Chapter 2. Conservation of Energy

Do you pronounce it Joule's to rhyme with schools,
Joule's to rhyme with Bowls,
or Joule's to rhyme with Scowls?
Whatever you call it, by Joule's,
or Joule's,
or Joule's, it's good! -- Advertising slogan of the Joule brewery. The name, and the corresponding unit of energy, are now usually pronounced so as to rhyme with “school.”

2.1 Energy

a / James Joule, 1818-1889. The son of a wealthy brewer, Joule was tutored as a young man by the famous scientist John Dalton. Fascinated by electricity, he and his brother experimented by giving electric shocks to each other and to the family's servants. Joule ran the brewery as an adult, and science was merely a serious hobby. His work on energy can be traced to his attempt to build an electric motor that would replace steam engines. His ideas were not accepted at first, partly because they contradicted the widespread belief that heat was a fluid, and partly because they depended on extremely precise measurements, which had not previously been common in physics.

b / Heat energy can be converted to light energy. Very hot objects glow visibly, and even objects that aren't so hot give off infrared light, a color of light that lies beyond the red end of the visible rainbow. This photo was made with a special camera that records infrared light. The man's warm skin emits quite a bit of infrared light energy, while his hair, at a lower temperature, emits less.

The energy concept

You'd probably like to be able to drive your car and light your apartment without having to pay money for gas and electricity, and if you do a little websurfing, you can easily find people who say they have the solution to your problem. This kind of scam has been around for centuries. It used to be known as a perpetual motion machine, but nowadays the con artists' preferred phrase is “free energy.”¹ A typical “free-energy” machine would be a sealed box that heats your house without needing to be plugged into a wall socket or a gas pipe. Heat comes out, but nothing goes in, and this can go on indefinitely. But an interesting thing happens if you try to check on the advertised performance of the machine. Typically, you'll find out that either the device is still in development, or it's back-ordered because so many people have already taken advantage of this Fantastic Opportunity! In a few cases, the magic box exists, but the inventor is only willing to demonstrate very small levels of heat output for short periods of time, in which case there's probably a tiny hearing-aid battery hidden in there somewhere, or some other trick.

Since nobody has ever succeeded in building a device that creates heat out of nothing, we might also wonder whether any device exists that can do the opposite, turning heat into nothing. You might think that a refrigerator was such a device, but actually your refrigerator doesn't destroy the heat in the food. What it really does is to extract some of the heat and bring it out into the room. That's why it has big radiator coils on the back, which get hot when it's in operation.

If it's not possible to destroy or create heat outright, then you might start to suspect that heat was a conserved quantity. This would be a successful rule for explaining certain processes, such as the transfer of heat between a cold Martini and a room-temperature olive: if the olive loses a little heat, then the drink must gain the same amount. It would fail in general, however.

Sunlight can heat your skin, for example, and a hot lightbulb filament can cool off by emitting light. Based on these observations, we could revise our proposed conservation law, and say that there is something called heatpluslight, which is conserved. Even this, however, needs to be generalized in order to explain why you can get a painful burn playing baseball when you slide into a base. Now we could call it heatpluslightplusmotion. The word is getting pretty long, and we haven't even finished the list.

Rather than making the word longer and longer, physicists have hijacked the word “energy” from ordinary usage, and give it a new, specific technical meaning. Just as the Parisian platinum-iridium kilogram defines a specific unit of mass, we need to pick something that defines a definite unit of energy. The metric unit of energy is the joule (J), and we'll define it as the amount of energy required to heat 0.24 grams of water from 20 to 21 degrees Celsius. (Don't memorize the numbers.)²

Note how only differences in temperature and energy appeared in the preceding example. In other words, we don't have to make any assumptions about whether there is a temperature at which all an object's heat energy is removed. Historically, the energy and temperature units were invented before it was shown that there is such a temperature, called absolute zero. There is a scale of temperature, the Kelvin scale, in which the unit of temperature is the same as the Celsius degree, but the zero point is defined as absolute zero. But as long as we only deal with temperature differences, it doesn't matter whether we use Kelvin or Celsius. Likewise, as long as we deal with differences in heat energy, we don't normally have to worry about the total amount of heat energy the object has. In standard physics terminology, “heat” is used only to refer to differences, while the total amount is called the object's “thermal energy.”This distinction is often ignored by scientists in casual speech, and in this book I'll usually use “heat” for either quantity.

We're defining energy by adding up things from a list, which we lengthen as needed: heat, light, motion, etc. One objection to this approach is aesthetic: physicists tend to regard complication as a synonym for ugliness. If we have to keep on adding more and more forms of energy to our laundry list, then it's starting to sound like energy is distressingly complicated. Luckily it turns out that energy is simpler than it seems. Many forms of energy that are apparently unrelated turn out to be manifestations of a small number of forms at the atomic level, and this is the topic of section 2.4.

Discussion Questions

c / As in figure b, an infrared camera distinguishes hot and cold areas. As the bike skids to a stop with its brakes locked, the kinetic energy of the bike and rider is converted into heat in both the floor (top) and the tire (bottom).

Logical issues

Another possible objection is that the open-ended approach to defining energy might seem like a kind of cheat, since we keep on inventing new forms whenever we need them. If a certain experiment seems to violate conservation of energy, can't we just invent a new form of invisible “mystery energy” that patches things up? This would be like balancing your checkbook by putting in a fake transaction that makes your calculation of the balance agree with your bank's. If we could fudge this way, then conservation of energy would be untestable --- impossible to prove or disprove.

Actually all scientific theories are unprovable. A theory can never be proved, because the experiments can only cover a finite number out of the infinitely many situations in which the theory is supposed to apply. Even a million experiments won't suffice to prove it in the same sense of the word “proof” that is used in mathematics. However, even one experiment that contradicts a theory is sufficient to show that the theory is wrong. A theory that is immune to disproof is a bad theory, because there is no way to test it. For instance, if I say that 23 is the maximum number of angels that can dance on the head of a pin, I haven't made a properly falsifiable scientific theory, since there's no method by which anyone could even attempt to prove me wrong based on observations or experiments.

Conservation of energy is testable because new forms of energy are expected to show regular mathematical behavior, and are supposed to be related in a measurable way to observable phenomena. As an example, let's see how to extend the energy concept to include motion.

d / A simplified drawing of Joule's paddlewheel experiment.

e / The heating of the tire and floor in figure c is something that the average person might have predicted in advance, but there are other situations where it's not so obvious. When a ball slams into a wall, it doesn't rebound with the same amount of kinetic energy. Was some energy destroyed? No. The ball and the wall heat up. These infrared photos show a squash ball at room temperature (top), and after it has been played with for several minutes (bottom), causing it to heat up detectably.

Kinetic energy

Energy of motion is called kinetic energy. (The root of the word is the same as the word “cinema” -- in French, kinetic energy is “énergie cinétique.”) How does an object's kinetic energy depend on its mass and velocity? Joule attempted a conceptually simple experiment on his honeymoon in the French-Swiss Alps near Mt. Chamonix, in which he measured the difference in temperature between the top and bottom of a waterfall. The water at the top of the falls has some gravitational energy, which isn't our subject right now, but as it drops, that gravitational energy is converted into kinetic energy, and then into heat energy due to internal friction in the churning pool at the bottom:

In the logical framework of this book's presentation of energy, the significance of the experiment is that it provides a way to find out how an object's kinetic energy depends on its mass and velocity. The increase in heat energy should equal the kinetic energy of the water just before impact, so in principle we could measure the water's mass, velocity, and kinetic energy, and see how they relate to one another.³

Although the story is picturesque and memorable, most books that mention the experiment fail to note that it was a failure! The problem was that heat wasn't the only form of energy being released. In reality, the situation was more like this:

f / A realistic drawing of Joule's apparatus, based on the illustration in his original paper. The paddlewheel is sealed inside the can in the middle. Joule wound up the two 13-kg lead weights and dropped them 1.6 meters, repeating this 20 times to produce a temperature change of only about half a degree Fahrenheit in the water inside the sealed can. He claimed in his paper to be able to measure temperatures to an accuracy of 1/200 of a degree.

What about the factor of 1/2 in front? It comes out to be exactly 1/2 by the design of the metric system. If we'd been using the old-fashioned British engineering system of units (which is no longer used in the U.K.), the equation would have been $K=(7.44\times {10}^{-2}\text{Btu}\cdot {\text{s}}^2/\text{slug}\cdot {\text{ft}}^2)m{v}^2$. The version of the metric system called the SI,⁵ in which everything is based on units of kilograms, meters, and seconds, not only has the numerical constant equal to 1/2, but makes it unitless as well. In other words, we can think of the joule as simply an abbreviation, 1 J=1 kg⋅m²/s². More familiar examples of this type of abbreviation are 1 minute=60 s, and the metric unit of land area, 1 hectare=10000 m².

Power

Power, P, is defined as the rate of change of energy, dE/dt. Power thus has units of joules per second, which are usually abbreviated as watts, 1 W=1 J/s. Since energy is conserved, we would have dE/dt=0 if E was the total energy of a closed system, and that's not very interesting. What's usually more interesting to discuss is either the power flowing in or out of an open system, or the rate at which energy is being transformed from one form into another. The following is an example of energy flowing into an open system.

Now here's an example of energy being transformed from one form into another.

g / A skateboarder rises to the edge of an empty pool and then falls back down.

h / The sum of kinetic plus gravitational energy is constant.

i / Two balls start from rest, and roll from A to B by different paths.

j / How much energy is required to raise the submerged box through a height Δ y?

k / A seesaw.

l / The biceps muscle is a reversed lever.

m / Discussion question C.

Gravitational energy

Gravitational energy, to which I've already alluded, is different from heat and kinetic energy in an important way. Heat and kinetic energy are properties of a single object, whereas gravitational energy describes an interaction between two objects. When the skater in figures g and h is at the top, his distance from the bulk of the planet earth is greater. Since we observe his kinetic energy decreasing on the way up, there must be some other form of energy that is increasing. We invent a new form of energy, called gravitational energy, and written U or U_g, which depends on the distance between his body and the planet. Where is this energy? It's not in the skater's body, and it's not inside the earth, either, since it takes two to tango. If either object didn't exist, there wouldn't be any interaction or any way to measure a distance, so it wouldn't make sense to talk about a distance-dependent energy. Just as marriage is a relationship between two people, gravitational energy is a relationship between two objects.

There is no precise way to define the distance between the skater and the earth, since both are objects that have finite size. As discussed in more detail in section 2.3, gravity is one of the fundamental forces of nature, a universal attraction between any two particles that have mass. Each atom in the skater's body is at a definite distance from each atom in the earth, but each of these distances is different. An atom in his foot is only a few centimeters from some of the atoms in the plaster side of the pool, but most of the earth's atoms are thousands of kilometers away from him. In theory, we might have to add up the contribution to the gravitational energy for every interaction between an atom in the skater's body and an atom in the earth.

For our present purposes, however, there is a far simpler and more practical way to solve problems. In any region of the earth's surface, there is a direction called “down,” which we can establish by dropping a rock or hanging a plumb bob. In figure h, the skater is moving up and down in one dimension, and if we did measurements of his kinetic energy, like the made-up data in the figure, we could infer his gravitational energy. As long as we stay within a relatively small range of heights, we find that an object's gravitational energy increases at a steady rate with height. In other words, the strength of gravity doesn't change much if you only move up or down a few meters. We also find that the gravitational energy is proportional to the mass of the object we're testing. Writing y for the height, and g for the overall constant of proportionality, we have

The number g, with units of joules per kilogram per meter, is called the gravitational field. It tells us the strength of gravity in a certain region of space. Near the surface of our planet, it has a value of about 9.8 $\text{J}/\text{kg}\cdot \text{m}$, which is conveniently close to 10 $\text{J}/\text{kg}\cdot \text{m}$ for rough calculations.

There are a couple of important things to note about this example. First, we were able to massage the equation so that it only involved Δy, rather than y itself. In other words, we don't need to worry about where y=0 is; any coordinate system will work, as long as the positive y axis points up, not down. This is no accident. Gravitational energy can always be changed by adding a constant onto it, with no effect on the final result, as long as you're consistent within a given problem.

The other interesting thing is that the mass canceled out: even if the skater gained weight or strapped lead weights to himself, his velocity at the bottom would still be 8 m/s. This isn't an accident either. This is the same conclusion we reached in section 1.2, based on the equivalence of gravitational and inertial mass. The kinetic energy depends on the inertial mass, while gravitational energy is related to gravitational mass, but since these two quantities are equal, we were able to use a single symbol, m, for them, and cancel them out.

We can see from the equation $v=\sqrt{-2g\mathrm{\Delta }y}$ that a falling object's velocity isn't constant. It increases as the object drops farther and farther. What about its acceleration? If we assume that air friction is negligible, the arguments in section 1.2 show that the acceleration can't depend on the object's mass, so there isn't much else the acceleration can depend on besides g. In fact, the acceleration of a falling object equals -g (in a coordinate system where the positive y axis points up), as we can easily show using the chain rule:

where I've calculated dv/dK as 1/(dK/dv), and dK/dU=-1 can be found by differentiating $K+U=\text{(constant)}$ to give dK+dU=0.

We can also check that the units of g, $\text{J}/\text{kg}\cdot \text{m}$, are equivalent to the units of acceleration,

and therefore the strength of the gravitational field near the earth's surface can just as well be stated as 10 m/s².

\myeqnspacing

The seesaw example demonstrates the principle of the lever, which is one of the basic mechanical building blocks known as simple machines. As discussed in more detail in chapters 3 and 4, the principle applies even when the forces involved aren't gravitational. (A rigorous definition of “force” is given in chapter 3.)

Note that although a lever makes it easier to lift a heavy weight, it also decreases the distance traveled by the load. By reversing the lever, we can make the load travel a greater distance, at the expense of increasing the amount of force required. The human muscular-skeletal system uses reversed levers of this kind, which allows us to move more rapidly, and also makes our bodies more compact, at the expense of brute strength. A piano uses reversed levers so that a small amount of motion of the key produces a longer swing of the hammer. Another interesting example is the hydraulic jack shown in figure n. The analysis in terms of gravitational energy is exactly the same as for the seesaw, except that the relationship between Δy₁ and Δy₂ is now determined not by geometry but by conservation of mass: since water is highly incompressible, conservation of mass is approximately the same as a requirement of constant volume, which can only be satisfied if the distance traveled by each piston is in inverse proportion to its cross-sectional area.

Discussion Questions

n / A hydraulic jack.

o / The surfaces are frictionless. The black blocks are in equilibrium.

p / Water in a U-shaped tube.

Equilibrium and stability

The seesaw in figure k is in equilibrium, meaning that if it starts out being at rest, it will stay put. This is known as a neutral equilibrium, since the seesaw has no preferred position to which it will return if we disturb it. If we move it to a different position and release it, it will stay at rest there as well. If we put it in motion, it will simply continue in motion until one person's feet hit the ground.

Most objects around you are in stable equilibria, like the black block in figure o/3. Even if the block is moved or set in motion, it will oscillate about the equilibrium position. The pictures are like graphs of y versus x, but since the gravitational energy U=mgy is proportional to y, we can just as well think of them as graphs of U versus x. The block's stable equilibrium position is where the function U(x) has a local minimum. The book you're reading right now is in equilibrium, but gravitational energy isn't the only form of energy involved. To move it upward, we'd have to supply gravitational energy, but downward motion would require a different kind of energy, in order to compress the table more. (As we'll see in section 2.4, this is electrical energy due to interactions between atoms within the table.)

A differentiable function's local extrema occur where its derivative is zero. A position where dU/dx is zero can be a stable (3), neutral (2), or unstable equilibrium, (4). An unstable equilibrium is like a pencil balanced on its tip. Although it could theoretically remain balanced there forever, in reality it will topple due to any tiny perturbation, such as an air current or a vibration from a passing truck. This is a technical, mathematical definition of instability, which is more restrictive than the way the word is used in ordinary speech. Most people would describe a domino standing upright as being unstable, but in technical usage it would be considered stable, because a certain finite amount of energy is required to tip it over, and perturbations smaller than that would only cause it to oscillate around its equilibrium position.

The domino is also an interesting example because it has two local minima, one in which it is upright, and another in which it is lying flat. A local minimum that is not the global minimum, as in figure o/5, is referred to as a metastable equilibrium.

q / A car drives over a cliff.

Predicting the direction of motion

Kinetic energy doesn't depend on the direction of motion. Sometimes this is helpful, as in the high road-low road example (p. 84, example 9), where we were able to predict that the balls would have the same final speeds, even though they followed different paths and were moving in different directions at the end. In general, however, the two conservation laws we've encountered so far aren't enough to predict an object's path through space, for which we need conservation of momentum (chapter 3), and the mathematical technique of vectors. Before we develop those ideas in their full generality, however, it will be helpful to do a couple of simple examples, including one that we'll get a lot of mileage out of in section 2.3.

Suppose we observe an air hockey puck gliding frictionlessly to the right at a velocity v, and we want to predict its future motion. Since there is no friction, no kinetic energy is converted to heat. The only form of energy involved is kinetic energy, so conservation of energy, ΔE=0, becomes simply ΔK=0. There's no particular reason for the puck to do anything but continue moving to the right at constant speed, but it would be equally consistent with conservation of energy if it spontaneously decided to reverse its direction of motion, changing its velocity to -v. Either way, we'd have ΔK=0. There is, however, a way to tell which motion is physical and which is unphysical. Suppose we consider the whole thing again in the frame of reference that is initially moving right along with the puck. In this frame, the puck starts out with K=0. What we originally described as a reversal of its velocity from v to -v is, in this new frame of reference, a change from zero velocity to -2v, which would violate conservation of energy. In other words, the physically possible motion conserves energy in all frames of reference, but the unphysical motion only conserves energy in one special frame of reference.

For our second example, we consider a car driving off the edge of a cliff (q). For simplicity, we assume that air friction is negligible, so only kinetic and gravitational energy are involved. Does the car follow trajectory 1, familiar from Road Runner cartoons, trajectory 2, a parabola, or 3, a diagonal line? All three could be consistent with conservation of energy, in the ground's frame of reference. For instance, the car would have constant gravitational energy along the initial horizontal segment of trajectory 1, so during that time it would have to maintain constant kinetic energy as well. Only a parabola, however, is consistent with conservation of energy combined with Galilean relativity. Consider the frame of reference that is moving horizontally at the same speed as that with which the car went over the edge. In this frame of reference, the cliff slides out from under the initially motionless car. The car can't just hover for a while, so trajectory 1 is out. Repeating the same math as in example 8 on p. 83, we have

in this frame of reference, where the stars indicate coordinates measured in the moving frame of reference. These coordinates are related to the ground-fixed coordinates (x,y) by the equations

where v is the velocity of one frame with respect to the other. We therefore have

in our original frame of reference. Eliminating t, we can see that this has the form of a parabola:

2.2 Numerical Techniques

a / Approximations to the brachistochrone curve using a third-order polynomial (solid line), and a seventh-order polynomial (dashed). The latter only improves the time by four milliseconds.

Engineering majors are a majority of the students in the kind of physics course for which this book is designed, so most likely you fall into that category. Although you surely recognize that physics is an important part of your training, if you've had any exposure to how engineers really work, you're probably skeptical about the flavor of problem-solving taught in most science courses. You realize that not very many practical engineering calculations fall into the narrow range of problems for which an exact solution can be calculated with a piece of paper and a sharp pencil. Real-life problems are usually complicated, and typically they need to be solved by number-crunching on a computer, although we can often gain insight by working simple approximations that have algebraic solutions. Not only is numerical problem-solving more useful in real life, it's also educational; as a beginning physics student, I really only felt like I understood projectile motion after I had worked it both ways, using algebra and then a computer program. (This was back in the days when 64 kilobytes of memory was considered a lot.)

In this section, we'll start by seeing how to apply numerical techniques to some simple problems for which we know the answer in “closed form,” i.e. a single algebraic expression without any calculus or infinite sums. After that, we'll solve a problem that would have made you world-famous if you could have done it in the seventeenth century using paper and a quill pen! Before you continue, you should read Appendix 1 on page 835 that introduces you to the Python programming language.

First let's solve the trivial problem of finding how much time it takes an object moving at speed \verb-v- to travel a straight-line distance \verb-dist-. This closed-form answer is, of course, \verb-dist/v-, but the point is to introduce the techniques we can use to solve other problems of this type. The basic idea is to divide the distance up into \verb-n- equal parts, and add up the times required to traverse all the parts. The following Python function does the job. Note that you shouldn't type in the line numbers on the left, and you don't need to type in the comments, either. I've omitted the prompts \verb->>>- and \verb-...- in order to save space.

import math
def time1(dist,v,n):
 x=0 # Initialize the position.
 dx = dist/n # Divide dist into n equal parts.
 t=0 # Initialize the time.
 for i in range(n):
 x = x+dx # Change x.
 dt=dx/v # time=distance/speed
 t=t+dt # Keep track of elapsed time.
 return t

>>> time1(1.0,1.0,10) # dist, v, n
0.99999999999999989

>>> time1(1.0,1.0,1) # dist, v, n
1.0

Now let's do an example where the answer isn't obvious to people who don't know calculus: how long does it take an object to fall through a height \verb-h-, starting from rest? We know from example 8 on page 83 that the exact answer, found using calculus, is $\sqrt{2h/g}$. Let's see if we can reproduce that answer numerically. The main difference between this program and the previous one is that now the velocity isn't constant, so we need to update it as we go along. Conservation of energy gives mgh=(1/2)mv²+mgy for the velocity v at height y, so $v=-\sqrt{-2g(h-y)}$. (We choose the negative root because the object is moving down, and our coordinate system has the positive y axis pointing up.)

import math
def time2(h,n):
 g=9.8 # gravitational field
 y=h # Initialize the height.
 v=0 # Initialize the velocity.
 dy = -h/n # Divide h into n equal parts.
 t=0 # Initialize the time.
 for i in range(n):
 y = y+dy # Change y. (Note dy<0.)
 v = -math.sqrt(2*g*(h-y)) # from cons. of energy
 dt=dy/v # dy and v are <0, so dt is >0
 t=t+dt # Keep track of elapsed time.
 return t

>>> time2(1.0,10) # h, n
0.35864270709233342

>>> time2(1.0,10000) # h, n
0.44846664060793945

A subtle point is that we changed \verb-y- in line 9, and then we calculated \verb-v- in line 10, which depends on \verb-y-. Since \verb-y- is only changing by a ten-thousandth of a meter with each step, you might think this wouldn't make much of a difference, and you'd be almost right, except for one small problem: if we swapped lines 9 and 10, then the very first time through the loop, we'd have \verb-v-=0, which would produce a division-by-zero error when we calculated \verb-dt-! Actually what would make the most sense would be to calculate the velocity at height \verb-y- and the velocity at height \verb-y+dy- (recalling that \verb-dy- is negative), average them together, and use that value of \verb-y- to calculate the best estimate of the velocity between those two points. Since the acceleration is constant in the present example, this modification results in a program that gives an exact result even for \verb-n-=1:

import math
def time3(h,n):
 g=9.8
 y=h
 v=0
 dy = -h/n
 t=0
 for i in range(n):
 y_old = y
 y = y+dy
 v_avg = -(math.sqrt(2*g*(h-y_old))+math.sqrt(2*g*(h-y)))/2.
 dt=dy/v_avg
 t=t+dt
 return t

>>> time3(1.0,1) # h, n
0.45175395145262565

Now we're ready to attack a problem that challenged the best minds of Europe back in the days when there were no computers. In 1696, the mathematician Johann Bernoulli posed the following famous question. Starting from rest, an object slides frictionlessly over a curve joining the point (a,b) to the point (0,0). Of all the possible shapes that such a curve could have, which one gets the object to its destination in the least possible time, and how much time does it take? The optimal curve is called the brachistochrone, from the Greek “short time.” The solution to the brachistochrone problem evaded Bernoulli himself, as well as Leibniz, who had been one of the inventors of calculus. The English physicist Isaac Newton, however, stayed up late one night after a day's work running the royal mint, and, according to legend, produced an algebraic solution at four in the morning. He then published it anonymously, but Bernoulli is said to have remarked that when he read it, he knew instantly from the style that it was Newton --- he could “tell the lion from the mark of his claw.”

Rather than attempting an exact algebraic solution, as Newton did, we'll produce a numerical result for the shape of the curve and the minimum time, in the special case of a=1.0 m and b=1.0 m. Intuitively, we want to start with a fairly steep drop, since any speed we can build up at the start will help us throughout the rest of the motion. On the other hand, it's possible to go too far with this idea: if we drop straight down for the whole vertical distance, and then do a right-angle turn to cover the horizontal distance, the resulting time of 0.68 s is quite a bit longer than the optimal result, the reason being that the path is unnecessarily long. There are infinitely many possible curves for which we could calculate the time, but let's look at third-order polynomials,

where we require c₃=(b-c₁a-c₂a²)/a³ in order to make the curve pass through the point (a,b). The Python program, below, is not much different from what we've done before. The function only asks for c₁ and c₂, and calculates c₃ internally at line 4. Since the motion is two-dimensional, we have to calculate the distance between one point and the next using the Pythagorean theorem, at line 16.

import math
def timeb(a,b,c1,c2,n):
 g=9.8
 c3 = (b-c1*a-c2*a**2)/(a**3)
 x=a
 y=b
 v=0
 dx = -a/n
 t=0
 for i in range(n):
 y_old = y
 x = x+dx
 y = c1*x+c2*x**2+c3*x**3
 dy = y-y_old
 v_avg = (math.sqrt(2*g*(b-y_old))+math.sqrt(2*g*(b-y)))/2.
 ds = math.sqrt(dx**2+dy**2) # Pythagorean thm.
 dt=ds/v_avg
 t=t+dt
 return t

As a first guess, we could try a straight diagonal line, y=x, which corresponds to setting c₁=1, and all the other coefficients to zero. The result is a fairly long time:

>>> a=1.
>>> b=1.
>>> n=10000
>>> c1=1.
>>> c2=0.
>>> timeb(a,b,c1,c2,n)
0.63887656499994161

>>> c1=0.
>>> c2=0.
>>> timeb(a,b,c1,c2,n)
0.59458339947087069

by using the program listed in appendix 2 on page 838 to search automatically for the optimal curve. The seventh-order approximation shown in the figure came from a straightforward extension of the same program.

2.3 Gravitational Phenomena

a / An ellipse is circle that has been distorted by shrinking and stretching along perpendicular axes.

b / An ellipse can be constructed by tying a string to two pins and drawing like this with a pencil stretching the string taut. Each pin constitutes one focus of the ellipse.

c / If the time interval taken by the planet to move from P to Q is equal to the time interval from R to S, then according to Kepler's equal-area law, the two shaded areas are equal. The planet is moving faster during time interval RS than it was during PQ, because gravitational energy has been transformed into kinetic energy.

Cruise your radio dial today and try to find any popular song that would have been imaginable without Louis Armstrong. By introducing solo improvisation into jazz, Armstrong took apart the jigsaw puzzle of popular music and fit the pieces back together in a different way. In the same way, Newton reassembled our view of the universe. Consider the titles of some recent physics books written for the general reader: The God Particle, Dreams of a Final Theory. When the subatomic particle called the neutrino was recently proven for the first time to have mass, specialists in cosmology began discussing seriously what effect this would have on calculations of the evolution of the universe from the Big Bang to its present state. Without the English physicist Isaac Newton, such attempts at universal understanding would not merely have seemed ambitious, they simply would not have occurred to anyone.

This section is about Newton's theory of gravity, which he used to explain the motion of the planets as they orbited the sun. Newton tosses off a general treatment of motion in the first 20 pages of his Mathematical Principles of Natural Philosophy, and then spends the next 130 discussing the motion of the planets. Clearly he saw this as the crucial scientific focus of his work. Why? Because in it he showed that the same laws of nature applied to the heavens as to the earth, and that the gravitational interaction that made an apple fall was the same as the as the one that kept the earth's motion from carrying it away from the sun.

Kepler's laws

d / As the earth and Mars revolve around the sun at different rates, the combined effect of their motions makes Mars appear to trace a strange, looped path across the background of the distant stars.

With Brahe's death, it fell to his former assistant Kepler to try to make some sense out of the volumes of data. After 900 pages of calculations and many false starts and dead-end ideas, Kepler finally synthesized the data into the following three laws:

\myindented{4mm}{ Kepler's elliptical orbit law: The planets orbit the sun in elliptical orbits with the sun at one focus.
Kepler's equal-area law: The line connecting a planet to the sun sweeps out equal areas in equal amounts of time.
Kepler's law of periods: The time required for a planet to orbit the sun, called its period, T, is proportional to the long axis of the ellipse raised to the 3/2 power. The constant of proportionality is the same for all the planets.}

Although the planets' orbits are ellipses rather than circles, most are very close to being circular. The earth's orbit, for instance, is only flattened by 1.7% relative to a circle. In the special case of a planet in a circular orbit, the two foci (plural of “focus”) coincide at the center of the circle, and Kepler's elliptical orbit law thus says that the circle is centered on the sun. The equal-area law implies that a planet in a circular orbit moves around the sun with constant speed. For a circular orbit, the law of periods then amounts to a statement that the time for one orbit is proportional to r^3/2, where r is the radius. If all the planets were moving in their orbits at the same speed, then the time for one orbit would simply depend on the circumference of the circle, so it would only be proportional to r to the first power. The more drastic dependence on r^3/2 means that the outer planets must be moving more slowly than the inner planets.

Our main focus in this section will be to use the law of periods to deduce the general equation for gravitational energy. The equal-area law turns out to be a statement on conservation of angular momentum, which is discussed in chapter 4. We'll demonstrate the elliptical orbit law numerically in chapter 3, and analytically in chapter 4.

e / A cannon fires cannonballs at different velocities, from the top of an imaginary mountain that rises above the earth's atmosphere. This is almost the same as a figure Newton included in his Mathematical Principles.

Circular orbits

Kepler's laws say that planets move along elliptical paths (with circles as a special case), which would seem to contradict the proof on page 88 that objects moving under the influence of gravity have parabolic trajectories. Kepler was right. The parabolic path was really only an approximation, based on the assumption that the gravitational field is constant, and that vertical lines are all parallel. In figure e, trajectory 1 is an ellipse, but it gets chopped off when the cannonball hits the earth, and the small piece of it that is above ground is nearly indistinguishable from a parabola. Our goal is to connect the previous calculation of parabolic trajectories, y=(g/2v²)x², with Kepler's data for planets orbiting the sun in nearly circular orbits. Let's start by thinking in terms of an orbit that circles the earth, like orbit 2 in figure e. It's more natural now to choose a coordinate system with its origin at the center of the earth, so the parabolic approximation becomes y=r-(g/2v²)x², where r is the distance from the center of the earth. For small values of x, i.e. when the cannonball hasn't traveled very far from the muzzle of the gun, the parabola is still a good approximation to the actual circular orbit, defined by the Pythagorean theorem, r²=x²+y², or $y=r\sqrt{1-x^2/r^2}$. For small values of x, we can use the approximation $\sqrt{1+ϵ}\approx 1+ϵ/2$ to find y≈r-(1/2r)x². Setting this equal to the equation of the parabola, we have g/2v²=(1/2r), or

The sun's gravitational field

We can now use the circular orbit condition $v=\sqrt{gr}$, combined with Kepler's law of periods, T∝r^3/2 for circular orbits, to determine how the sun's gravitational field falls off with distance.⁷ From there, it will be just a hop, skip, and a jump to get to a universal description of gravitational interactions.

The velocity of a planet in a circular orbit is proportional to r/T, so

If gravity behaves systematically, then we can expect the same to be true for the gravitational field created by any object, not just the sun.

There is a subtle point here, which is that so far, r has just meant the radius of a circular orbit, but what we have come up with smells more like an equation that tells us the strength of the gravitational field made by some object (the sun) if we know how far we are from the object. In other words, we could reinterpret r as the distance from the sun.

f / The gravitational energy U=-Gm₁m₂/r graphed as a function of r.

g / Cavendish's original drawing of the apparatus for his experiment, discussed in example 14. The room was sealed to exclude air currents, and the motion was observed through telescopes sticking through holes in the walls.

h / A simplified drawing of the Cavendish experiment, viewed from above. The rod with the two small masses on the ends hangs from a thin fiber, and is free to rotate.

i / The Pioneer 10 space probe's trajectory from 1974 to 1992, with circles marking its position at one-year intervals. After its 1974 slingshot maneuver around Jupiter, the probe's motion was determined almost exclusively by the sun's gravity.

Gravitational energy in general

We now want to find an equation for the gravitational energy of any two masses that attract each other from a distance r. We assume that r is large enough compared to the distance between the objects so that we don't really have to worry about whether r is measured from center to center or in some other way. This would be a good approximation for describing the solar system, for example, since the sun and planets are small compared to the distances between them --- that's why you see Venus (the “evening star”) with your bare eyes as a dot, not a disk.

The equation we seek is going to give the gravitational energy, U, as a function of m₁, m₂, and r. We already know from experience with gravity near the earth's surface that U is proportional to the mass of the object that interacts with the earth gravitationally, so it makes sense to assume the relationship is symmetric: U is presumably proportional to the product m₁m₂. We can no longer assume ΔU∝Δr, as in the earth's-surface equation ΔU=mgΔy, since we are trying to construct an equation that would be valid for all values of r, and g depends on r. We can, however, consider an infinitesimally small change in distance dr, for which we'll have dU=m₂g₁dr, where g₁ is the gravitational field created by m₁. (We could just as well have written this as dU=m₁g₂dr, since we're not assuming either mass is “special” or “active.”) Integrating this equation, we have

where we're free to take the constant of integration to be equal to zero, since gravitational energy is never a well-defined quantity in absolute terms. Writing G for the constant of proportionality, we have the following fundamental description of gravitational interactions:

Let's interpret this. First, don't get hung up on the fact that it's negative, since it's only differences in gravitational energy that have physical significance. The graph in figure f could be shifted up or down without having any physical effect. The slope of this graph relates to the strength of the gravitational field. For instance, suppose figure f is a graph of the gravitational energy of an asteroid interacting with the sun. If the asteroid drops straight toward the sun, from A to B, the decrease in gravitational energy is very small, so it won't speed up very much during that motion. Points C and D, however, are in a region where the graph's slope is much greater.

As the asteroid moves from C to D, it loses a lot of gravitational energy, and therefore speeds up considerably.

By the way, Pioneer 10 and 11 have shown some behavior that some scientists have interpreted as indicating small deviations from the expected -Gm₁m₂/r behavior of the gravitational energy. For more information, see the Wikipedia article "Pioneer anomaly."

The gravitational field

We got the energy equation U = -Gm₁m₂/r by integrating g∝ 1/r² and then inserting a constant of proportionality to make the proportionality into an equation. The opposite of an integral is a derivative, so we can now go backwards and insert a constant of proportionality in g∝ 1/r² that will be consistent with the energy equation:

This kind of inverse-square law occurs all the time in nature. For instance, if you go twice as far away from a lightbulb, you receive 1/4 as much light from it, because as the light spreads out, it is like an expanding sphere, and a sphere with twice the radius has four times the surface area. It's like spreading the same amount of peanut butter on four pieces of bread instead of one --- we have to spread it thinner.

Discussion Questions

j / A spherical shell of mass M interacts with a pointlike mass m.

k / The gravitational energy of a mass m at a distance s from the center of a hollow spherical shell of mass.

l / The actual trajectory of the Apollo 11 spacecraft, A, and the straight-line trajectory, B, assumed in the example.

The shell theorem

Newton's great insight was that gravity near the earth's surface was the same kind of interaction as the one that kept the planets from flying away from the sun. He told his niece that the idea came to him when he saw an apple fall from a tree, which made him wonder whether the earth might be affecting the apple and the moon in the same way. Up until now, we've generally been dealing with gravitational interactions between objects that are small compared to the distances between them, but that assumption doesn't apply to the apple. A kilogram of dirt a few feet under his garden in England would interact much more strongly with the apple than a kilogram of molten rock deep under Australia, thousands of miles away. Also, we know that the earth has some parts that are more dense, and some parts that are less dense. The solid crust, on which we live, is considerably less dense than the molten rock on which it floats. By all rights, the computation of the total gravitational energy of the apple should be a horrendous mess. Surprisingly, it turns out to be fairly simple in the end. First, we note that although the earth doesn't have the same density throughout, it does have spherical symmetry: if we imagine dividing it up into thin concentric shells, the density of each shell is uniform.

Second, it turns out that a uniform spherical shell interacts with external masses as if all its mass was concentrated at its center.

\mythmhdr{The shell theorem} The gravitational energy of a uniform spherical shell of mass M interacting with a pointlike mass m outside it equals -GMm/s, where s is the center-to-center distance. If mass m is inside the shell, then the energy is constant, i.e. the shell's interior gravitational field is zero.

\mythmhdr{Proof} Let b be the radius of the shell, h its thickness, and ρ its density. Its volume is then V=(area)(thickness)=4πb²h, and its mass is M=ρV=4πρb²h. The strategy is to divide the shell up into rings as shown in figure j, with each ring extending from θ to θ+dθ. Since the ring is infinitesimally skinny, its entire mass lies at the same distance, r, from mass m. The width of such a ring is found by the definition of radian measure to be w=bdθ, and its mass is dM=(ρ)(circumference)(thickness)(width)= (ρ)(2πb sin θ)(h)(bdθ)=2πb²hsinθdθ. The gravitational energy of the ring interacting with mass m is therefore

The integral has a mixture of the variables r and θ, which are related by the law of cosines,

and to evaluate the integral, we need to get everything in terms of either r and dr or θ and dθ. The relationship between the differentials is found by differentiating the law of cosines,

and since sinθdθ occurs in the integral, the easiest path is to substitute for it, and get everything in terms of r and dr:

This was all under the assumption that mass m was on the outside of the shell. To complete the proof, we consider the case where it's inside. In this case, the only change is that the limits of integration are different:

The two results are equal at the surface of the sphere, s=b, so the constant-energy part joins continuously onto the 1/s part, and the effect is to chop off the steepest part of the graph that we would have had if the whole mass M had been concentrated at its center. Dropping a mass m from A to B in figure k releases the same amount of energy as if mass M had been concentrated at its center, but there is no release of gravitational energy at all when moving between two interior points like C and D. In other words, the internal gravitational field is zero. Moving from C to D brings mass m farther away from the nearby side of the shell, but closer to the far side, and the cancellation between these two effects turns out to be perfect. Although the gravitational field has to be zero at the center due to symmetry, it's much more surprising that it cancels out perfectly in the whole interior region; this is a special mathematical characteristic of a 1/r interaction like gravity.

The following example applies the numerical techniques of section 2.2.

import math
def tmoon(vi,ri,rf,n):
 bigg=6.67e-11 # gravitational constant
 me=5.97e24 # mass of earth
 mm=7.35e22 # mass of moon
 rm=3.84e8 # earth-moon distance
 r=ri
 v=vi
 dr = (rf-ri)/n
 e=-bigg*me/ri-bigg*mm/(rm-ri)+.5*vi**2
 t=0
 for i in range(n):
 u_old = -bigg*me/r-bigg*mm/(rm-r)
 k_old = e - u_old
 v_old = math.sqrt(2.*k_old)
 r = r+dr
 u = -bigg*me/r-bigg*mm/(rm-r)
 k = e - u
 v = math.sqrt(2.*k)
 v_avg = .5*(v_old+v)
 dt=dr/v_avg
 t=t+dt
 return t

>>> re=6.378e6 # radius of earth
>>> rm=1.74e6 # radius of moon
>>> ri=re+3.363e5 # re+initial altitude
>>> rf=3.8e8-rm # earth-moon distance minus rm
>>> vi=1.083e4 # initial velocity
>>> tmoon(vi,ri,rf,1000)/3600. # convert seconds to hours
59.654047441976552

>>> vi=1.2e4
>>> tmoon(vi,ri,rf,1000)/3600.
18.177752636111677

Evidence for repulsive gravity

Until recently, physicists thought they understood gravity fairly well. Einstein had modified Newton's theory, but certain characteristrics of gravitational forces were firmly established. For one thing, they were always attractive. If gravity always attracts, then it is logical to ask why the universe doesn't collapse. Newton had answered this question by saying that if the universe was infinite in all directions, then it would have no geometric center toward which it would collapse; the forces on any particular star or planet exerted by distant parts of the universe would tend to cancel out by symmetry. More careful calculations, however, show that Newton's universe would have a tendency to collapse on smaller scales: any part of the universe that happened to be slightly more dense than average would contract further, and this contraction would result in stronger gravitational forces, which would cause even more rapid contraction, and so on.

When Einstein overhauled gravity, the same problem reared its ugly head. Like Newton, Einstein was predisposed to believe in a universe that was static, so he added a special repulsive term to his equations, intended to prevent a collapse. This term was not associated with any attraction of mass for mass, but represented merely an overall tendency for space itself to expand unless restrained by the matter that inhabited it. It turns out that Einstein's solution, like Newton's, is unstable. Furthermore, it was soon discovered observationally that the universe was expanding, and this was interpreted by creating the Big Bang model, in which the universe's current expansion is the aftermath of a fantastically hot explosion.⁹ An expanding universe, unlike a static one, was capable of being explained with Einstein's equations, without any repulsion term. The universe's expansion would simply slow down over time due to the attractive gravitational forces. After these developments, Einstein said woefully that adding the repulsive term, known as the cosmological constant, had been the greatest blunder of his life.

This was the state of things until 1999, when evidence began to turn up that the universe's expansion has been speeding up rather than slowing down! The first evidence came from using a telescope as a sort of time machine: light from a distant galaxy may have taken billions of years to reach us, so we are seeing it as it was far in the past. Looking back in time, astronomers saw the universe expanding at speeds that ware lower, rather than higher. At first they were mortified, since this was exactly the opposite of what had been expected. The statistical quality of the data was also not good enough to constute ironclad proof, and there were worries about systematic errors. The case for an accelerating expansion has however been nailed down by high-precision mapping of the dim, sky-wide afterglow of the Big Bang, known as the cosmic microwave background. Some theorists have proposed reviving Einstein's cosmological constant to account for the acceleration, while others believe it is evidence for a mysterious form of matter which exhibits gravitational repulsion. The generic term for this unknown stuff is “dark energy.”

As of 2008, most of the remaining doubt about the repulsive effect has been dispelled. During the past decade or so, astronomers consider themselves to have entered a new era of high-precision cosmology. The cosmic microwave background measurements, for example, have measured the age of the universe to be 13.7 ± 0.2 billion years, a figure that could previously be stated only as a fuzzy range from 10 to 20 billion. We know that only 4% of the universe is atoms, with another 23% consisting of unknown subatomic particles, and 73% of dark energy. It's more than a little ironic to know about so many things with such high precision, and yet to know virtually nothing about their nature. For instance, we know that precisely 96% of the universe is something other than atoms, but we know precisely nothing about what that something is.

m / The WMAP probe's map of the cosmic microwave background is like a “baby picture” of the universe.

2.4 Atomic Phenomena

a / A vivid demonstration that heat is a form of motion. A small amount of boiling water is poured into the empty can, which rapidly fills up with hot steam. The can is then sealed tightly, and soon crumples.

Variety is the spice of life, not of science. So far this chapter has focused on heat energy, kinetic energy, and gravitational energy, but it might seem that in addition to these there is a bewildering array of other forms of energy. Gasoline, chocolate bars, batteries, melting water --- in each case there seems to be a whole new type of energy. The physicist's psyche rebels against the prospect of a long laundry list of types of energy, each of which would require its own equations, concepts, notation, and terminology. The point at which we've arrived in the study of energy is analogous to the period in the 1960's when a half a dozen new subatomic particles were being discovered every year in particle accelerators. It was an embarrassment. Physicists began to speak of the “particle zoo,” and it seemed that the subatomic world was distressingly complex. The particle zoo was simplified by the realization that most of the new particles being whipped up were simply clusters of a previously unsuspected set of fundamental particles (which were whimsically dubbed quarks, a made-up word from a line of poetry by James Joyce, “Three quarks for Master Mark.”) The energy zoo can also be simplified, and it's the purpose of this section to demonstrate the hidden similarities between forms of energy as seemingly different as heat and motion.

b / Random motion of atoms in a gas, a liquid, and a solid.

c / The spring's energy is really due to electrical interactions among atoms.

Heat is kinetic energy.

What is heat really? Is it an invisible fluid that your bare feet soak up from a hot sidewalk? Can one ever remove all the heat from an object? Is there a maximum to the temperature scale?

The theory of heat as a fluid seemed to explain why colder objects absorbed heat from hotter ones, but once it became clear that heat was a form of energy, it began to seem unlikely that a material substance could transform itself into and out of all those other forms of energy like motion or light. For instance, a compost pile gets hot, and we describe this as a case where, through the action of bacteria, chemical energy stored in the plant cuttings is transformed into heat energy. The heating occurs even if there is no nearby warmer object that could have been leaking “heat fluid” into the pile.

An alternative interpretation of heat was suggested by the theory that matter is made of atoms. Since gases are thousands of times less dense than solids or liquids, the atoms (or clusters of atoms called molecules) in a gas must be far apart. In that case, what is keeping all the air molecules from settling into a thin film on the floor of the room in which you are reading this book? The simplest explanation is that they are moving very rapidly, continually ricocheting off of the floor, walls, and ceiling. Though bizarre, the cloud-of-bullets image of a gas did give a natural explanation for the surprising ability of something as tenuous as a gas to exert huge forces.

The experiment shown in figure a, for instance, can be explained as follows. The high temperature of the steam is interpreted as a high average speed of random motions of its molecules. Before the lid was put on the can, the rapidly moving steam molecules pushed their way out of the can, forcing the slower air molecules out of the way. As the steam inside the can thinned out, a stable situation was soon achieved, in which the force from the less dense steam molecules moving at high speed balanced against the force from the more dense but slower air molecules outside. The cap was put on, and after a while the steam inside the can began to cool off. The force from the cooler, thin steam no longer matched the force from the cool, dense air outside, and the imbalance of forces crushed the can.

This type of observation leads naturally to the conclusion that hotter matter differs from colder in that its atoms' random motion is more rapid. In a liquid, the motion could be visualized as people in a milling crowd shoving past each other more quickly. In a solid, where the atoms are packed together, the motion is a random vibration of each atom as it knocks against its neighbors.

We thus achieve a great simplification in the theory of heat. Heat is simply a form of kinetic energy, the total kinetic energy of random motion of all the atoms in an object. With this new understanding, it becomes possible to answer at one stroke the questions posed at the beginning of the section. Yes, it is at least theoretically possible to remove all the heat from an object. The coldest possible temperature, known as absolute zero, is that at which all the atoms have zero velocity, so that their kinetic energies, K=(1/2)mv², are all zero. No, there is no maximum amount of heat that a certain quantity of matter can have, and no maximum to the temperature scale, since arbitrarily large values of v can create arbitrarily large amounts of kinetic energy per atom.

The kinetic theory of heat also provides a simple explanation of the true nature of temperature. Temperature is a measure of the amount of energy per molecule, whereas heat is the total amount of energy possessed by all the molecules in an object.

There is an entire branch of physics, called thermodynamics, that deals with heat and temperature and forms the basis for technologies such as refrigeration. Thermodynamics is discussed in more detail in chapter 5, and I've provided here only a brief overview of the thermodynamic concepts that relate directly to energy.

d / All these energy transformations turn out at the atomic level to be due to changes in the distances between atoms that interact electrically.

e / This figure looks similar to the previous ones, but the scale is a million times smaller. The little balls are the neutrons and protons that make up the tiny nucleus at the center of a uranium atom. When the nucleus splits (fissions), the source of the kinetic energy is partly electrical and partly nuclear.

All energy comes from particles moving or interacting.

If I stretch the spring in figure c and then release it, it snaps taut again. The creation of some kinetic energy shows that there must have been some other form of energy that was destroyed. What was it?

We could just invent a new type of energy called “spring energy,” study its behavior, and call it quits, but that would be ugly. Are we going to have to invent a new forms of energy like this, over and over? No: the title of this book doesn't lie, and physics really is fundamentally simple. As shown in figure d, when we bend or stretch an object, we're really changing the distances between the atoms, resulting in a change in electrical energy. Electrical energy isn't really our topic right now --- that's what most of the second half of this book is about --- but conceptually it's very similar to gravitational energy. Like gravitational energy, it depends on 1/r, although there are some interesting new phenomena, such as the existence of both attraction and repulsion, which doesn't occur with gravity because gravitational mass can't be negative. The real point is that all the apparently dissimilar forms of energy in figure d turn out to be due to electrical interactions among atoms. Even if we wish to include nuclear reactions (figure e) in the picture, there still turn out to be only four fundamental types of energy:

\myindented{6mm}{ kinetic energy (including heat)
gravitational energy
electrical and magnetic energy
nuclear energy }

Astute students have often asked me how light fits into this picture. This is a very good question, and in fact it could be argued that it is the basic question that led to Einstein's theory of relativity as well as the modern quantum picture of nature. Since these are topics for the second half of the book, we'll have to be content with half an answer at this point. For now, we may think of light energy as a form of kinetic energy, but one calculated not according to (1/2)mv² but by some other equation. (We know that (1/2)mv² would not make sense, because light has no mass, and furthermore, high-energy beams of light do not differ in speed from low-energy ones.)

Discussion Questions

2.5 Oscillations

a / The spring has a minimum-energy length, 1, and energy is required in order to compress or stretch it, 2 and 3. A mass attached to the spring will oscillate around the equilibrium, 4-13.

b / Three functions with the same curvature at x=0.

c / The amplitude would usually be defined as the distance from equilibrium to one extreme of the motion, i.e. half the total travel.

d / Example 22.

e / Water in a U-shaped tube.

Let's revisit the example of the stretched spring from the previous section. We know that its energy is a form of electrical energy of interacting atoms, which is nice conceptually but doesn't help us to solve problems, since we don't know how the energy, U, depends on the length of the spring. All we know is that there's an equilibrium (figure a/1), which is a local minimum of the function U. An extremely important problem which arises in this connection is how to calculate oscillatory motion around an equilibrium, as in a/4-13. Even if we did special experiments to find out how the spring's energy worked, it might seem like we'd have to go through just as much work to deal with any other kind of oscillation, such as a sapling swinging back and forth in the breeze.

Surprisingly, it's possible to analyze this type of oscillation in a very general and elegant manner, as long as the analysis is limited to small oscillations. We'll talk about the mass on the spring for concreteness, but there will be nothing in the discussion at all that is restricted to that particular physical system. First, let's choose a coordinate system in which x=0 corresponds to the position of the mass where the spring is in equilibrium, and since interaction energies like U are only well defined up to an additive constant, we'll simply define it to be zero at equilibrium:

Since x=0 is an equilibrium, U(x) must have a local minimum there, and a differentiable function (which we assume U is) has a zero derivative at a local minimum:

There are still infinitely many functions that could satisfy these criteria, including the three shown in figure b, which are x²/2, x²/2(1+x²), and (e^3x+e^-3x-2)/18. Note, however, how all three functions are virtually identical right near the minimum. That's because they all have the same curvature. More specifically, each function has its second derivative equal to 1 at x=0, and the second derivative is a measure of curvature. We write k for the second derivative of the energy at an equilibrium point,

Physically, k is a measure of stiffness. For example, the heavy-duty springs in a car's shock absorbers would have a high value of k. It is often referred to as the spring constant, but we're only using a spring as an example here. As shown in figure b, any two functions that have U(0)=0, dU/dx=0, and d²U/dx²=k, with the same value of k, are virtually indistinguishable for small values of x, so if we want to analyze small oscillations, it doesn't even matter which function we assume. For simplicity, we'll just use U(x)=(1/2)kx² from now on.

Now we're ready to analyze the mass-on-a-spring system, while keeping in mind that it's really only a representative example of a whole class of similar oscillating systems. We expect that the motion is going to repeat itself over and over again, and since we're not going to include frictional heating in our model, that repetition should go on forever without dying out. The most interesting thing to know about the motion would be the period, T, which is the amount of time required for one complete cycle of the motion. We might expect that the period would depend on the spring constant, k, the mass, m, and and the amplitude, A, defined in figure c.¹⁰

In examples like the brachistochrone and the Apollo 11 mission, it was generally necessary to use numerical techniques to determine the amount of time required for a certain motion. Once again, let's dust off the \verb-time3- function from page 91 and modify it for our purposes. For flexibility, we'll define the function U(x) as a separate Python function. We really want to calculate the time required for the mass to come back to its starting point, but that would be awkward to set up, since our function works by dividing up the distance to be traveled into tiny segments. By symmetry, the time required to go from one end to the other equals the time required to come back to the start, so we'll just calculate the time for half a cycle and then double it when we return the result at the end of the function. The test at lines 16-19 is necessary because otherwise at the very end of the motion we can end up trying to take the square root of a negative number due to rounding errors.

import math
def u(k,x):
 return .5*k*x**2

def osc(m,k,a,n):
 x=a
 v=0
 dx = -2.*a/n
 t=0
 e = u(k,x)+.5*m*v**2
 for i in range(n):
 x_old = x
 v_old = v
 x = x+dx
 kinetic = e-u(k,x)
 if kinetic<0. :
 v=0.
 print "warning, K=",kinetic,"<0"
 else :
 v = -math.sqrt(2.*kinetic/m)
 v_avg = (v+v_old)/2.
 dt=dx/v_avg
 t=t+dt
 return 2.*t

>>> osc(1.,1.,1.,100000)
warning, K= -1.43707268307e-12 <0
6.2831854132667919

The first thing to notice is that with this particular set of inputs (m=1 kg, k=1 J/m², and A=1 m), the program has done an excellent job of computing 2π=6.2831853…. This is Mother Nature giving us a strong hint that the problem has an algebraic solution, not just a numerical one. The next interesting thing happens when we change the amplitude from 1 m to 2 m:

>>> osc(1.,1.,2.,100000)
warning, K= -5.7482907323e-12 <0
6.2831854132667919

With these hints, it seems like we start looking for an algebraic solution. For guidance, here's a graph of x as a function of t, as calculated by the \verb-osc- function with \verb-n-=10.

This looks like a cosine function, so let's see if a x=Acos(ωt+δ) is a solution to the conservation of energy equation --- it's not uncommon to try to “reverse-engineer” the cryptic results of a numerical calculation like this. The symbol ω=2π/T (Greek omega) is a standard symbol for the number of radians per second of oscillation, and the phase angle δ is to allow for the possibility that t=0 doesn't coincide with the beginning of the motion. The energy is

According to conservation of energy, this has to be a constant. Using the identity sin²+cos²=1, we can see that it will be a constant if we have mω²=k, or $\omega =\sqrt{k/m}$, i.e. $T=2\pi \sqrt{m/k}$. Note that the period is independent of amplitude.

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Homework Problems

b / Problem 16.

c / Problem 17.

d / Problem 18.

e / Problem 19.

f / Problem 27.

g / Problem 32.

h / Problem 35.

i / Problem 36.

j / Problem 37.

a / Problem 4.