Transcript Document

Relativity
PHY232
Remco Zegers
[email protected]
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
An argument between Newton and Einstein
 Newton is standing along a road. Einstein is passing by in a car driving
with a speed of 10 m/s. They later meet in a pub, and Newton
congratulates Einstein with achieving a speed of 10 m/s. Modest
Einstein answers that he was not moving with 10 m/s at all: he thought
Newton was moving with a speed of 10 m/s, in the opposite direction
and congratulates him. A long argument follows, which is only settled
after many beers. What did they agree on in the end?
 a) Newton was right that the car was moving at a speed of 10 m/s
 b) Einstein was right that Newton was moving at a speed of 10 m/s
 c) neither were right
 d) both were right
 e) that they had too many beers
For the record: Newton: 1643 – 1727 Einstein: 1879-1955; the above
encounter is entirely fictional…unless…see later
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relative motion
70 mph
20 mph
Which train is having what speed, relative to what?
In what ways can one distinguish between the 2?
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If light is a wave…
…then what is oscillating?
waves in water: water molecules move up and down
Transverse waves
sound:
molecules in air move to and fro
(no sound in vacuum!)
Longitudinal waves
light waves: ???
let’s call it the ‘ether’
Transverse or Longitudinal??
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properties of ‘ether’
• It must be a very tenuous gas (doesn’t interfere with
our movements)
• It must be very rigid (speed of light is very large)
… possible???
against common sense, but…
we need something to measure motion against.
Something must be the absolute frame of reference
to measure motion against:
• earth?
• sun?
• Galaxy?
• ETHER!
(mid 19th century)
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Measuring the ether wind…
Our postulates:
• So now we have ether as our fixed reference frame
• Light is a transverse oscillation of the ether material
light
vearth
vlight,ether
ether
vlight,measured on earth=vlight,ether+vearth
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Michelson Morley experiment
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ether & Michelson-Morley
ether wind
c: speed of light
v: ether wind speed
v
tf=L/(c+v) + L/(c-v) = 2Lc/(c2-v2)
tm=L/vup + L/vdown

vup=vdown= (c2-v2)
c
vup

tm=2L/ (c2-v2)
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ether wind
ether wind?
c: speed of light
v: ether wind speed
1
tf/tm=c/(c2-v2)=
2
v
1 2
c
if v<c then tf>tm
corresponding virtual path length ratio (use x=vt)
1
df/dm=
v2
1 2
c
Path length difference creates
interference pattern!
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ether wind?
If ether would exist…
tf>tm
tf=tm
The interference pattern would change when setup is rotated!
DOES NOT HAPPEN!!! so ether does not exist
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Wow!!
There is no ether, so there is no way to determine
absolute motion
All motion must be considered relative
to some object arbitrarily taken at rest.
Any reference object (reference frame)
can be taken: no object/frame is more at
rest than other
Einstein
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Einstein’s postulates
1. All laws of physics are the same in all frames of
reference moving with constant velocity relative to
each other (whatever you try to measure, the result
is the same independent on the frame of reference)
2. The speed of light is constant (2.9979245x108 m/s)
in all inertial reference frames.
Postulates of Special relativity: no acceleration
involved.
A massive object cannot move faster than the speed of
light
This has some important (and weird) consequences…
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train paradox
lightning strikes two ends of a fast moving train simultaneously, when
seen by an observer standing along the railroad. However, the person
in the train sees the light come in at different times and thus thinks the
times were different. Who is right?
a) observer along the railroad
b) observer in the train
c) neither
d) both
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simultaneous?
Newton: ‘Absolute, true and mathematical time, of itself
and from nature, flows equably without relation to anything
external’ (time
is absolute)
Einstein: Simultaneity is not an absolute concept but one that
depends on the state of motion of the observer.
(time is relative)
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Time dilation
d
vtB
A is sitting in a train moving with velocity
v. She shines a light toward a mirror on
the roof (height d). The reflected light
takes tA=2d/c to get back to A.
An observer (B) outside the train sees A
moving with v. The distance traveled by
the train until the reflected light returns
is vtB
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ctB/2
d
vtB/2
Time dilation
 ct B   vt B 
2

 
 d
 2   2 
2d
2d
t B 
t A 
c
v2
c 1 2
c
t A
1
t B 
 t A

2
2
v
v
1 2
1 2
c
c
2
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Time dilatation
A clock moving past an observer at speed v runs more
slowly than an identical clock at rest with respect to
the observer by a factor of 1/.
v
1

c

1 
2
tp: proper time is the time interval between two events
as measured by an observer who sees the two events
occur at the same positions. Or in other words, is in the same frame
of reference
Note: this equation can only be applied in a frame of reference that is
not accelerating (else Einstein’s postulates of special relativity do not
hold)
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question
 A person on a far-away planet X takes 5 hours to read a
book. If earth moves with a velocity of 0.9 times the speed
of light relative to the planet, how long does the reader
take to finish the book when viewed by an observer on
earth?
Step 1) In which frame is the proper time measured?
a) in the frame of the reader on planet X
b) in the frame of the person on earth
The ‘events’, starting and finishing the book occur at the
same place for the person on planet X, so he measured
the proper time
Step 2) Use the time dilation equation:
t=(5 hours)/(1-[0.9c/c]2)=11.47 hours
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twin paradox
 Two identical twins, Newton and Einstein, live on earth and are both
20 years old. Einstein decides to settle on planet Y. He travels there
by spaceship with an average speed of 0.95 times the speed of light.
As measured by a clock in his spaceship it takes 5 years. Upon
arrival he feels homesick and returns immediately at the same
average speed. Which of the following is correct?
 a) Upon his return, Einstein has aged 10 years and Newton has aged
x10 years=10 years/(1-[0.95c/c]2)=32 years
 b) Upon his return, Einstein has aged 10 years and Newton has aged
10/ years=10 years/(1-[0.95c/c]2)=3.1 years
 c) both are 20+32=52 years old
 d) both are 20+10=30 years old
The proper time is in the frame of Einstein. Note that since Einstein is
at certain times accelerating/decelerating, the time dilation equation
cannot be applied by Einstein, even though from his point of view
Newton is traveling at 0.95c.
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Lorentz-Fitzgerald length contraction
A person A measures the
length of a rope to be 10 m
Lp=10 m
(we call this the proper length
Lp, proper meaning that the
rope is in the same reference
frame as the person).
A space ship passes by with v=0.9c. How long is the rope
according to a person B in the space ship?
1. According to A, it takes the ship t=Lp/v to get from one
end of the rope to the other
2. For B, this time is reduced to tp= t/ (dilation)
proper time because in frame of B
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Lorentz-Fitzgerald length contraction
L=4.4 m
Lp=10 m
3. So, according to B, the length of the rope is:
L=vtp= vt/=Lp/=Lp(1-v2/c2)
L=Lp x 0.44 = 4.4 meter
The length of an object measured in a frame moving with
respect to the object is less than the proper length.
L=Lp/ 
Length contraction only takes place along direction of motion
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slow
question
L=Lp/ 
When passing by very slowly (v<<c) the length of a space ship
is 10 m as observed from the ground. What is its length
(as observed from the ground) if the ship has a velocity of
0.9c (=2.27) ?
a) L=Lp/=10/2.27=4.4 meter c) L=Lp=10m
b) L=Lpx=10x2.27=22.7 meter
L=Lp/=10/2.27=4.4 meter
The proper length is now in
the frame of the ship!
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loncapa
now do questions 1,2,3,4,5 of set 10
note: a light year is the distance traveled by light
in one year. It is ‘just’ another measure of distance
like miles or kilometers.
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Relativistic addition of velocities
 A person is walking in a moving train. The train moves with a speed of
10 m/s to the right, and the person walks with a speed of 2 m/s to the
right, relative to the moving train. You are standing on a platform in
the station. The speed of the person, from your point of view is:
 a) 8 m/s
b) 10 m/s
c) 12 m/s
• A spaceship is passing by you with a velocity of 0.8c. It shoots a
rocket in the same direction as the moving ship, which according to the
pilot of the ship, has a velocity of 0.6c. What is the velocity of the rocket
from your point of view?
a) v=0.2c b) v=0.8c c) v>0.8c but v<c d) v=c e) v=1.4c
if we simply add 0.6c+0.8c, the velocity is larger than c,
which is impossible.
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Relativistic addition of velocities
 frame d is moving in the +x direction relative to frame b
with a velocity vdb. The velocity of an object a is
measured in frame d to be vad. Then the above equation
gives the velocity vab of a in the frame b.
 Note that if vad and vdb are small, vab=vad+vdb which is the
common equation for relative motion.
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Relativistic addition of velocities
• A spaceship is passing by you with a velocity of 0.8c. It shoots a
rocket in the same direction as the moving ship, which according to the
pilot of the ship, has a velocity of 0.6c. What is the velocity of the rocket
from your point of view?
a: rocket
d: frame of reference of spaceship
b: your frame of reference
vad=0.6c
vdb=0.8c
vab=(0.6c+0.8c)/(1+0.6cx0.8c/c2)=0.946c
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Doppler effect: a non-moving source
v
source
f=v/
you

The velocity v (say of light or sound) is fixed
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doppler effect: a source moving towards you
the distance between
the wave front is
shortened
vsource vsound vsource
   


f
f
f
vsource
source
v
f 
you


vsound
f 
 vsound  vsource



prime’: received observable
The frequency becomes larger: wavelength smaller
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doppler effect: a source moving away from you
the distance between
the wave front becomes longer
vsource vsound vsource
   


f
f
f
vsource
you
v
f 

source

vsound
f 
 vsound  vsource



prime’: received observable
vsource negative
The frequency becomes lower: wavelength higher
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applications of the doppler effect: speed
radar
 v  vobserver
f   f 
 v  vsource


v

f
vv

approachin g car 


 

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lon-capa
now do question 6,7,10 from set 10.
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relativistic energy and momentum
 We have seen that Newtonian laws for motion do not hold at
relativistic energies.
 The equations for momentum, energy and kinetic energy must also
be modified.
momentum:
 an important conclusion by Einstein was that energy and mass are
equivalent:
 The total energy of an object is given by:
 The kinetic energy is the total energy of an object minus its rest mass
(energy):
(use in question 12)
 By combining:
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relativistic protons
v

c

1
1  2
 a proton (rest mass of 938 MeV/c2) is accelerated over a potential
difference of 2x107 V.
 what are: a) the kinetic energy of the proton b) the velocity of the
proton c) the total energy of the proton.
a) the energy given by the potential is qV=1.6x10-19 x 2x107 J=3.2x10-12 J
Note: 1eV = 1.6x10-19 J so kinetic energy is also: 2x107 eV=20 MeV
b)
so =1+Ekin/(mc2)=1+Ekin/Erest=1+20/938=1.021
=(1-1/2)=0.2 so v=0.2c
c) E=Ekin+Erest=20 MeV+938 MeV=958 MeV = 1.53x10-10 J
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lon-capa
now do questions 8, 9, 11, 12 of loncapa 10.
Note: for question 9: you need a calculator that
can handle many digits. If yours doesn’t do that
(like mine) one option is the standard windows
calculator, but be sure to set it to ‘scientific view’.
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Einstein’s General relativity
person in a free-falling elevator
feels similar (weightless) to
a person in a rocket far away
from any planet (gravitational
field)
person in an accelerating rocket
feels similar (same weight) as
a person standing on a planet
The force of gravity is the acceleration you feel when you
move through space-time…
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space-time
This has as a consequence that a ray of light would bend
in a gravitational field (observed!!).
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postulates of general relativity
1. All the laws in nature have the same form for
observers in any frame of reference (accelerated or
not).
2. In the vicinity of any given point, a gravitational field
is equivalent to an accelerated frame of reference
without a gravitational field (principle of equivalence).
The gravitational effect at a certain point is given
by the so-called curvature of space time…
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curvature of space time…
Masses produce a curvature in space-time (which would
otherwise be flat). Smaller masses (earth) follow the
curvature of larger masses (sun).
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really??
Strong gravity makes time run slower…
boulder (Colorado)
1 mile
Washington
Atomic clocks have an error of 1 in 1014 (1 s in 3 million year)
After 1 day, an atomic clock in Boulder runs faster by 15 ns
(15x10-9s) than an atomic clock in Washington. This
difference is 17 times larger than the error!!
Gravity is slightly different: time is different!
Important for satellites (GPS systems!!)
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