Relativity with stationary rod

[latentcorpse]

An observer in S' sees a stationary rod lying in the x'-y' plane and making an angle $\phi'$ with the x' axis. Show that
$\tan{\phi'}=\frac{1}{\gamma} \tan{\phi}$

where $\phi$ is the angle the rod makes with the x-axis according to an observer in S.

so far I have $\tan{\phi'} = \frac{y'}{x'}=\frac{y}{\gamma(x-vt)}=\frac{1}{\gamma} \frac{y}{x-vt}$

if i could just get rid of that vt, i'd be there but i can't seem to get it to go away...

 

[dx]

I assume the rod is moving in the x direction. In the S frame, let the length in the x direction be L_x and it's length in the y direction be L_y. In the S' frame, the length in the y direction will be the same i.e. L'_y = L_y, and the length in the x direction will be shorter by a factor of √(1 - v²), i.e. L'_x = √(1 - v²) L_x. So we get

tan θ = L_y / L_x

tan θ' = L'_y / L'_x = ^L_y/_{Lx√(1 - v²)}

 

[dx]

Your mistake was that you took tan θ to be y/x. If the coordinates of the tips of the rod are (x₁, y₁) and (x₂, y₂), then tan θ will be ∆y/∆x where ∆x = x₂ - x₁ and ∆y = y₂ - y₁. You should apply the Lorentz transform to both (x₁, y₁) and (x₂, y₂) to obtain ∆x' and ∆y'. tan θ' will then be ∆y'/∆x'.