Michelson Morley Experiment
Michelson Morley Experiment
Relativity and Astrophysics
Lecture 09
Terry Herter
Outline
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Worked Problems
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Spacelike and Timelike intervals
Simultaneity Example
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Problem L-2, page 112 (due Friday/Monday)
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Sample problem L-2 (pg. 106-107)
Will hand back on Monday if you hand it in on Friday
Prelim
Wednesday, Sept. 23
Closed book and notes
Will cover material through Friday (9/18)
Should know
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3-12: Michelson-Morley Experiment
Spacetime interval, fundamental postulates of Special Relativity,
simultaneity and other issues raise by relativity
Will have both qualitative and quantitative questions
Lorentz Transformation equations will be provided (if needed)
Michelson Morley Experiment
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Michelson Morley Experiment
Michelson-Morley Exp. (Prob. 3-12)

Suppose we have a plane flying from point A
to point B with velocity c

Let the distance from A to B be L, then the time
to travel back and forth with no wind is
Traveling along wind direction
tn  2 L c

For a wind blowing with velocity, v, blowing
from A to B, the travel time from A to B (with
the wind) is
t AB  L c  v 

L
A
c
c
B
v
Flying back against the wind, the travel time is
t BA  L c  v 

Summing gives the total time
t ABA  t AB  t BA 

t ABA 
2c
L
L
L 2 2

c v
cv cv
2L
1
c 1 v2 / c2
Note that this is highly non-linear, if
v > c you can’t fly back from B to a!
Relative to the no wind time, we have
t ABA
2c c
1
L 2 2

tn
c  v 2L 1  v 2 c 2
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Problem 3-12b

Now suppose the plane flies from A to C with
the wind perpendicular to the direction of
motion. The air speed is still c
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Traveling  to wind direction
But the plane has to fly into the wind with a
velocity component equal to v or it will go off
track, that is, c  v
L
y

Let cx be the velocity component in the x
direction (A-C direction), then
A
c 2  cx2  c y2  c x2  v 2

The travel time from A to C is
t AC  L cx  L

C
v
c2  v2
The return trip takes the same time, so that
t ACA  t AC  tCA 

c
c
2L
t ACA 
2L
1
c 1 v2 / c2
c2  v2
Relative to the no wind time, we have
Note, you may not make if v > c
you can’t get from A to C or C to A!
1
t ACA
c
2L


2
2
2
2
tn
c  v 2L 1  v c
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Michelson Morley Experiment
Problem 3-12c
 to wind
t ABA
 to wind
2L
1

c 1 v2 / c2
t ACA 
2L
1
c 1 v2 / c2
Let us compare the two round trip times for ABA and ACA
Assuming v << c, we can us the expansion
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
1  z n  1  nz
if
z  1
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The for moving parallel to the wind (ABA)
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2L  v 2 
1  
c  c 2 
And for perpendicular to the wind (ACA)
2L 
v2 
1  2 
t ACA 
c  2c 
t ABA 
Thus ACA takes less time and the time difference is:
2L v 2
2L is round trip distance
t  t ACA  t ABA 
c 2c 2
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Problem 3-12d
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Suppose a South Pole station send out 4
planes travel as shown to weather
stations located 300 km away.
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What is the wind speed?
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Assume that the wind is blowing
horizontally in the diagram
We have from our previous slide
t 
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300 km
The planes are traveling at 300 km/hr
The time difference between the first and
last plane returning is 4 seconds
2L v 2
c 2c 2
1/ 2
 c
v  c  t 
 L

South Pole station with by weather
stations located 300 km away in
locations shown.
Thus
1/ 2
 4 300 
v  300 

 3600 300 
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 10 km/hr
The wind is blowing along the path which
takes the longest time. The direction is
not known with additional info.
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Michelson Morley Experiment
Problem 3-12e
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Michelson and Morley designed an
experiment to look for a change in the speed
of light due to motion (of the Earth) relative to
the “ether”
They used an “interferometer which divided an
incoming light beam into two parts
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The total path length (2L) was 22m for their
experiment
The orbital speed of the Earth is 30 km/sec
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eye
22m
1  30 10
2L v


3 108 m/sec 2  3 108
c 2c 2
3
2
Corrector to get
same optical
path length
telescope
We have for the time difference
t 
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mirror
Numbers
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beam
splitter
The beams travel in perpendicular directions,
until they hit a mirror and retrace their steps
The returning beams “interfere” with one
another
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mirror



2
Michelson interferometer used in
Michelson-Morley experiment
So that
t  3.7 10 16 sec
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Which is indeed very, very short
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Problem 3-12f
mirror
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Since this timing is too short to measure (even
by today’s standards, they used a Na light
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Thus for a period (1 cycle)
T
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beam
splitter
Emits at specific wavelength, 589 nm
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mirror
1 
589  109 m


 2  10 15 sec
f c
3  108 m/sec
The experiment was done to 1/100 of this
value (1 dark fringe to another)
Corrector to get
same optical
path length
telescope
texp  2 10 17 sec
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Now Michelson and Morley rotated the table
so they were looking for twice the effect
(computed in part e)
texp
2  10 17 sec
1


2  3.7 10 16 sec 37
t
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The velocity accuracy relative to Earth motion
is (extra factor of 2 for rotation):
2L v 2
t  2
c 2c 2
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eye
Michelson interferometer used in
Michelson-Morley experiment.
Round trip path, 2L = 22 m. Earth
speed 30 km/sec
1/ 2
 3 108

c

 vm  
2 10 17 
texp   
22
2
L




1/ 2
Michelson Morley Experiment
 5 km/ sec
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Michelson Morley Experiment
Problem 3-12g
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No effect was detected with the experiment
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Does the Michelson-Morley experiment by itself
disprove that light is propagated through an ether?
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A null result – the experiment could not tell that the Earth is
moving through an “ether”
Not quite – but it is close. One could have length contraction
(only) of the interferometer along the direction of motion
This could compensate for the a slower travel time by the
light (photons) making the arm shorter.
This looks like relativity but it is not, the speed of light is not
a constant in this case
A modified experiment with arms of unequal length
(see problem 3-13) still shows there is not motion
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Requires both length contraction and time dilation for no
fringe shift (as predicted by Relativity).
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Spacelike Intervals
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In lab frame we see two events (event A and event B) separated by a
distance x (= xB – xA) and time t (= tB – tA)
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Can we find a rocket frame in which the events are simultaneous?
Let the coordinates for event A be zero in both frames. Then we have
xA = x′A = 0
tA = t′A = 0
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xB = known
tB = known
x′B = unknown
t′B = 0
We want the speed vrel of the rocket such that t′B = 0. We use the
Lorentz transformation equations
t B   t B  vrel  xB

t B  vrel xB
 vrel 
t′B = 0
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vrel is the required speed of the rocket so that t′B = 0.
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tB
xB
Note that since vrel < 1, we must have xB > t
Thus t2B - x2B < 0 (the spacetime interval is spacelike)
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Michelson Morley Experiment
Timelike Intervals

In lab frame we see two events (event A and event B) separated by a
distance x (= xB – xA) and time t (= tB – tA)


Can we now find a rocket frame in which the events occur at the same
location?
Let the coordinates for event A be zero in both frames. Then we have
xA = x′A = 0
tA = t′A = 0

xB = known
tB = known
x′B = 0
t′B = unknown
We want the speed vrel of the rocket such that x′B = 0. We use the
Lorentz transformation equations
xB   xB  vrel  t B

xB  vrel t B
 vrel 
x′B = 0

Again vrel is the required speed of the rocket so that x′B = 0.
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xB
tB
Note that since vrel < 1, we must have xB < t
Thus t2B - x2B > 0 (the spacetime interval is timelike)
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