Physics Tutorial: Relativistic Transformation of Velocity and the Relativistic Doppler Effect

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In this Physics tutorial, you will learn:

  • What is the relationship between velocities measured in two different inertial systems of reference?
  • What is the relationship between values of light wavelength measured in two different inertial systems of reference?
  • The same for frequencies of light waves
  • How does Relativistic Doppler Effect helps in understanding how the universe works?

Introduction

So far, we have discussed a number of new concepts that are applied in relativistic events. These concepts are very helpful in finding a relationship between the velocities of particles measured in different inertial frames of reference. Certainly, we must find the same value for the speed of light in all inertial systems considered. This element will act as a touchstone for the accuracy of this new theory.

We will see in this tutorial that in relativistic events, we cannot use anymore the classical formulae for velocity. In addition, we will recall the Doppler Effect of light waves discussed in tutorial 12.7 (The Doppler Effect) but now applying a relativistic approach. Obviously, the findings when using these two approaches must be the same.

Relativistic Transformation of Velocity

Let's suppose we are inside a wagon moving due right at velocity V along the horizontal tracks, as shown in the figure below.

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

On the left side of wagon, there is a source B1 which emits material particles and a receiver R of light signals. In addition, there is a clock at the same position, which measures the time intervals of events occurring inside the wagon. On the right side of wagon, there is a light source B2, which has a special property: it emits a signal towards the receiver R when hit by any material particle coming from B1. The receiver then records the signal and the clock shows the time of this event. In other words, the particle emitted from B1 reaches B2 and the light emitted from B2 reaches R (where there is also the source B1 and the clock).

Let's denote by Δt' the time elapsed since the emission of particle until the light signal arrival to the receiver (calculated through the clock connected to the wagon, i.e. in the inertial system S'). If we denote by L' the length of wagon when measured from inside (in S'), we have

∆t' = L'/v' + L'/c

where v' is the velocity of the particles emitted from B1 when considered from the system S' connected to the wagon.

Now let's consider the event from the viewpoint of an observer outside the wagon, at rest to the Earth (in the inertial S therefore). We denote by Δt1 the time elapsed from the particle's emission to its arrival at B2. During this time, the particle has moved by L + V · Δt1, where L is the length of wagon measured from outside it. Hence, if we denote by v the velocity of particles emitted by B1 when measured from outside the wagon (in S), we obtain

v ∙ ∆t1 = L + V ∙ ∆t1

Now, let's denote by Δt2 the time needed for the light signal emitted from B2 to reach the receiver R when considered from the system S connected to the Earth. Using the same reasoning as above, we obtain

c ∙ ∆t2 = L - V ∙ ∆t2

The negative sign is because light signal moves in the opposite direction to the wagon's motion. From the two above formulae, we obtain

∆t1 = L/v - v and ∆t2 = L/c + V

The total time of event measured by the clock outside the wagon (in S) is

∆t = ∆t1 + ∆t2 = L/v - v + L/c + V

Now we have both Δt' and Δt expressed in terms of L, L', v and V. We already know from the previous tutorial the relationship between Δt' and Δt as well as those between L' and L:

∆t' = √1 - V2/c2 ∙ ∆t and L = √1 - V2/c2 ∙ L'

Using these formulae, we find:

∆t' = L'/v' + L'/c
= L' ∙ (c + v' )/c ∙ v'
= √1 - V2/c2 ∙ ∆t
= √1 - V2/c2 ∙ (L/v - v + L/c + V)

Thus,

L' ∙ (c + v' )/c ∙ v' = √1 - V2/c2L ∙ (c + V) + L(v - v)/(v - v) ∙ (c + V)
L' ∙ (c + v' )/c ∙ v' = √1 - V2/c2L ∙ (c + v)/(v - v) ∙ (c + V)

Substituting

L = √1 - V2/c2 ∙ L'

in the right side of the above equation, we obtain

L' ∙ (c + v' )/c ∙ v' = √1 - V2/c21 - V2/c2 ∙ L' ∙ (c + v)/(v - v) ∙ (c + V)
(c + v' )/c ∙ v' = 1 - V2/c2(c + v)/(v - v) ∙ (c + V)

After doing some easy (but long) mathematical transformations, we obtain

v = v' + V/1 + v' ∙ V/c2

This formula gives the relationship between velocities of the same particle when considered from two inertial frames of reference S and S'. We will find again this formula in the next tutorial using another method.

For V << c (as usually occurs in daily life), we obtain the classical relationship between velocities v = v' (as expected). If we use light signals instead of material particles in the above experiment (i.e. if instead of v' we use c), we obtain

v = c + V/1 + c ∙ V/c2
= c + V/1 + V/c
= c + V/1 + c+ V/c
= c

This result (v = c) for the light signal when viewed from outside the wagon (from S) is correct and expected as the speed of light is equal in all inertial frames of reference.

Example 1

An atomic nucleus flies through a long laboratory tunnel at 0.5c. At a certain point during the flight, this particle emits a "daughter" particle, which moves in the direction of "parent" particle at 0.7c relative to it. What is the speed of daughter particle measured from the system connected to the laboratory (at rest to the ground)?

Solution 1

Using the relativistic formula of velocities relationship, we have

v = v' + V/1 + v' ∙ V/c2

where >v' = 0.5cV = 0.7c, thus,

v = 0.5c + 0.7c/1 + 0.5c ∙ 0.7c/c2
= 1.2c/1 + 0.35c2/c2
= 1.2c/1.35
= 0.89c

If we used the classical formula, we would obtain 0.5c + 0.7c = 1.2c for the value of v. This result would contradict the Special Theory of Relativity, which says (rightfully) that nothing can exceed the speed of light.

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.

Summary

In relativistic events we cannot use the formulae of classical physics as they give wrong results. Using the relativistic approach, we find the following relationship between velocities of the same particle when considered from two inertial frames of reference S and S':

v = v' + V/1 + v' ∙ V/c2

where v is the velocity of the moving particle when viewed from the system S at rest, v' the velocity of particle when viewed from the system connected to the source, V is the moving velocity of system S' relative to S and c is the speed of light in vacuum.

For V << c (as usually occurs in daily life), we obtain the classical relationship between velocities v = v' (as expected). If we use light signals instead of material particles in the above experiment (i.e. if we put c instead of v'), we obtain v = c.

The light waves 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).

The formula that relates the frequency of original and observed light during red shift (when the star or galaxy is moving away from Earth) is

f = f' ∙ 1-V/c/1 + V/c

while that of blue shift (when the star or galaxy is approaching the Earth) is

f = f' ∙ 1 + V/c/1-V/c

where V is the speed of the moving star, f is the observable frequency and f' is the original frequency of light.

Physics Revision Questions for Relativistic Transformation of Velocity and the Relativistic Doppler Effect

1. A particle is moving at 0.8c due right and at a certain point it explodes in two equal daughter particles where one of them moves in the original direction at 0.7c relative to the parent particle. What is the velocity of daughter particle when viewed from a stationary position to the Earth surface?

  1. 0.96c
  2. 1.5c
  3. 0.1c
  4. 0.75c

Correct Answer: A

2. A helium atom is in a galaxy travelling at a velocity of 4.78 × 106 m/s away from the Earth. An astronomer on Earth observes a frequency of 6.52 × 1014 Hz emitted from the helium atom. What frequency of light is really emitted from this helium atom?

  1. 5.18 × 1014 Hz
  2. 4.48 × 1014 Hz
  3. 6.22 × 1014 Hz
  4. 6.82 × 1014 Hz

Correct Answer: C

3. An astronomer is observing a remote galaxy from Earth. Measurements show that the light waves emitted by helium atoms, which normally have a wavelength of 600 nm, are observed from Earth at the new wavelength 500 nm. How fast is the galaxy moving towards or away from us?

  1. 1.66 × 109 m/s
  2. 3.3 × 109 m/s
  3. 3 × 107 m/s
  4. 1.8 × 107 m/s

Correct Answer: D

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