11.3
Mass and Energy: E mc 2
In the previous section, we found that the definition of momentum has to be modified
for relativistic speeds if we are to continue to have a conservation-of-momentum law.
It similarly turns out that if we are to continue to have a conservation-of-energy law,
we must revise the definition of energy.
When a force is applied to an object, work is done on it, increasing its kinetic energy.
In Section 4.2 you learned that
work done EK
EKf EKi
1
1
mvf2 mvi2
2
2
1
work done m(vf2 vi2)
2
We assumed that the object’s mass remained constant and that any energy transferred
to increase its kinetic energy resulted in an increase in speed only. Classically, we would
expect this relation to continue to be valid as long as the force continues to act—the
speed simply increases without bound. But this is not the case. As speeds become relativistic, the classical treatment of energy must be re-examined.
Einstein suggested that the total (relativistic) energy associated with an object of rest
mass m, moving at speed v relative to an inertial frame is
mc 2
Etotal v2
1 2
c
If the object happens to be at rest in this inertial frame, then v is zero and the total (relativistic) energy is simply
Erest mc 2
conservation of mass–energy
the principle that rest mass and
energy are equivalent
This is Einstein’s famous E mc 2. He proposed two things: rest mass is a form of energy
that is associated with all massive objects, and there might be forces or interactions in
nature that could transform mass into the more familiar types of energy or vice versa.
In classical mechanics, mass and energy are conserved separately. In special relativity,
these conservation laws are generalized to become one combined law, the conservation
of mass-energy.
Relativistic kinetic energy is the extra energy an object with mass has as a result of its
motion:
Etotal Erest EK
EK Etotal Erest
mc 2
EK mc 2
v2
1 c2
580 Chapter 11
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Section 11.3
For an object at rest in a given inertial frame (v 0), its total energy in that frame is
not zero but equals its rest energy, which is convertible to other forms of energy.
There are many instances when this conversion of mass into energy has been observed:
•
•
•
The energy E released by the fission of uranium in a nuclear reactor is accompanied by a decrease m in the rest mass of the reactants. Precise measurements
E
of both have shown that m = .
c2
0
When the p meson (a subatomic particle) decays, it completely disappears,
losing all of its rest mass. In its place appears electromagnetic radiation having an
equivalent energy. (We will return to the p 0 meson in Chapter 13.)
DID YOU
KNOW
?
Rest Mass and Energy Units
Physicists normally express the
rest masses of subatomic particles
in energy units. For example, the
rest mass me of an electron is
stated as 511 keV.
When hydrogen atoms react in the interior of the Sun, in a process called fusion,
the tremendous energy released, E, is accompanied by a corresponding loss of
E
In fact, the Sun is losing mass by this process
rest mass, m, given by m = .
c2
at a rate of 4.1 109 kg/s.
Only with the technological advances of the twentieth century did physicists begin
to examine the atomic and subatomic systems where matter-into-energy conversions
and energy-into-matter conversions occur. Previously, mass and energy were seen to be
conserved separately. It is believed that the equation E mc 2 applies to most processes,
although the changes are usually far too small to be measured.
For example, the heat of combustion of coal is approximately 3.2 107 J/kg. If 1.0 kg
of coal is burned, what change would the release of energy in light and warmth make in
the mass of the coal compared with the mass of the products of combustion? Using
E (m)c 2 for an original rest mass of 1.0 kg of coal:
E
m c2
3.2 107 J
(3.0 108 m/s)2
m 3.6 1010 kg
As a percentage of the coal’s original mass,
3.6 1010 kg
100% 0.000 000 036%
1.0 kg
As you can see, this is an insignificant loss in mass, not detectable with even the best
electronic balance.
However, this equation is a powerful theoretical tool that makes it possible to predict
the amount of energy available from any process that results in a decrease in mass. For
example, let us now calculate the energy available from the complete conversion of that
same 1.0 kg of coal. Again, using E (m)c 2 and m 1.0 kg:
E (m)c 2
(1.0 kg)(3.0 108 m/s)2
E 9.0 1016 J
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Einstein’s Special Theory of Relativity 581
This is the same amount of energy that results from the combustion of approximately
three billion kilograms of coal. It is no wonder that scientists are searching for methods
of converting mass into energy. Even in the best nuclear fission reactors, only a fraction of one percent of the reacting mass is converted to energy, but the energy yield is
enormous. One hope for solving the world’s energy problems is the discovery of methods
of converting some small fraction of normal mass into energy, without radioactive
byproducts.
SAMPLE problem 1
If the 0.50 kg mass of a ball at rest were totally converted to another form of energy, what
would the energy output be?
Solution
m 0.50 kg
E ?
E (m)c 2
(0.50 kg)(3.00 108 m/s)2
E 4.5 1016 J
The energy equivalent is 4.5 1016 J, or approximately half the energy emitted by the Sun
every second.
SAMPLE problem 2
In subatomic physics, it is convenient to use the electron volt, rather than the joule, as the
unit of energy. An electron moves at 0.860c in a laboratory. Calculate the electron’s rest
energy, total energy, and kinetic energy in the laboratory frame, in electron volts.
Solution
m 9.11 1031 kg
v 0.860c
Erest ?
Etotal ?
EK ?
Erest mc 2
(9.11 1031 kg)(3.00 108 m/s)2
Erest 8.199 1014 J
Converting to electron volts:
1.60 1019 J 1 eV
1 eV
(8.199 1014 J) 5.12 105 eV
1.60 1019 J
582 Chapter 11
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Section 11.3
This result is still a large number, so it is convenient to express the energy in mega electron volts (MeV), giving the result 0.512 MeV.
Reference books, however, show that the accepted rest energy of an electron is
0.511 MeV. Where did we go wrong? Our calculation assumed a value of 3.00 108 m/s for
c. As noted earlier, a more precise value for c is 2.997 924 58 108 m/s. Using this value for
c and taking the (measured) rest mass of an electron to be 9.109 389 1031 kg, we find
the rest energy of an electron to be 0.511 MeV, in agreement with the accepted value.
The total energy is given by the relationship
mc 2
Etotal v2
1 2
c
But
mc 2
0.511 MeV, therefore
0.511 MeV
Etotal (0.860c)2
1 c2
Etotal 1.00 MeV
The total energy of the electron is 1.00 MeV.
Etotal EK + Erest
EK Etotal Erest
1.00 MeV 0.511MeV
EK 0.489 MeV
The kinetic energy of the electron is 0.489 MeV.
Practice
Understanding Concepts
1. Calculate the rest energy of a proton in joules and in mega electron volts.
(mp 1.67 1027 kg)
2. The rest energy of a small object is 2.00 102 MJ. Calculate its rest mass.
3. A proton moves with a speed of 0.950c through a laboratory.
(a) Calculate the total energy of the proton in the laboratory frame.
(b) Calculate the kinetic energy of the proton in the laboratory frame.
Give your answers in mega electron volts.
4. An electron is at rest in an inertial frame. Calculate the work needed to accel-
1. 1.50 1010 J; 939 MeV
2. 2.22 109 kg
3. (a) 3.00 103 MeV
(b) 2.07 103 MeV
4. 4.99 1013 J
5. 1.09 102 kg
6. 2.31 1030 kg
erate it to a speed of 0.990c.
5. The total annual energy consumption of Canada is about 9.80 Answers
1018 J.
How
much mass would have to be totally converted to energy to meet this need?
6. The rest mass energies of a proton and a neutron are 938.3 MeV and 939.6 MeV,
respectively. What is the difference in their rest masses in kilograms?
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Einstein’s Special Theory of Relativity 583
SUMMARY
•
•
•
•
•
Mass and Energy: E mc 2
Rest mass is the mass of an object at rest.
Only the rest mass is used in relativistic calculations.
Einstein’s famous mass–energy equivalence equation is E mc 2.
mc 2
The total energy of a particle is given by Etotal .
v2
1 2
c
Rest mass is a form of energy that is convertible into other more common and
usable forms, for example, thermal energy.
Section 11.3 Questions
Understanding Concepts
1. Explain how the equation E 6. A hypothetical particle has a rest energy of 1.60 MeV and
mc 2
is consistent with the
law of conservation of mass–energy.
2. Calculate how much energy can be produced by the com-
plete annihilation of 1.0 kg of mass.
3. Earth, of mass 5.98 1024 kg, revolves around the Sun at
an average speed of 2.96 104 m/s. How much mass, if
converted into energy, could accelerate Earth from rest to
that speed?
4. A nuclear generating plant produces electric power at an
average rate of 5.0 GW, by converting a portion of the rest
mass of nuclear fuel. How much fuel is converted to energy
in one year, if the generation of electric power from the
annihilated mass is 100% efficient? (Hint: Recall that a
machine that delivers 1 W of power is delivering energy at
the rate of 1 J/s.)
5. Calculate the energy required to accelerate a proton from
rest to 0.90c. (mp 1.673 1027 kg)
584 Chapter 11
(in a certain inertial frame) a total energy of 3.20 MeV.
(a) Calculate its rest mass.
(b) Calculate its kinetic energy in the given frame.
Making Connections
7. How much would the rest energy in question 2 cost at the
typical utility price of $0.15/kWh?
8. It is reasonable to assume that 4 L of gasoline produces
1.05 108 J of energy and that this energy is sufficient to
operate a car for 30.0 km. An Aspirin tablet has a mass of
325 mg. If the Aspirin could be converted completely into
thermal energy, how many kilometres could the car go on a
single tablet?
9. Deuteron, a heavy hydrogen nucleus ( 21H), consists of a
proton and a neutron. Its rest energy is 1875.6 MeV. How
much energy is liberated, as kinetic energy and as a
gamma ray, when a deuteron is created from a separate
proton (a particle of rest mass energy 938.3 MeV) and a
neutron (a particle of rest mass energy 939.6 MeV)?
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