Slide show Notes_06.ppt:Relativistic Doppler Effect

You have learned about the Doppler Effect (DE) in Physics 212, right?Perhaps even earlier, at high school, or from your own reading – because DE plays such an important role in our life (e.g., police uses DE to catch speed limit violators!) and in science (e.g., from studies of the so-called “Doppler shift” astronomers have learned about the expanding Universe).

The basic mechanism of DE for sound waves (i.e., at which the DE material in Ph212 primarily focused) is different than that for light waves: in the DE forsound waves the medium (such as air, water) plays a crucial role – but in thelight propagation no medium is involved! Also, which is quite obvious, all relativistic effects (time dilation, length contraction) are negligibly small in theDE for sound and other mechanical waves.

However, just for refreshing your memory, it is worth to begin with a briefoverview of the DE for sound waves.

Let’s start with a somewhat simplified model. The transmitter (loudspeaker) emits a sound wave, and at some distance from itThere is a receiver (e.g., a human ear, in the picture below symbolizedby the question-mark-like shape).

We can think of the receiving process that each time a sound wave “crest”reaches the receiver, it produces a PING! Many such “pings” heard in a second give us the impression of a continuous sound (the real mecha-nism of hearing is surely more complicated, but the basic physics in our model is OK (actually, for analyzing the basic principles of DE, instead of considering a continuous wave, one can think of a sound signal in the form of a sequence of short sound signals: ping!-ping!-ping!-ping-ping!

Let’s refresh our memory: DE for sound waves

Well, here is the same animation as in the preceding slide, but slowed down – note that each PING! occurs exactly at the momentwhen a “crest” (i.e., a maximum) reaches the “ear”.

Now, consider two observers, of which one is stationary, and the other moves towards the sound source with constant speed:

Note that the moving observer registers more Pings! per time unit than the stationary one – it means that she/he registers a higher frequency.Question: how would the frequency change if the second observer moved away from the sound source?

Yes, of course – then the frequency would be lower,

The same as before, but slowed down a bit: note that in both cases the PING! Appears precisely at the moment a consecutive maximum reaches the ear:

Now, let’s talk about the relativistic Doppler effect.

Now, the signal transmitted by one observer, and received by another, is a light wave. It makes a signifant difference compared to the situationin Doppler effect with sound waves: there is no medium, and the velocity of the wave (i.e., of light) is the same for both observers. And the the “transmitter” and “receiver” move relative to each other with such a speed that relativistic effects have to be taken into account.

We will consider the following situation: the “transmitter” is in the frameO that moves away with speed -u (meaning: to the left) from the frame O’in which the “receiver” located. At some moment the “transmitter” startsto broadcast a light wave. On the next slide, you will see an animation.The position of O at the moment it starts transmitting will be indicated by a marker, and another marker will show the position of O at the momentThe light signal reaches O’.

The animation is repeated several times, and then it stops:

On the next slide, we will perform some calculations.

'tu 'tc

A total of N waves sent out from O

't – the time that elaped between the beginning of transmission and the moment the wave-front reached the observer O’ , measured in the O’ frame.

For the O’ observer, the N waves sent out from the O source are

stretched over the distance )(''' cuttctu

Hence, the wavelength λ’ according to the O’ observer is:

N

uct )(''

Denote the time registered in the O frame between the beginning oftransmission and the moment the wave-front reached O’ as Δt0 ,

And the frequency of the signal for the O observer as ν .The frequency can be thought of as the number of waves sent out in a time unit. Therefore, the total number of waves emitted is

0tN By combining the two equations, we obtain:

0

)(''

t

uct

The general relation between the wavelength and frequency of a lightwave is: wavelength = (speed of light)/(frequency). Therefore, the wavelength and the frequency the O’ observer registers are related as:

''

c

After equating this with the result for the same wavelength at the bottomof the preceding slide, and some simple algebra, we obtain:

cut

t

/1

1

'' 0

Now, we can use the time dilation formula:

220 /1' cutt

cu

cu

cucu

cucu

cu

cu

cu

cu

/1

/1

)/1)(/1(

)/1)(/1(

/1

/1

/1

/1'

2222

cu

cu

/1

/1'

Let’s compare the relativistic DE with the classical DE for sound waves:

The general formula for the Doppler frequency shift of sound waves is:

air). still to(relative source theof speed theis

air), still to(relativeobserver theof speed theis

sound; of speed theis here where'

s

o

s

o

u

u

cuc

uc

If we consider an analogous situation as before, then only the source moves:

c

u

cuuc

c

u

s

ss

o

1/1

1'

then ;0put we, therefore

1

for valid

11

1

We used:

Comparison of relativistic and classical DE, continued:

Now let’s use the equation we have derived for the relativistic frequencyshift. Let’s assume that the source speed u is small compared with thespeed of light; then, we can use the same approximation as we have used in the preceding slide:

c

ucu

cu

cu1/1

/1

/1' 2

It is the same formula that we obtained a moment ago for sound waves.However, for source or observer speeds comparable with the speed of light one can no longer use the same formula as for sound waves.

But are there any such situations that we can observe?

The answer is YES! Due to the expansion of Universe, distant galaxies are moving away from our galaxy with such speeds that we have to use the exact formula.

An important thing to remember:

The Doppler shift in the frequency of light waves arriving from distantgalaxies is one of the main sources of our knowledge of the Universe.

The light arriving from distant galaxies is shifted toward lower frequencies.This is called “the reddening of galaxies”.

How do we know that the frequency is lower? Well, all stars emit certaincharacteristic “spectral lines”, the frequency of which is well known.One of such lines is “the blue line of hydrogen”, with wavelength λ= 434 nm.

Suppose that in the light from a distant galaxy the same line has a wavelength of λ’= 600 nm – such light is no longer blue, but red (therefore, the term “reddening”).

Question: what is the “receding speed” u of that galaxy?

31.0 434600434600

/1600/1434

/1

/1

nm 434nm 600 hence, ; :use We

2222

22

cuc

u

cucu

cu

cuccc

Quick quiz:

Find the % errorin the value of u

obtained using theclassical formulafor the Dopplerfrequency shift.

The amazing TWIN PARADOX

There are two twins, Amelia and Casper.

Casper stays on Earth

Amelia takes off in a spaceshipand goes to a distant star...

Casper is very happy when she comes back...But he is now an old man...

...whereas for Amelia, due to the timedilation, the calendar advanced slower,And she is still young...

Is this story consistent with the relativity theory?

Twin Paradox, continued.

Twin Paradox, continued (2).

But someone may say: I see the whole thing differently.The relativity theory says that all systems are equivalent, right? So, from Amelia’sviepoint, it is Casper who takes off with the entire planet. ...

Therefore, it is the Casper’s clock that will“tick” slower than Amelia’s clock, and it is Casper who will be younger when theymeet again!

Twin Paradox, continued (3).

Who is right?!

I suggest that we resolve this dilemma by

DEMOCRATIC MEANS

Let’s vote!

Who thinks that Casper will be older, raise your hand!

And now, who thinks that Amelia will be older, raise your hand!

Twin Paradox, continued (4).The problem is not trivial! It evoked a heated discussion shortly after

the publication of Einstein’s 1905 paper, in which it was mentioned for the first time.

Than, there was another vigorous discussion in the 1950s – probably,because it was a period of America’s great fascination with SF literature.

Only after a few years of debate, and after the publication of about 40 articles on that topic in various scientific journals, someone got the

right idea how to solve the problem – namely by employing THE DOPPLER EFFECT

In order to explain how the Doppler Effect can help, “let’s do the numbers”.

Suppose that Amelia’s travel destination is a star 12 light-years away fromEarth (i.e., light from this star has to travel 12 years until it reaches Earth).Next, suppose that Amelia’s spacecraft travels with a speed of 0.6c .So, it takes 20 years Earth time for Amelia to get to the star, and 20 years totravel back to Earth.

But due to time dilation, in Amelia’s frame only

yrs 168.0

yrs20

64.0

yrs20

/)6.0(1

yrs 2022

cc

will elapse on her way to the star,and another 16 years on her way back to Earth

Twin Paradox, continued (5).Suppose that Amelia takes off exactly on a day that is hers and Casper’s birthday. And suppose that before Amelia’s departurethe two siblings make the following decision: on his every next birthday (Earth time) Casper will send Amelia a light signal. An on every birthday of hers – spacecraft time – she will send Caspera similar light signal.

Well , so the frequency of the light signals Casper sends out is:

1yr1 yr 1

1 isfrequency signal sCasper' that so

1 T

Using the formula for relativistic frequency shift, we find the frequencyν’ with which Amelia receives Casper’s signals:

(her time) years every two signal one :meanswhich

,yr 5.06.1

4.0yr 1

/1

/1' 11-

cu

cu

During the 16-year flight to the star, Amelia will thus receive

8 Casper’s light signals

Twin Paradox, continued (6).

However, on her way back to Earth, Amelia’s spacecraft speed is

) (minus but ,longer No uuu It does make a difference!

Because now the frequency with which she receives Casper’s signals is:

(her time)year singleevery signals TWO :meanswhich

,yr 24.0

6.1yr 1

/1

/1' 11-

cu

cu

thusduring her 16-year flight back to Earth, Amelia will receive

32 Casper’s light signalsDuring the entire trip she will thus receive

a total of 8+32=40 Casper’s signals

Simple “Minkowski diagrams” illustrating the the Doppler Effect-based explanation of the Twin Paradox