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Physics Lesson 18.4.1 - Relativistic Transformation of Velocity

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Welcome to our Physics lesson on Relativistic Transformation of Velocity, this is the first 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.

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.

You have reached the end of Physics lesson 18.4.1 Relativistic Transformation of Velocity. 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.

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18.4Relativistic Transformation of Velocity and the Relativistic Doppler Effect
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18.4.1Relativistic Transformation of Velocity
18.4.2The Relativistic Doppler Effect

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  3. Relativity Revision Notes: Relativistic Transformation of Velocity and the Relativistic Doppler Effect. Print the notes so you can revise the key points covered in the physics tutorial for Relativistic Transformation of Velocity and the Relativistic Doppler Effect
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  6. Continuing learning relativity - read our next physics tutorial: Lorentz Transformations

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