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Physics 25 Chapter 28 Special Relativity Dr. Alward
Video Lecture 1: Mass-Energy Equivalence
Video Lecture 2: Relativistic Energies
Video Lecture 3: Time Dilation
Video Lecture 4: Length Contraction
Video Lecture 5: Relative Velocity
Mass-Energy Equivalence
According to Albert Einstein’s Special Theory of Relativity, energy added to an object
causes its mass to increase; similarly, all or part of the mass of an object can undergo a
transformation into energy (heat, light, sound, potential energy). This “equivalence” of
mass and energy is expressed in the following equation:
Example:
Gain in energy causes a gain in mass, as the
example below illustrates:
600 J of light energy Q is absorbed by a 10-
kg object. What is the new mass of the
object?
Δm = Q /c2
= 600 /(3.0 x 108)2
= 6.67 x 10-15 kg
m = 10.000 000 000 000 067 kg
Δm = Q /c2
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The most important example of a conversion of mass into energy occurs in
the interior of the sun where two different isotopes of hydrogen,
“deuterium” (1H2) and “tritium” (1H
3), combine (“fuse”) to form helium.
The mass of the product (the helium nucleus) is less
than the mass of the reactants (the hydrogen
isotopes) by the amount 3 x 10-29 kg:
Δm = -3 x 10-29 kg
Q = (Δm) c2
= (-3 x 10-29) (3 x 108)2
= -2.70 x 10-12 J
The created energy is electromagnetic energy that’s radiated away.
The electron’s “antiparticle” is the positron. If they encounter each
other, they will annihilate each other and, in their place will appear
electromagnetic energy.
The reverse can occur:
Under the right conditions, electromagnetic energy can disappear, and
in its place will appear the equivalent amount of mass in the form of
an electron-positron pair.
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Total Energy
According to Einstein, the total energy of an object
moving with speed v is:
E = mc2 / (1 - β2)1/2
β = v/c, where c = 3.00 x 108 m/s.
Note that when the object is at rest, β = 0, and E = mc2,
which is called the “rest-mass energy.”
Relativistic Kinetic Energy
Subtract the rest portion (mc2) of the energy from the
total energy, and what’s left is the motion part of the
energy kinetic energy.
K = E - mc2
= mc /(1-β2)1/2 - mc2
Note that if v = 0, then K = 0, as expected for an object
that’s not moving.
We often refer to this kinetic energy as the “relativistic”
kinetic energy to alert the reader to the fact that this
kinetic energy is the true kinetic energy equation, while
the other kinetic energy equation, ½ mv2, is called the
“classical” kinetic energy.
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Example:
(a) What is the rest-mass energy of an
electron (in MeV)?
Electron Mass: 9.11 x 10-31 kg
E = mc2
= 9.11 x 10-31 (3.0 x 108)2
= 8.20 x 10-14 J
= 8.20 x 10-14 J / 1.6 x 10-19 J/eV
= 0.51 x 106 eV
= 0.51 MeV
(b) What the total energy of an electron
moving at 0.9995 c?
β = 0.9995
E = mc2 / (1-β2)1/2
= 0.51 / (1 - 0.99952)1/2
= 16.13 MeV
(c) What is its relativistic kinetic energy?
K = E - mc2
= 16.13 - 0.51
= 15.62 MeV
(d) What is the kinetic energy using the
“classical” equation?
K = ½ mv2
= ½ (9.11 x 10-31)(0.9995 x 3.0 x 108)2
= 4.10 x 10-14 J
= 4.10 x 10-14 J/ 1.6 x 10-19 J/eV
= 0.26 x 106 eV
= 0.26 MeV
At very high speeds the classical kinetic
energy equation is wildly inaccurate, as we
see above, but at lower speeds the classical
kinetic energy equation is extremely accurate,
as we will see on the next page.
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Example:
Calculate and compare the classical and relativistic kinetic
energies (in joules) of a proton traveling at the “non-
relativistic” speed, v = 0.01c (β = 0.01).
v = 0.01 (3.0 x 108)
= 3.0 x 106 m/s
Classical: K = ½ mv2
= ½ (1.67 x 10-27) (3.0 x 106)2
= 7.515 x 10-15 J
Relativistic:
mc2 = 1.67 x 10-27 (3.0 x 108)2
= 1.503 x 10-10 J
K = mc2 / (1- β2)1/2 - mc2
= 1.503 x 10-10 (1 - 0.012)-1/2 - 1.503 x 10-10
= 7.516 x 10-15 J
Even at speeds as large as 3.0 x 106 m/s, the classical
kinetic energy equation is still accurate to within one-
hundredth of one percent.
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Classical versus Relativistic Kinetic Energy
Prove: At small speeds v, the classical kinetic energy equation (K = ½ mv2)
predicts kinetic energies as accurately as does the relativistic kinetic energy
equation.
The “binomial theorem”:
If x << 1, then
(1 - x)n 1 – nx
For example, let x = 0.0001:
(1 – 0.0001)2 = 0.9998
Compare to 1.0000 – 2 (0.0001) = 0.9998
Proof:
The role of x is played by β2, while the role of n is played by -½:
(1 - β2)-1/2 = 1 - (-½) β2
= 1 + ½ β2 if β << 1
K = mc2 [ (1 - β2)-1/2 - 1]
= mc2 [1 + ½ β2 - 1]
= mc2 (½ β2)
= mc2 (½) (v2/c2)
= ½ mv2 if v << c
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Example:
How much work (in MeV) would need to be done on a
proton to increase its speed from βo = 0.60 to β = 0.70?
Mass of proton: m = 1.67 x 10-27 kg
mc2 = 1.67 x 10-27 (3.0 x 108)2
= 1.50 x 10-10 J
= 1.50 x 10-10 J / 1.6 x 10-19 J/eV
= 9.39 x 108 eV
= 939 x 106
= 939 MeV
K = mc2 / (1- β2)1/2 - mc2
= 939 / (1 - 0.702)1/2 - 939
= 375.9 MeV
Ko = mc2 / (1- βo2)1/2 - mc2
= 939 / (1 - 0.602)-1/2 - 939
= 234.8 MeV
Use the work-kinetic energy theorem:
W = K - Ko
= 375.9 - 234.8
= 141.1 MeV
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Time Dilation
The time required for an event to occur depends on the
observer’s motion relative to the event.
Observers moving relative to an event
measure a longer time than do
observers stationary relative to the
event.
--Albert Einstein, 1905
Times measured by observers at rest relative to the event
are called the “proper time.”
To = “proper time”
Times measured by observers moving relative to the event
are called the “dilated time.”
T = “dilated time”
= To / (1 - β2)1/2
Note: The denominator is less than 1.0, so T > To.
Note: the word “proper” does not imply it’s the “real” time,
and that the times observed by moving observers are
somehow flawed because of measurement errors or illusion.
The dilated time measured by the moving observer is just as
real for that observer as it is for the observer who is at rest
relative to the event.
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Example A:
An observer at rest relative to an ice cube says it took
eight minutes for the ice to melt.
Proper Time: To = 8.0 minutes
How long does an observer, moving at a speed of four-
tenths the speed of light say it took for the ice to melt?
β = 0.40
T = To / (1 - β2)1/2
= 8.00 / (1 - 0.402)1/2
= 8.73 minutes
Dilated Time: 8.73 minutes
Example B:
A runner completes a journey in 2.00 minutes,
according to her. Does she measure the dilated time, or
the proper time?
The runner is located wherever the event (the running)
is occurring, so she measures the proper time.
To = 2.00 minutes
Suppose a passenger in a spaceship flying overhead at
0.80 c observes the runner. How long will he say it
took the runner to complete her run?
T = 2.00 /(1 - 0.802)1/2
= 3.33 minutes
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Example:
A hovering drone in a hypothetical
universe observes an automobile
drive at speed v = 40 m/s across town
in 14 minutes. The automobile
driver says the trip took 12.0
minutes; what is the speed of light in
that universe?
The event is the trip; it is occurring
wherever the driver is, so she
measures the proper time.
The drone’s location relative to the automobile is constantly changing, i.e., the drone
moving relative to the event, so it measures the dilated time, T = 14.0 minutes.
To = 12.0 minutes
T = 14.0 minutes
Assume Einstein’s Special Theory holds true in the hypothetical universe.
T = To / (1 - β2)1/2
14.0 = 12.0 / (1 - β2)1/2
β = 0.515
40/c = 0.515
c = 77.67 m/s
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Muons
Muons are unstable particles that live for only 2.2 x 10-6 s
before decaying—according to observers stationary
relative to the muon. This is the proper time.
If muons are moving relative to observers, they live
longer—according to the moving observers.
For example, muons are created when protons streaming
toward Earth from the sun collide with molecules in the
upper atmosphere; these muons move at β = 0.998
toward the ground. A hypothetical rider atop the muon
would say that the muon after creation would travel the
following distance through the atmosphere before
decaying:
Distance = Speed x Time
= (0.998)(3.0 x 108) (2.2 x 10-6)
= 659 m
Earth observers measure the dilated time:
T = To / (1 - β2)1/2
= (2.2 x 10-6) /(1 - 0.9982)1/2
= 3.48 x 10-5 s
Earth observers say the muon in fact travels this many
meters before dying:
(0.998) (3.0 x 108 m/s) (3.48 x 10-5 s) = 10, 419 m
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Length Contraction
L = Lo ( 1 - β2)1/2
Lo = Length measured by an observer at rest relative to the object
= “the proper length”
L = Length measured by an observer moving relative to the object
= “the contracted length”
Both lengths are correct for their respective observers.
Contraction occurs in the dimension that is along the direction of motion.
Below we show an object moving at various speeds to the right, along
the x-axis. The narrowing of the object’s width occurs along the
object’s x-axis.
Note: the word “proper” does not imply that the length measured by the
stationary observer is the only “real” length, and all other measurement are
somehow flawed because of errors by the observer, faulty equipment, or an
illusion. The length measured by the moving observer is just as real for that
observer as it is for the observer who is at rest relative to the object.
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Distance-Speed-Time Calculations Involving Light-Years
One Light-Year (LY) is the distance light will travel in one year. For example, if
an object is traveling at one-tenth the speed of light, it will take ten years to travel
1.0 LY. The table below further illustrates the point we wish to make:
Let D = Distance in LY
D (in LY) t (in Y)
1.0 1.0 1.0
1.0 0.2 5.0
10.0 0.4 25.0
20.0 0.5 40.0
t = D/
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Example
A passenger on a spaceship leaving Earth at = 0.80 travels to a star and says
its distance from Earth is 8.0 LY. The passenger is moving relative to the path
connecting Earth to the star, so this distance is the contracted length, L:
L = 8.0 LY
(a) How long (in years) does the passenger say it took the spaceship to reach
the star?
The passenger is always located where the event (the traveling) is occurring
(wherever she is), so she measures the proper time, To. The passenger
calculates the travel time To (in years, Y) as shown below:
To = L/
= 8.0 / 0.80
= 10 Y
(b) How long (in years) does the Earth observer say the spaceship took to reach
the star?
The Earth observer is moving relative to where the event (the traveling) is
occurring, so he measures the dilated time, T:
T = 10 /(1 – 0.802)1/2
= 16.67 Y
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Example:
A stationary observer measures length of a path to be 400 m. The observer is at
rest relative to the path, so he measures the proper length:
Lo = 400 m
A person runs along that path at a speed of 0.70 c:
v = 0.70 (3.0 x 108 m/s)
= 2.1 x 108 m/s
(a) What does the runner say is the length of the path?
The runner is moving relative to the path, so he measures the contracted length:
L = Lo (1 - β2)1/2
= 400 (1 - 0.702)1/2
= 286 m
(b) How much time does the runner say it took to complete the run?
Time = Distance / Speed
= 286 m / 2.1 x 108 m/s
= 1.36 x 10-6 s
The event in question is the running; the runner is located wherever the running
is happening, do the runner is at rest relative to the event and therefore
measures the proper time:
To = 1.36 x 10-6 s
(c) Calculate the runtime according to the stationary observer using two
different methods.
1. Time = Distance / Speed
= 400 m / 2.1 x 108 m/s
= 1.90 x 10-6 s
2. T = To / (1 - β2)1/2
= 1.36 x 10-6 / (1 - 0.702)1/2
= 1.90 x 10-6 s
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Example:
The catcher on a baseball team says the path
connecting home plate to first base has a
length of 27.40 m. He is at rest relative to the
path, so he measures the proper length, Lo.
Lo = 27.40 m
The catcher observes a runner traveling from
home plate to first base at a speed of 0.90 c.
v = 0.90 (3.0 x 108)
= 2.7 x 108 m/s
(a) What does the catcher say is the time it
takes the runner to reach first base?
Time = Distance/Speed:
= 27.40 m /2.7 x 108 m/s
= 1.01 x 10-7 s
The event, which is the running, is taking
place at the runner’s location, so the runner
measures the proper time.
The catcher is moving relative to the event’s
location (the runner’s location), so the catcher
measures the dilated time:
T = 1.01 x 10-7 s
(b) What does the runner say is the time it
takes to reach first base?
The runner measures the proper time:
T = To / (1 - β2)1/2
1.01 x 10-7 = To / (1 - 0.902)1/2
To = 4.40 x 10-8 s
(c) What does the runner say is the length
of the path?
The runner is moving relative to the path, so
he measures the contracted length:
L = Lo (1 - β2)1/2
= 27.40 (1 - 0.902)1/2
= 11.94 m
(d) What does the runner say her speed is?
Runner Measures Distance = 11.94 m
Runner Measures Time = 4.40 x 10-8 s
Speed = Distance / Time
= (11.94 m) / (4.40 x 10-8 s)
= 2.7 x 108 m/s
Note that this is the same speed the catcher
measures. The two persons disagree about
distance between home plate and first base,
and they disagree about the time it takes the
runner to run from home plate to first base,
but they agree about the runner’s speed,
i.e., they agree about the ratio of distance
and time. The catcher says the distance is
longer, while the time is longer; the runner
says the distance is shorter, while the time
is shorter, but the ratios of distance over
times-the speeds---are the same.
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Classical Relative Velocities
Classical relative velocities are velocities that
are negligibly small compared to the speed of
light.
Let the velocity of an Object A relative to an
Earth observer be VAO and the velocity of Object
B relative to the Earth observer be VBO.
Objects moving toward Earth are negative, while
objects moving away from Earth are positive.
The velocity of Object A relative to Object B is
VAB = VAO - VBO
Example: VAO = 20 m/s and VBO = -30 m/s
VAB = 20 - (-30)
= 50 m/s
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Relativistic Relative Velocities
At very great velocities--greater than about one-hundredth the
speed of light, Newtonian physics, i.e., non-relativistic physics, is
no longer reliable.
The equation that is valid for relativistic, as well as classical,
velocities is shown below for Objects A and B moving relative to
an observer on Earth.
AB = (AO - BO) / (1 - AO BO)
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Example:
Spaceship A is moving at a velocity 0.50 c relative to
observers on Earth, as it chases Spaceship B, which is moving
relative to Earth at 0.30 c.
(a) What is the velocity of A as seen by Spaceship B?
AB = (AO - BO) / (1 - AO BO)
= (0.50 - 0.30 ) / [1 - (0.50) (0.30)]
= 0.24
(b) Spaceship A is emitting light whose frequency is
fA = 6.0 x 1014 Hz. What is fB, the frequency seen by
Spaceship B?
Recall the Doppler Effect for Light equation discussed in
Chapter 24:
fo = fs (1 ± )
Here, the role of the observer is played by Spaceship B, while
the source is Spaceship A.
fB = fA (1 ± )
Spaceship A is approaching Spaceship B, so we use the
positive sign to ensure that the observed frequency is greater
than the broadcast frequency:
fB = 6.00 x 1015 (1 + 0.24)
= 7.44 x 1015 Hz