Relativistic Doppler effect .

The basic mechanism of DE for sound waves is different than that for light waves.

Lets look at light waves straight away since you are familiar with sound waves.

DE: Doppler Effect

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 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.

'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 light wave 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 bottom of 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”).

31.0 434600434600

/1600/1434

/1

/1

nm 434nm 600 hence, ; :use We

2222

22

cuc

u

cucu

cu

cuccc

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

Quick quiz:Find the % errorin the value of u

obtained using theclassical formulafor the Dopplerfrequency shift.