Lecture 2. Relativistic Kinematics, part II

Outline:

Length Contraction

Relativistic Velocity Addition

Relativistic Doppler Effect

“Red shift” in the Universe

Relativistic effects: length contraction

K0

K

Question : how long does the signal take to complete the round trip?

mirr

or

An observer in the car’s rest RF : 00 2

xt

c

An observer on the ground :

1 21 2 1 2

1 2 2 2

1 1 2

x V t x V tt t t t t

c cx x c

t t t x xc V c V c V c V c V

These intervals are related by the time dilation formula:0

2 21 /

tt

V c

2 20 1 /x x V c

“Moving objects are shortened in the direction of

motion”

2 202 2

22 1 /x c

V c xc c V

0x

0t - the proper time interval

V

Length Contraction (cont’d)

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).

01x

observer

V

K0

K

02x

2 1 2 10 0 2 12 1 2 2

2 21 1

x x V t t x xx x

V Vc c

0 0

2

21

x Vtx

Vc

0

2

21

x Vtx

Vc

Comment It’s easier to write L.Tr. for the “proper” length interval in the right-hand side:

Of course, the same result follows directly from L.Tr.:

2 1t t

Proper length L0 : the length of an object measured in its rest RF ( ). 0 0

0 2 1x x x

- the end positions are measured simultaneously in K

2

0 21V

L Lc

- moving objects are contracted in the direction of

their motion

0 02 1 2 1 2 1 2 1x x x x V t t x x

0L LCompare:

Length contraction (cont’d)

2

0 21V

L Lc

- moving objects are contracted in the direction of their motion

/V c

0/L L

1

1

20 Contraction occurs only in the direction of relative

motion of RFs!

VK

K’

disc at rest the same disc as seen by observer K’

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.

ct ct x

x x ct

y y

z z

Recapitulation: decay of cosmic-ray muons

82.994 10 / 0.998 0.998v m s c N0– the number of muons generated at high altitude

N – the number of muons measured in the sea-level lab

60 2.2 10t s In the muon’s rest frame

By ignoring relativistic effects (wrong!), we get the decay length:

6 80

0 0

2.2 10 3 10 / 660

20,000exp exp 30

660

L t c s m s m

N N N

In fact, the decay length is much greater, the muons can be detected even at the sea level!

~20 km

Because of the time dilation, in the RF of the lab observer the muon’s lifetime is:

6 835 10 3 10 / 10.5L s m s km

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)

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.

60

235 10

1

tt s

altit

ude

0 0

20,000exp exp 2

10,500N N N

Decay of cosmic-ray muons in the muon’s RF

We can re-interpret this situation in terms of the length contraction:

The life-time in the rest frame:

2 40 1 2 10 0.063 1260L L m m

82.994 10 / 0.998 0.998v m s c

60 2.2 10t s

becomes comparable with the muon life-time.

Let’s reconsider the same situation, but now our observer moves with the muon (the muon’s rest IRF)

The travel time

N0– the number of muons generated at high altitude

N – the number of muons measured in the sea-level lab

~20 km

altit

ude

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.

In the muon’s rest frame, the distance to the Earth (~20 km in the Earth’s RF) is significantly shortened:

68

12604 10

3 10 /

mt s

m s

Problems1. 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?

V

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 ???

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:

K K’V

2' 1 /L L V c L

and the time of travel is 21 /'

'L V cL

tV V

2 222 2

2 2 2

/1 / ' 1 ' 0.864

/ '

L V V V L cL V c Vt t

c c c c L c t

4.3 years

According to earth observers:/ 4.3

5/ 0.864

L L c yrt yrV V c

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.

0

21

tt

2 3 2 6 2 601 10 1 10 1 10t

t

66 10

1 10 1 0.99999952

211 1 .....

2!n n n

n

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?

1 1 11 1 1 1

2 1 21 12

and so on…

Relativistic Velocity Addition

Speed of light is the largest speed in nature, no body nor any signal can travel with the speed greater than c.

IRF K: a particle moves a distance dx in a time dt

2 1

2 1

x xv

t t

observer

V

K

K’

v

IRF K’: a particle moves a distance dx’ in a time dt’

1 1 1'x x Vt

1 1,x t 2 2,x t2 1

2 1

' ''

' '

x xv

t t

2 2 2'x x Vt

21 1 1' /t t V c x

22 2 2' /t t V c x

2 1 2 1

22 1 2 1

2

'/ 1

x x V t t v Vv

vVt t V c x xc

2

'1

v Vv

vVc

“+” – anti-parallel

“-” - parallel,v

V

, 'v V c v v V

2

'1

c Vv c v c

cVc

- Galilean velocity addition

Problems

2

1.2' 0.88

1 0.361

v V cv c

vVc

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?

K’

K Galilean velocity addition: ' 1.2v v V c 0.6V c

Relativistic velocity addition:V

0.6v 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?

1 0.75v c2 0.5v c

3 0.33v c3 2' 0.83v v V v v c

K’

K

3 23 2

3 2

5 / 6' 0.711 / 1 1 / 6

v v cv c

v v c

IRF K (rocket): 0.6v cIRF K’ (Earth) moves with respect to K with

IRF K (gun) IRF K’ (Earth)

Yes: 0.83c > 0.75c

No: 0.71c < 0.75cSolve the same problem using IRF K’’ (getaway car).

Doppler Effect for Sound

0 0

1 /

1 /s s

s s

v v v vf f f

v V V v

f0 – the frequency of sound in the rest frame of the source

observersource of

sound

air (the medium where the waves propagate)

vV

v – the speed of an observer with respect to air

V – the speed of the source of sound with respect to air

v

V

v “+” observer moves toward the source “-” observer moves away from the source

“-” source moves toward the observer “+” source moves away from the observer

V

f – the frequency of sound heard by an 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!):

1. Transverse Doppler effect

lightwavefronts observer

The origin of the transverse Doppler effect is time dilation, this is a pure relativistic effect, no counterpart in classical mechanics.

2 20

0

1 11 1f f

T T

V

0T - the period of oscillations of the e.-m. field in the rest

RF of the source K (the “proper” time interval)

0 01/f T

T - the period of oscillations in the RF of the moving observer

1/f Tf is always smaller than f0 – “red shift” (shift to lower frequencies)

K

K’

Longitudinal Doppler Effect for Light

V is the velocity of the relative motion of an observer with respect to the light source.

The most frequent encounter with Doppler effect in light (microwave): police radar speed

detectors (relativistic effects are negligible)

1VT V

T T Tc c

0 0 02

1 1 11

1 11T T T f f

V V

light observer

V

K

K’

10

an extra time needed for the next wave front to reach an observer

0

21 /

TT

V c

20

- the same time dilation as in the case of the transverse Doppler Effect

- “red shift”

The light source and the observer move away from each other.

The light source and the observer approach each other.

0

1

1f f

- “blue shift” (shift toward higher frequencies)

2

0 0 0

11 1 1

1f f f f

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?

0

1

1astf f

00 0

0

'1 1' 1 1 1

1 1

f ff f f

f

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)

0

1 1'

1 1astf f f

46

9

5 107.7 10

6.5 10

Hz

Hz

667.7 10

3.85 10 1.1 /2 2 2

c km s

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: 0V H d

V - the observed velocity of the galaxy away from usH0 - Hubble's "constant" (units: s-1)d - the distance to the galaxy (1 Megaparsec=3106 light-yrs)

Most recent measurements of H0 ~ 71 ± 2 (km/s)/Mpc. Hubble’s constant gives us the age of the Universe 0:

90 01/ 13.8 10H yr 18 1

0

/70 2.3 10km s

H sMpc

tnow

R

the horizon of visibility = infinite

red shift

c0

Extreme “red shifts”: quasars and CMBR

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).

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. PenziasNobel 1978

Mather, Smoot, Nobel 2006

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 (!)

Currently, the energy of the CMB photons is “red shifted” to ~ 3K (f = f0/1000 !).