- #1

- 7

- 0

## Homework Statement

A rod of proper length L0 oriented parallel to the x axis moves with speed u along the x axis in S. What is the length measured by an observer in S'?

## Homework Equations

Relativistic relative velocity:

For an object moving with speed u in S, its speed u' in S' = (u - v)/(1 - (u*v)/(c^2)) where v is the speed of S' relative to S.

Lorentz transformations:

x' = γ(x - vt)

t' = γ(t - xv/(c^2))

## The Attempt at a Solution

The answer to the problem is given, but I think it's been heavily algebraically manipulated and I'm not sure if the answer I got matches. I want to check if the way I did it makes sense:

Consider two events in S: when the right end of the rod crosses the y-axis and when the left end of the rod crosses the y-axis. Looking at the x and t coordinates only (since the rod's movement is parallel to the x-axis)

event #1: (x,t) = (0,0)

event #2: (x,t) = (0, L0/u)

Now the length of the rod = Δt * u = L0.

The length L' can be found similarly as Δt' * u' = L'

t1' = γ(t1 - x1*v/(c^2)) = 0

t2' = γ(t2 - x2*v/(c^2)) = γ*t2 = γ*L0/u

So Δt' = t2' - t1' = γ*L0/u - 0 = γ*L0/u

Using the relativistic relative velocity equation, u' = (u - v)/(1 - u*v/(c^2))

So L' = γ*L0/u * (u - v)/(1 - u*v/(c^2))

The answer given is L' = L0 * sqrt[(c^2 - v^2)(c^2 - u^2)]/(c^2 - u*v)

These answers could potentially match up, but I couldn't get them to using algebraic manipulations, but I might just not know how they manipulated it/not know some trick they used. Does the way I solved the problem make sense, though?