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Physics Lesson 18.4.2 - The Relativistic Doppler Effect

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Welcome to our Physics lesson on The Relativistic Doppler Effect, this is the second lesson of our suite of physics lessons covering the topic of Relativistic Transformation of Velocity and the Relativistic Doppler Effect, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

The Relativistic Doppler Effect

In Section 12, tutorial 12.7, we have explained the Doppler Effect illustrating it with examples from sound waves. The classical formula of Doppler Effect is

f = v ± vd/v ∓ vs ∙ f0

where v is the speed of wave, vs is the speed of source and vd is the speed of detector, f0 is the original frequency emitted by the source and f is the observed (detected) frequency by the receiver. The plus or minus signs are used to show whether the source is approaching the detector or moving away from it.

It must be highlighted the fact that the effect of frequency change in Doppler Effect of mechanical waves (including sound waves) has nothing to do with the motion of source and detector relative to each other but because they are moving in the medium of wave's propagation.

The light waves however, do not require any material medium to propagate. Therefore, any possible Doppler Effect is simply determined by the movement of observer relative to the source (for light vd = vs = 0 because they are too small to be considered and moreover, there is no material medium of propagation in vacuum; the cosmic ether does not exist).

To calculate the Doppler Effect of light, we consider a light source O' moving in the positive direction of the X-axis of an observer located at origin O of a fixed system of reference S connected to the ground, as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Relativistic Transformation of Velocity and the Relativistic Doppler Effect

Let's suppose that in the instant t = t' = 0 (which represent the time origins for the inertial systems S and S' where S is the system connected to the stationary observer and S' is the system connected to the light source), the source starts emitting light rays represented in the figure through the curved waveforms. The source emits N waves in a time interval Δt'. Obviously, all waves are emitted from the same point for an observer O' part of the system S' that moves together with the source, despite the fact that for the observer O in the system S (connected to the ground) they are both moving at velocity V.

Let's denote by Δt the time interval in which the observer O has detected the N waves emitted from the source. It is clear that the total distance d travelled by the Nth wave during this time interval is

d = c ∙ ∆t + V ∙ ∆t

because not only the light waves are moving at c but the source itself is moving at velocity V as well. Thus, since wavelength is calculated by dividing the total distance by the number of waves, we have

λ = d/N
= c ∙ ∆t + V ∙ ∆t/N
= (c + V) ∙ ∆t/N

On the other hand, we have for the number of waves N:

N = f' ∙ ∆t' = f ∙ ∆t

where f' is the frequency of light waves measured from O' connected to S' and f is the frequency of light waves measured from O connected to S (we expect a lower frequency when measured from O' because the source is moving in the same direction to the observer). Thus, substituting N in the previous equation, we obtain for the wavelength λ:

λ = (c + V) ∙ ∆t/(f' ∙ ∆t'

From the equation of waves c = λ · f, we can write

f' = c/λ'

where λ' is the wavelength of light signals when measured from O'. Thus, we obtain

λ = (c + V) ∙ ∆t ∙ λ'/c ∙ ∆t'

We already know the time dilation formula found in the previous tutorial:

∆t' = √1 - V2/c2 ∙ ∆t

Substituting it in the last equation, we obtain

λ = (c + V) ∙ λ'/c ∙ √1 - V2/c2
= c ∙ (1 + V/c) ∙ λ'/c ∙ √(1 - V/c) ∙ (1 + V/c)
= λ' ∙ 1 + V/c/1 - V/c

Hence, since λ = c / f and λ' = c / f', we have

c/f = c/f'1 + V/c/1-V/c

and therefore, we obtain the formula that relates the frequencies of light wave measured from the two inertial systems of reference S and S' (at their origins O and O'):

f = f' ∙ 1-V/c/1 + V/c

The last equation represents the formula of Doppler Effect of light. We have introduced this formula in tutorial 12.7 "The Doppler Effect" but without explaining how it was obtained. We simply substituted the term V/c by β and explained the phenomena of red-shift and blue-shift observed when a star is approaching us or moving away from us at high speed, which are purely relativistic phenomena.

We see that f < f' when the source is moving away from the observer (as occurs in our example). Hence, the colours of visible light observed from a star which moves away from us point towards red, as this colour has the lowest frequency of visible light (red-shift). Since this phenomenon is observed only in remote parts of the universe, scientists concluded that the universe is expanding. The opposite can also occur (a star can be moving towards us - a phenomenon called "blue shift"; in this case, the observed frequency of light increases and the rays emitted by the star points towards blue - one of the most powerful colours of visible light). The formula of light frequency during blue shift therefore is

f = f' ∙ 1 + V/c/1-V/c

As a conclusion, we can say that this apparently modest formula (the Doppler Effect formula for light) has been used to understand how the universe works and how did it start. Thus, since the universe is expanding, it is logical that in its early days the whole universe has been concentrated in a very small region until it exploded due to the high energy accumulated. This explosion is called "Big Bang", which we will discuss in the next sections of this course.

Example 2

A remote galaxy is observed from Earth. Measurements show that the light waves emitted by hydrogen atoms, which normally have a wavelength of 434 nm, are observed at Earth at the new wavelength 600 nm (from red they look blue). How fast is the galaxy moving towards or away from us?

Solution 2

We have the following clues:

λ' = 434 nm = 434 × 10-9 m = 4.34 × 10-7 m
λ = 600 nm = 600 × 10-9 m = 6 × 10-7 m
(c = 3 × 108 m/s)
V = ?

We must find the corresponding frequencies first:

f' = c/λ'
= 3 × 108 m/s/4.34 × 10-7 m
= 6.91 × 1014 Hz

and

f = c/λ
= 3 × 108 m/s/6 × 10-7 m
= 5 × 1014 Hz

Hence, applying the Doppler Effect formula for light when the source is moving towards us (because there is a blue shift), we obtain

f = f' ∙ 1 + V/c/1-V/c
f2 = f'21 + V/c/1-V/c
f2 ∙ (1-V/c) = f'2 ∙ (1 + V/c)
f2-f2V/c = f'2 + f'2V/c
f2-f'2 = (f'2 + f2 ) ∙ V/c
V = f2 - f'2/f'2 + f2 ∙ c
= (6.91 × 1014 Hz)2-(5 × 1014 Hz)2/(5 × 1014 Hz)2 + (6.91 × 1014 Hz)2 ∙ (3 × 108 m/s)
= (47.75 - 25) × 1028/(25 + 47.75) × 1028 ∙ (3 × 108 m/s)
= 22.75/72.75 ∙ (3 × 108 m/s)
= 0.313 ∙ (3 × 108 m/s)
= 0.313 c

This value is about 94 000 km/s, a very high speed.

You have reached the end of Physics lesson 18.4.2 The Relativistic Doppler Effect. There are 2 lessons in this physics tutorial covering Relativistic Transformation of Velocity and the Relativistic Doppler Effect, you can access all the lessons from this tutorial below.

More Relativistic Transformation of Velocity and the Relativistic Doppler Effect Lessons and Learning Resources

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18.4Relativistic Transformation of Velocity and the Relativistic Doppler Effect
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18.4.1Relativistic Transformation of Velocity
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