In a relativistically moving train, a metal rod is dropped from the ceiling, parallel to the floor. To stay in the realm of special relativity, let's say it's moving down at constant velocity rather than falling under gravity. In the reference frame of a stationary platform watching the train zoom by, does the rod stay parallel to the ground as it falls? The following calculation is Pete's attempt at a solution, and it concludes that the rod is not parallel to the ground in the platform frame.

$$S = $$ rest frame of the rod

$$S' = $$ rest frame of the train

$$S'' = $$ rest frame of the platform

$$v = (0, u)$$ is the velocity of $$S'$$ relative to $$S$$

$$v' = (V, 0)$$ is the velocity of $$S''$$ relative to $$S'$$

$$\gamma = 1/\sqrt{1-u^2/c^2}$$

$$\gamma' = 1/\sqrt{1-V^2/c^2}$$

The rod has two ends, A and B. In the rod frame, we place them at (x,y) = (0,0) and (1,0) respectively.

In $$S$$:

$$\begin{align}

A &= (x_A, \ y_A) \\

&= (0, \ 0)

\end{align}$$

$$\begin{align}

B &= (x_B, \ y_B) \\

&= (1, \ 0)

\end{align}$$

The angle of the rod with the x-axis at time t is:

$$\begin{align}

\tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\

&= 0

\end{align}$$

A &= (x_A, \ y_A) \\

&= (0, \ 0)

\end{align}$$

$$\begin{align}

B &= (x_B, \ y_B) \\

&= (1, \ 0)

\end{align}$$

The angle of the rod with the x-axis at time t is:

$$\begin{align}

\tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\

&= 0

\end{align}$$

In $$S'$$:

$$\begin{align}

A' &= (x_A'(t'), \ y_A'(t')) \\

&= (x_A, \ \frac{y_A}{\gamma} - ut') \\

&= (0, \ -ut')

\end{align}$$

$$\begin{align}

B' &= (x_B'(t'), \ y_B'(t')) \\

&= (x_B, \ \frac{y_B}{\gamma} - ut') \\

&= (1, \ -ut')

\end{align}$$

The angle of the rod with the x-axis at time t' is:

$$\begin{align}

\tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\

&= 0

\end{align}$$

A' &= (x_A'(t'), \ y_A'(t')) \\

&= (x_A, \ \frac{y_A}{\gamma} - ut') \\

&= (0, \ -ut')

\end{align}$$

$$\begin{align}

B' &= (x_B'(t'), \ y_B'(t')) \\

&= (x_B, \ \frac{y_B}{\gamma} - ut') \\

&= (1, \ -ut')

\end{align}$$

The angle of the rod with the x-axis at time t' is:

$$\begin{align}

\tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\

&= 0

\end{align}$$

In $$S''$$:

$$\begin{align}

A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\

&= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\

&= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\

&= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right)

\end{align}$$

$$\begin{align}

B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\

&= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\

&= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\

&= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right)

\end{align}$$

The angle of the rod with the x-axis at time t'' is:

$$\begin{align}

\tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\

&= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\

&= \frac{uV\gamma'^2}{\gamma c^2}

\end{align}$$

A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\

&= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\

&= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\

&= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right)

\end{align}$$

$$\begin{align}

B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\

&= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\

&= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\

&= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right)

\end{align}$$

The angle of the rod with the x-axis at time t'' is:

$$\begin{align}

\tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\

&= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\

&= \frac{uV\gamma'^2}{\gamma c^2}

\end{align}$$