Physics Lesson 18.5.4 - Lorentz Transformation of Velocity

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

Lorentz Transformation of Velocity

Lorentz transformations of spacetime coordinates allow us find the relativistic formulae of velocity transformations for a particle moving and observed in two inertial systems S and S'. The particle's motion can be random; we will discuss only for the instantaneous velocity of particle in the two systems, as defined in Kinematics section.

Let's suppose a particle is at the point x, y, z of the system S in the instant t (this can be thought as an event). The same event is characterized by the coordinates x', y', z' of the system S'. After a short time interval Δt has elapsed in S, the particle (event) has the new coordinates x + Δx, y + Δy, z + Δz in the time t + Δt. This can be thought as the second event, which has the new spacetime coordinates x' + Δx', y' + Δy', z' + Δz', t' + Δt' in S'. If the event occurs only according X or X', based on the Lorentz transformation of spacetime coordinates we have:

x' = x - V ∙ t/√1 - V²/c²
y' = y
z' = z
t' = t - x ∙ V/c²/√1 - V²/c²

and

x' + ∆x' = (x + ∆x) - V ∙ (t + ∆t)/√1 - V²/c²
y' + ∆y' = y + ∆y
z' + ∆z' = z + ∆z
t' + ∆t' = (t + ∆t)-(x + ∆x) ∙ V/c²/√1 - V²/c²

Subtracting the first set of equations from the second, we have

(x' + ∆x' ) - x' = (x + ∆x) - V ∙ (t + ∆t)/√1 - V²/c² - x - V ∙ t/√1 - V²/c²
(y' + ∆y' ) - y' = (y + ∆y) - y
(z' + ∆z' ) - z' = (z + ∆z) - z
(t' + ∆t') - t' = (t + ∆t) - (x + ∆x) ∙ V/c²/√1 - V²/c² - t - x ∙ V/c²/√1 - V²/c²

After doing the operations, we obtain

∆x' = ∆x - V ∙ ∆t/√1 - V²/c²
∆y' = ∆y
∆y' = ∆y
∆t' = ∆t - V/c² ∙ ∆x/√1 - V²/c²

Dividing each of the first three equations by the fourth one, we obtain:

∆x'/∆t' = ∆x - V ∙ ∆t/∆t - V/c² ∙ ∆x
∆y'/∆t' = ∆y/∆t - V/c² ∙ ∆x ∙ √1 - V²/c²
∆z'/∆t' = ∆z/∆t - V/c² ∙ ∆x ∙ √1 - V²/c²

Dividing the numerator and denominator in the right side of each equation by Δt, we obtain

∆x'/∆t' = ∆x/∆t - V ∙ ∆t/∆t/∆t/∆t - V/c² ∙ ∆x/∆t
∆y'/∆t' = ∆y/∆t ∙ √1 - V²/c²/∆t/∆t - V/c² ∙ ∆x/∆t
∆z'/∆t' = ∆z/∆t ∙ √1 - V²/c²/∆t/∆t - V/c² ∙ ∆x/∆t

∆x'/∆t' = ∆x/∆t - V/1 - V/c² ∙ ∆x/∆t
∆y'/∆t' = ∆y/∆t ∙ √1 - V²/c²/1 - V/c² ∙ ∆x/∆t
∆z'/∆t' = ∆z/∆t ∙ √1 - V²/c²/1 - V/c² ∙ ∆x/∆t

Assuming the time intervals in the two systems as very small (Δt' → 0 and Δt → 0), i.e. taking the above rates as limits, we find the components of particle's velocity in the two inertial systems S' and S:

v_x' = v_x - V/1 - V ∙ v_x/c²
v_y' = v_y ∙ √1 - V²/c²/1 - V ∙ v_x/c²
v_z' = v_z ∙ √1 - V²/c²/1 - V ∙ v_x/c²

If we consider a one dimensional light ray (for example a light ray emitted by a laser) moving only in the X (X') direction, we have v_x = c, v_y = 0 and v_z = 0 (in S). Thus, we find for the velocity in S' (giving that V << c):

v_x' = c - V/1 - V ∙ c/c² = c - V/1 - V/c = c/1 = c
v_y' = 0 ∙ √1 - V²/c²/1 - V ∙ v_x/c² = 0
v_z' = 0 ∙ √1 - V²/c²/1 - V ∙ v_x/c² = 0

Therefore, the observer in S' measures the same velocity c in the positive direction of X' while the velocity of light in the other directions is zero (as expected).

The Lorentz transformations for velocity converge with the classical formulae for V << c or for c → ∞. This is obvious given their structure.

Example 3

A spaceship leaves the Earth at v₁ = 0.9c and at a certain point, it emits a spatial module at v₂ = 0.7c relative to itself. What is the velocity of module when viewed from Earth?

Solution 3

Clues:

v₁ = V = 0.9c
v₂ = v'x = 0.7c
v_x = ?

Using the equation for Lorentz transformation of velocity

v_x' = v_x - V/1 - V ∙ v_x/c²

we obtain after rearranging and substituting the given values:

v_x' - V ∙ v_x/c² ∙ v_x' = v_x -V
v_x + V ∙ v_x'/c² ∙ v_x = v_x' + V
v_x ∙ (1 + V ∙ v_x'/c² ) = v_x' + V
v_x = v_x' + V/1 + V ∙ v_x'/c²
= 0.7c + 0.6c/1 + 0.6c ∙ 0.7c/c²
= 1.3c/1 + 0.42
= 1.3c/1.42
= 0.915c

This result is reasonable as it is smaller than c. If we used the classical approach, the calculations would give v = 0.9c + 0.7c = 1.6c - a wrong result which contradicts the findings of the Special Theory of Relativity discovered by Einstein.

You have reached the end of Physics lesson 18.5.4 Lorentz Transformation of Velocity. There are 4 lessons in this physics tutorial covering Lorentz Transformations, you can access all the lessons from this tutorial below.

More Lorentz Transformations Lessons and Learning Resources

Relativity Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
18.5	Lorentz Transformations
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
18.5.1	The Spacetime Lorentz Transformations
18.5.2	Galilean Transformations as Limit of Lorentz Transformations
18.5.3	Quick Recap of Dilation and Contraction
18.5.4	Lorentz Transformation of Velocity

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