ch28 Relativity.pptx

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
Chapter 28
Special Relativity

28.1 Events and Inertial Reference Frames
An event is a physical “happening” that occurs at a certain place and
time.
To record the event, each observer uses a reference frame that consists of a
coordinate system and a clock.
Each observer is at rest relative to her own reference frame.
An inertial reference frame is one in which Newton’s law of inertia is
valid.

28.1 Events and Inertial Reference Frames
In this example, the event is the space shuttle lift off.

28.2 The Postulates of Special Relativity
THE POSTULATES OF SPECIAL RELATIVITY
1. The Relativity Postulate. The laws of physics are the same
in every inertial reference frame.
2. The Speed of Light Postulate. The speed of light in a vacuum,
measured in any inertial reference frame, always has the same value
of c, no matter how fast the source of light and the observer are
moving relative to one another.

28.2 The Postulates of Special Relativity

28.3 The Relativity of Time: Time Dilation
TIME DILATION
A light clock

Suppose two identical clocks are built.
One is kept on earth, and the other is
placed aboard a spacecraft that travels
at a constant velocity relative to the
earth
The astronaut is at rest with respect to the
clock on the spacecraft and,
therefore, sees the light pulse move
along the up/down path shown in top
diagram.
• According to the astronaut, the time interval ∆𝑡0 required for the light to
follow this path is the distance 2D divided by the speed of light c; ∆𝑡0 =
2𝐷
𝑐
• An earth-based observer, however, does not measure ∆𝑡0 as the time
interval between these two events.
• Since the spacecraft is moving, the earth-based observer sees the light pulse
follow the diagonal path shown in red in part b of the drawing.
• This path is longer than the up/down path seen by the astronaut.
• But light travels at the same speed c for both observers, in accord with the
speed-of-light postulate

An observer on the earth sees the light
pulse travel a greater distance
between ticks.
The time interval ∆𝑡 that the earth-based
observer measures in Figure b can be
determined as follows
• While the light pulse travels from the
source to the detector, the spacecraft
moves a distance 2𝐿 = 𝑣∆𝑡 to the right,
where v is the speed of the spacecraft
relative to the earth.
• From the drawing it can be seen that the light pulse travels a total
diagonal distance of 2s during the time interval∆𝑡
Applying the Pythagorean theorem, we find that
But the distance 2s is also equal to the speed of light times the time
interval∆𝑡, so that

And
Therefore
solving for ∆𝑡 we get
But
2𝐷
𝐶
= ∆𝑡0, where ∆𝑡0 is the time interval between successive “ticks” of the
spacecraft’s clock as measured by the astronaut.
Substituting we get the equation for TIME DILATION given by

2
2
1 c
v
t
t o




Time dilation
The symbols in this formula are defined as follows:

PROPER TIME INTERVAL
The time interval measured at rest with respect to the clock is
called the proper time interval.
In general, the proper time interval between events is the time
interval measured by an observer who is at rest relative to the
events.
Proper time interval
o
t

Check your understanding 1 to 3 page 876

Example 1 Time Dilation
The spacecraft is moving past the earth at a constant speed of
0.92 times the speed of light. The astronaut measures the time
interval between ticks of the spacecraft clock to be 1.0 s. What is
the time interval that an earth observer measures?
2
2
1 c
v
t
t o




 
s
6
.
2
92
.
0
1
s
0
.
1
2



c
c

28.4 The Relativity of Length: Length Contraction
The shortening of the distance between two points is one
example of a phenomenon known as length contraction.
Length contraction
2
2
1
c
v
L
L o 


Example 4 The Contraction of a Spacecraft
An astronaut, using a meter stick that is at rest relative to a cylindrical
spacecraft, measures the length and diameter to be 82 m and 21 m
respectively. The spacecraft moves with a constant speed of 0.95c
relative to the earth. What are the dimensions of the spacecraft,
as measured by an observer on earth.

    m
26
95
.
0
1
m
82
1
2
2
2




 c
c
c
v
L
L o
Diameter stays the same.

28.5 Relativistic Momentum
Relativistic
momentum
2
2
1 c
v
mv
p



28.6 The Equivalence of Mass and Energy
THE TOTAL ENERGY OF AN OBJECT
Total energy
of an object
Rest energy
of an object
2
2
2
1 c
v
mc
E


2
mc
Eo 
o
E
E 

KE
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,
E = E0 + KE
Kinetic energy
of an object

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.
• From the equation of relativistic momentum, make
𝑝
𝑣
the subject of formula
• Substitute
𝑝
𝑣
on the equation for energy
2
2
2
1 c
v
mc
E


• Substitute
𝑣
𝑐
on the equation for energy
Solving the equation for energy 𝐸2
we get

Why the Speed of Light in a Vacuum Is the Ultimate Speed
One of the important consequences of the theory of special relativity is that
objects with mass cannot reach the speed c of light in a vacuum
Consider Equation,
The equation which gives the kinetic energy of an object moving at a
speed v
As v approaches the speed of light c, the denominator
Approaches zero
Hence, the kinetic energy becomes infinitely large
However, the work–energy theorem 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.
Thus, c represents the ultimate speed.

Problems 1 to 20
 
0
2 2
1 /
t
t
v c

 


The dilated time interval t is related to the proper time interval t0 by
 
2 2
0 / 1 /
t t v c
    . Solving this equation for the speed v of the spacecraft yields
When you measure your breathing rate, you count N = 8.0 breaths during a proper time interval
of Δt0 = 1.0 minutes, and in so doing you determine a rate of
R0 = N/Δt0 = (8.0 breaths)/(1.0 minute) = 8.0 breaths/minute. When measured by
monitors on the earth, the N = 8.0 breaths occur in a dilated time interval Δt that is related
to the proper time interval by 0
2
2
1
t
t
v
c

 

The breathing rate R measured by a monitor on the earth, then, is given by
N
R
t


(1)

5. REASONING The observer is moving with respect to the oscillating object. Therefore, to
the observer, the oscillating object is moving with a speed of v = 1.90 × 108
m/s, and the
observer measures a dilated time interval for the period of oscillation. To determine this
dilated time interval t = Tdilated, we must use the time-dilation equation:
where t0 is the proper time interval, as measured in the reference frame to which the fixed end
of the spring is attached. The proper time interval is the period T of the oscillation as
given by Equations

Example 8 The Sun is Losing Mass
The sun radiates electromagnetic energy at a rate of 3.92x1026W.
What is the change in the sun’s mass during each second that it
is radiating energy? What fraction of the sun’s mass is lost during
a human lifetime of 75 years.

  
 
kg
10
36
.
4
s
m
10
3.00
s
0
.
1
s
J
10
92
.
3 9
2
8
26
2








c
E
m o
  
12
30
7
9
sun
10
19
.
5
kg
10
99
.
1
yr)
75
(
s/yr
10
16
.
3
s
kg
10
36
.
4








m
m

• The total momentum of the man/woman system is conserved,
since no net external force acts on the system.
• Therefore, the final total momentum pm + pw must equal the initial total
momentum, which is zero.
• As a result, pm = – pw
   
w w
m m
2 2
m w
–
1– / 1– /
m v
m v
v c v c

m –2.0 m/s
v 
Solving for vm, we find

According to the work-energy theorem, he work that must be done on the electron to accelerate it from rest to a speed of
0.999c is equal to the kinetic energy of the electron when it is moving at 0.999c.
2
2 2
–31 8 2 –12
2 2
1
–1
KE
1– ( / )
1
–1
(9.11 10 kg)(3.00 10 m/s) 1.8 10 J
1– (0.999 ) /
mc
v c
c c
 
 

 
 
 
 
    
 
 
• The mass m of the aspirin is related to its rest energy E0
• Since it requires 1.1  108 J to operate the car for twenty miles, we can
calculate the number of miles that the car can go on the energy that is equivalent to
the mass of one tablet
• The number N of miles the car can go on one aspirin tablet is
       
  
   
2
0
8 8
2
6 8
6
8
1.1 10 J / 20.0 mi 1.1 10 J / 20.0 mi
325 10 kg 3.0 10 m/s
5.3 10 mi
1.1 10 J / 20.0 mi
E mc
N

 
 
 
  


• According to the work-energy theorem, the change in the kinetic energy equals the
work W done to accelerate the electron. Since the electron starts from rest, its initial
kinetic energy is zero, so its final kinetic energy KE equals the work done.
This conclusion is valid for either nonrelativistic or relativistic speeds. The given potential
difference has a magnitude of 7
1.8 10 V
V
   and is related to the work done
according to 0
/
V W q
 
The relativistic kinetic energy of the electron is 2
2 2
1
KE 1
1 /
mc
v c
 
 
 
 

 
Recognizing that the electron’s final kinetic energy KE equals the work W done
to accelerate the electron, we have KE W

for the speed v, we proceed as follows

The total energy for each particle is given by
  
2
–31 8
2
–14
2 2
9.11 10 kg 3.00 10 m/s
9.5 10 J
0.50
1– 1–
mc
E
v c
c c
 
   
   
   
   
The annihilation energy is twice this value, so
13
2 1.9 10 J
E E 
   
.

Conceptual Example 9 When is a Massless Spring Not Massless?
The spring is initially unstrained and assumed to be massless. Suppose
that the spring is either stretched or compressed. Is the mass of
the spring still zero, or has it changed? If the mass has changed, is
the mass change greater for stretching or compressing?

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