates of a particular point in space, x, y, and z. The Δr vector, whose components are Δ x, Δ y, and Δ z, can then be used to represent motion that starts at one point and ends at another. Adding two Δ r vectors is interpreted as a trip with two legs: by computing the Δ r vector going from point A to point B plus the vector from B to C, we find the vector that would have taken us directly from A to C.

Calculations with magnitude and direction

If you ask someone where Las Vegas is compared to Los Angeles, she is unlikely to say that the Δ x is 290 km and the Δ y is 230 km, in a coordinate system where the positive x axis is east and the y axis points north. She will probably say instead that it's 370 km to the northeast. If she was being precise, she might specify the direction as 38° counterclockwise from east. In two dimensions, we can always specify a vector's direction like this, using a single angle. A magnitude plus an angle suffice to specify everything about the vector. The following two examples show how we use trigonometry and the Pythagorean theorem to go back and forth between the x-y and magnitude-angle descriptions of vectors.

The following example shows the correct handling of the plus and minus signs, which is usually the main cause of mistakes by students.

Addition of vectors given their components

The easiest type of vector addition is when you are in possession of the components, and want to find the components of their sum.

Addition of vectors given their magnitudes and directions

In this case, you must first translate the magnitudes and directions into components, and the add the components.

Graphical addition of vectors

Often the easiest way to add vectors is by making a scale drawing on a piece of paper. This is known as graphical addition, as opposed to the analytic techniques discussed previously.

o / Example 61.

Even when we don't intend to do an actual graphical calculation with a ruler and protractor, it can be convenient to diagram the addition of vectors in this way, as shown in figure p. With Δr vectors, it intuitively makes sense to lay the vectors tip-to-tail and draw the sum vector from the tail of the first vector to the tip of the second vector. We can do the same when adding other vectors such as force vectors.

Unit vector notation

When we want to specify a vector by its components, it can be cumbersome to have to write the algebra symbol for each component:

A more compact notation is to write

where the vectors , , and , called the unit vectors, are defined as the vectors that have magnitude equal to 1 and directions lying along the x, y, and z axes. In speech, they are referred to as “x-hat,” “y-hat,” and “z-hat.”

A slightly different, and harder to remember, version of this notation is unfortunately more prevalent. In this version, the unit vectors are called , , and :

Applications to relative motion, momentum, and force

Vector addition is the correct way to generalize the one-dimensional concept of adding velocities in relative motion, as shown in the following example:

Force is a vector, and we add force vectors when more than one force acts on the same object.

By the way, the word “pressure” that I threw around casually in the preceding example has a precise technical definition: force per unit area. The SI units of pressure are N/m², which can be abbreviated as pascals, 1 Pa = 1 N/m². Atmospheric pressure is about 100 kPa. By applying the equation F_g+F_f = 0 to the top and bottom surfaces of a cubical volume of fluid, one can easily prove that the difference in pressure between two different depths is Δ P=ρ gΔ y. (In physics, “fluid” can refer to either a gas or a liquid.)

Pressure is discussed in more detail in chapter 5.

Discussion Questions

3.4.4 Calculus with vectors

Differentiation

In one dimension, we define the velocity as the derivative of the position with respect to time, and we can think of the derivative as what we get when we calculate Δ x/Δ t for very short time intervals. The quantity Δ x=x_f-x_i is calculated by subtraction. In three dimensions, x becomes r, and the Δr vector is calculated by vector subtraction, Δr=r_f-r_i. Vector subtraction is defined component by component, so when we take the derivative of a vector, this means we end up taking the derivative component by component,

or

All of this reasoning applies equally well to any derivative of a vector, so for instance we can take the second derivative,

or

A counterintuitive consequence of this is that the acceleration vector does not need to be in the same direction as the motion. The velocity vector points in the direction of motion, but by Newton's second law, a=F/m, the acceleration vector points in the same direction as the force, not the motion. This is easiest to understand if we take velocity vectors from two different moments in the motion, and visualize subtracting them graphically to make a Δv vector. The direction of the Δv vector tells us the direction of the acceleration vector as well, since the derivative dv/d t can be approximated as Δv/Δ t. As shown in figure x/1, a change in the magnitude of the velocity vector implies an acceleration that is in the direction of motion. A change in the direction of the velocity vector produces an acceleration perpendicular to the motion, x/2.

Although I've relegated the results a=ω² r=|v|²/r to an example because they are a straightforward corollary of more general principles already developed, they are important and useful enough to record for later use.

These results are counterintuitive as well. Until Newton, physicists and laypeople alike had assumed that the planets would need a force to push them forward in their orbits. Figure y may help to make it more plausible that only an inward force is required. A forward force might be needed in order to cancel out a backward force such as friction, z, but the total force in the forward-backward direction needs to be exactly zero for constant-speed motion.

When you are in a car undergoing circular motion, there is also a strong illusion of an outward force. But what object could be making such a force? The car's seat makes an inward force on you, not an outward one. There is no object that could be exerting an outward force on your body. In reality, this force is an illusion that comes from our brain's intuitive efforts to interpret the situation within a noninertial frame of reference. As shown in figure aa, we can describe everything perfectly well in an inertial frame of reference, such as the frame attached to the sidewalk. In such a frame, the bowling ball goes straight because there is no force on it. The wall of the truck's bed hits the ball, not the other way around.

Integration

An integral is really just a sum of many infinitesimally small terms. Since vector addition is defined in terms of addition of the components, an integral of a vector quantity is found by doing integrals component by component.

Discussion Questions

3.4.5 The dot product

How would we generalize the mechanical work equation d E=F d x to three dimensions? Energy is a scalar, but force and distance are vectors, so it might seem at first that the kind of “magic-wand” generalization discussed on page 198 failed here, since we don't know of any way to multiply two vectors together to get a scalar. Actually, this is Nature giving us a hint that there is such a multiplication operation waiting for us to invent it, and since Nature is simple, we can be assured that this operation will work just fine in any situation where a similar generalization is required.

How should this operation be defined? Let's consider what we would get by performing this operation on various combinations of the unit vectors , , and . The conventional notation for the operation is to put a dot, ⋅, between the two vectors, and the operation is therefore called the dot product. Rotational invariance requires that we handle the three coordinate axes in the same way, without giving special treatment to any of them, so we must have and . This is supposed to be a way of generalizing ordinary multiplication, so for consistency with the property 1×1=1 of ordinary numbers, the result of multiplying a magnitude-one vector by itself had better be the scalar 1, so . Furthermore, there is no way to satisfy rotational invariance unless we define the mixed products to be zero, ; for example, a 90-degree rotation of our frame of reference about the z axis reverses the sign of , but rotational invariance requires that produce the same result either way, and zero is the only number that stays the same when we reverse its sign. Establishing these six products of unit vectors suffices to define the operation in general, since any two vectors that we want to multiply can be broken down into components, e.g., . Thus by requiring rotational invariance and consistency with multiplication of ordinary numbers, we find that there is only one possible way to define a multiplication operation on two vectors that gives a scalar as the result.¹⁷ The dot product has all of the properties we normally associate with multiplication, except that there is no “dot division.”

The result of example 72 is very useful. It gives us a way to find the angle between two vectors if we know their components. It can be used to show that the dot product of any two perpendicular vectors is zero. It also leads to a nifty proof that the dot product is rotationally invariant --- up until now I've only proved that if a rotationally invariant product exists, the dot product is it --- because angles and lengths aren't affected by a rotation, so the right side of the equation is rotationally invariant, and therefore so is the left side.

I introduced the whole discussion of the dot product by way of generalizing the equation d E=Fd x to three dimensions. In terms of a dot product, we have

(If that step seemed like black magic, try writing it out in terms of components.) If the force is perpendicular to the motion, as in figure ae, then the work done is zero. The pack horse is doing work within its own body, but is not doing work on the pack.

A good application of the dot product is to allow us to write a simple, streamlined proof of separate conservation of the momentum components. (You can skip the proof without losing the continuity of the text.) The argument is a generalization of the one-dimensional proof on page 128, and makes the same assumption about the type of system of particles we're dealing with. The kinetic energy of one of the particles is (1/2)mv⋅v, and when we transform into a different frame of reference moving with velocity u relative to the original frame, the one-dimensional rule v→ v+u turns into vector addition, v→ v+u. In the new frame of reference, the kinetic energy is (1/2)m(v+u)⋅(v+u). For a system of n particles, we have

As in the proof on page 128, the first sum is simply the total kinetic energy in the original frame of reference, and the last sum is a constant, which has no effect on the validity of the conservation law. The middle sum can be rewritten as

so the only way energy can be conserved for all values of u is if the vector sum of the momenta is conserved as well.

3.4.6 Gradients and line integrals (optional)

This subsection introduces a little bit of vector calculus. It can be omitted without loss of continuity, but the techniques will be needed in our study of electricity and magnetism, and it may be helpful to be exposed to them in easy-to-visualize mechanical contexts before applying them to invisible electrical and magnetic phenomena.

In physics we often deal with fields of force, meaning situations where the force on an object depends on its position. For instance, figure af could represent a map of the trade winds affecting a sailing ship, or a chart of the gravitational forces experienced by a space probe entering a double-star system. An object moving under the influence of this force will not necessarily be moving in the same direction as the force at every moment. The sailing ship can tack against the wind, due to the force from the water on the keel. The space probe, if it entered from the top of the diagram at high speed, would start to curve around to the right, but its inertia would carry it forward, and it wouldn't instantly swerve to match the direction of the gravitational force. For convenience, we've defined the gravitational field, g, as the force per unit mass, but that trick only leads to a simplification because the gravitational force on an object is proportional to its mass. Since this subsection is meant to apply to any kind of force, we'll discuss everything in terms of the actual force vector, F, in units of newtons.

If an object moves through the field of force along some curved path from point r₁ to point r₂, the force will do a certain amount of work on it. To calculate this work, we can break the path up into infinitesimally short segments, find the work done along each segment, and add them all up. For an object traveling along a nice straight x axis, we use the symbol d x to indicate the length of any infinitesimally short segment. In three dimensions, moving along a curve, each segment is a tiny vector . The work theorem can be expressed as a dot product, so the work done along a segment is F⋅dr. We want to integrate this, but we don't know how to integrate with respect to a variable that's a vector, so let's define a variable s that indicates the distance traveled so far along the curve, and integrate with respect to it instead. The expression F⋅dr can be rewritten as |F| |dr| cosθ, where θ is the angle between F and dr. But |dr| is simply d s, so the amount of work done becomes

Both F and θ are functions of s. As a matter of notation, it's cumbersome to have to write the integral like this. Vector notation was designed to eliminate this kind of drudgery. We therefore define the line integral

as a way of notating this type of integral. The `C' refers to the curve along which the object travels. If we don't know this curve then we typically can't evaluate the line integral just by knowing the initial and final positions r₁ and r₂.

The basic idea of calculus is that integration undoes differentiation, and vice-versa. In one dimension, we could describe an interaction either in terms of a force or in terms of an interaction energy. We could integrate force with respect to position to find minus the energy, or we could find the force by taking minus the derivative of the energy. In the line integral, position is represented by a vector. What would it mean to take a derivative with respect to a vector? The correct way to generalize the derivative d U/d x to three dimensions is to replace it with the following vector,

called the gradient of U, and written with an upside-down delta¹⁸ like this, ∇ U. Each of these three derivatives is really what's known as a partial derivative. What that means is that when you're differentiating U with respect to x, you're supposed to treat y and z and constants, and similarly when you do the other two derivatives. To emphasize that a derivative is a partial derivative, it's customary to write it using the symbol ∂ in place of the differential d's. Putting all this notation together, we have

The gradient looks scary, but it has a very simple physical interpretation. It's a vector that points in the direction in which U is increasing most rapidly, and it tells you how rapidly U is increasing in that direction. For instance, sperm cells in plants and animals find the egg cells by traveling in the direction of the gradient of the concentration of certain hormones. When they reach the location of the strongest hormone concentration, they find their destiny. In terms of the gradient, the force corresponding to a given interaction energy is F=-∇ U.

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Homework Problems

In problems 27-31, analyze the forces using a table in the format shown in section 3.2.6. Analyze the forces in which the italicized object participates.

i / Problem 50.

k / Problem 57.

p / Problem 71.