Einstein’s Postulates
• In 1905 Albert Einstein published a paper on the
electrodynamics of moving bodies. In this paper, he
postulated that absolute motion can not be detected by
any experiment. That is, there is no ether. The reference
frame connected with earth is considered to be at rest
and the velocity of the light will be the same in any
direction. His theory of special relativity can be derived
from two postulates:
• Postulate 1: Absolute uniform motion can not be
detected.
• Postulate 2: The speed of light is independent of
the motion of the source.
Terminology of Theory of Special
Relativity
• Event. Such as collisions, creation, or annihilation of
particles. Events are absolute. If an event happens in one
reference frame (RF), it also occurs in any other. TSR
considers only point events (having negligible extension in
space)
• In a given RF, coordinates of an event are given by four
coordinates (t,x,y,z). An event then can be represented by
a point in 4D space-time (Minkovsky space).
Interval
• Interval. Interval is a mathematical form of second
postulate. In two RF’s coordinates on an event are
different.
• Consider two events (event 1 and 2): an emission and
absorption of a photon. In one RF (S), coordinates are
(t1 , x1 , y1 , z1) and (t2 , x2 , y2 , z2) . We have:
(x 2− x 1 )2 +( y 2 − y 1) 2+( z 2 − z 1 )2= c 2 (t 2 − t 1)2
• In other RF (S') the coordinates of these two events are
(t1 ', x1' , y1' , z1' ) and (t2' , x2' , y2' , z2').
(x ' 2 − x ' 1) +( y ' 2 − y ' 1) +( z' 2− z' 1) = c (t ' 2 − t ' 1 )
2
2
2
2
2
The interval between events 1 and 2 is
s12 = c t2  t1   x2  x1    y2  y1   z2  z1 
2
2
2
2
2
When s12= 0 in one RF, then s'12= 0 in any other RF.
The interval is zero only if the two events are happened at the
speed of light. Zero interval is also called light-like interval.
Interval
For light-like intervals we proved that s12=0 in all RFs.
Consider a general case: two infinitely close events 1 and
2. Consider three RFs: S, S', and S''.
S' moves with respect to S with speed v1, S'' moves with
respect to S' with speed v2, and S'' moves with respect to S
with speed v. The events are separated by intervals ds,
ds', and ds'' in corresponding RFs. For example, in S:
ds = c 2 dt 2  dx 2  dy 2  dz 2
ds, ds' and ds'' are related to each other by a function of
velocity, say a(v2), that can only depend on the magnitude of
their relative velocities.
Interval
ds, ds' and ds'' are related to each other by a function, say
a(v2), that can only depend on the magnitude of their
relative velocities:
 
ds' = a v12 ds;
 
ds' ' = a v22 ds' ;
 
ds' ' = a v 2 ds
a (v 22) ds'= a (v 2 )ds ;
a (v 22)a (v 12) ds= a (v 2) ds
a (v 22)a (v 12)= a (v 2 )
Interval
The equation we obtained
a (v 22)a (v 12)= a (v 2 )
does not depend on angles between v1, v2, and v → a(v2) is a
constant. Since a2=a → a=1 → ds=ds' → s=s'+c. For light-like
interval s=s' → c=0.
Therefore for any two events the interval
s12 = c t2  t1   x2  x1    y2  y1   z2  z1 
2
2
2
2
2
is the same in all RFs.
s=s'
This is the mathematical form of the postulate on the constant
speed of light.
The Lorentz Transformation
Galilean transformations:
x = x’ + vt’, y = y’, z = z’, t = t’
The inverse transformations are
x’ = x – vt,
y’ = y, z’ = z, t’ = t
These equations are consistent with experimental observations as
long as v is much less than c. They lead to the familiar classical
addition law for velocities. If a particle has velocity ux = dx/dt in
frame S, its velocity in frame S’ is
dx dx d
dx
u x 

 ( x  v t) 
 v  ux  v
dt dt dt
dt
from here we have
d u x d u x
ax 

 a x
dt
dt
The Lorentz Transformation
It should be clear that the Galilean transformation
is not consistent with Einstein’s postulates of special
relativity.
If light moved along the x axis with speed ux’ = c
in S’, these equations imply that the speed in S is ux = c
+ v rather than ux=c, which is not consistent with
Einstein’s postulates and experiment.
The classical transformation equations must
therefore be modified .
The Lorentz Transformation
We assume that the relativistic transformation
equation for x is the same as the classical equation
except for a constant multiplier on the right side:
x   ( x' vt' )
where γ is a constant that can depend on v and c but
not on coordinates. The inverse transformation in this
case
x'   ( x  vt)
Lorentz Transformation
The transformation described by these
equations is called the Lorentz transformation.
Lorentz’s equations replace the flawed Galileo
transformation equations in relating the
measurements of two different observers in uniform
motion relative to each other.
Lorentz Transformation
x'  γ x  ut 
y'  y
z'  z
 xu 
t'  γ  t  2 
c 

1
γ
1β2
u
β
c
1
γ
2
u 
1  
 
c
Addition of Velocities
We start by taking the derivative with respect to t' of
the first Lorentz transformation equation:
 dx
dx'
dt 
 γ u 
dt'
dt' 
 dt'
From Calculus, we have that:
 dx dt
dx'
dt 
 γ
u 
dt'
dt' 
 dt dt'
 dx dt
dx'
dt 
 γ
u 
dt'
dt' 
 dt dt'
dx'
 dx
 dt
 γ   u
dt'
 dt
 dt'
Using the definition of velocity, we have:
dt
v '  γ v  u 
dt'
Differentiating the fourth Lorentz equation with
respect to t, we obtain:
dt'  dt u dx 

 γ   2
dt
 dt c dt 
dt' 
u
 γ 1  2
dt
 c
2
 γ
v  
 c2
c2  u v




dt
c
1


2
dt' γ(c  u v) γ1 u v 

2 
c 

We now substitute this result into our velocity equation
to obtain:
 vu 
vu
c
v'

u v  c  βv 
1
2
c
where v is velocity of object seen by unprimed observer, v '
is velocity of object seen by primed observer, u is velocity
of the primed observer as seen by unprimed observer.
EXERSICE: Show that both observers agree on the
speed of light.
SOLUTION: We determine the speed v' as seen by the
primed observer for a beam of light v = c seen by the
unprimed observer. Using the velocity addition formula
we have:
cu cu
v'

cu
u
1
1
2
c
c
cu 
cc
 

c

u


Relative Speeds of Cosmic Rays
Suppose that two cosmic ray protons approach Earth
from opposite directions as shown in Fig.
The speeds relative to the Earth measured to be v1=0.6c
and v2=-0.8c. What is Earth’s velocity relative to each
proton, and what is the velocity of each proton relative to
the other?
Relative Speeds of Cosmic Rays
Consider each particle and the Earth to be inertial reference frames
S’, S’’, and S with they respective x-axis parallel. With this
arrangements v1=u1x=0.6c and v2=u2x= -0.8c. Thus, the speed of Earth
measured in S’ is v ’Ex= -0.6c and the speed of Earth measured in S’’ is
v ’’Ex=0.8c.
To find the speed of proton 2 with respect to proton 1, we apply
equation:
vu
v '
uv
1
c2
to compute the speed of the particle 2 in S’:
'
u2 x
0.8c  (0.6c)
1.4c


 0.95c
2
1.48
1  [(0.6c)(0.8c)] / c
Relative Speeds of Cosmic Rays
The observer in S’’:
0.6c  (0.8c)
1.4c
u 

 0.95c
2
1  (0.6c)( 0.8c) / c
1.48
''
1x
Lorentz-Fitgerald (Length) Contraction
Fitzgerald's length contraction is now a direct consequence of
the Lorentz transformation. Consider two different observers in
relative motion who measure the length of a box as shown below:
S
S’
x1
observer sees box moving
at u
observer sees stationary
box of length L 0  x 2'  x 1'
At a single instant of time t, the unprimed observer measures
the box's length in S as
L0  x2'  x1'
while the observer in S’ measures the box's length as
L  x2  x1
'
Using the Lorentz transformation equation for x', we have that
'
'




x2
x1
'
L    ut2     ut1 

 

L
'
x x
'
2

'
1

ut  ut


x x
'
2

'
1
Therefore, we have that
L
'
L0

Lorentz-Fitzgerald (Length) Contraction
'
L  Lo
u
1  
c
2
where L’ is the improper length
Lo is the proper length
u is the relative speed between the
observers
Length Contraction
Warning: Your everyday use of English terms can get you
in trouble in this material. The term "proper" is not
meant to indicate that its the "correct" or "right" answer.
Two observers might measure the length of a train as 10
m when they see the train as stationary. If one observer
rides in the train as it moves down the track, he will
measure the train's length as 10 m while an observer
standing by the track sees the train shorter than 10 m.
Both Observers Are CORRECT!!! The question "What is
the length of the train?" is meaningless!! You must
specify the frame. There is no such thing as absolute
distance anymore!
Time Dilation
Another interesting aspect of relativity is
that moving clocks always run slower.
This is not an artifact of the clock! It is a
consequence of the nature of time itself
according to Einstein.
All clocks will behave this way including
your biological processes!
Time Dilation
Let us consider a "Light Clock" in which one unit
of time corresponds to the time it takes a light pulse
to travel between two meters a distance L apart. We
will place the clock on a train traveling at speed u
with the unprimed observer.
Mirror
L
Light Path As Seen By Train Observer
Time Dilation
For the unprimed observer on the train, the
time it takes the light pulse to make a single
tick is given by
L
t
c
Time Dilation
The primed observer standing by the railroad track
sees the train and clock pass with a speed of u. Thus,
the path of the light beam appears to follow the
diagonal path shown below:
ct'
L
u t'
u
Time Dilation
From the Pythagorean theorem, t' is given by
c t' 
2
 u t'   L
2
2
c t'  u t'  L
2
c
2
2
2

2
 u t'  L
2
2
2
L
t'  2

2
c u
2

2

2
2
L
2

 u  
2
c 1  


c




Time Dilation
L
t' 
c
u 
1  
c
2
L
γ
c
We now plug in our results for the unprimed observer
and we find that
t'  γ t
Thus, our observers do not agree upon time!
Time Dilation Equation
T 
To
u
1  
c
2
where To is the proper time
T’ is the improper time
Time Dilation as a Function of Speed
Two events occur at the same point x’ at times t’1 and t’2 in frame S’,
which is traveling at speed u relative to frame S. (a) What is the
spatial separation of these events in frame S? (b) What is the
temporal separation of these events in frame S?
Two events occur at the same point x’ at times t’1 and t’2 in frame
S’, which is traveling at speed u relative to frame S. (a) What is
the spatial separation of these events in frame S? (b) What is the
temporal separation of these events in frame S?
The spatial separation in S is x2 - x1, where x2 and x1 are the
coordinates of the events in S, which can be found using Lorentz
transformation for position:
x   ( x  vt )
'
'
x   ( x  vt )
'
'
(a) The spatial separation in S is Δx = x2 – x1:
x1   ( x '  vt1' )
x2   ( x '  vt2' )
x  x2  x1   ( x '  vt2' )   ( x '  vt1' ) 
 v(t 2'  t1' ) 
v(t 2'  t1' )
v2
1 2
c
x 
v(t  t )
'
2
'
1
2
v
1 2
c
Using the time dilation formula
T  T '
(b)
t  t  t   (t  t )
'
2
t 
'
1
(t  t )
'
2
'
1
2
v
1 2
c
'
2
'
1
Simultaneity
We saw that proper time is the time interval between two events
that occur at the same point in some reference frame.
REMEMBER: In each frame there is a clock at each point in space, and the time
of an event in a given frame is measured by a clock at that point.
However, in another reference frame moving relative to the first,
the same two events occur in different places, so two clocks are
needed to record the times.
The time of each
event is measured on a different clock, and the interval is found by
subtraction. The procedure requires that the clocks be
synchronized.
Synchronized clocks:
Two clocks that are synchronized in one reference frame
typically are not synchronized in any other frame moving
relative to the first frame.
Simultaneous events:
Two events that are simultaneous in one reference frame
typically are not simultaneous in other frame that is moving
relative to the first.
We will thus define two events to be simultaneous in an
inertial reference frame if the light signal from the events
reach an observer halfway between them at the same time
as recorded by a clock at that location, called a local clock.
Inertial reference frame formed from a lattice of measuring rods with a clock at
each intersection. The clocks are all synchronized using a reference clock. The
measuring rods could all have different length as required by the scale and
precision of the measurements being considered. The three space dimensions are
the clock positions. The fourth spacetime dimension, time, is shown by readings
on the clocks.
An astronomer on Earth observes a meteoroid in the southern sky
approaching the Earth at a speed of 0.800c. At the time of its
discovery the meteoroid is 20.0 ly from the Earth. Calculate (a) the
time interval required for the meteoroid to reach the Earth as
measured by the Earthbound astronomer, (b) this time interval as
measured by a tourist on the meteoroid, and (c) the distance to the
Earth as measured by the tourist.
An astronomer on Earth observes a meteoroid in the southern sky
approaching the Earth at a speed of 0.800c. At the time of its
discovery the meteoroid is 20.0 ly from the Earth. Calculate (a) the
time interval required for the meteoroid to reach the Earth as
measured by the Earthbound astronomer, (b) this time interval as
measured by a tourist on the meteoroid, and (c) the distance to the
Earth as measured by the tourist.
(a)
The 0.8c and the 20 ly are measured in the Earth frame, so
in this frame,
.
x 20 ly 20 ly 1c
t 

 25.0 yr
v 0.8c
0.8c 1 ly yr
(b) We see a clock on the meteoroid moving, so we do not
measure proper time; that clock measures proper time.
t tp
tp 
t


25.0 yr
1 1-v2 c2
 25.0 yr 1 0.82  25.0 yr 0.6  15.0 yr
An astronomer on Earth observes a meteoroid in the southern sky
approaching the Earth at a speed of 0.800c. At the time of its
discovery the meteoroid is 20.0 ly from the Earth. Calculate (a) the
time interval required for the meteoroid to reach the Earth as
measured by the Earthbound astronomer, (b) this time interval as
measured by a tourist on the meteoroid, and (c) the distance to the
Earth as measured by the tourist.
(c) Method one: We measure the 20 ly on a stick stationary in
our frame, so it is proper length. The tourist measures it to be
contracted to
Lp
20 ly
20 ly
L


 12.0 ly

1 1 0.82 1.667
Method two: The tourist sees the Earth approaching at
 0.8 ly
yr15 yr  12.0 ly
Classical Doppler Shift
Anyone who has watched auto racing on TV is
aware of the Doppler shift.
As a race car approaches the camera, the sound of
its engine increases in pitch (frequency).
After the car passes the camera, the pitch of its
engine decreases.
We could use this pitch to determine the relative
speed of the car. This technique is used
in many real world applications including ultrasound
imaging, Doppler radar, and to determine the motion
of stars.