Lorentz Transformation with Two Moving Frames

[trevor51590]

Homework Statement

Essentially, a particle is moving downward from the top of a tower at 0.98c, and the tower is moving up at 0.98c. I am to find the apparent height of the tower from the perspective of the particle

Homework Equations

Δx=γ(Δx′+vΔt′)

Δt=γ(Δt′+vΔx′/c2)


Δx′=γ(Δx−vΔt)

Δt′=γ(Δt−vΔx/c2)

The Attempt at a Solution

I've tried to solve multiple ways and I think I just am getting the concept wrong and could use a hint-

Essentially I figured if both frames are moving at .98c, then the apparent movement of the particle would be 2(.98c) if the tower frame would be fixed, and that I could solve it as a Galilean. The way I was told to solve it was by modifying the Lorentz formulae and frankly I'm having a brain fart figuring out my frames. I was also told c cannot be over 1, which made my original method not work.

 

[BruceW]

Brain fart sounds uncomfortable...

So in the question, they are saying that in some reference frame the particle is moving down at 0.98c and the tower is moving up at 0.98c.

Or are they saying that in the tower's reference frame the particle is moving down at 0.98c and in the particle's reference frame, the tower is moving up at 0.98c?

If it is the second case, then obviously the relative speed between the tower and the particle is 0.98c.

If it is the first case, the velocities do not add so simply as you assumed. So it would not be 2(.98c). You need to use the equation for addition of velocities in special relativity.