Transcript CHAPTER 2: Special Theory of Relativity

Physics 334
Modern Physics
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CHAPTER 4
Special Theory of Relativity 1
4-1 Foundations of Special Relativity
The Experimental Basis of Relativity
Einstein’s Postulates
Albert Michelson
4-2 Relationship between Space and
(1852-1931)
Time
The Lorentz Transformation
Time Dilation and Length Contraction It was found that there was
The Doppler Effect
no displacement of the
The Twin Paradox and Other Surprises interference fringes, so that
the result of the experiment
was negative and would,
therefore, show that there is
still a difficulty in the theory
itself…
- Albert Michelson, 1907
Newtonian (Classical) Relativity
Newton’s laws of motion must be implemented with respect to (relative to)
some reference frame.
y
y’
z’
x’
z
x
A reference frame is called an inertial frame if Newton’s laws are
valid in that frame.
Such a frame is established when a body, not subjected to net
external forces, moves in rectilinear motion at constant velocity.
Difference Between Inertial and Non-Inertial
Reference Frame
Newtonian Principle of Relativity
If Newton’s laws are valid in one reference frame, then they are also
valid in another reference frame moving at a uniform velocity relative to
the first system.
This is referred to as the Newtonian principle of relativity or Galilean
invariance.
If the axes are also parallel, these frames are said to be Inertial
Coordinate Systems
The Galilean Transformation
For a point P:
In one frame S: P = (x, y, z, t)
In another frame S’: P = (x’, y’, z’, t’)
The Inverse Relations
x  x  vt
y  y
z  z
t  t
1. Parallel axes
2. S’ has a constant relative
velocity (here in the x-direction)
with respect to S.
3. Time (t) for all observers is a
Fundamental invariant, i.e.,
it’s the same for all inertial
observers.
x  x  vt 
y  y
z  z
t  t
A need for ether
In Maxwell’s theory, the speed of
light, in terms of the permeability
and permittivity of free space, was
given by:
v
Thus the velocity of light is constant
Aether was proposed as an
absolute reference system in which
the speed of light was this constant
and from which other
measurements could be made.
Properties of Aether:
Low density
Elasticity
Transverse waves
Galilean transformation
Maxwell’s equations are not invariant under
Galilean transformations.
The Michelson-Morley experiment was
an attempt to show the existence of
aether.
Michelson-Morley experiment
Michelson and Morley realized that
the earth could not always be
stationary with respect to the
aether. And light would have a
different path length and phase shift
depending on whether it
propagated parallel and antiparallel or perpendicular to the
aether.
Perpendicular
propagation
Supposed
velocity of earth
through the
aether
Parallel and
anti-parallel
propagation
Michelson-Morley
Experimental Analysis
Exercise 4-1: Show that the time difference between path
differences after 90° rotation is given by:
v2
2  t  t   2 L 3
c
Recall that the phase shift is
w times this relative delay:
v2
2w L 3
c
or:
L v2
4
 c2
The Earth’s orbital speed is: v = 3 × 104 m/s , and the interferometer
size is: L = 1.2 m, So the time difference becomes: 8 × 10−17 s, which,
for visible light, is a phase shift of: 0.2 rad = 0.03 periods
The Michelson interferometer should’ve
revealed a fringe shift as it was rotated with
respect to the aether velocity. MM expected
0.4 of the width of a fringe, and could only
see 0.01 equal to the uncertainty in the
measurement.
Interference fringes showed
no change as the
interferometer was rotated.
Thus, aether seems not to exist!
Einstein’s Postulates
Albert Einstein was only two years old
when Michelson and Morley reported
their results.
At age 16 Einstein began thinking
about Maxwell’s equations in moving
inertial systems.
In 1905, at the age of 26, he
published his startling proposal:
the Principle of Relativity.
It nicely resolved the Michelson and
Morley experiment (although this
wasn’t his intention and he
maintained that in 1905 he wasn’t
aware of Michelson and Morley’s
work…)
Albert Einstein (1879-1955)
It involved a fundamental
new connection between
space and time and that
Newton’s laws are only an
approximation.
Einstein’s Two Postulates
With the belief that Maxwell’s equations must be valid in
all inertial frames, Einstein proposed the following
postulates:
The principle of relativity: The laws of physics are the same in all
inertial reference frames.
The constancy of the speed of light: The speed of light in a
vacuum is equal to the value c, independent of the motion of
the source.
Relativity of Simultaneity
In Newtonian physics, we
previously assumed that t’ = t.
With synchronized clocks, events
in S and S’ can be considered
simultaneous.
Einstein realized that each system
must have its own observers with their
own synchronized clocks and meter
sticks.
Events considered simultaneous in S
may not be in S’.
Also, time may pass more slowly in
some systems than in others.
The constancy of the speed of light
Lorentz Transformation
Exercise 4-2: The equations for a spherical wavefronts in S is
x2+y2+z2=c2t2 , Show that the equation for the spherical wavefronts in
S’ cannot be x’2+y’2+z’2=c2t’2 in the Galilean transformation.
Exercise 4-3: Show that x’ = g (x – vt) so that x = g’ (x’ + vt’) , yields
the g factoid
1
g
v2
1 2
c
and that for small velocities
1 v2
g  1 2
2c
Lorentz Transformation
Exercise 4-4: Use x’ = g (x – vt) and x = g’ (x’ + vt’) , to find t’ = g (t – v x
/c2)
Exercise 4-5: Use x’ = g (x – vt) and t’ = g (t – v x /c2) to show that the
equations for spherical wave fronts in S and S’ are the same.
Lorentz Transformation Equations
A more symmetrical form:
  v/c
g
1
1 v / c
2
2
Properties of g
Recall that  = v / c < 1 for all observers.
g equals 1 only when v = 0.
In general:
Graph of g vs. :
(note v < c)
The complete Lorentz Transformation
x 
x  vt
1  v2 / c 2
y  y
x
z  z
t 
t  vx / c 2
1  v2 / c 2
x  vt 
1  v2 / c 2
y  y
z  z
t
t   vx / c 2
1  v2 / c 2
If v << c, i.e., β ≈ 0 and g ≈ 1, yielding the familiar Galilean transformation.
Space and time are now linked, and the frame velocity cannot exceed c.
Relativistic Velocity Transformation
Exercise 4-6: Suppose a shuttle takes off quickly from a space ship
already traveling very fast (both in the x direction). Imagine that the
space ship’s speed is v, and the shuttle’s speed relative to the
space ship is u’. What will the shuttle’s velocity (u) be in the rest
frame?
ux  v
dx
g (dx  v dt )
ux 


2
dt g [dt   (v/c ) dx] 1+ ux v/c 2
uy
dy
dy
uy 


2
dt g [dt   (v/c ) dx] g (1+ ux v/c 2 )
uz
dz
dz
uz 


2
dt g [dt   (v/c ) dx] g (1+ ux v/c 2 )
The Inverse Lorentz Velocity Transformations
If we know the shuttle’s velocity in the rest frame, we can calculate it
with respect to the space ship. This is the Lorentz velocity
transformation for u’x, u’y , and u’z. This is done by switching primed
and unprimed and changing v to –v:
ux  v
dx
ux 

dt 1  ux v/c 2
uy
dy
uy 

2
dt g (1  ux v/c )
uz
dz
uz 

dt g (1  ux v/c 2 )
Lorentz velocity transformation
Example: As the outlaws escape in their really fast getaway ship at
3/4c, the police follow in their pursuit car at a mere 1/2c, firing a
bullet, whose speed relative to the gun is 1/3c. Question: does the
bullet reach its target a) according to Galileo, b) according to
Einstein?
vpg = 1/2c
police
vbp = 1/3c
vog = 3/4c
bullet
vpg = velocity of police relative to ground
vbp = velocity of bullet relative to police
vog = velocity of outlaws relative to ground
outlaws
Galileo’s addition of velocities
In order to find out whether justice is met, we need to compute
the bullet's velocity relative to the ground and compare that with
the outlaw's velocity relative to the ground.
In the Galilean transformation, we simply add the bullet’s velocity
to that of the police car:
vbg  vbp  v pg  vbg  13 c  12 c  65 c
Therefore,
5
6
c  34 c  justice is served!
Einstein’s addition of velocities
Due to the high speeds involved, we really must relativistically
add the police ship’s and bullet’s velocities:
ux  v
ux 
1  ux v c2
 vbg 
5
7
1 
vbg 
1
3
1
3
vbp  v pg
1  vbp v pg c 2
c  12 c
5

7c
2
1
c  2 c / c
c  34 c  justice is not served!
Gedanken
(Thought)
experiments
It was impossible to achieve
the kinds of speeds
necessary to test his ideas
(especially while working in
the patent office…), so
Einstein used Gedanken
experiments or Thought
experiments.
Young Einstein
The complete Lorentz Transformation
x 
x  vt
1  v2 / c 2
y  y
z  z
t 
t  vx / c 2
1  v2 / c 2
x
Length
contraction
1  v2 / c 2
y  y
z  z
Simultaneity
problems
t
Time
dilation
x  vt 
t   vx / c 2
1  v2 / c 2
If v << c, i.e., β ≈ 0 and g ≈ 1, yielding the familiar Galilean transformation.
Space and time are now linked, and the frame velocity cannot exceed c.
Time Dilation and Length Contraction
More very interesting consequences of the Lorentz Transformation:
Time Dilation:
Clocks in S’ run slowly with respect to stationary clocks in S.
Length Contraction:
Lengths in S’ contract with respect to the same lengths in
stationary S.
We must think about how we
measure space and time.
In order to measure an object’s length in space,
we must measure its leftmost and rightmost
points at the same time if it’s not at rest.
If it’s not at rest, we must ask someone
else to stop by and be there to help out.
In order to measure an event’s duration in time,
the start and stop measurements can occur at
different positions, as long as the clocks are
synchronized.
If the positions are different, we must ask someone
else to stop by and be there to help out.
Proper Time
To measure a duration, it’s best to use
what’s called Proper Time.
The Proper Time, t, is the time between
two events (here two explosions) occurring
at the same position (i.e., at rest) in a system as measured by a
clock at that position.
Same location
Proper time measurements are in some sense the most fundamental
measurements of a duration. But observers in moving systems, where
the explosions’ positions differ, will also make such measurements.
What will they measure?
Time Dilation and Proper Time
Frank’s clock is stationary in S where two explosions occur.
Mary, in moving S’, is there for the first, but not the second.
Fortunately, Melinda, also in S’, is there for the second.
Melinda
Mary
S’
Mary and Melinda
are doing the best
measurement that
can be done.
Each is at the right
place at the right
time.
S
Frank
If Mary and
Melinda are
careful to time and
compare their
measurements,
what duration will
they observe?
Time Dilation
Mary and Melinda measure the times for the two explosions in system S’ as t’1
and t’2 . By the Lorentz transformation:
(t2  t1 )  (v c)( x2  x1 )
'
'
'
t  t  t 
2
2
2 1
1 v c
This is the time interval as measured in the frame S’.
This is not proper time due to the motion of S’: x1  x2.
Frank, on the other hand, records x2 – x1 = 0 in S with a (proper)
time: t = t2 – t1, so we have:
'
t 
t
1  v2 c2
 gt  gt
Time Dilation
1) ∆t ’ > ∆t:(γ >1) the time measured
between two events at different
positions is greater than the time
between the same events at
one position: this is time dilation.
2) The events do not occur at the same space and time
coordinates in the two systems.
3) System S requires 1 clock and S’ requires 2 clocks for the
measurement.
4) Because the Lorentz transformation is symmetrical, time
dilation is reciprocal: observers in S see time travel
faster than for those in S’. And vice versa!
Time Dilation Example: Reflection
Mirror
L
Mirror
v∆t/2
S’
v
Mary
S
Frank
Exercise 4-7: Show that the event in its rest frame (S’) occurs
faster than in the frame that’s moving compared to it (S).
Fred
Time stops for a light wave
Because:
t
t' 
1 v c
2
2
 gt  gt
And, when v approaches c:
1
1 v c
2
2

For anything traveling at the speed of light:
'
t  
In other words, any finite interval at rest appears infinitely long at the
speed of light.
Proper Length
When both endpoints of an
object (at rest in a given frame)
are measured in that frame,
the resulting length is called
the Proper Length.
We’ll find that the proper length is the largest length observed.
Observers in motion will see a contracted object.
Lp
Length Contraction
Frank Sr., at rest in system S, measures the length of his
somewhat bulging waist:
← Proper length
Lp = xr  xℓ
Frank Sr.
Now, Mary and Melinda S’, measure it, too, making simultaneous
measurements (t’l = t’r ) of the left, x’l , and the right x’r endpoints
Frank Sr.’s measurement in terms of Mary’s and Melinda’s:
'
( xr'  x' )  v(tr'  t ' )
L
l
l 
L p  xr  x 
 g L'
l
1  v2 c2
1  v2 c2
where Mary’s and Melinda’s measured length is:
Lp
'
L
 L p 1  v2 c2
g
L  xr  x
Moving objects
appear thinner!
Length contraction is also reciprocal.
So Mary and Melinda see Frank Sr. as thinner than he is in his
own frame.
But, since the Lorentz transformation is symmetrical, the effect
is reciprocal: Frank Sr. sees Mary and Melinda as thinner by a
factor of g also.
Length contraction is also known as Lorentz contraction.
Also, Lorentz contraction does not occur for the transverse
directions, y and z.
Lorentz
Contraction
v = 10% c
v = 80% c
A fastmoving
plane at
different
speeds.
v = 99% c
v = 99.9% c
Experimental Verification of Time Dilation
Cosmic Ray Muons: Muons are produced in the upper atmosphere in
collisions between ultra-high energy particles and air-molecule nuclei.
But they decay (lifetime = 1.52 ms) on their way to the earth’s surface:
N  N0e
t /t
No relativistic correction
With relativistic correction
Top of the atmosphere
Now time dilation says
that muons will live
longer in the earth’s
frame, that is, t will
increase if v is large.
And their average
velocity is 0.98c!
Detecting muons to see time dilation
At 9000 m it takes muons (9000/0.998c ~ 30 μs) about 15 lifetimes to
reach earth. If No = 108 and t = 15τ, N ~ 31 muons should reach earth.
From relativistic
approach, the distance
traveled is only 600m at
that speed in 1 lifetime (2
μs) and therefore N =
3.68 x 107
Experiments have
confirmed this relativistic
prediction
Space-time Invariants
This is a quantity that is invariant under Lorentz transformation. It is defined
in the following way;
(∆s)2 = (c2∆t2) - [∆x2 + ∆y2 + ∆z2]
The quantity Δs2 between two events is invariant (the same) in any inertial
frame.
Δs is known as the space-time interval between two events.
There are three possibilities for Δs2:
Δs2 = 0: Δx2 = c2 Δt2, and the two events can be connected only by a light signal.
The events are said to have a light-like separation.
Δs2 > 0: Δx2 > c2 Δt2, and no signal can travel fast enough to connect the two
events. The events are not causally connected and are said to have a
space-like separation.
Δs2 < 0: Δx2 < c2 Δt2, and the two events can be causally connected. The interval
is said to be time-like.
Space-time
When describing events in
relativity, it’s convenient to
represent events with a
space-time diagram.
In this diagram, one spatial
coordinate x, specifies
position, and instead of time t,
ct is used as the other
coordinate so that both
coordinates will have
dimensions of length.
Space-time diagrams were
first used by H. Minkowski in
1908 and are often called
Minkowski diagrams. Paths
in Minkowski space-time are
called world-lines.
Particular Worldlines
Stationary
observers live
on vertical lines.
A light wave has
a 45º slope.
Worldline is the record of the particle’s travel
through spacetime, giving its speed (1/slope) and
acceleration (=1/rate of change of slope).
The Light Cone
The past, present, and future are easily identified in space-time
diagrams. And if we add another spatial dimension, these regions
become cones.
The Doppler Effect
The Doppler effect for sound yields an
increased sound frequency as a source
such as a train (with whistle blowing)
approaches a receiver and a decreased
frequency as the source recedes.
Christian Andreas Doppler
(1803-1853)
A similar change in sound frequency occurs when the source is fixed
and the receiver is moving.
But the formula depends on whether the source or receiver is moving.
The Doppler effect in sound violates the principle of relativity because
there is in fact a special frame for sound waves. Sound waves depend
on media such as air, water, or a steel plate in order to propagate. Of
course, light does not!
Waves from a source at rest
Viewers at rest
everywhere
see the waves
with their
appropriate
frequency and
wavelength.
Recall the Doppler Effect
A receding
source yields a
red-shifted
wave, and an
approaching
source yields a
blue-shifted
wave.
A source
passing by
emits bluethen redshifted waves.
The Relativistic Doppler Effect
So what happens when we throw in Relativity?
Exercise 4-8: Consider a source of light (for example,
a star) in system S’ receding from a receiver (an
astronomer) in system S with a relative velocity v.
Show that the frequency can be obtained from
1 
f 
f0
1 
Where f0 is the proper frequency
Exercise 4-9: What would be the frequency if the
source was approaching?
Exercise 4-10: Use the results from exercise 8 and 9
to deduce the expressions for non-relativistic
velocities.
v∆t
c∆t
Using the Doppler shift to sense rotation
The Doppler shift has a zillion uses.
Using the Doppler shift to sense rotation
Example: The Sun rotates at the equator once in about 25.4 days. The
Sun’s radius is 7.0x108m. Compute the Doppler effect that you would
expect to observe at the left and right limbs (edges) of the Sun near
the equator for the light of wavelength  = 550 nm = 550x10-9m (yellow
light). Is this a redshift or a blueshift?
“Aether Drag”
Exercise 4-12: In 1851, Fizeau measured the degree to which light
slowed down when propagating in flowing liquids.
Fizeau found experimentally:
1

u  c / n  1  2  v
 n 
This so-called “aether drag” was considered evidence for the aether
concept.
Derive this equation from velocity addition equations.
Lorentz-FitzGerald Contraction
Exercise 4-13: Lorentz and FitzGerald, proposed that the null test
of Michelson Morley’s experiment can be explained by using the
concept of length contraction to explain equal path lengths and
zero phase shift. Show that this proposition can work.
The Twin Paradox
The Set-up
Mary and Frank are twins. Mary, an astronaut,
leaves on a trip many lightyears (ly) from the
Earth at great speed and returns; Frank
decides to remain safely on Earth.
The Problem
Frank knows that Mary’s clocks measuring her age must run slow, so
she will return younger than he. However, Mary (who also knows
about time dilation) claims that Frank is also moving relative to her,
and so his clocks must run slow.
The Paradox
Who, in fact, is younger upon Mary’s return?
The Twin-Paradox Resolution
Frank’s clock is in an inertial system during the entire trip. But Mary’s
clock is not. As long as Mary is traveling at constant speed away from
Frank, both of them can argue that the other twin is aging less rapidly.
But when Mary slows down to turn around, she leaves her original
inertial system and eventually returns in a completely different
inertial system.
Mary’s claim is no longer valid,
because she doesn’t remain
in the same inertial system.
Frank does, however, and
Mary ages less than Frank.
t
x
Twin Paradox
Exercise 4-14: A clock is placed in a satellite that orbits Earth
with a period of 108 min. (a) By what time interval will this clock
differ from an identical clock on Earth after 1 year? (b) How
much time will have passed on Earth when the two clocks differ
by 1.0 s? (Assume special relativity applies and neglect general
relativity.)