# 02_SR1x

Mechanics

Nov 14, 2013 (4 years and 6 months ago)

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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

Time Dilation

Length Contraction

Energy, E = mc
2

1

+
v
2
)

ALL have been observed and
verified experimentally.

Frames of Reference

Reference Frame:

coordinate system in which

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

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

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,

†

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

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