Why Does a Moving Rod Appear Inclined in Different Reference Frames?

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Homework Statement

A $S'$ frame moves with velocity $v\hat{e_{x}}$ with respect to a S frame.



There's a rod with length $L$, parallel to the $x'$ axis moving with velocity $u \hat{e_{y'}}$.



Show that the rod seems inclined and what's the inclination angle?

Relevant Equations

$t = \gamma (t' + vx')$

$x = \gamma(x' + vt')$

$t' = \gamma (t - vx)$

$x' = \gamma(x - vt)$

Ateempt of solution:

There are two key coordinates in this scenario, the leftmost tip of the rod, which in $S'$ is $C_{0} = (t', 0, ut',0)$ and the opposite tip
$C_{1} = (t', L,ut',0)$

An angle $\phi$ could be found through a relationship such as $tan(\phi) = \frac{ \Delta x}{ \Delta y}$

For $\Delta x$ I have
$x_{0} = \gamma ( x_{0'} + vt') = \gamma vt'$ and

$x_{1} = \gamma (x_{1'} + vt') = \gamma (L + vt')$.

$\Delta x = \gamma (L + vt' - vt') = \gamma L$

For $\Delta y$ I'm not sure how to calculate its coordinates in $S$. Should I use the velocity composition formula? The lorentz transformations would be trivially applied then?

Edit:
I tried to deduce a composition for this case in the following manner

In $S'$, $u_{s'} = \frac {\Delta y'}{\Delta t'}$

I can write these for $S$ to find $\frac{\Delta y}{\Delta t}$

$u_{s'} = \frac {\Delta y}{\gamma (\Delta t - v \Delta x)}$ and dividing by $\Delta t$

$u_{s'} = \frac{ u_{s}}{\gamma (1 - v u_{s})}$

Solving for $u_{s}$ I get $u_{s} = \frac{ u_{s'} \gamma}{1+ \gamma u_{s'} v}$

So to find the $y$ coordinates do I use

$y = \gamma_{u_{s}} (y' + u_{s} t')$

but this is wrong because all these parameters are the same for both $C_{0'}$ and $C_{1'}$, what would imply $y_{0} = y_{1}$ hence $\Delta y = 0$

 

[PeroK]

The rod is parallel to the $x'$ axis. You have it parallel to the $y'$ axis.

 

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The rod is parallel to the x′ axis. You have it parallel to the y′ axis.

Thanks, I'll edit the post