Lorentz Transformation of y-velocity

[muffinbottoms]

Homework Statement

A Particle moves with uniform speed V'y = Δy'/Δt' along the y'-axis of the rocket frame. Transform Δy' and Δt' to laboratory displacements Δx, Δy, and Δt using the Lorentz transformation equations. Show that the x-component and the y-component of the velocity of this particle in the laboratory frame are given by the expressions ... (under relevant equations)

Homework Equations

Vx = V rel
Vy = Vy'(1-Vrel^2)^.5

The Attempt at a Solution

Okay so the textbook i got this problem from is lacking in both directions and example problems. This is what I have so far..

x = x' because the particle is moving along the y-axis
z=z'

Δt = vγy' + γt
Δx = x'
Δy= γy' + Vγt'

 

[muffinbottoms]

or is it that
t' = -Vrelγy+γt = γ(-Vrel(y)+t)
y' = γy-Vrelγt = γ(y-Vrel(t))

how should i continue?

 

[Chestermiller]

Homework Statement

A Particle moves with uniform speed V'y = Δy'/Δt' along the y'-axis of the rocket frame. Transform Δy' and Δt' to laboratory displacements Δx, Δy, and Δt using the Lorentz transformation equations. Show that the x-component and the y-component of the velocity of this particle in the laboratory frame are given by the expressions ... (under relevant equations)

Homework Equations

Vx = V rel
Vy = Vy'(1-Vrel^2)^.5

The Attempt at a Solution

Okay so the textbook i got this problem from is lacking in both directions and example problems. This is what I have so far..

x = x' because the particle is moving along the y-axis
z=z'

Δt = vγy' + γt
Δx = x'
Δy= γy' + Vγt'

The relevant equations are not correct. If the particle is moving relative to the rocket, and the rocket is moving relative to the laboratory, then you have to use the relativistic Velocity Addition Theorem to get the velocity of the particle relative to the laboratory.

 

[cryora]

I was able to get the given expressions by assuming that the rocket moves only along the x-axis in the laboratory frame with relative speed v_rel, as suggested by v_x = v_rel, though the fault is on the book for not mentioning that specifically, thus forcing you to assume that the relative velocity could be in any direction.

 

[HalfLostAndSearching]

To find the x-component of the velocity in the lab frame, we can use the Lorentz transformation equation for velocity, Vx = Vrel, where Vrel is the relative velocity between the rocket frame and the lab frame. Since the particle is moving along the y'-axis in the rocket frame, its velocity in the lab frame will only have an x-component, which is equal to the relative velocity between the two frames.

To find the y-component of the velocity in the lab frame, we can use the Lorentz transformation equation for y-velocity, Vy = Vy'(1-Vrel^2)^.5. This equation takes into account the time dilation and length contraction effects of special relativity.

Substituting the given values for V'y and Vrel into the equation, we get:

Vy = (Δy'/Δt')(1-Vrel^2)^.5

= (Δy'/Δt')(1-v^2)^.5

= (Δy'/Δt')(1-(v^2/c^2))^.5

= (v/c)(Δy'/Δt') [since v<<c]

= γy' + Vγt' [using the Lorentz transformation equations for time and length]

Therefore, the expressions for the x-component and y-component of the velocity in the lab frame are:

Vx = Vrel = 0 [since the particle is moving along the y-axis in the rocket frame]

Vy = γy' + Vγt' = (v/c)(Δy'/Δt') + (v/c)(Δt'/Δt') = (v/c)(Δy' + Δt')/Δt'

This shows that the x-component of the velocity is zero and the y-component of the velocity is dependent on the ratio of the particle's displacement and time in the rocket frame relative to the speed of light. This is consistent with the principles of special relativity, where the velocity of an object is dependent on the observer's frame of reference.