28.6

The Equivalence of Mass and Energy

| The Total Energy of an Object

One of the most astonishing results of special relativity is that mass and energy are equivalent, in the sense that a gain or loss of mass can be regarded equally well as a gain or loss of energy. Consider, for example, an object of mass m traveling at a speed v. Einstein showed that the total energy E of the moving object is related to its mass and speed by the following relation:

| (28.4)

To gain some understanding of Equation 28.4, consider the special case in which the object is at rest. When v = 0 m/s, the total energy is called the rest energy E0, and Equation 28.4 reduces to Einstein’s now-famous equation:

| (28.5)

The rest energy represents the energy equivalent of the mass of an object at rest. As Example 6 shows, even a small mass is equivalent to an enormous amount of energy.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 6 | | | The Energy Equivalent of a Golf Ball

A 0.046-kg golf ball is lying on the green, as Figure 28-9 illustrates. If the rest energy of this ball were used to operate a 75-W light bulb, how long would the bulb remain lit?

Figure 28-9 The rest energy of a golf ball is sufficient to keep a 75-W light bulb burning for an incredibly long time (see Example 6).

Reasoning The average power delivered to the light bulb is 75 W, which means that it uses 75 J of energy per second. Therefore, the time that the bulb would remain lit is equal to the total energy used by the light bulb divided by the energy per second (i.e., the average power) delivered to it. This energy comes from the rest energy of the golf ball, which is equal to its mass times the speed of light squared.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value
Mass of golf ball | m | 0.046 kg
Average power delivered to light bulb |  | 75 W
Unknown Variable |  |
Time that light bulb would remain lit | t | ?

Modeling the Problem

| Step 1 Power
| The average power is equal to the energy delivered to the light bulb divided by the time t (see Section 6.7 and Equation 6.10b), or . In this case the energy comes from the rest energy E0 of the golf ball, so . Solving for the time gives Equation 1 at the right. The average power is known, and the rest energy will be evaluated in Step 2.

| (1)

| Step 2 Rest Energy
| The rest energy E0 is the total energy of the golf ball as it rests on the green. If the golf ball’s mass is m, then its rest energy is

| (28.5)

where c is the speed of light in a vacuum. Both m and c are known, so we substitute this expression for the rest energy into Equation 1, as indicated at the right.

Solution Algebraically combining the results of each step, we have

The time that the light bulb would remain lit is

Expressed in years , this time is equivalent to

Related Homework: Problems 26, 28

When an object is accelerated from rest to a speed v, the object acquires kinetic energy in addition to its rest energy. The total energy E is the sum of the rest energy E0 and the kinetic energy KE, or E = E0 + KE. Therefore, the kinetic energy is the difference between the object’s total energy and its rest energy. Using Equations 28.4 and 28.5, we can write the kinetic energy as

| (28.6)

This equation is the relativistically correct expression for the kinetic energy of an object of mass m moving at speed v.
Equation 28.6 looks nothing like the kinetic energy expression introduced in Section 6.2—namely, (Equation 6.2). However, for speeds much less than the speed of light (), the relativistic equation for the kinetic energy reduces to , as can be seen by using the binomial expansion* to represent the square root term in Equation 28.6:

Suppose v is much smaller than c—say v = 0.01c. The second term in the expansion has the value , while the third term has the much smaller value . The additional terms are smaller still, so if , we can neglect the third and additional terms in comparison with the first and second terms. Substituting the first two terms into Equation 28.6 gives

which is the familiar form for the kinetic energy. However, Equation 28.6 gives the correct kinetic energy for all speeds and must be used for speeds near the speed of light, as in Example 7.

| Example 7 | | | A High-Speed Electron

An electron is accelerated from rest to a speed of v = 0.9995c in a particle accelerator. Determine the electron’s (a) rest energy, (b) total energy, and (c) kinetic energy in millions of electron volts or MeV. Reasoning and Solution

a. | The electron’s rest energy is

| (28.5)

Since , the electron’s rest energy is

b. | The total energy of an electron traveling at a speed of v = 0.9995c is

| (28.4)

c. | The kinetic energy is the difference between the total energy and the rest energy:

| (28.6)

For comparison, if the kinetic energy of the electron had been calculated from , a value of only 0.26 MeV would have been obtained.

Since mass and energy are equivalent, any change in one is accompanied by a corresponding change in the other. For instance, life on earth is dependent on electromagnetic energy (light) from the sun. Because this energy is leaving the sun (see Figure 28-10), there is a decrease in the sun’s mass. Example 8 illustrates how to determine this decrease.

Figure 28-10 The sun emits electromagnetic energy over a broad portion of the electromagnetic spectrum. These photographs were obtained using that energy in the indicated regions of the spectrum. (Top, Mark Marten/NASA/Photo Researchers; bottom, Dr. Leon Golub/Photo Researchers)

| Example 8 | | | The Sun is Losing Mass

The sun radiates electromagnetic energy at the rate of . (a) What is the change in the sun’s mass during each second that it is radiating energy? (b) The mass of the sun is . What fraction of the sun’s mass is lost during a human lifetime of 75 years?
Reasoning Since 1 W = 1 J/s, the amount of electromagnetic energy radiated during each second is . Thus, during each second, the sun’s rest energy decreases by this amount. The change in the sun’s rest energy is related to the change in its mass by , according to Equation 28.5.
Solution

a. | For each second that the sun radiates energy, the change in its mass is

Over 4 billion kilograms of mass are lost by the sun during each second.
b. | The amount of mass lost by the sun in 75 years is

Although this is an enormous amount of mass, it represents only a tiny fraction of the sun’s total mass:

Any change in the rest energy of a system causes a change in the mass of the system according to . It does not matter whether the change in energy is due to a change in electromagnetic energy, potential energy, thermal energy, or so on. Although any change in energy gives rise to a change in mass, in most instances the change in mass is too small to be detected. For instance, when 4186 J of heat is used to raise the temperature of 1 kg of water by 1 C°, the mass changes by only . Conceptual Example 9 illustrates further how a change in the energy of an object leads to an equivalent change in its mass.

| Conceptual Example 9 | | | When is a Massless Spring not Massless?

Figure 28-11a shows the top view of a spring lying on a horizontal table. The spring is initially unstrained and assumed to be massless. Suppose that the spring is either stretched or compressed by an amount x from its unstrained length, as part b of the drawing shows. Is the mass of the spring still zero, or has it changed? And, if the mass has changed, is the change greater, smaller, or the same when the spring is stretched rather than compressed?

Figure 28-11 (a) This spring is unstrained and assumed to have no mass. (b) When the spring is either stretched or compressed by an amount x, it gains elastic potential energy and, hence, mass.

Reasoning and Solution Whenever a spring is stretched or compressed, its elastic potential energy changes. In Section 10.3 we showed that the elastic potential energy of an ideal spring is equal to , where k is the spring constant and x is the amount of stretch or compression. Consistent with the theory of special relativity, any change in the total energy of a system, including a change in the elastic potential energy, is equivalent to a change in the mass of the system. Thus, the mass of a strained spring is greater than that of an unstrained (and massless) spring. Furthermore, since the elastic potential energy depends on x2, the increase in mass of the spring is the same whether it is compressed or stretched, provided the magnitude of x is the same in both cases.
The increase is exceedingly small because it is equal to the elastic potential energy divided by c2 and c2 is so large.
Related Homework: Conceptual Questions 8, 9, 10

It is also possible to transform matter itself into other forms of energy. For example, the positron (see Section 31.4) has the same mass as an electron but an opposite electrical charge. If these two particles of matter collide, they are completely annihilated, and a burst of high-energy electromagnetic waves is produced. Thus, matter is transformed into electromagnetic waves, the energy of the electromagnetic waves being equal to the total energies of the two colliding particles. The medical diagnostic technique known as positron emission tomography or PET scanning depends on the electromagnetic energy produced when a positron and an electron are annihilated (see Section 32.6).
The transformation of electromagnetic waves into matter also happens. In one experiment, an extremely high-energy electromagnetic wave, called a gamma ray (see Section 31.4), passes close to the nucleus of an atom. If the gamma ray has sufficient energy, it can create an electron and a positron. The gamma ray disappears, and the two particles of matter appear in its place. Except for picking up some momentum, the nearby nucleus remains unchanged. The process in which the gamma ray is transformed into the two particles is known as pair production.

| The Relation Between Total Energy and Momentum

It is possible to derive a useful relation between the total relativistic energy E and the relativistic momentum p. We begin by rearranging Equation 28.3 for the momentum, to obtain

With this substitution, Equation 28.4 for the total energy becomes

Using this expression to replace v/c in Equation 28.4 gives

Solving this expression for E2 shows that

| (28.7)

| The Speed of Light is the Ultimate Speed

One of the important consequences of the theory of special relativity is that objects with mass cannot reach the speed of light in a vacuum. Thus, the speed of light represents the ultimate speed. To see that this speed limitation is a consequence of special relativity, consider Equation 28.6, which gives the kinetic energy of a moving object. As v approaches the speed of light c, the term in the denominator approaches zero. Hence, the kinetic energy becomes infinitely large. However, the work-energy theorem (Section 6.2) tells us that an infinite amount of work would have to be done to give the object an infinite kinetic energy. Since an infinite amount of work is not available, we are left with the conclusion that objects with mass cannot attain the speed of light c.

Check Your Understanding 3

Consider the same cup of coffee in the following four situations. (a) It is hot (95 °C) and sits on a table at sea level. (b) It is cold (60 °C) and sits on the same table at sea level. (c) It is hot (95 °C) and sits on a table in a cottage in the mountains. (d) It is cold (60 °C) and sits on a table in a cottage in the mountains. In which situation does the cup of coffee have the greatest and the smallest mass? (The answers are given at the end of the book.)
Background: Mass and energy are equivalent. A gain or loss of energy in any form is equivalent to a gain or loss of mass.
For similar questions (including calculational counterparts), consult Self-Assessment Test 28.2, which is described at the end of Section 28.7.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.