Transcript Slide 1

Theory of Special Relativity
Theory of Special Relativity
Einstein is recognised for:
 encompassing all of the known results
 restating the laws of relativity and the
required modifications for the laws of Physics
to be consistent
Theory of Special Relativity
Correspondence Principle
Correspondence Principle

Any new theory proposed must agree with an
older theory in terms of correct predictions
which the theory gave.

In this case the theory of special relativity
must agree with classical mechanics at low
speeds.
Correspondence Principle


This requirement guided Einstein in
developing the transformations which are the
foundation of Special Relativity.
It is a good idea to verify that these equations
reduce to those from classical physics.
Einstein’s Postulates (1905)
The postulates of special relativity are:
1) The laws of Physics are the same in all
inertial reference frames.
2) The speed of light in a vacuum is c in all
inertial reference frames.
Einstein’s Postulates (1905)


Thus there is no ether (or rather no need for
an absolute frame) as there is no preferred
reference frame.
The Galilean transform was replaced by a
Lorentz transformation.
Consequences of Special Relativity


Several phenomena are predicted as a result
of these formulations which conflict with our
everyday experiences.
Specifically an effect on length and time.
Consequences of Special Relativity
For example:

there is no absolute length

or absolute time

events at different locations occurring
simultaneously in one frame are not
simultaneous in another frame.
Definitions
Proper frame: the reference frame in which the
observed body is at rest.
Proper length: the length in the frame where
the object is at rest wrt the observer.
Proper time: the time interval recorded by a
clock attached to the observed body.
Consequences of Special Relativity

We will be looking at three particular
phenomena: time dilation, length contraction
and simultaneity.
Consequences of Special
Relativity
Time Dilation
Time Dilation

To illustrate how time is altered consider the
case where there is a mirror attached to the
ceiling of a car moving with a velocity v.
S′
S
V
Time Dilation

An observer in the car flashes a light, the path of
which is shown below.
D

The diagram shows the path as seen by an
observer in the car.
Time Dilation
D

Time taken for the ray to be reflected to observer is
t 0 
2 D0
c
Time Dilation
D


t 0 
2 D0
c
The time measured by this observer is the proper
time since only one clock is needed to measure the
time.
The length is the proper length since the observer is
at rest wrt the mirror.
Time Dilation

An observer outside the car will see a moving mirror.
ct
2
ct
2
A
Do
B
v t

As the car travels a distance v t, the light moves a
distance of c t.
Time Dilation
ct
2
ct
2
A
Do
B
v t


The time taken for the light to be reflected back is t
Two clocks measured the time, one at A and the
other at B
Time Dilation
ct
2
ct
2
A
Do
B
v t


The relationship between t and t 0 must be found.t
First using Pythagoras we determine an expression
for t containing D0 :
 ct 
 vt 
2

 
  D0
 2 
 2 
2
2
Time Dilation
 ct 
 vt 
2

 
  D0
 2 
 2 
2
2


Simplifying we get c 2  v 2 t 2  4D02
However t 0 
Therefore
c
2
2 D0
 2D0  ct 0
c

 v 2 t 2  c 2 t02
c2
2
 t  2


t
0
c  v2
2
 t 


1
1
2
v
c2
 t 0 (Time dilation)
Time Dilation
This can be written as t   t 0
 
1
1
2
v
c2
(Lorentz Factor)
the quantity v c is often represented by
Therefore  
1
1  2

Features of Time Dilation


The time measured in the stationary frame is
longer than that measured in the moving
frame.
Moving clocks run slower!! This effect is
called time dilation.
Experimental Evidence of
Time Dilation
Decaying Muons
Decaying Muons



Muons are particles produced by cosmic
radiation
have a life of 2.2s .
If they move at a speed of 0.99c the max.
distance a muon can travel is 600 m .
Decaying Muons




Muons are particles produced by cosmic
radiation
have a life of 2.2s.
If they move at a speed of 0.99c the max.
distance a muon can travel is 600 m .
But muons traveled4800 m.
Decaying Muons

How can the muons travel 4800 m ?
Decaying Muons

How can the muons travel 4800 m ?

Answer
Time Dilation
Decaying Muons



Consider the two reference frames – the
earth frame and the muon reference frame.
The time measured by an observer is the
proper time since only one clock (attached to
the muons) is required.
However an observer in the earth frames
needs to clocks!!
The Earth Frame


Because of time dilation the lifetime of a
muon is longer as measured by an observer
in the earth frame.
by a factor   7.1
The Earth Frame



Because of time dilation the lifetime of a
muon is longer as measured by an observer
in the earth frame.
by a factor   7.1
decay time in the earth frame,t  16s
The Earth Frame




Because of time dilation the lifetime of a
muon is longer as measured by an observer
in the earth frame.
by a factor   7.1
decay time in the earth frame,t  16s
Hence available time  0.99c  16s  4800m
The Muon Frame


In the moving frame the muons see a shorter
length to be travelled 600 m .
Achievable in their lifetime.
Consequences of Special
Relativity
Length Contraction
Length Contraction


length in a moving frame of reference will
appear contracted in the direction of the
motion.
NB: The proper length is the length measured
in which the object is at rest wrt the frame of
reference.
Length Contraction
v
E
Consider a spaceship moving with a velocity v
between two stars. Two observers, one on earth at
rest with respect to the stars and the other on the
ship measure the distance.
Length Contraction
Earth
frame


L0
Since at rest wrt the stars he measures a
proper length L0 .
L0
From his frame the time for the trip is t 
v
Length Contraction
v
Ship
frame



L
For him the star moving towards him a
velocity v.
Therefore measures a shorter distance.
t
He measures a time t 0 
and distance

L  vt 0
Length Contraction
v
Ship
frame

L
t
L0
Substituting for t0 we get L  v
L



1
L0
1 v2 c2


L  L0 1  v
2
c
2

1
2
(Length contraction)
Features of Length Contraction


A moving observer measures a contracted
length in the direction of motion.
There is no contraction perpendicular to the
motion.
Consequences of Special
Relativity
Simultaneity
Simultaneity

Unlike our everyday experiences, two events
which are simultaneous in one frame are in
general not simultaneous in another.
Simultaneity
The following thought experiment highlights this.
v
O′
O
A


B
A train is moving with a velocity v to the right when
its ends are struck by lightning.
Two observers at O and O  both points midway
between the two ends record the occurrence via light
reaching them.
Stationary Frame


The light from the strikes at A and B reach
the observer O at the same time.
Since the distance travelled by the light is the
same, he concludes that the events were
simultaneous.
Train Frame

By the time the light reaches O , the observer
has moved (and hence his point of reference)
tA
tB
v
O′
O
A
B
t′B < t′A
Train Frame



Since the speed of light is constant, light from
B′ reaches the observer first.
Hence he concludes that the events are not
simultaneous.
The lightning strikes the right side first.
Twin Paradox
Twin Paradox

Famous thought experiment in special
relativity.
Twin Paradox


Famous thought experiment in special
relativity.
Experiment involves two identical twins
Twin Paradox


Famous thought experiment in special
relativity.
Experiment involves two identical twins – one
stays on earth and the other journeys in a
spaceship traveling near the speed of light to
a star 30lys away.
Twin Paradox



Famous thought experiment in special
relativity.
Experiment involves two identical twins – one
stays on earth and the other journeys in a
spaceship traveling near the speed of light to
a star 30lys away.
On reaching the star he immediately returns
to earth at the same speed.
Twin Paradox



Experiment involves two identical twins – one
stays on earth and the other journeys in a
spaceship traveling near the speed of light to
a star 30lys away.
On reaching the star he immediately returns
to earth at the same speed.
He returns to find his brother 80 yrs while he
only aged 10 yrs.
Twin Paradox




Experiment involves two identical twins – one
stays on earth and the other journeys in a
spaceship traveling near the speed of light to
a star 30lys away.
On reaching the star he immediately returns
to earth at the same speed.
He returns to find his brother 80 yrs while he
only aged 10 yrs.
Thus the traveler appears younger than the
brother who was at rest.
Twin Paradox


From the frame of the brother on earth, he
was at rest while his brother travelled at high
speed.
While according to his brother he was at rest
in his frame and it was brother who sped
away from him and then returned.
Twin Paradox

The “apparent” paradox is who was travelling
at high speed and therefore should have
aged?

The view by both brother appears symmetric
and hence the paradox.
Twin Paradox

When this hypothetical situation was first
stated it appeared that there was a conflict.
Twin Paradox (Solution)

The problem is that trip is not symmetrical as
stated.
Twin Paradox (Solution)


The problem is that trip is not symmetrical as
stated.
The brother on spaceship experiences a
series of accelerations and decelerations as
he leaves and returns to earth.
Twin Paradox (Solution)



The problem is that trip is not symmetrical as
stated.
The brother on spaceship experiences a
series of accelerations and decelerations as
he leaves and returns to earth.
Therefore he is not always in an inertial
frame!!
Twin Paradox (Solution)




The problem is that trip is not symmetrical as
stated.
The brother on spaceship experiences a
series of accelerations and decelerations as
he leaves and returns to earth.
Therefore he is not always in an inertial
frame!!
Hence the laws of special relativity are not
valid in his frame.
Twin Paradox (Solution)


The other brother has remained in an inertial
frame.
Therefore his measurements are correct.
Twin Paradox (Solution)

Therefore the brother on the spaceship is
indeed younger.
Time Dilation Example

The period of a pendulum is measured to be
3s in the inertial frame of the pendulum. What
is the period when measured by an observer
moving at a speed of 0.95c wrt the
pendulum?
Time Dilation Example


The period of a pendulum is measured to be
3s in the inertial frame of the pendulum. What
is the period when measured by an observer
moving at a speed of 0.95c wrt the
pendulum?
1
Solution: t   t 0  
2
v
1 2
c
Time Dilation Example


The period of a pendulum is measured to be
3s in the inertial frame of the pendulum. What
is the period when measured by an observer
moving at a speed of 0.95c wrt the
pendulum?
1
1

 3.2
Solution: t   t 0  
2
2
v
1 2
c
1  0.9
Time Dilation Example


The period of a pendulum is measured to be
3s in the inertial frame of the pendulum. What
is the period when measured by an observer
moving at a speed of 0.95c wrt the
pendulum?
1
1

 3.2
Solution: t   t 0  
2
2
v
1 2
c
 t  3.2  3  9.6s
1  0.9
Length Contraction Example

An observer on earth sees a spaceship at an
altitude of 435m moving downward towards
the earth with a speed of 0.97c. What is the
altitude of the spaceship as measured by an
observer on the spaceship?
Length Contraction Example


An observer on earth sees a spaceship at an
altitude of 435m moving downward towards
the earth with a speed of 0.97c. What is the
altitude of the spaceship as measured by an
observer on the spaceship?
L0
Solution: L 

Length Contraction Example


An observer on earth sees a spaceship at an
altitude of 435m moving downward towards
the earth with a speed of 0.97c. What is the
altitude of the spaceship as measured by an
observer on the spaceship?
L0
Solution: L 


L  4351  0.9
 106 m
2

1
2
Simultaneity Example

A relativistic snake of proper length 100cm is
moves with speed v=0.6c to the right across a
table. A boy holding two hatchets 100cm apart
plans to bounce them simultaneously so that the
left hatchet lands immediately behind the
snake’s tail.
The boy sees the length of the snake contracted
so that the hatchets miss the snake. The snake
sees a contracted distance between the
hatchets so it is cut. Who is correct?
Simultaneity Example



Solution: A problem on simultaneity!
Boy’s Frame
The length of the snake = 80cm
L
L0

1
5
; 

2
4
1  0.6
100cm
L
 80cm
5
4
Simultaneity Example


Boy’s Frame
Let us looks at the time which the two events
occur.
vxL 


tL    tL  2   0
c 

Since xL=tL=0
vxR  5 
0.6c 100cm 


tR    tR  2    0 

2
c  4
c


 t R  2.5ns
Simultaneity Example

In S′ the two hatchets do not fall
simultaneously.