Transcript Lecture. Photoelectric Effect

Lecture 2. Relativistic Kinematics, part II
Outline:
Length Contraction
Relativistic Velocity Addition
Relativistic Doppler Effect
“Red shift” in the Universe
Relativistic effects: length contraction
Question : how long does the signal take to complete the round trip?
x0
K0
mirror
V
K
An observer in the car’s rest RF :
t0  2
x0
c
t0
- the proper time interval
x  V t1
x  V t2
t2 
c
c
x
x
1 
2c
 1
t1 
t2 
t  x 



x

c V
c V
 c V c V 
 c2  V 2 
t0
t 
These intervals are related by the time dilation formula:
1  V 2 / c2
An observer on
the ground :
t  t1  t2
x0
 2c 
2
 1  V 2 / c 2 x  2
2 
c
 c V 
t1 
x  x0 1  V / c
2
2
“Moving objects
are shortened in
the direction of
motion”
Length Contraction (cont’d)
Of course, the same result follows directly from L.Tr.:
K0
x10
Proper length L0 : the length of an object measured in its rest RF (
An observer in the RF K moving with respect to the RF K0
with the velocity V directed parallel to the meter stick,
measures its length. In order to do that, he/she finds two
points x1 and x2 in his/her RF that would simultaneously
coincide with the ends of the moving stick (t1 =t2).
x20
V
K
Comment It’s easier to write L.Tr. for the “proper” length
interval in the right-hand side:
observer
x
t2  t1
x x
0
2
0
1
x0  x20  x10
x2  x1
V
1 2
c
V
1 2
c
Compare:
x0 
V2
1 2
c
x  Vt
V2
1 2
c
- the end positions are measured simultaneously in K
 x  x   V  t2  t1  
 2 1
2
x 0  Vt 0
2
L  L0 1 
2
V
c2
- moving objects
are contracted in
the direction of
their motion
x20  x10   x2  x1  V t2  t1   x2  x1
L  L0
).
Length contraction (cont’d)
V2
L  L0 1  2
c
- moving objects are
contracted in the direction
of their motion
10 To observe this effect, the relative speed of the reference
frames should be large. For the fastest spacecraft, the speed
is ~10-4c, and the effect is of an order of 10-8.
1
L / L0
 V / c
20 Contraction occurs only
in the direction of relative
motion of RFs!
ct     ct   x 
x    x   ct 
y  y
z  z
K
V
disc at rest
K’
the same disc as seen
by observer K’
1
Recapitulation: decay of cosmic-ray muons
Muon – an electrically charged unstable elementary particle with a rest energy ~ 207 times
greater than the rest energy of an electron. The muon has an average half-life of 2.2 10-6 s.
Muons are created at high altitudes due to collisions of fast cosmic-ray particles (mostly
protons) with atoms in the Earth atmosphere. (Most cosmic rays are generated in our
galaxy, primarily in supernova explosions)
N0– the number of muons
generated at high altitude
v  2.994 108 m / s  0.998c   0.998
In the muon’s rest frame
t0  2.2 106 s
By ignoring relativistic effects (wrong!), we get the decay length:
~20 km
altitude
L  t0  c  2.2 106 s  3 108 m / s  660m
N – the number of
muons measured in
the sea-level lab
 20,000 
N  N0 exp  
  N0 exp  30 
660


In fact, the decay length is much greater, the muons can be
detected even at the sea level!
Because of the time
dilation, in the RF of the
lab observer the muon’s
lifetime is:
t 
t0
 35 106 s
1  2
L  35 106 s  3 108 m / s  10.5km
 20,000 
N  N0 exp  
  N0 exp  2 
10,500


Decay of cosmic-ray muons in the muon’s RF
Let’s reconsider the same situation, but now our observer moves with the muon
(the muon’s rest IRF)
v  2.994 108 m / s  0.998c   0.998
N0– the number of muons
generated at high altitude
We can re-interpret this situation in terms of the
length contraction:
The life-time in the rest frame:
In the muon’s rest frame, the distance to the Earth (~20 km in
the Earth’s RF) is significantly shortened:
altitude
~20 km
t0  2.2 106 s
L  L0 1   2  2 104 m  0.063  1260m
The travel time
N – the number of
muons measured in
the sea-level lab
t 
1260m
6

4

10
s
8
3 10 m / s
becomes comparable with the muon life-time.
Thus, again, there is a considerable number of muons
(the same as we’ve calculated in the lab RF) that can be
detected at the sea level.
Problems
1. The nearest star to the Earth is Proxima Centauri, 4.3 light-years away.
- at what constant speed must a spacecraft travel from the Earth if it is to reach
the star in 2.5 years, as measured by travelers on the spacecraft?
- how long does this trip take according to earth observers?
K’
K
Consider two IRFs, K (the Earth) and K’ (the
rest RF of the spacecraft). By astronaut's
reckoning (K’), the distance to the star is
contracted:
V
L '  L 1  V / c 
L
2
2
L'

V
V
2
2
2
2
 L   V  V 
 Vt '   1  2      t '
c  c  c
and the time of travel is
L 1  V / c 
L 1  V / c 
t'
According to earth observers:
t
2
4.3 years
V

c
L/c
 L / c    t '
2
2
 0.864
L L / c 4.3 yr


 5 yr
V V / c 0.864
2. Consider a disc at rest. We know that the “circumference/diameter”
ratio is . Now the disc rotates around its center. If one applies the
Lotentz length contraction to the disc, the result would be puzzling: the
circumference “shrinks” while the diameter (which is normal to the
velocity) remains intact, so “circumference/diameter”   ! What’s
going on ???
V
Problem
Imagine an alien spaceship traveling so fast that it crosses our galaxy (whose rest
diameter is 100,000 light-years) in only 100 years of spaceship time. Observers at rest in
the galaxy would say that this is possible because the ship’s speed  is so close to 1 that
the proper time it measures between its entry into and departure from the galaxy is much
shorter than the galaxy-frame coordinate time (~100,000 ly) between those events. Find
the exact value of the speed  that the aliens must have to cross the galaxy in 100 years.
t 
t0
1  2
t0
1  
 103
t
2
  1  10
6
1   2  106
 2  1  106
106
 1
 0.9999995
2
n  n  1 2
  .....
1     1  n 
2!

1
1
1

1   1
 1 

 1
2
1 
2
1 1 
2
n
and so on…
How does it look to the aliens? To them, their clocks are running normally, but the galaxy,
which moves backward relative to them at speed  1, is Lorentz contracted. What is the
galaxy’s size by aliens’ reckoning?
Relativistic Velocity Addition
K
x1 , t1
observer
v
x2 , t2
K’
x1 '    x1  Vt1 
x2 '    x2 Vt2 
t1 '   t1  V / c 2  x1 
t2 '   t2  V / c 2  x2 
v,V  c v '  v  V
c V
v  c v' 
c
cV
1 2
c
V
IRF K: a particle moves a
distance dx in a time dt
IRF K’: a particle moves a
distance dx’ in a time dt’
x2  x1
v
t2  t1
x2 ' x1 '
v' 
t2 ' t1 '
  x2  x1   V  t2  t1  
v V
v' 

2
  t2  t1   V / c   x2  x1   1  vV
c2
v V
“+” – anti-parallel
v' 
v, V
vV
“-” - parallel
1 2
c
- Galilean velocity addition
Speed of light is the largest speed in
nature, no body nor any signal can
travel with the speed greater than c.
Problems
1. A person on a rocket traveling at 0.6c (with respect to the Earth RF) observes a meteor
passing him at a speed he measures as 0.6c. How fast is the meteor moving with respect
to the Earth?
v  0.6c IRF K (rocket): v  0.6c
IRF K’ (Earth) moves with respect to K with
K
Galilean velocity addition:
V
Relativistic velocity addition:
K’
V  0.6c
v '  v  V  1.2c
v V
1.2c
v' 

 0.88c
vV 1  0.36
1 2
c
2. As the outlaws escape in their getaway car, which goes 3/4c, the police officer fires a bullet
from the pursuit car, which only goes 1/2c. The muzzle velocity of the bullet (relative to the
gun) is 1/3c. Does the bullet reach its target (a) according to Galileo, (b) according to
Einstein?
K
IRF K (gun)
v3  0.33c
v2  0.5c
K’
v '  v  V  v3  v2  0.83c
v1  0.75c
IRF K’ (Earth)
Solve the same problem using IRF K’’ (getaway car).
Yes: 0.83c > 0.75c
v3 ' 
v3  v2
5 / 6c

 0.71c
2
1  v3v2 / c 1  1/ 6
No: 0.71c < 0.75c
Doppler Effect for Sound
air (the medium where the waves propagate)
v
V
observer
source of
sound
vv – the speed of an observer with respect to air
V – the speed of the source of sound with respect to air
f0 – the frequency of sound in the rest frame of the source
f – the frequency of sound heard by an observer
f  f0
vs  v
1  v / vs
 f0
vs  V
1  V / vs
v
“+” observer moves toward the source
“-” observer moves away from the source
V
“-” source moves toward the observer
“+” source moves away from the observer
Transverse Doppler Effect for Light
Doppler effect for light - a change in the observed light frequency due to a relative
motion of the light source and an observer (no special RF associated with the medium
where light propagates!):
light
K
wave
1. Transverse Doppler effect
observer
V
fronts
T0 - the period of oscillations
of the e.-m. field in the rest
K’
RF of the source K
(the “proper” time interval)
T - the period of oscillations in the RF of the moving observer
f0  1/ T0
f  1/ T
f 
1 1

1   2  f0 1   2
T T0
f is always smaller than f0
– “red shift” (shift to lower
frequencies)
The origin of the transverse Doppler effect is time dilation, this is a pure
relativistic effect, no counterpart in classical mechanics.
Longitudinal Doppler Effect for Light
10
The light source and the observer move away from each other.
K
observer
light
K’
V
V is the velocity of the relative
motion of an observer with respect
to the light source.
VT
 V

T T 
 T 1  
c
 c
T
an extra time needed for the next
wave front to reach an observer
T   T0 1   
20
1 
 T0
2
1 
1 
1
1 
1 
1  V / c 
2
- the same time dilation as in the
case of the transverse Doppler Effect
f   f0
1 
1 
The light source and the observer approach each other.
f   f0
T0
- “red shift”
V  V
- “blue shift” (shift toward higher frequencies)
The most frequent encounter with Doppler
effect in light (microwave): police radar speed
detectors (relativistic effects are negligible)
f   f0
1 
 
1 
1  f0
1   
2
 f0 1   
Problem
A spaceship approaches an asteroid and sends out a radio signal with proper frequency
6.5x109 Hz. The signal bounces off the asteroid’s surface and returns shifted by 5x104 Hz.
What is the relative speed of the spaceship and the asteroid?
In this situation, there Doppler shift occurs twice.
Firstly, the original frequency is received by an
asteroid as
Secondly, the spaceship receives the
reflected signal with the frequency
(the asteroid is the “secondary”
source of light)
f '  f ast
1   
f ' f0
1 
f ' f0  f0 
 1 
 1
 1 
1 
f0
1   
1 
f ast  f0
1 
 1  
1 
 f0 

1 
1




5 104 Hz
6


7.7

10
6.5 109 Hz

 7.7 106

 
 3.85 106 c  1.1km / s
2  2
2
Hubble’s Law (1929)
The Universe expands: the larger the distance to an object, the larger the
(relative) speed. By measuring the red shift of (identifiable) spectral lines,
one can calculate the recessional speed of the light source with respect to
the Earth’s observer.
According to Hubble's Law, there is a direct proportionality (at least at not
too large distances) between the velocity and the distance to the source:
V  H0  d
V - the observed velocity of the galaxy away from us
H0 - Hubble's "constant" (units: s-1)
d - the distance to the galaxy (1 Megaparsec=3106 light-yrs)
Most recent measurements of H0 ~ 71 ± 2 (km/s)/Mpc.
Hubble’s constant gives us the age of the Universe 0:
H 0  70
R
km / s
 2.3 1018 s 1
Mpc
 0  1/ H0  13.8 109 yr
c0
now
the horizon of
visibility = infinite
red shift
t
Extreme “red shifts”: quasars and CMBR
Quasars, very bright objects (like 100-10,000 our Galaxies)
of a very small size (10-4 of our Galaxy size), believed to be
supermassive black holes in the nuclei of distant galaxies.
Distance: (2-10)109 light-years [~ (0.8-3)103 Mpc].
Doppler shift: f/f ~0.1-6.4 (!)
Cosmic Microwave Background Radiation (CMBR) In the
standard Big Bang model, the radiation is decoupled from
the matter in the Universe about 300,000 years after the Big
Bang, when the temperature dropped to the point where
neutral atoms form (T~3000K). At this moment, the Universe
became transparent for the “primordial” photons. This
radiation is coming from all directions and its spectrum is
quite distinct from the radiation from stars and galaxies).
Currently, the energy of the CMB photons is “red shifted” to ~ 3K (f = f0/1000 !).
The sub-mm/THz range contains ~ half of the total luminosity of the Universe and
98% of all the photons emitted since the Big Bang.
R. Wilson
A. Penzias
Nobel 1978
Mather, Smoot, Nobel 2006