Special Relativity
Galilean Relativity
Galileo Galilei (
ㄶ〰1
Mechanical laws
of physics are
same
for
every inertial observer
. By observing the
outcome of mechanical experiments, one
cannot distinguish
a state of
rest from
a
state of
constant velocity
.
Special Relativity
Albert Einstein (1905)
No experiment
(whether mechanical or
electromagnetic or optical or whatever)
can distinguish
a state of absolute
rest
from uniform straight

line motion
.
What’s the big difference???
Special Relativity
Einstein PREDICTED these effects
at v
挺
Time Dilation
Length Contraction
Energy, E = mc
2
Addition of velocities
⡶
1
+
v
2
)
ALL have been observed and
verified experimentally.
Frames of Reference
Reference Frame:
coordinate system in which
measurements are made.
Does not need to be fixed to
an object in the system.
Lab frame
: fixed with respect to Earth’s
surface. Convenient origin: center of circle.
Particle frame
: view as though we were
"riding" on the particle and making
measurements. The Earth spins around the
particle in this frame.
Both
frames of reference
are
experiencing acceleration
(~0 for lab frame)
and
are
called
non

inertial frames
.
v
Example: Particle in Uniform Circular Motion
Inertial Frames of Reference
A frame where a particle, under the influence of no
net external forces, travels in a straight line at
constant speed (includes v = 0!). (From Newton’s law
of Inertia).
HUH!!??
Any frame travelling at constant v in an inertial frame is also
inertial.
Therefore,
by approximating the earth’s surface as an
inertial frame, a coordinate system attached to a bus moving at
constant velocity,
v’
, is also inertial.
A frame moving at constant velocity…
Galilean Transformations
Lets say that
at t = 0
, the two frames coincide, that is
x =
x
´
= 0, y = y
´
= 0, z = z
´
= 0 at t = t
´
= 0
.
At time t = t
1
we expect:
Consider
two
inertial frames
:
frame
S’
moves with
constant velocity, , with respect to frame S.
S
x
z
y
S
x
z
y
vt
1
t =0
t =t
1
v
z’
S’
x’
y’
v
z’
S’
x’
y’
v
Galilean Transformations:
Position
Consider the
coordinate transformation for
the
horizontal position of a particle at rest in S
´
(x
1
´
, y
1
´
,
z
1
´
)
to frame S(x
1
, y
1
, z
1
).
S
x
z
y
vt
1
t =t
1
z’
S’
x’
y’
v
x
1
’
x
1
x
1
=
x
1
´
+ vt
1
x’ = x
–
vt
y’ = y, z’ = z and t’ = t
Simplify by
dropping
subscripts
Galilean Transformations:
Velocity and Acceleration
For
velocities
we
use symbol u
to avoid confusion
u
x
’ = dx’/dt’ = d(x
–
vt)/dt = dx/dt
–
v = u
x
–
v.
u
y
’ = dy’/dt’ = dy/dt = u
x
u
z
’ = dz’/dt’ = dz/dt = u
z
Acceleration :
a’ =du’/dt = d(u

v)/dt = du/dt
–
dv/dt = du/dt = a
u’ = u
–
v
OR
u = u’+v
0
Accelerations are the same in all inertial frames
.
Galileo’s contribution to physics.
Children playing catch on a bus make the same throws as if
they were on the ground if they are same distance apart.
a’ = a
Read pages
5

6 of text
Galilean Relativity
Postulate:
The laws of MECHANICS are the same in all
inertial frames.
Result:
Galilean transformations shown above
Prediction:
Measurement of
Speed of light
waves, c,
depends on
the
observer’s velocity
.
v
v+c
Ground

based observer
c
Ship

based observer
Problems with Galilean Relativity
Maxwell
’s equations
predicted c = 1/(
0
0
)
½
=
2.9979 m/s,
for the speed of light in the vacuum. It
was
believed
that
this speed would vary
according the frame of reference
. The
above
value
could
only
be
measured
in a
special frame
of absolute rest
with respect to the medium
believed to propagate light, the ether.
Earth’s speed
in orbit around the sun (
~30 km/s
) was the
fastest available speed
for testing dependence of c upon
relative motion of earth in “ether wind”.
30 km/s
Measuring the speed of Light
Fizeau’s apparatus 1849
Not accurate enough
Measuring the speed of Light
Michelson

Morley
apparatus
1887
Measuring the speed of Light
Michelson

Morley
apparatus
1887
Result from Michelson

Morely
Interometry
Compared velocity of light beams at right angles
Unless the ether happens to be moving at the same
velocity as earth,
one of the two perpendicular beams
SHOULD HAVE BEEN affected by the “ether wind”.
Astonished scientists found c was constant!!!
A
repeat experiment
months later to ensure that ether
velocity did not happen to match the earth’s velocity
during the previous run
confirmed the original result
.
30 km/s
Michelson

Morely Aftermath
Did scientists abandon the concept of the ether?
NO! Not really see:
http://www

groups.dcs.st

and.ac.uk/~history/HistTopics/Special_relativity.html
They invented miraculous properties for the ether and
went on using the correct Lorentz (which we will see later)
transforms as “fudges” until Einstein proved these were
physically correct.
What is light moving relative to???
Einstein’s Postulates
1.
All of the laws of
PHYSICS are
the SAME for every
observer
in an inertial frame.
•
This one was easy to accept
–
but only worked
using Lorentz transformations (which nobody
understood) for electromagnetic phenomena
2.
The speed of light is a constant, independent of
the motion of the source.
•
This one sounded crazy (given beliefs of the time)
but had been proven by experiments
Together
these postulates form the
basis
for the
Special Theory
of Relativity
which generated amazing results which demolished
the understanding of space and time as separate things.
Events and Observers
An Event
is something that happens at a particular
point in space and at a particular instant in time. A
hockey game is not an event. E.g. a one hour hockey
is NOT an event, a face

off is. Events do not “belong”
in any reference frame.
An Observer
is a suitable recording device (e.g. a
person) who does exist in a particular reference frame.
Relativity of Time
Time
is
measured
using
a
clock
with a regular repeated
motion.
As
a thought experiment,
consider a “light
clock” at rest in
frame
S
´
which moves at v relative to
S
source/detector
S
´
(x
´
,y
´
)
v
y
´
x
´
mirror
d
x
0
´
Observer
at
rest in
frame S’
observes
that
light pulse takes,
T
´
= 2d/c
to leave the source,
bounce off the mirror and
return to the detector
Relativity of Time
However, according to an observer at rest in S
source/detector
S(x,y)
v
y
x
mirror
d
x
0
x
0
+v
T
In S
the
light travels farther
than 2d (as in S
´
). Light speed,
c, is
the same
.
Therefore
T
> 2d/c =
T
´
.
T
T
´
because c limits
the maximum velocity with which information is carried.
Proper Time
x
0
S(x,y)
v
y
x
d
x
0
+v
T
For
any object moving at constant velocity
relative in
S, we
can always define a frame
S
´
where the
object is
at rest
. In S
´
the beginning and end of the time interval
(i.e. the emission and detection of the light pulses) are
both at the same position.
x
´
S
´
(x
´
,y
´
)
v
y
´
d
x
0
´
This is called the
“one position” or
“proper” time,
T
0
.
For the time
measured in ANY
OTHER inertial
frame the observer
is at two different
positions at
endpoints. These
times are called
“two position”
times,
吮
†
T
0
T
Time Dilation
S(x,y)
v
y
x
d
v
T
From earlier,
in S
´
´
= 2d/c =
T
0
AB
=(d
2
+ (v
T/2)
2
)
1/2
= BC
T
=(AB + BC)/c
= (2/c) (d
2
+ (v
T/2)
2
)
1/2
(
T)
2
= (4/c
2
) (d
2
+ (v
T/2)
2
)
(
T)
2
= (4d
2
+ (v
T)
2
)
/c
2
(
T)
2
c
2
= 4d
2
+ v
2
(
T)
2
Time Dilation
S(x,y)
v
y
x
d
v
T
From earlier,
in S
´
´
= 2d/c =
T
0
(
T)
2
(c
2

v
2
)= 4d
2
c
2d
/c
v
1
1
v
c
2d
ΔT
2
2
2
2
T
=
T
0
2
2
/c
v
1
1
where…
Time Dilation
Since, 0
瘠
1
v = 0
㴠1
v
Time dilation effects have been observed on highly accurate
clocks carried aboard jet liners. They have also been
observed in Muon decay.
Lorentz Transformations
S
x
z
y
t =t
1
y’ = y, z’ = z and t’
t
So far, we have only considered time effects the
special case where x = 0. Consider the following:
x =
(
x’ + vt’)
x’ =
(
x
–
vt)
x =
(
x’ + vt’)
x’ =
(
x
–
vt)
Einstein’s 2
nd
postulate means
these “light sphere”
will be the same size
in S and S
’
z’
S’
x’
y’
v
t’ =t
1
’
z’
S’
x’
y’
v
t’ =t
1
’
Lorentz Transformations
x =
(
x’ + vt’)
x’ =
(
x
–
vt)
x =
(
(
x
–
vt) + vt’)
v
x
1
t
t
2
2
Exercise
Lorentz Transformations
S
x
z
y
t =t
1
z’
S’
x’
y’
v
t’ =t
1
’
x
2
+ y
2
+ z
2
= c
2
t
2
x’
2
+ y’
2
+ z’
2
= c
2
t’
2
x’ =
(
x
–
vt)
v
x
1
t
t
2
2
What is
?
Lorentz Transformations
x
2
+ y
2
+ z
2
= c
2
t
2
x’
2
+ y’
2
+ z’
2
= c
2
t’
2
x’ =
(
x
–
vt)
v
x
1
t
t
2
2
2
2
2
2
2
v
x
1
t
2
2
c
2
2
z
y
vt
x
y’ = y
z’ = z
Lorentz Transformations
2
2
4
2
2
2
v
x
1
v
xt
1
t
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
c
c
c
z
y
t
v
xvt
x
Lorentz Transformations
x
2
+ y
2
+ z
2
= c
2
t
2
v
xt
1
t
v
1
2
2
2
2
4
2
2
2
2
2
2
2
2
2
2
2
2
c
xvt
v
c
z
y
c
x
2
2
2
2
2
1
0
Comparing to:
Equation of coefficients gives three conditions…
c
2
Lorentz Transformations
1
2
2
2
2
2
4
v
1
2
c
Any one of those conditions could be used to
derive
. Using the x
2
coefficient:
2
2
2
1
2
2
v
1
2
c
Lorentz Transformations
Any one of those conditions could be used to
derive
. Using the x
2
coefficient:
1
2
2
2
v
1
2
c
2
2
v
2
c
c
2
2
2
2
c
c
2
2
v
2
2
2
v/c
v
2
1
1
c
c
As
before
Lorentz Transformations
So now we have
2
v/c
1
1
x’ =
(
x
–
vt)
y’ = y
z’ = z
t’ =
(t
–
vx/c
2
)
Exercise
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