Transcript A Quantum Journey

Special Relativity
An Introduction To ‘High Speed’ Physics
What Is Light?
There were two contradicting theories as to the nature of light:
Newton – light is corpuscular
Huygens – light is a wave
18th Century
19th Century
20th Century
Newton must be right!?
diffraction/interference
wave-particle duality
If light is a wave, through what does it propagate?
The Aether
Space is permeated by an invisible lumineferous aether
(light-bearing medium)
Medium through which light can propagate
The Earth must be moving relative to the aether
So light will travel faster or slower, depending on the orientation
Differences can be determined by experiment
Test for the existence of the aether
The Michelson-Morley Experiment
The Michelson-Morley Experiment
Test for the presence of an aether using an interferometer
B
A is a half-silvered mirror
B/C are mirrors
O is a detector
speed c – v
C
incoming light
t ACA
t ABA
d
d
2d 
v2



1 2
cv cv
c 
c
2d

c


1  v 

c 2 

2
 12




1
t ACA  t ABA
v ms-1 relative
to the aether
A
speed c + v
O
d
There is a phase difference between the two beams
Light at O should be phase-shifted, but no phase shift was observed
Maxwell’s Predictions
Electric and magnetic fields interact
changing E-Field
E-Field
B-Field
changing B-Field
Accelerating charges produce EM waves
Maxwell’s equations predict that these waves
propagate through a vacuum at a constant speed
c
1
 00
What Has Gone Wrong?
Consider a train:
u
ms-1
The resultant velocity of the person is u + v
v ms-1
Galilean velocity
transformation
The resultant velocity of the light is c not c + v
Is Maxwell wrong? Are Michelson and Morley’s results wrong?
Or is Galileo wrong?
Einstein’s Postulates
We can now state two postulates:
1) The laws of physics are the same in all inertial frames of
reference
2) The speed of light in a vacuum is the same in all inertial
frames of reference
But what is an inertial frame of reference?
Frames Of Reference
A frame of reference is the coordinate system of an observer
y
y
y
v ms-1
x
z
x
z
stationary frame
a ms-2
x
z
frame moving at
constant velocity v
inertial frames
accelerating frame
indistinguishable from
a gravitational field
The Galilean Transformation
Consider two inertial frames, S and S’
S’ is moving at velocity v away from S, and vy = vz = 0 ms-1
y’
y
S
S’
v ms-1
x’
x
z
z’
object in
frame S’
The object has the same y- and z-coordinates in both frames
The time measured at any instant is the same in both frames
The x-coordinate is constant in S’, but changes in S
The Galilean Transformation
At a time t, the x-axis of S’is a distance vt from the x-axis of S
So the x-coordinate in S is the x-coordinate in S’+ vt
All other coordinates are unchanged
This transformation between frames can be written as:
x  x '  vt
y  y'
z  z'
The transformation from S’to S
x '  x  vt
Galilean Transformation
t  t'
The transformation from S to S’
y'  y
z'  z
t'  t
A New Transformation…
The Galilean transformation contradicts Einstein’s second postulate
We need to derive a new transformation, with the properties:
The speed of light must be a constant
It must tend to the Galilean transformation for low velocities
We are not assuming that time is absolute
So we need to be careful when referring to time
An event has both space and time coordinates
This can be written as a four-vector (x, y, z, t)
A Thought Experiment
Alice is on a long train journey, and is rather bored
d
She decides to build a clock using her mirror and a torch
What is the time interval between a pulse leaving and
returning to the torch?
2d
t 
c
A Thought Experiment
Bob is standing on platform 9¾ and watches Alice in the train
d
l
l
v ms-1
vt’
What is the time interval t’ in Bob’s frame of reference? 9 ¾
l
 vt ' 
2
d 

 2 
2
2l 2
 vt ' 
2
t ' 

d 

c
c
 2 
2
ct
2  ct 
 vt ' 
d
 t ' 

 

2
c  2 
 2 
2
2
t ' 
t
v2
1 2
c
The Lorentz Factor
The factor g is the Lorentz Factor
g (v ) 
For v << c, g(v) = 1
As v tends to c, g(v) tends to infinity
g
1
0
c
Speed /
ms-1
1
v2
1 2
c
Time Dilation
g
1
v2
1 2
c
t '  gt
We have this relationship, but what does it mean?
If a body is travelling slowly w.r.t. an observer (g = 1) time
intervals are the same for the body and the observer
If a body is travelling fast w.r.t. an observer (g >> 1) time
intervals appear longer to the observer
If you are in a spacecraft travelling close to c, time will pass
normally for you, but will speed up around you
Notice, therefore, that photons do not age
The Twin Paradox
Once upon a time there were two twins…
Bill
Ben
Ben goes on a journey into space, but Bill stays on Earth
When Ben returns, which of the twins is oldest?
Bill thinks he will be older, as Ben travelled very fast away from him
Ben thinks he will be older, as Bill travelled very fast away from him
Who is right?
Bill is older, because he stayed in the same inertial frame, but Ben
had to accelerate in the rocket
The viewpoints are not identical
Length Contraction
How do lengths appear in a different frame?
Similar derivation as for time dilation
There is no change in the directions perpendicular to travel
In the direction of travel, we can show that:
g
1
1
2
v
c2
1
L  g L0
So, for a body travelling with v close to c, relative to an observer,
the body will appear shorter to the observer, in the direction of v
How do you measure your velocity?
It is meaningless to have an absolute velocity
Velocity can only be measured relative to another body
But what about length contraction and time dilation?
If lengths are shortened, then can you measure the change?
No – the ruler is length-contracted as well
Similarly, clocks slow down, so changes in time can’t be measured
But doesn’t relativity define c as a maximum absolute speed?
Not quite – this is a maximum relative speed, as light has the
same speed (c) relative to any frame
The Lorentz Transformation
We can now derive a relativistic transformation
y’
y
S
S’
v ms-1
x’
x
z’
z
this length is contracted
in the frame S
The y- and z-coordinates will be the same in both frames, as before
From the Galilean transformation, x = x’+ vt
But in frame S, x’is length-contracted to g-1x’
x'
x  vt 
g
x'  g x  vt 
The Lorentz Transformation
To get the time transformation is a little trickier
Notice that the transformations are linear
x'
x  vt 
g
x
x'  vt '
g
x'  g x  vt 
x
g( x  vt )  vt '
g
Lorentz time transformation
vx 

t '  g t  2 
c 

The Lorentz Transformation
We can now state the full Lorentz Transformation:
x'  g x  vt 
y'  y
z'  z
vx 

t '  g t  2 
c 

x '  x  vt
v << c
y'  y
z'  z
t'  t
This satisfies the conditions for the transformation
Can We Go Faster Than Light?
From Newton’s Second Law, F = ma (constant mass)
So if we provide a continuous force, we can achieve v > c
But momentum must be conserved
p  mv
p  gm0 v
relativistic momentum
m0 is the rest mass
The g factor in effect increases the mass, as v increases
A greater force is needed to provide the same acceleration
To reach the speed of light, an infinite force would be required
Conclusions
We have looked at the theory of Special Relativity
This resolved the conflict between Newtonian mechanics, and
Maxwell’s equations
It is a ‘special’ theory, as it doesn’t consider accelerations
This is dealt with in General Relativity:
Acceleration and gravity are equivalent
Geometrical interpretation of gravitation
Much more difficult theory!
Derivation left as an exercise to the student 