Physics 6C
Special Relativity
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Two Main Ideas – The Postulates of Special Relativity
• Light travels at the same speed in all inertial reference frames.
• Laws of physics yield identical results in all inertial reference frames.
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Two Main Ideas – The Postulates of Special Relativity
• Light travels at the same speed in all inertial reference frames.
• Laws of physics yield identical results in all inertial reference frames.
Inertial reference frames refer to observers moving
at constant velocity with respect to each other.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Two Main Ideas – The Postulates of Special Relativity
• Light travels at the same speed in all inertial reference frames.
• Laws of physics yield identical results in all inertial reference frames.
Inertial reference frames refer to observers moving
at constant velocity with respect to each other.
If there is a nonzero acceleration, the frames are not inertial, and we
would need to use General Relativity. Way too much math for this
course – sorry…
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Two Main Ideas – The Postulates of Special Relativity
• Light travels at the same speed in all inertial reference frames.
• Laws of physics yield identical results in all inertial reference frames.
Inertial reference frames refer to observers moving
at constant velocity with respect to each other.
If there is a nonzero acceleration, the frames are not inertial, and we
would need to use General Relativity. Way too much math for this
course – sorry…
The relationship between what is seen in the two reference frames is
found via the Lorentz Transformation. We will see the following factor
in all of our equations:
1

2
1  v2
c
Here v is the relative speed of the frames, and c is the speed of light.
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The main results of Special Relativity are the following:
1) Time Dilation - if an object is moving, an observer will measure times to
be longer (compared to the frame of the object itself)
t 
t0
2
1  v2
c
 t    t0
t0 refers to the object at rest in its own frame
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
The main results of Special Relativity are the following:
1) Time Dilation - if an object is moving, an observer will measure times to
be longer (compared to the frame of the object itself)
t 
t0
2
1  v2
c
 t    t0
t0 refers to the object at rest in its own frame
2) Length Contraction – if an object is moving, an observer will measure
lengths to be shorter in the direction of motion (compared to the frame of
the object itself)
L  L0 1 
v2
c2
L 
L0

L0 refers to the object at rest in its own frame
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
The main results of Special Relativity are the following:
1) Time Dilation - if an object is moving, an observer will measure times to
be longer (compared to the frame of the object itself)
t 
t0
2
1  v2
c
 t    t0
t0 refers to the object at rest in its own frame
2) Length Contraction – if an object is moving, an observer will measure
lengths to be shorter in the direction of motion (compared to the frame of
the object itself)
L  L0 1 
v2
c2
L 
L0

L0 refers to the object at rest in its own frame
3) Addition of velocities is more complicated than in the non-relativistic
case. At low speeds, we just add or subtract the relative velocities and it
works fine, but near the speed of light we need to be more careful.
Here’s a formula:
v2 3 
v2 1  v1 3
1
v 21 v13
c2
v13 is the relative speed between the frames, and v21 and
v23 are the velocities of the object in each frame. These
formulas are tricky to use, so practice several examples.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
The main results of Special Relativity are the following:
1) Time Dilation - if an object is moving, an observer will measure times to
be longer (compared to the frame of the object itself)
t 
t0
2
1  v2
c
 t    t0
t0 refers to the object at rest in its own frame
2) Length Contraction – if an object is moving, an observer will measure
lengths to be shorter in the direction of motion (compared to the frame of
the object itself)
L  L0 1 
v2
c2
L 
L0

L0 refers to the object at rest in its own frame
3) Addition of velocities is more complicated than in the non-relativistic
case. At low speeds, we just add or subtract the relative velocities and it
works fine, but near the speed of light we need to be more careful.
Here’s a formula:
v2 3 
v2 1  v1 3
1
v 21 v13
c2
v13 is the relative speed between the frames, and v21 and
v23 are the velocities of the object in each frame. These
formulas are tricky to use, so practice several examples.
4) Energy and Mass are equivalent (Erest=m0c2). We can also get formulas
for relativistic momentum and total energy.

p

mv
2
1  v2
c
Etotal 
m0c2
1
v2
c2
 K  Eres t
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http://www.youtube.com/watch?v=JQnHTKZBTI4
http://www.youtube.com/watch?v=C2IGMLYyA7U&NR=1
http://www.youtube.com/watch?v=PpFZpqlWiF0
Visual demonstrations of special relativity.
Prepared by Vince Zaccone
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Here’s a problem:
In the year 2084, a spacecraft flies over Moon Station III at a speed of 0.800c. A scientist on
the moon measures the length of the moving spacecraft to be 140 m. The spacecraft later
lands on the moon, and the same scientist measures the length of the now stationary
spacecraft. What value does she get?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
In the year 2084, a spacecraft flies over Moon Station III at a speed of 0.800c. A scientist on
the moon measures the length of the moving spacecraft to be 140 m. The spacecraft later
lands on the moon, and the same scientist measures the length of the now stationary
spacecraft. What value does she get?
Use the length contraction formula with L=140m and v=.8c. We are looking for L0.
L  L0 1 
v2
c2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
In the year 2084, a spacecraft flies over Moon Station III at a speed of 0.800c. A scientist on
the moon measures the length of the moving spacecraft to be 140 m. The spacecraft later
lands on the moon, and the same scientist measures the length of the now stationary
spacecraft. What value does she get?
Use the length contraction formula with L=140m and v=.8c. We are looking for L0.
L  L0 1 
v2
c2
140m  L 0 1 
.8c 2
c2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
In the year 2084, a spacecraft flies over Moon Station III at a speed of 0.800c. A scientist on
the moon measures the length of the moving spacecraft to be 140 m. The spacecraft later
lands on the moon, and the same scientist measures the length of the now stationary
spacecraft. What value does she get?
Use the length contraction formula with L=140m and v=.8c. We are looking for L0.
L  L0 1 
v2
c2
140m  L 0 1 
140m  L 0 0.6 
.8c 2
Notice that c2 cancels out. This usually happens
when you use speeds written in terms of c.
c2
L 0  233m
Our result is consistent with the concept of length contraction.
The ship is measured to be shorter when it is moving.
Prepared by Vince Zaccone
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Here’s a problem:
Inside a spaceship flying past the earth at ¾ the speed of light, a pendulum is swinging.
a) If each swing takes 1.5 s as measured by an astronaut performing an experiment inside
the spaceship, how long will the swing take as measured by a person at mission control on
earth who is watching the experiment?
b) If each swing takes 1.5 s as measured by a person at mission control on earth, how long
will the swing take as measured by an astronaut inside the spaceship?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Inside a spaceship flying past the earth at ¾ the speed of light, a pendulum is swinging.
a) If each swing takes 1.5 s as measured by an astronaut performing an experiment inside
the spaceship, how long will the swing take as measured by a person at mission control on
earth who is watching the experiment?
b) If each swing takes 1.5 s as measured by a person at mission control on earth, how long
will the swing take as measured by an astronaut inside the spaceship?
We will be using the time dilation formula. Notice the difference between part a) and part b)
– In part a) the time as measured on the spaceship is given. This is Δt0 because the
pendulum is at rest relative to the ship.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Inside a spaceship flying past the earth at ¾ the speed of light, a pendulum is swinging.
a) If each swing takes 1.5 s as measured by an astronaut performing an experiment inside
the spaceship, how long will the swing take as measured by a person at mission control on
earth who is watching the experiment?
b) If each swing takes 1.5 s as measured by a person at mission control on earth, how long
will the swing take as measured by an astronaut inside the spaceship?
We will be using the time dilation formula. Notice the difference between part a) and part b)
– In part a) the time as measured on the spaceship is given. This is Δt0 because the
pendulum is at rest relative to the ship.
t 
t 0
2
1  v2
c
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Inside a spaceship flying past the earth at ¾ the speed of light, a pendulum is swinging.
a) If each swing takes 1.5 s as measured by an astronaut performing an experiment inside
the spaceship, how long will the swing take as measured by a person at mission control on
earth who is watching the experiment?
b) If each swing takes 1.5 s as measured by a person at mission control on earth, how long
will the swing take as measured by an astronaut inside the spaceship?
We will be using the time dilation formula. Notice the difference between part a) and part b)
– In part a) the time as measured on the spaceship is given. This is Δt0 because the
pendulum is at rest relative to the ship.
t 
t0
2
1  v2
c
 t 
1.5s
.75c 
1
2
c2
 t  2.3s
The people on earth measure a longer
(dilated) time for each swing, as expected.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Inside a spaceship flying past the earth at ¾ the speed of light, a pendulum is swinging.
a) If each swing takes 1.5 s as measured by an astronaut performing an experiment inside
the spaceship, how long will the swing take as measured by a person at mission control on
earth who is watching the experiment?
b) If each swing takes 1.5 s as measured by a person at mission control on earth, how long
will the swing take as measured by an astronaut inside the spaceship?
We will be using the time dilation formula. Notice the difference between part a) and part b)
– In part a) the time as measured on the spaceship is given. This is Δt0 because the
pendulum is at rest relative to the ship.
t 
t0
2
1  v2
c
 t 
1.5s
.75c 
1
2
c2
 t  2.3s
The people on earth measure a longer
(dilated) time for each swing, as expected.
Part b) uses the same formula, but now we are given Δt instead.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Inside a spaceship flying past the earth at ¾ the speed of light, a pendulum is swinging.
a) If each swing takes 1.5 s as measured by an astronaut performing an experiment inside
the spaceship, how long will the swing take as measured by a person at mission control on
earth who is watching the experiment?
b) If each swing takes 1.5 s as measured by a person at mission control on earth, how long
will the swing take as measured by an astronaut inside the spaceship?
We will be using the time dilation formula. Notice the difference between part a) and part b)
– In part a) the time as measured on the spaceship is given. This is Δt0 because the
pendulum is at rest relative to the ship.
t 
t0
2
1  v2
c
 t 
1.5s
.75c 
1
2
 t  2.3s
c2
The people on earth measure a longer
(dilated) time for each swing, as expected.
Part b) uses the same formula, but now we are given Δt instead.
t 
t0
2
1  v2
c
 1.5s 
t0
 t0  1.0s
.75c 
Again the people on earth measure a longer time
1
c2
because the clock is moving relative to them.
2
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Here’s a problem:
Two particles are created in a high-energy accelerator and move off in opposite directions.
The speed of one particle, as measured in the laboratory, is 0.65c, and the speed of each
particle relative to the other is 0.95c. What is the speed of the second particle, as measured
in the laboratory?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Two particles are created in a high-energy accelerator and move off in opposite directions.
The speed of one particle, as measured in the laboratory, is 0.65c, and the speed of each
particle relative to the other is 0.95c. What is the speed of the second particle, as measured
in the laboratory?
Pictures will probably help here
(the laboratory is object #3)
v2 3 
v2 1  v1 3
1
v 21 v13
c2
v23=?
2
1
v13=0.65c
This is what you see in the laboratory.
Particle 1 is moving at 0.65c, and
Particle 2 is moving the other direction.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Two particles are created in a high-energy accelerator and move off in opposite directions.
The speed of one particle, as measured in the laboratory, is 0.65c, and the speed of each
particle relative to the other is 0.95c. What is the speed of the second particle, as measured
in the laboratory?
Pictures will probably help here
(the laboratory is object #3)
v2 3 
v2 1  v1 3
1
v 21 v13
c2
v23=?
2
1
v13=0.65c
This is what you see in the laboratory.
Particle 1 is moving at 0.65c, and
Particle 2 is moving the other direction.
v21=-0.95c
2
1
This is the same scenario in the reference frame
of Particle 1. Particle 2 is moving away at -0.95c,
and Particle 1 is at rest in its own frame.
(If we were to put the laboratory in this picture
we would see it moving to the left at -0.65c.)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Two particles are created in a high-energy accelerator and move off in opposite directions.
The speed of one particle, as measured in the laboratory, is 0.65c, and the speed of each
particle relative to the other is 0.95c. What is the speed of the second particle, as measured
in the laboratory?
Pictures will probably help here
(the laboratory is object #3)
v2 3 
v
v2 1  v1 3
1
v 21 v13
c2
2
1
v13=0.65c
This is what you see in the laboratory.
Particle 1 is moving at 0.65c, and
Particle 2 is moving the other direction.
 0.95c  0.65c
1
0.65c 0.95c 
c2
 0.3c
1  0.6175
v  0.78c
v
v23=?
v21=-0.95c
2
1
This is the same scenario in the reference frame
of Particle 1. Particle 2 is moving away at -0.95c,
and Particle 1 is at rest in its own frame.
(If we were to put the laboratory in this picture
we would see it moving to the left at -0.65c.)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a problem:
Two particles are created in a high-energy accelerator and move off in opposite directions.
The speed of one particle, as measured in the laboratory, is 0.65c, and the speed of each
particle relative to the other is 0.95c. What is the speed of the second particle, as measured
in the laboratory?
Pictures will probably help here
(the laboratory is object #3)
v2 3 
v
v2 1  v1 3
1
v 21 v13
c2
v23=?
2
1
v13=0.65c
This is what you see in the laboratory.
Particle 1 is moving at 0.65c, and
Particle 2 is moving the other direction.
 0.95c  0.65c
1
0.65c 0.95c 
c2
 0.3c
1  0.6175
v  0.78c
v
So in the lab, Particle 2 looks like it is
moving to the left at speed 0.78c.
v21=-0.95c
2
1
This is the same scenario in the reference frame
of Particle 1. Particle 2 is moving away at -0.95c,
and Particle 1 is at rest in its own frame.
(If we were to put the laboratory in this picture
we would see it moving to the left at -0.65c.)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
The sun produces energy by nuclear fusion reactions, in which matter is converted to energy.
The rate of energy production is 3.8 x 1026 Watts. How many kilograms of mass does the sun
convert to energy each second?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
The sun produces energy by nuclear fusion reactions, in which matter is converted to energy.
The rate of energy production is 3.8 x 1026 Watts. How many kilograms of mass does the sun
convert to energy each second?
We only need to use Einstein’s E=mc2 for this one.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
The sun produces energy by nuclear fusion reactions, in which matter is converted to energy.
The rate of energy production is 3.8 x 1026 Watts. How many kilograms of mass does the sun
convert to energy each second?
We only need to use Einstein’s E=mc2 for this one.

s 2
3.8  1026 J  m  3  108 m
Remember, a Watt is a Joule per second
m  4.2  109kg
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