Travelling near the speed of light

Question

Suppose that there is a long straight board held horizontal to the ground, and it is released from rest. If I were to travel at 0.9c past the board (from one end to the other), would I perceive both ends of the board to fall simultaneously? Or will I see one end of the board reach the ground faster than the other?

Answer 1

If both ends are released simultaneously in your frame, they will hit the ground simultaneously in your frame. If they are released simultaneously in the ground frame, they will hit the ground simultaneously in the ground frame, hence not in yours.

Answer 2

In order to tackle this problem, one would first model the dimensions of the board at time t relative to the observer. Then one would model the propagation of light bouncing off the board and reaching the observers eye.

In step 1, we obviously have length contraction of the board which acts in one dimension since the downwards speed is non relativistic.

In step 2, we must consider that light emitted from the back of the board will take ever so slightly longer than light emitted from the front of the board to reach the observer's eye. Therefore the back of the board will be slightly higher in the image formed. However I see this effect as negligible since the downwards speed is small. One would instead notice an elongation of the board compared to what is expected by length contraction

Answer 3

When you say "would I perceive both ends to fall simultaneously" there are actually two possible interpretations of the time you "perceive" an event: when light literally hits your eye, and what you calculate the original event to have happened after adjusting for the time the light traveled. It is the latter time that is usually of interest. But for completeness there are 4 events in spacetime whose coordinates we can calculate:

(1) The back of the board begins to fall ($B_{fall}$)

(2) The front of the board begins to fall ($F_{fall}$)

(3) Light from the back of the board falling hits your eye ($B_{eye}$)

(4) Light from the front of the board falling hits your eye ($F_{eye}$)

Suppose in the rest frame of the Earth the board has length $L$ and we set the origin at the point in spacetime where the back of the board begins to fall. So the "Earth" coordinates are $(t,x) = (0, 0)$ for $B_{fall}$, and $(0, L)$ for $F_{fall}$.

Let's assume the spaceship is traveling at speed $v$ in the positive x direction, and use the same spacetime point as Earth did as the origin. To calculate spaceship coordinates from Earth coordinates we use the Lorentz transformation $(t', x') = (\gamma(t-vx), \gamma(x-vt))$ where $\gamma=1/\sqrt{1-v^2}$ (for convenience we are using units where c=1, so length is measured in "light seconds"). In the spaceship coordinates $B_{fall}$ is still $(0, 0)$, but $F_{fall}$ is now $(\gamma(-vL), {\gamma}L)$, which means the times for $F_{fall}$ and $B_{fall}$ are different.

You could do a similar calculation for the events $B_{eye}$ and $F_{eye}$, but you'll have to know exactly where your eye is relative to the board.