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 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 mechanism 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
u  t '
c  t '
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
u  t 'c  t '  t ' (u  c)
Hence, the wavelength λ’ according to the O’ observer is:
t ' ( c  u )
'
N
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
N    t0
By combining the two equations, we obtain:
t ' ( c  u )
'
  t 0
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:
t0
1
 ' 

t ' 1  u / c
Now, we can use the time dilation formula:
t0  t ' 1  u / c
2
2
1 u / c
1 u / c
 ' 

1 u / c
1 u / c
(1  u / c)(1  u / c)

(1  u / c)(1  u / c)
2
2
1 u / c

1 u / c
1 u / c
 ' 
1 u / c
2
2
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:
c  uo
 '
c  us
where here c is the speed of sound;
uo is the speed of the observer (relative to still air),
u s is the speed of the source (relative to still air).
If we consider an analogous situation as before, then only the source moves:
therefore , we put uo  0; then
c
1
 us 
 ' 

  1  
c  us
1  us / c
c

We used:
1
 1 
1 
valid for
  1
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:
1 u / c
 ' 

1 u / c
1  u / c 
2
 u
  1  
 c
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?
c
c
1 u / c


600 nm 434 nm 1  u / c
434 2  1  u / c   600 2  1  u / c 
We use :  
c

; hence,

600 2  434 2  600 2  434 2 
u
c
 u  0.31c
Quick quiz:
Find the % error
in the value of u
obtained using the
classical formula
for the Doppler
frequency shift.
The amazing TWIN PARADOX
There are two twins, Amelia and Casper.
Casper stays on Earth
Amelia takes off in a spaceship
and goes to a distant star...
Twin Paradox, continued.
Casper is very happy
when she comes back...
But he is now an old man...
...whereas for Amelia, due to the time
dilation, the calendar advanced slower,
And she is still young...
Is this story consistent with the relativity theory?
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’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!
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 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
20 yrs
1  (0.6c) / c
2
2

20 yrs 20 yrs

 16 yrs
0.8
0.64
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 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:
1
1

so that Casper' s signal frequency is  
 1yr 1
T
1 yr
Using the formula for relativistic frequency shift, we find the frequency
ν’ with which Amelia receives Casper’s signals:
1 u / c
0.4
-1
 ' 
 1 yr 
 0.5 yr 1 ,
1 u / c
1.6
which means : one signal every two years (her time)


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
No longer u, but  u (minus u )
It does make a difference!
Because now the frequency with which
she receives Casper’s signals is:
1 u / c
1.6
-1
 ' 
 1 yr 
 2 yr 1 ,
1 u / c
0.4
which means : TWO signals every single year (her time)


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