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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 for

sound waves the medium (such as air, water) plays a crucial role – but in the

light propagation no medium is involved! Also, which is quite obvious, all

relativistic effects (time dilation, length contraction) are negligibly small in the

DE for sound and other mechanical waves.

However, just for refreshing your memory, it is worth to begin with a brief

overview of the DE for sound waves.

Let’s refresh our memory: DE for sound waves

Let’s start with a somewhat simplified model. The transmitter

(loudspeaker) emits a sound wave, and at some distance from it

There is a receiver (e.g., a human ear, in the picture below symbolized

by 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!

Well, here is the same animation as in the preceding slide, but

slowed down – note that each PING! occurs exactly at the moment

when 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 situation

in 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 frame

O that moves away with speed -u (meaning: to the left) from the frame O’

in which the “receiver” located. At some moment the “transmitter” starts

to 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 moment

The light signal reaches O’.

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

On the next slide, we will perform some calculations.

A total of N waves sent out from O

–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

Hence, the wavelengthλ’ according to the O’ observer is:

Denote the time registered in the O frame between the beginning of

transmission 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

By combining the two equations, we obtain:

The general relation between the wavelength and frequency of a light

wave is: wavelength = (speed of light)/(frequency). Therefore, the

wavelength and the frequency the O’ observer registers are related as:

After equating this with the result for the same wavelength at the bottom

of the preceding slide, and some simple algebra, we obtain:

Now, we can use the time dilation formula:

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:

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

We used:

Comparison of relativistic and classical DE, continued:

Now let’s use the equation we have derived for the relativistic frequency

shift. Let’s assume that the source speed u is small compared with the

speed of light; then, we can use the same approximation as we have

used in the preceding slide:

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 distant

galaxies 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 certain

characteristic “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?

Quick quiz:

Find the % error

in the value of u

obtained using the

classical formula

for the Doppler

frequency shift.

There are two twins, Amelia and Casper.

Casper stays on Earth

Amelia takes off in a spaceship

and goes to a distant star...

...whereas for Amelia, due to the time

dilation, the calendar advanced slower,

And she is still young...

Casper is very happy

when she comes back...

But he is now an old man...

Is this story consistent with the relativity theory?

But someone may say: I see the whole thing differently.

The relativity theory says that all systems are

equivalent, right? So, from Amelia’s

viepoint, 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 they

meet again!

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!

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 from

Earth (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 to

travel back to Earth.

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

will elapse on her way to the star,

and another 16 years on her

way back to Earth

Suppose that Amelia takes off exactly on a day that is hers and

Casper’s birthday. And suppose that before Amelia’s departure

the 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 Casper

a similar light signal.

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

Using the formula for relativistic frequency shift, we find the frequency

ν’ with which Amelia receives Casper’s signals:

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

8 Casper’s light signals

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

It does make a difference!

Because now the frequency with which

she receives Casper’s signals is:

thus

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

32 Casper’s light signals

During 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

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