Special Relativity

94 Relativistic Energy

Learning Objectives

  • Calculate the total energy of an object moving at relativistic speeds.
  • Calculate the relativistic kinetic energy of a moving object.
  • Describe the concept of rest energy and explain how mass can be converted into other forms of energy.
  • Explain why no object with mass can be accelerated to the speed of light.
  • Describe the relationship between relativistic energy and relativistic momentum.
The National Spherical Torus Experiment (NSTX), an experimental fusion reactor used to investigate controlled nuclear fusion.
Figure 94.1. The National Spherical Torus Experiment (NSTX) was an experimental fusion reactor designed to study controlled nuclear fusion. During fusion, a very small amount of mass is converted into an enormous amount of energy according to Einstein's mass-energy relationship. Understanding this process requires the concepts of relativistic energy introduced in this chapter. (Credit: Princeton Plasma Physics Laboratory.)

One of the most famous equations in all of science is Einstein's equation

[latex]E=mc^2.[/latex]

This simple expression reveals a profound idea: mass itself is a form of energy. Although we often think of mass and energy as separate physical quantities, special relativity shows that they are fundamentally related and can be converted into one another under the appropriate conditions.

This principle explains the enormous amounts of energy released during nuclear reactions. In both nuclear fission and nuclear fusion, the total mass of the products is slightly smaller than the total mass of the original nuclei. The "missing" mass has not disappeared—it has been transformed into other forms of energy, including kinetic energy, electromagnetic radiation, and heat.

A tokamak, such as the experimental fusion reactor shown in Figure 94.1, is designed to harness this process by fusing light nuclei into heavier ones. If controlled fusion becomes practical on a large scale, it could provide a powerful source of clean energy while producing far less long-lived radioactive waste than conventional nuclear fission reactors.

The law of conservation of energy remains one of the most fundamental principles in physics. Classical physics teaches that energy cannot be created or destroyed, only transformed from one form into another. Special relativity extends this idea by recognizing that mass must also be included as part of the total energy of a system. As a result, the combined quantity of mass and energy is conserved in every inertial reference frame.

Like relativistic momentum, relativistic energy is defined so that the laws of physics remain identical for all inertial observers, satisfying Einstein's first postulate of special relativity. This revised definition successfully explains phenomena that classical physics cannot, including nuclear reactions, particle-antiparticle annihilation, and the tremendous energy produced inside stars.

Healthcare Connection

The conversion of mass into energy is fundamental to several medical technologies. Radioisotopes used in nuclear medicine undergo radioactive decay, releasing energy that can be detected to produce diagnostic images in procedures such as positron emission tomography (PET). Likewise, radiation therapy relies on energetic particles and photons whose behavior is accurately described only by relativistic energy and momentum. Although the amount of mass converted during these processes is extraordinarily small, the released energy is sufficient for both medical imaging and cancer treatment.

In this chapter, we develop the relativistic expressions for total energy and kinetic energy, explore the meaning of rest energy, and examine why no object with mass can ever reach the speed of light. We will also discover how energy and momentum are linked by one of the most important equations in modern physics, providing the foundation for understanding nuclear reactions, particle physics, and many applications in medicine and astrophysics.

Total Energy and Rest Energy

One of Einstein's most important achievements was showing that the law of conservation of energy remains valid in special relativity, provided that energy is defined differently than in classical physics. Just as momentum required a relativistic correction, energy must also include the effects of motion at speeds approaching the speed of light.

Einstein's first postulate states that the laws of physics are identical in every inertial reference frame. Therefore, the law of conservation of energy must hold for every observer, regardless of their state of motion. To satisfy this requirement, the total energy of a moving object includes a relativistic factor known as the Lorentz factor.

Total Energy

The total relativistic energy of an object is

[latex]E=\gamma mc^2,[/latex]

where

  • [latex]m[/latex] is the object's rest mass,
  • [latex]c[/latex] is the speed of light in a vacuum,
  • [latex]v[/latex] is the object's speed relative to the observer, and
  • [latex]\gamma[/latex] is the Lorentz factor,
    [latex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.[/latex]

Total energy includes both the object's energy of motion and the energy associated with its mass. As the object's speed increases, the Lorentz factor becomes larger, causing the total energy to increase rapidly.

Unlike classical mechanics, the total energy of an object is not zero when the object is at rest. When an object is stationary relative to an observer, its velocity is zero, so

[latex]v=0 \qquad\Rightarrow\qquad \gamma=1.[/latex]

Substituting this into the total energy equation gives a remarkable result: even a stationary object possesses energy.

Rest Energy

The energy associated with an object's mass while it is at rest is called its rest energy.

[latex]E_0=mc^2.[/latex]

This equation is the correct form of Einstein's famous mass-energy relationship. It demonstrates that every object possesses an enormous amount of energy simply because it has mass, even if it is completely motionless.

The implications of this equation are profound. If energy is added to an object—for example, by compressing a spring, charging a battery, or heating a material—the object's total energy increases. According to special relativity, this also means its mass increases slightly. Conversely, if an object releases energy, its mass decreases by a corresponding amount. Although these changes are usually far too small to measure in everyday situations, they become measurable during nuclear reactions.

Today, mass-energy equivalence forms the foundation of nuclear physics, particle physics, astrophysics, and many medical technologies. It explains the energy produced by stars, nuclear power plants, radioactive decay, and particle-antiparticle annihilation.

Healthcare Connection

Mass-energy conversion plays an essential role in nuclear medicine. Radioactive isotopes used in PET imaging and cancer therapy release energy because tiny amounts of mass are converted into radiation during nuclear decay. Although the mass difference is extraordinarily small, Einstein's equation predicts that even these minute changes produce enough energy for medical imaging and therapeutic applications.

Example 94.1: Calculating Rest Energy of a Small Mass

Problem

Calculate the rest energy of a mass of 1.00 g.

Strategy

Although one gram is a very small amount of matter, Einstein's equation shows that even tiny masses contain enormous amounts of stored energy. First convert the mass to kilograms, then substitute it into the rest-energy equation.

Solution

  1. Identify the known quantities.
    • [latex]m=1.00\times10^{-3}\ \text{kg}[/latex]
    • [latex]c=3.00\times10^8\ \text{m/s}[/latex]
  2. Identify the unknown.[latex]E_0[/latex]
  3. Choose the appropriate equation.
    [latex]E_0=mc^2[/latex]
  4. Substitute the known values.
    [latex]\begin{aligned} E_0 &=(1.00\times10^{-3}\,\text{kg}) (3.00\times10^8\,\text{m/s})^2\\ &=9.00\times10^{13}\, \text{kg}\cdot\text{m}^2/\text{s}^2 \end{aligned}[/latex]
  5. Express the answer in joules.Since
    [latex]1\ \text{kg}\cdot\text{m}^2/\text{s}^2=1\ \text{J},[/latex]

    the rest energy is

    [latex]E_0=9.00\times10^{13}\ \text{J}.[/latex]

Discussion

This result illustrates one of the most astonishing predictions of modern physics: an extremely small amount of matter contains an enormous amount of energy. The rest energy of just one gram of matter is approximately [latex]9\times10^{13}[/latex] joules, comparable to roughly twice the energy released by the Hiroshima atomic bomb.

Fortunately, this energy is not normally accessible. In everyday chemical reactions, such as burning fuel or charging a battery, only tiny fractions of the total rest energy are involved. Significant conversion of mass into usable energy occurs primarily in nuclear reactions, such as fission inside nuclear power plants and fusion inside the Sun.

The enormous value of [latex]mc^2[/latex] arises because the speed of light is extremely large, and squaring it produces an even larger number. Consequently, even a very small amount of mass corresponds to a tremendous quantity of energy.

Today, the practical applications of mass-energy conversion are well established. Nuclear power plants generate electricity by converting a tiny fraction of nuclear mass into energy through fission, while nuclear fusion powers the Sun and other stars. Although the same physical principles also underlie nuclear weapons, the focus here is on the scientific understanding of how mass and energy are related and on the many peaceful applications that have emerged from Einstein's theory.

When Einstein proposed the relationship between mass and energy in the early twentieth century, its implications were revolutionary. Radioactivity had been discovered only a few years earlier, and scientists were puzzled by the tremendous amount of energy emitted by certain radioactive materials. Einstein suggested that this energy originated from the conversion of a very small amount of mass into radiation during nuclear processes. Because the mass change is extraordinarily small, it was beyond the experimental capabilities of the time to measure directly.

Over the following decades, increasingly precise experiments confirmed Einstein's prediction. Today we know that every nuclear reaction—including radioactive decay, nuclear fission, and nuclear fusion—involves small changes in mass that are accompanied by corresponding changes in energy, exactly as predicted by the equation

[latex]E=mc^2.[/latex]
The Sun produces energy through nuclear fusion, while a nuclear power plant produces electricity through nuclear fission. Both processes convert a small amount of mass into energy.
Figure 94.2. Both the Sun (a) and nuclear power plants (b) obtain energy by converting a small fraction of mass into energy. In the Sun, hydrogen nuclei fuse to form helium, whereas nuclear reactors release energy through the fission of heavy nuclei such as uranium. Although the nuclear processes are different, both obey Einstein's mass-energy relationship. (Credits: (a) NASA/Goddard Space Flight Center, Scientific Visualization Studio; (b) U.S. Government.)

Today, physicists view mass as one form of energy rather than as a completely separate physical quantity. This insight transformed our understanding of nature. It explains how stars shine for billions of years, why radioactive materials emit energetic particles, why Earth's interior remains hot, and how nuclear technologies can produce such enormous amounts of energy from relatively small amounts of matter.

Stored Energy and Potential Energy

Mass-energy equivalence applies not only to nuclear reactions but also to every form of stored energy. Whenever energy is added to an object—whether by charging a battery, compressing a spring, stretching an elastic band, or heating a material—that energy becomes part of the object's total energy. Consequently, the object's rest mass also increases, although by an extremely small amount.

This idea may seem surprising because, in everyday life, we usually treat mass as constant. Classical physics even regarded the conservation of mass as a fundamental law. Special relativity shows that this is actually an approximation that works extremely well whenever the energy changes are small compared with the enormous value of [latex]mc^2[/latex].

Why don't we notice these changes in mass? The answer is that the increase is far too small to measure with ordinary instruments. The following example illustrates just how tiny the effect is for an everyday object.

Example 94.2: How Charging a Battery Changes Its Mass

Problem

A car battery can deliver 600 A·h of charge at a potential difference of 12.0 V.

  1. Calculate the increase in the battery's rest mass when it is charged from completely discharged to fully charged.
  2. Determine the percentage increase in mass if the battery has a mass of 20.0 kg.

Strategy

The electrical energy stored in the battery is equal to the electrical potential energy it can later deliver:

[latex]PE_{\rm elec}=qV.[/latex]

The total charge is obtained from

[latex]q=It.[/latex]

Once the stored energy is known, Einstein's equation relates that energy to the increase in mass:

[latex]\Delta E=\Delta mc^2.[/latex]

Solution (a)

  1. Known quantities
    • [latex]It=600\ \text{A}\cdot\text{h}[/latex]
    • [latex]V=12.0\ \text{V}[/latex]
    • [latex]c=3.00\times10^8\ \text{m/s}[/latex]
  2. Convert ampere-hours to coulombs.
    [latex]q=(600\ \text{A}\cdot\text{h}) \left(\frac{3600\ \text{s}}{1\ \text{h}}\right) =2.16\times10^6\ \text{C}[/latex]
  3. Calculate the stored electrical energy.
    [latex]PE=qV =(2.16\times10^6)(12.0) =2.592\times10^7\ \text{J}[/latex]
  4. Calculate the corresponding increase in mass.
    [latex]\Delta m=\frac{PE}{c^2} =\frac{2.592\times10^7}{(3.00\times10^8)^2} =2.88\times10^{-10}\ \text{kg}[/latex]

Solution (b)

[latex]\begin{aligned} \%\ \text{increase} &=\frac{\Delta m}{m}\times100\%\\ &=\frac{2.88\times10^{-10}} {20.0}\times100\%\\ &=1.44\times10^{-9}\% \end{aligned}[/latex]

Discussion

The battery becomes slightly heavier when it is charged because the stored electrical energy contributes to its total mass. However, the increase is only about three ten-billionths of a kilogram, corresponding to less than one-billionth of one percent of the battery's total mass.

This change is far too small to detect with ordinary laboratory balances, which explains why mass appears to remain constant in everyday situations. Only in nuclear reactions, where much larger amounts of energy are released, do the associated changes in mass become large enough to measure directly.

Healthcare Connection

Medical devices such as implantable pacemakers, infusion pumps, and portable defibrillators store electrical energy in batteries. According to special relativity, a fully charged battery has a slightly greater mass than the same battery after it has been discharged. Although the difference is far too small to affect clinical practice or to be measured with ordinary instruments, it illustrates that all forms of stored energy—including electrical, chemical, and mechanical energy—contribute to the total mass of a system.

Kinetic Energy and the Ultimate Speed Limit

Kinetic energy is the energy associated with motion. In classical mechanics, it is given by the familiar expression

[latex]KE_{\text{class}}=\frac{1}{2}mv^2.[/latex]

This equation accurately describes objects moving much slower than the speed of light. However, as an object's speed becomes a significant fraction of [latex]c[/latex], classical mechanics no longer provides accurate predictions. To describe high-speed motion correctly, we must use the relativistic form of kinetic energy.

Relativistic kinetic energy can be derived from the work-energy theorem, which states that the net work done on an object equals the increase in its kinetic energy. If an object begins from rest, then

[latex]W_{\rm net}=KE.[/latex]

Initially, the object possesses only its rest energy,

[latex]E_0=mc^2.[/latex]

After work is done on the object, its total energy becomes

[latex]E=\gamma mc^2.[/latex]

The work performed therefore equals the increase in total energy:

[latex]\begin{aligned} W_{\rm net} &=E-E_0\\ &=\gamma mc^2-mc^2\\ &=(\gamma-1)mc^2. \end{aligned}[/latex]

Since the work done on the object becomes its kinetic energy, we obtain the relativistic expression for kinetic energy.

Relativistic Kinetic Energy

The kinetic energy of an object moving at relativistic speeds is

[latex]KE_{\rm rel}=(\gamma-1)mc^2.[/latex]

This equation behaves exactly as we expect when the object is at rest. If

[latex]v=0,[/latex]

then

[latex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=1,[/latex]

and therefore

[latex]KE_{\rm rel}=0.[/latex]

Although the relativistic expression looks very different from the classical equation, it actually reduces to the familiar result at low speeds. When

[latex]v\ll c,[/latex]

the Lorentz factor can be approximated using a binomial expansion:

[latex]\gamma\approx1+\frac12\frac{v^2}{c^2}.[/latex]

Subtracting one from both sides gives

[latex]\gamma-1\approx\frac12\frac{v^2}{c^2}.[/latex]

Substituting this approximation into the relativistic kinetic energy equation yields

[latex]\begin{aligned} KE_{\rm rel} &=\left(\frac12\frac{v^2}{c^2}\right)mc^2\\ &=\frac12mv^2\\ &=KE_{\rm class}. \end{aligned}[/latex]

This result is reassuring because it demonstrates that special relativity contains classical mechanics as a limiting case. Whenever an object's speed is much smaller than the speed of light, relativistic effects become negligible, and the familiar Newtonian equations remain valid.

The behavior at very high speeds is far more remarkable. As an object's velocity approaches the speed of light, the Lorentz factor grows without bound. Consequently, the relativistic kinetic energy also increases without limit.

In other words, every additional increase in speed requires more energy than the previous one. Near the speed of light, enormous amounts of energy produce only tiny increases in velocity.

This leads to one of the most important conclusions of special relativity:

The Ultimate Speed Limit

No object with mass can ever be accelerated to the speed of light.

As an object's speed approaches [latex]c[/latex], its relativistic kinetic energy approaches infinity. Reaching the speed of light would therefore require an infinite amount of work and an infinite supply of energy—both of which are physically impossible.

This conclusion has been confirmed repeatedly by experiments using modern particle accelerators. Electrons and protons can be accelerated to speeds extremely close to the speed of light, often exceeding 99.9999% of [latex]c[/latex], but they never actually reach it. Instead, the additional energy supplied by the accelerator increases the particles' relativistic energy far more than it increases their speed.

Healthcare Connection

Medical linear accelerators (LINACs) used in radiation therapy routinely accelerate electrons to relativistic speeds before they strike a target and produce high-energy X-rays for cancer treatment. Engineers must use relativistic equations—not classical ones—to calculate the particles' kinetic energy accurately. If classical mechanics were used instead, the energy of the therapeutic radiation beam would be significantly underestimated, leading to incorrect dose calculations.

The Speed of Light

No object with mass can attain the speed of light.

The speed of light is therefore the ultimate speed limit for every particle or object that has mass. No matter how much energy is supplied, its speed can only approach [latex]c[/latex]; it can never equal or exceed it.

This conclusion is consistent with relativistic velocity addition. Even when two speeds are combined, the resulting speed remains less than [latex]c[/latex]. Both the relativistic kinetic-energy equation and the existence of this universal speed limit have been confirmed in many experiments.

As an object's speed becomes closer to the speed of light, additional energy produces progressively smaller increases in speed. The object's kinetic energy continues to rise, but its velocity approaches [latex]c[/latex] only asymptotically.

Graph comparing relativistic and classical kinetic energy as speed increases. The two curves are nearly identical at low speeds, but relativistic kinetic energy rises much more steeply and approaches infinity as speed approaches the speed of light.
Figure 94.3. Relativistic and classical kinetic energy agree closely at low speeds. At high speeds, however, the relativistic kinetic energy increases much more rapidly. As [latex]v[/latex] approaches [latex]c[/latex], the relativistic kinetic energy approaches infinity, showing why an object with mass cannot reach the speed of light.

Example 94.3: Comparing Relativistic and Classical Kinetic Energy

Problem

An electron moves at a speed of

[latex]v=0.990c.[/latex]
  1. Calculate its relativistic kinetic energy in MeV.
  2. Calculate the classical kinetic energy at the same speed and compare the two results.

Use an electron mass of

[latex]m=9.11\times10^{-31}\ \text{kg}.[/latex]

Strategy

Because the electron is moving at 99.0% of the speed of light, its motion is strongly relativistic. We must therefore use

[latex]KE_{\rm rel}=(\gamma-1)mc^2.[/latex]

We first calculate the Lorentz factor [latex]\gamma[/latex]. We then calculate the classical result using

[latex]KE_{\rm class}=\frac12mv^2[/latex]

to determine how seriously classical physics underestimates the electron's kinetic energy.

Solution (a): Relativistic Kinetic Energy

  1. Calculate the Lorentz factor.
    [latex]\begin{aligned} \gamma &=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ &=\frac{1}{\sqrt{1-(0.990)^2}}\\ &=7.0888 \end{aligned}[/latex]
  2. Calculate the relativistic kinetic energy.
    [latex]\begin{aligned} KE_{\rm rel} &=(\gamma-1)mc^2\\ &=(7.0888-1) (9.11\times10^{-31}\,\text{kg}) (3.00\times10^8\,\text{m/s})^2\\ &=4.99\times10^{-13}\ \text{J} \end{aligned}[/latex]
  3. Convert joules to megaelectron volts.Using
    [latex]1\ \text{MeV}=1.60\times10^{-13}\ \text{J},[/latex]

    we obtain

    [latex]\begin{aligned} KE_{\rm rel} &=(4.99\times10^{-13}\,\text{J}) \left( \frac{1\ \text{MeV}} {1.60\times10^{-13}\,\text{J}} \right)\\ &=3.12\ \text{MeV}. \end{aligned}[/latex]

Solution (b): Classical Kinetic Energy

[latex]\begin{aligned} KE_{\rm class} &=\frac12mv^2\\ &=\frac12 (9.11\times10^{-31}\,\text{kg}) (0.990)^2 (3.00\times10^8\,\text{m/s})^2\\ &=4.02\times10^{-14}\ \text{J} \end{aligned}[/latex]

Converting to MeV gives

[latex]\begin{aligned} KE_{\rm class} &=(4.02\times10^{-14}\,\text{J}) \left( \frac{1\ \text{MeV}} {1.60\times10^{-13}\,\text{J}} \right)\\ &=0.251\ \text{MeV}. \end{aligned}[/latex]

Comparison

[latex]\frac{KE_{\rm rel}}{KE_{\rm class}} = \frac{3.12}{0.251} =12.4.[/latex]

Discussion

The classical calculation underestimates the electron's kinetic energy by more than a factor of 12. At a speed of [latex]0.990c[/latex], the classical approximation is no longer valid.

This example also illustrates why it becomes increasingly difficult to accelerate a particle as its speed approaches [latex]c[/latex]. The accelerator continues to transfer energy to the particle, but that energy produces only a small additional increase in speed. Modern particle accelerators can easily give electrons energies of several MeV, and much larger facilities can produce electron energies of many GeV.

It is sometimes said that the extra energy “increases the mass” of the particle. In modern treatments, it is clearer to keep the particle's rest mass constant and describe the increase in terms of total energy, kinetic energy, and relativistic momentum.

Accelerating particles to speeds even closer to [latex]c[/latex] remains scientifically valuable. At very high energies, collisions can convert kinetic energy into new particles. By studying the particles created in these collisions, physicists investigate the internal structure of matter and the fundamental interactions of nature.

This principle is also important in medical physics. High-energy electrons, protons, and heavier ions are accelerated and directed toward tissue in radiation therapy. Their energies determine how deeply they penetrate, how they interact with matter, and how much energy they deposit in a tumor.

Healthcare Connection

Clinical radiation-therapy systems use particle energies ranging from a few MeV for electrons to hundreds of MeV for protons. At these energies, especially for electrons, relativistic calculations are essential. The kinetic energy determines the radiation's penetration depth and contributes to the dose delivered to tissue, so accurate energy calculations are central to safe treatment planning.

Aerial view of the Fermi National Accelerator Laboratory showing the large circular accelerator ring used to accelerate charged particles to extremely high energies.
Figure 94.4. The Fermi National Accelerator Laboratory (Fermilab), located near Batavia, Illinois, operated one of the world's largest particle accelerators. The Tevatron accelerated protons and antiprotons to energies approaching 1 TeV (one trillion electron volts), allowing physicists to study the fundamental building blocks of matter. Although the Tevatron ceased operation in 2011, particle accelerators remain indispensable tools for modern physics and medicine. (Credit: Fermilab, Reidar Hahn.)

Relativistic Energy and Momentum

In classical mechanics, momentum and kinetic energy are closely related. Since the momentum of an object is

[latex]p=mv,[/latex]

its kinetic energy can also be written as

[latex]KE_{\rm class} =\frac{p^2}{2m} =\frac{(mv)^2}{2m} =\frac12mv^2.[/latex]

Special relativity preserves the connection between energy and momentum, but the relationship becomes more general. Combining the relativistic definitions of total energy and momentum leads to one of the most important equations in modern physics:

[latex]E^2=(pc)^2+(mc^2)^2.[/latex]

This equation links three fundamental properties of a particle:

  • [latex]E[/latex], its total relativistic energy,
  • [latex]p[/latex], its relativistic momentum, and
  • [latex]m[/latex], its rest mass.

Unlike the classical expression, this equation is valid for every particle, regardless of its speed.

Notice what happens when an object is at rest. Since its momentum is zero,

[latex]p=0,[/latex]

the energy equation becomes

[latex]E=mc^2,[/latex]

which is precisely the expression for rest energy introduced earlier in this chapter.

As the object accelerates, its momentum increases, causing its total energy to increase as well. At extremely high speeds, the momentum term becomes much larger than the rest-energy term, so that

[latex](pc)^2\gg(mc^2)^2.[/latex]

In this limit, the total energy is well approximated by

[latex]E\approx pc.[/latex]

This approximation is commonly used when describing highly energetic particles produced in cosmic rays and particle accelerators.

Massless Particles

The energy-momentum relationship also explains the behavior of particles with zero rest mass. Setting

[latex]m=0[/latex]

reduces the equation to

[latex]E=pc,[/latex]

or equivalently,

[latex]p=\frac{E}{c}.[/latex]

This relationship applies to photons, the particles that make up electromagnetic radiation, including visible light, X-rays, and gamma rays. Although photons have no rest mass, they still carry both energy and momentum.

Special relativity further predicts that every massless particle must travel at exactly the speed of light. Unlike particles with mass, photons cannot travel slower or faster than [latex]c[/latex]; their speed is always the same in a vacuum.

Healthcare Connection

The equation [latex]E=pc[/latex] is fundamental to medical imaging and radiation therapy. X-rays and gamma rays used in diagnostic imaging, nuclear medicine, and cancer treatment are photons. Although they have no rest mass, they carry energy and momentum that determine how they interact with human tissue and deposit radiation dose.

Problem-Solving Strategy for Relativity

  1. Determine whether relativistic effects are important. Calculate or estimate the Lorentz factor,
    [latex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.[/latex]

    If [latex]\gamma\approx1[/latex], classical equations are usually sufficient.

  2. Identify the unknown quantity. Decide whether you are solving for energy, momentum, velocity, time, length, or another relativistic quantity.
  3. List the known information. Pay particular attention to the object's speed relative to the observer.
  4. Understand the physical situation before calculating. Determine which reference frame is being used and which observer measures quantities such as proper time or proper length.
  5. Select the appropriate relativistic equation. Use the chapter summary to identify the correct relationship for the problem.
  6. Avoid excessive rounding during intermediate calculations. Relativistic quantities are often sensitive to small numerical differences.
  7. Evaluate whether the answer is physically reasonable. Check that no object with mass exceeds the speed of light and that the result is consistent with the physical situation.

Check Your Understanding

A photon produces an electron-positron pair. If the electron moves at a speed of [latex]0.992c[/latex], what is its relativistic kinetic energy?

Answer

First calculate the Lorentz factor:

[latex]\gamma= \frac{1}{\sqrt{1-(0.992)^2}} =7.91.[/latex]

Then use the relativistic kinetic energy equation:

[latex]\begin{aligned} KE_{\rm rel} &=(\gamma-1)mc^2\\ &=(7.91-1) (9.11\times10^{-31}\,\text{kg}) (3.00\times10^8\,\text{m/s})^2\\ &=5.67\times10^{-13}\ \text{J}. \end{aligned}[/latex]

Section Summary

  • Special relativity extends the law of conservation of energy by recognizing that mass and energy are interchangeable.
  • The total relativistic energy of an object is
    [latex]E=\gamma mc^2.[/latex]
  • An object at rest possesses rest energy,
    [latex]E_0=mc^2,[/latex]

    demonstrating that mass itself is a form of stored energy.

  • Adding energy to a system increases its mass slightly, while releasing energy decreases its mass. These changes are usually too small to measure except in nuclear processes.
  • The relativistic kinetic energy is
    [latex]KE_{\rm rel}=(\gamma-1)mc^2.[/latex]

    At low speeds, this expression reduces to the classical equation [latex]\frac12mv^2[/latex].

  • No object with mass can reach the speed of light because doing so would require an infinite amount of energy.
  • Total relativistic energy and momentum are related by
    [latex]E^2=(pc)^2+(mc^2)^2.[/latex]
  • For highly relativistic particles,
    [latex]E\approx pc.[/latex]
  • For massless particles such as photons,
    [latex]E=pc,[/latex]

    and they always travel at the speed of light.

Conceptual Questions

  1. How does special relativity modify the classical laws of conservation of mass and conservation of energy?
  2. Suppose a sealed pot of water cools without gaining or losing any molecules. What happens to the mass of the water as it cools? Would this change be measurable in practice? Explain your reasoning.
  3. Consider the following thought experiment. An inflated balloon is placed on a highly sensitive scale outdoors early in the morning and remains there throughout the day. Assuming no air enters or leaves the balloon, would its measured mass change as the temperature changes? Discuss the physical reasons and the practical challenges involved in making this measurement.
  4. The fuel in a nuclear reactor loses a measurable amount of mass as energy is released. Does the combined mass of coal and oxygen decrease when coal is burned in a conventional power plant? If so, why is this effect much more difficult to observe?
  5. Objects with mass cannot travel faster than the speed of light. Does a similar upper limit exist for an object's momentum or its total energy? Explain.
  6. Light always travels at the speed of light in a vacuum. Does this mean light has mass? Explain your answer using the concepts introduced in this chapter.
  7. An Earth-based telescope projects a laser beam onto the Moon. By rotating the telescope, the illuminated spot can sweep across the Moon's surface faster than the speed of light. Does this violate special relativity? Explain why or why not. (Hint: The light itself travels from Earth to the Moon, not across the Moon's surface.)

Problems

  1. What is the rest energy of an electron, given its mass is [latex]9.11\times10^{-31}\ \text{kg}[/latex]? Give your answer in joules and MeV.
  2. Find the rest energy in joules and MeV of a proton, given its mass is [latex]1.67\times10^{-27}\ \text{kg}[/latex].
  3. If the rest energies of a proton and a neutron (the two constituents of nuclei) are 938.3 and 939.6 MeV respectively, what is the difference in their masses in kilograms?
  4. The Big Bang that began the universe is estimated to have released [latex]10^{68}\ \text{J}[/latex] of energy. How many stars could half this energy create, assuming the average star’s mass is [latex]4.00\times10^{30}\ \text{kg}[/latex]?
  5. A supernova explosion of a [latex]2.00\times10^{31}\ \text{kg}[/latex] star produces [latex]1.00\times10^{44}\ \text{J}[/latex] of energy. (a) How many kilograms of mass are converted to energy in the explosion? (b) What is the ratio [latex]\Delta m/m[/latex] of mass destroyed to the original mass of the star?
  6. (a) Using data from Table 7.1, calculate the mass converted to energy by the fission of 1.00 kg of uranium. (b) What is the ratio of mass destroyed to the original mass, [latex]\Delta m/m[/latex]?
  7. (a) Using data from Table 7.1, calculate the amount of mass converted to energy by the fusion of 1.00 kg of hydrogen. (b) What is the ratio of mass destroyed to the original mass, [latex]\Delta m/m[/latex]? (c) How does this compare with [latex]\Delta m/m[/latex] for the fission of 1.00 kg of uranium?
  8. There is approximately [latex]10^{34}\ \text{J}[/latex] of energy available from fusion of hydrogen in the world’s oceans. (a) If [latex]10^{33}\ \text{J}[/latex] of this energy were utilized, what would be the decrease in mass of the oceans? Assume that 0.08% of the mass of a water molecule is converted to energy during the fusion of hydrogen. (b) How great a volume of water does this correspond to? (c) Comment on whether this is a significant fraction of the total mass of the oceans.
  9. A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. (a) If all the lost mass is converted into the electron’s kinetic energy, find [latex]\gamma[/latex] for the electron. (b) What is the electron’s velocity?
  10. A [latex]\pi[/latex]-meson is a particle that decays into a muon and a massless particle. The [latex]\pi[/latex]-meson has a rest mass energy of 139.6 MeV, and the muon has a rest mass energy of 105.7 MeV. Suppose the [latex]\pi[/latex]-meson is at rest and all of the missing mass goes into the muon’s kinetic energy. How fast will the muon move?
  11. (a) Calculate the relativistic kinetic energy of a 1000-kg car moving at 30.0 m/s if the speed of light were only 45.0 m/s. (b) Find the ratio of the relativistic kinetic energy to classical.
  12. Alpha decay is nuclear decay in which a helium nucleus is emitted. If the helium nucleus has a mass of [latex]6.80\times10^{-27}\ \text{kg}[/latex] and is given 5.00 MeV of kinetic energy, what is its velocity?
  13. (a) Beta decay is nuclear decay in which an electron is emitted. If the electron is given 0.750 MeV of kinetic energy, what is its velocity? (b) Comment on how the high velocity is consistent with the kinetic energy as it compares to the rest mass energy of the electron.
  14. A positron is an antimatter version of the electron, having exactly the same mass. When a positron and an electron meet, they annihilate, converting all of their mass into energy. (a) Find the energy released, assuming negligible kinetic energy before the annihilation. (b) If this energy is given to a proton in the form of kinetic energy, what is its velocity? (c) If this energy is given to another electron in the form of kinetic energy, what is its velocity?
  15. What is the kinetic energy in MeV of a [latex]\pi[/latex]-meson that lives [latex]1.40\times10^{-16}\ \text{s}[/latex] as measured in the laboratory, and [latex]0.840\times10^{-16}\ \text{s}[/latex] when at rest relative to an observer, given that its rest energy is 135 MeV?
  16. Find the kinetic energy in MeV of a neutron with a measured life span of 2065 s, given its rest energy is 939.6 MeV, and rest life span is 900 s.
  17. (a) Show that [latex]\frac{(pc)^2}{(mc^2)^2}=\gamma^2-1[/latex]. This means that at large velocities [latex]pc\gg mc^2[/latex]. (b) Is [latex]E\approx pc[/latex] when [latex]\gamma=30.0[/latex], as for the astronaut discussed in the twin paradox?
  18. One cosmic ray neutron has a velocity of [latex]0.250c[/latex] relative to the Earth. (a) What is the neutron’s total energy in MeV? (b) Find its momentum. (c) Is [latex]E\approx pc[/latex] in this situation? Discuss in terms of the equation given in part (a) of the previous problem.
  19. What is [latex]\gamma[/latex] for a proton having a mass energy of 938.3 MeV accelerated through an effective potential of 1.0 TV (teravolt) at Fermilab outside Chicago?
  20. (a) What is the effective accelerating potential for electrons at the Stanford Linear Accelerator, if [latex]\gamma=1.00\times10^5[/latex] for them? (b) What is their total energy (nearly the same as kinetic in this case) in GeV?
  21. (a) Using data from Table 7.1, find the mass destroyed when the energy in a barrel of crude oil is released. (b) Given these barrels contain 200 liters and assuming the density of crude oil is [latex]750\ \text{kg/m}^3[/latex], what is the ratio of mass destroyed to original mass, [latex]\Delta m/m[/latex]?
  22. (a) Calculate the energy released by the destruction of 1.00 kg of mass. (b) How many kilograms could be lifted to a 10.0 km height by this amount of energy?
  23. A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?
  24. Suppose you use an average of [latex]500\ \text{kW}\cdot\text{h}[/latex] of electric energy per month in your home. (a) How long would 1.00 g of mass converted to electric energy with an efficiency of 38.0% last you? (b) How many homes could be supplied at the [latex]500\ \text{kW}\cdot\text{h}[/latex] per month rate for one year by the energy from the described mass conversion?
  25. (a) A nuclear power plant converts energy from nuclear fission into electricity with an efficiency of 35.0%. How much mass is destroyed in one year to produce a continuous 1000 MW of electric power? (b) Do you think it would be possible to observe this mass loss if the total mass of the fuel is [latex]10^4\ \text{kg}[/latex]?
  26. Nuclear-powered rockets were researched for some years before safety concerns became paramount. (a) What fraction of a rocket’s mass would have to be destroyed to get it into a low Earth orbit, neglecting the decrease in gravity? (Assume an orbital altitude of 250 km, and calculate both the kinetic energy (classical) and the gravitational potential energy needed.) (b) If the ship has a mass of [latex]1.00\times10^5\ \text{kg}[/latex] (100 tons), what total yield nuclear explosion in tons of TNT is needed?
  27. The Sun produces energy at a rate of [latex]4.00\times10^{26}\ \text{W}[/latex] by the fusion of hydrogen. (a) How many kilograms of hydrogen undergo fusion each second? (b) If the Sun is 90.0% hydrogen and half of this can undergo fusion before the Sun changes character, how long could it produce energy at its current rate? (c) How many kilograms of mass is the Sun losing per second? (d) What fraction of its mass will it have lost in the time found in part (b)?
  28. Unreasonable Results
    A proton has a mass of [latex]1.67\times10^{-27}\ \text{kg}[/latex]. A physicist measures the proton’s total energy to be 50.0 MeV. (a) What is the proton’s kinetic energy? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
  29. Construct Your Own Problem
    Consider a highly relativistic particle. Discuss what is meant by the term “highly relativistic.” (Note that, in part, it means that the particle cannot be massless.) Construct a problem in which you calculate the wavelength of such a particle and show that it is very nearly the same as the wavelength of a massless particle, such as a photon, with the same energy. Among the things to be considered are the rest energy of the particle (it should be a known particle) and its total energy, which should be large compared to its rest energy.
  30. Construct Your Own Problem
    Consider an astronaut traveling to another star at a relativistic velocity. Construct a problem in which you calculate the time for the trip as observed on the Earth and as observed by the astronaut. Also calculate the amount of mass that must be converted to energy to get the astronaut and ship to the velocity travelled. Among the things to be considered are the distance to the star, the velocity, and the mass of the astronaut and ship. Unless your instructor directs you otherwise, do not include any energy given to other masses, such as rocket propellants.
  31. Critical Thinking
    A space rock with a length of 1,000.0 m is moving through space at exactly 0.6 c. (a) If the space rock is moving toward an observer, what is the contracted length observed? (b) If the space rock is moving away from the observer, what is the contracted length observed? (c) Can the object reach the speed of light? (d) If the object were to stop in the observer’s reference frame, would it be observed to have proper length?

Glossary

Total energy
The total energy of an object moving at any speed, including both its rest energy and its kinetic energy. It is given by

[latex]E=\gamma mc^2,[/latex]

where

[latex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.[/latex]
Rest energy
The energy associated with an object's mass when it is at rest. It is given by

[latex]E_0=mc^2.[/latex]
Relativistic kinetic energy
The kinetic energy of an object moving at relativistic speeds. It is given by

[latex]KE_{\rm rel}=(\gamma-1)mc^2,[/latex]

where

[latex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.[/latex]
definition

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.