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Module 6
Aberration and Doppler Shift of Light
General Relationships
The term aberration used here means deviation. If a light source is moving relative to an observer then the observed path of
the light deviates from the emitted path. Consider a light source at the origin O’ in frame S’ as shown in Figure 6-1(a). It
emits a plane light wave along a path that forms an angle ’ with respect to the x’-axis. For an observer in frame S, the light
source is moving at speed v as shown in Figure 6-1(b). The observer sees the light traveling in a direction that forms angle 
with respect to the x-axis. The two angles are different; aberration has occurred.
Doppler shift means a frequency shift (and corresponding wavelength shift) of the light. This phenomenon is common to all
waves emitted by sources in relative motion with an observer. You probably are familiar with the Doppler shift of sound
waves. The frequency and wavelength of the light measured by the observer in frame S of Figure 6-1(b) are different than
the frequency and wavelength emitted by the source. Note the different wavelengths between consecutive wave fronts in the
figure.
S’
S
y
y’
l’
l
r’
r
’

x’
x
v
(a)
(b)
Fig. 6-1
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Aberration and Doppler Shift of Light
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Our goal in this module is to find how the observed angle  depends on the emitted angle ’ and how the observed
frequency f depends on the emitted frequency f’. To do this, let us write down a standard form for the electric field of the
harmonic plane wave in frame S’:
E ' (r ' , t ' )  Eo' cos( k ' r '' t ' )
(6.1)
Recall from your earlier studies of waves that E’o is the amplitude of the electric field wave, k’ is the wave number, and ’
is the angular frequency. The wave number is related to the wavelength l’ by
k '  2 / l'
(6.2)
and the angular frequency is related to the linear frequency by
'  2f '
(6.3)
We assume that the light is traveling in free space so that its speed is c. The speed can be expressed as
c  l' f '  ' / k '
(6.4)
Note from figure Figure 6-1(a) that we can write
x'  r ' cos '
y '  r ' sin '
(6.5)
or, equivalently,
x' cos '  r ' cos 2 '
y ' sin '  r ' sin 2 '
Module 6
Aberration and Doppler Shift of Light
(6.6)
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Adding the two equations (6.6) gives
r '  x' cos ' y ' sin '
(6.7)
Substituting this expression for r’ into Eq. (6.1) gives an alternate expression for the electric field of the light wave in S’,
E ' (r ' , t ' )  Eo' cos( k ' x' cos ' k ' y ' sin '' t ' )
(6.8)
We expect that the observer in frame S will see a plane wave traveling at c of the form
E (r , t )  Eo cos( kx cos   ky sin   t )
(6.9)
where k = 2 / l ,  = 2 f , and c = l f =  / k.
We now transform the electric field from frame S’ to S. This amounts to transforming x’, y’, and t’ in Eq. (6.8) with the
Lorentz transformation equations, Eqs. (3.16). (The Lorentz equations are written in the margin in case you have forgotten
them.) The transformations yield
x'   ( x  vt)
2
(6.10)
E (r , t )  Eo cos[ k '  ( x  vt) cos ' k ' y sin ''  (t  vx / c )]
y'  y
z'  z
The terms inside the cosine argument can be rearranged to give
E (r , t )  Eo cos[( k '  cos '' v / c 2 ) x  k ' y sin '('   k ' v cos ' )t ]
Module 6
Aberration and Doppler Shift of Light
t '   (t  vx / c 2 )
(6.11)
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We now make a direct comparison between Eqs. (6.11) and (6.9). Each is a valid expression for the light wave observed in
frame S. The only way that the two expressions can be equal for any values of x, y, or t is for the respective coefficients of
these coordinates to be equal. Equating the x and y coefficients gives, respectively,
k cos   k '  cos '' v / c 2
(6.12)
k sin   k ' sin '
(6.13)
and
We now divide Eq. (6.13) by Eq. (6.12) to find
tan  
sin 
k ' sin '

cos  k '  cos '' v / c 2
(6.14)
We make one more simplification. From Eq. (6.4), ’ = k’c so we can write the light aberration transformation equation
as
tan  
sin '
 (cos 'v / c)
(6.15)
Of course, we can swap primed and unprimed variables and replace v with –v to obtain the inverse transformation equation,
tan ' 
Module 6
sin 
 (cos   v / c)
Aberration and Doppler Shift of Light
(6.16)
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We have the light aberration equation but what about the Doppler shift? As you may have guessed, we find the frequency
equation by equating the t coefficients from Eqs. (6.11) and (6.9),
  '   k ' v cos '
(6.17)
'
v cos '
c
(6.18)
 v

  ' 1  cos ' 
 c

(6.19)
Replacing k’ with ’/ c gives
  '  
or
Since the angular frequencies are simply 2 times the linear frequencies, we can express the frequency transformation
equation as
 v

(6.20)
f  f ' 1  cos ' 
 c

with the corresponding inverse transform of
 v

f '  f 1  cos  
 c

(6.21)
The observed wavelength can be compared to the emitted wavelength using these frequency equations and the fact that
c  lf  l ' f '
Module 6
Aberration and Doppler Shift of Light
(6.22)
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Special Cases of Doppler Shift
Let us look at three special cases of the Doppler shift of light. The first two cases show the so-called longitudinal Doppler
effect and the last case is the transverse Doppler effect.
First, consider the case where the light source moves directly towards the observer so that the observer measures an angle of
 = 0 for the light. The situation is shown in Figure 6-2. Using Eq. (6.21) with  = 0 gives
f '  f  (1  v / c)
(6.23)
By substituting Eq. (3.15) for the gamma factor this becomes
f ' f
(1  v / c)
1- v2 / c2
 f
(1  v / c)
(1  v / c)(1  v / c)
 f
1 v / c
 f
1 v / c
cv
cv
(6.24)
Solving for f gives our final result,
f  f'
cv
cv
Note that the observed frequency of the light wave is higher than the
source frequency, a result that is consistent with sound waves. The
observed wavelength is therefore shorter that the source wavelength,
i.e. the light is blue-shifted.
(6.25)
S
y
v
x
Fig. 6-2
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Aberration and Doppler Shift of Light
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Next, consider the case where the light source moves directly away from the observer so that the observer measures an angle
of  = 180 for the light. The situation is shown in Figure 6-3. Using Eq. (6.21) with  = 180 gives
f '  f  (1  v / c)
(6.26)
By again substituting Eq. (3.15) for the gamma factor and simplifying we arrive at the final result of
f  f'
cv
cv
Note that the observed frequency of the light wave is
lower than the source frequency in this case. The
observed wavelength is therefore longer that the source
wavelength, i.e. the light is red-shifted.
(6.27)
S
y
v
x
Fig. 6-3
The last case we examine is when the observed angle is 90. It is impossible for the observed angle to have this value for
more than an instant if the source moves with a constant velocity to the right as we have been considering. The angle would
be 90 only when the source is directly below the observer. But let the source move in a circle with the observer at the
center of the circle as shown in Figure 6-4. In that case, the observed angle is always 90 for the light that reaches the
observer. Remember that the angle  is measured from the x-axis but that it is the direction of the relative velocity vector v
that determines the x-axis.
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Aberration and Doppler Shift of Light
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Using Eq. (6.21) with  = 90 gives
f ' f 
(6.28)
f  f '/ 
(6.29)
so the observed frequency is
The observed light is red-shifted; its frequency is observed to be lower
than the source frequency by a factor of gamma. Unlike the longitudinal
Doppler shift, this transverse shift is not predicted by classical
mechanics at all. In the classical picture, since the distance between the
source and observer is constant, there is no bunching up or stretching out
of the wave fronts. However, because of time dilation, there still is a
frequency shift.
v
v
The source records the proper time since the wave fronts always leave
the same position for the source. Suppose it sends out N wave fronts in a
proper time interval of tp. The source frequency is then f’ = N / tp.
This time interval is dilated for the observer by a factor of  so that the N
wave fronts take a longer time of tp to be recorded. The observed
frequency is f = N / tp or f = f’ /  which agrees with the result we
found in Eq. (6.29).
This red-shift has been observed in the radiation coming from particles
traveling in circles in accelerators. Its detection is another piece of
evidence supporting the special theory of relativity.
Module 6
Aberration and Doppler Shift of Light
v
Fig. 6-4
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An Example of Light Aberration
The observed aberration of star light serves as a nice example of light aberration. Based on the observed orientations of
their telescopes in viewing certain stars at different times, astronomers knew that the light was undergoing aberration. The
mechanism for the aberration was not understood. Newtonian relativity predicts an aberration but only if one adheres to the
straight Newtonian view that an observer can see light travel faster than c. As was discussed in Module 2, this view
disregards the results of the Michelson-Morley experiment. In addition, the predicted aberration only works if the relative
speed between the light source and observer is small. Special relativity predicts the aberration for any relative speed
between light source and observer and, as we know, explains the Michelson-Morley experiment with its principle of the
constancy of the speed of light.
Consider a star that is directly overhead and that emits light straight down to the Earth as shown in Figure 6-5(a). In its
frame of S’ the emitted angle ’ is 270 or 3/2. Because of the motion of the Earth along its orbit, there is relative motion
between the Earth and the star. If the Earth is going to the left at its orbital speed of v ~ 10-4c then it appears to the
astronomer that the star is traveling to the right at speed v as shown in Figure 6-5(b). The angle of the light observed by the
astronomer in frame S is  and it is different than 3/2. The telescope must be tilted slightly from the vertical by the angle .
Let’s calculate this tilt angle .
S’

y’
v
S
y
’ = 3 / 2

x’
x
(a)
(b)
Fig. 6-5
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Aberration and Doppler Shift of Light
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We use Eq. (6.15), the light aberration equation, to find
sin( 3 / 2)
c

[cos(3 / 2)  v / c] v
tan  
(6.30)
Since v/c ~ 10-4, the gamma factor is essentially unity so
c
v
tan  
(6.31)
Referring to Figure 6-5, note that  = ’ +  = 3/2 + . Using this fact and a trig identity, we can write
 3

tan   tan       cot 
 2

(6.32)
The angle  is very small. In the small angle approximation,
so that
cot  
cos  1

sin  
(6.33)
tan  
1

(6.34)
Equating Eqs. (6.31) and (6.34) gives us a tilt angle of
  v/c
(6.35)
Thus, the tilt angle of the telescope is approximately 10-4 radians or about 21 seconds of arc. While this angle is small, it is
measurable.
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Aberration and Doppler Shift of Light
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