Bob, in frame , is observing the moving plank of Exercise 38. He quickly fabricates a wall, fixed in his frame, that has a hole of length L and that is slanted at angle$\mathit{\theta }$, such that the plank will completely fill the hole as it passes through. This occurs at the instant t=0. According to Anna, moving with the plank, the plank is of course not of length L, but of length${\mathit{L}}_{\mathbf{0}}$. Moreover, because Bob’s wall moves relative to her. Anna sees a hole that is less than L in length, a plank longer than L is headed toward a hole shorter than . Can the plank pass through the hole according to Anna? If so, at what time(s)? Explain.

Consider a Bob, in frame S, is observing the moving plank. He quickly fabricates a wall, fixed in his frame, that has a hole of length L and that is slanted at angle $\theta$,

Consider the Anna, moving with the plank, the plank is of course not of length , but of length ${\mathrm{L}}_0$. Anna sees a hole that is less than L in length, a plank longer than L is headed toward a hole shorter than L.

Write the formula of time lag between these two events.

$\mathbf{∆}\mathbf{t}\mathbf{=}\mathbf{\gamma }\left(∆t^{\text{'}}+\frac{v∆x^{\text{'}}}{c^2}\right)$ …… (1)

Here, $\mathbf{∆}{\mathbf{t}}^{\mathbf{\text{'}}}$is time lag between two legs,c is moving speed.

As a matter of fact, Anna, the top of the board will go through the wall before the bottom. To acquire a descriptive analysis of an issue resembling this one, I could suggest going back and reading problem five of this chapter. The solution to these issues is that all physical phenomena remain constant across all frames of reference. Thus, if the plank can pass through the wall in Bob's frame (S), it can also pass through any other frame, such as Anna's frame (S). In this case, assuming that both the top and bottom of the plank would simultaneously pass through the wall in Bob's frame, a few arguments might be made to investigate the issue.

• Since the vertical distance is the same in both frames, Anna will regard the wall from her frame as being contracted in the direction of motion, causing it to be titled with a wider angle than that seen by Bob (review the previous problem to see the mathematical proof). However, the angle of the plank will be less in Anna's frame than it appears to Bob. This leads to the conclusion that, according to Bob's assertion, the two occurrences are not synced together since the top of the plank will pass the wall in Anna's frame before the bottom of the plank.
• Remembering that the contraction only takes place in the direction of motion provides another defence. As a result, in both frames, the vertical distance will remain constant. Knowing this, one may contend that no framing allows the board to strike the wall.

To demonstrate that if the top and bottom of the plank pass concurrently in Bob's frame, it cannot be synchronised in Anna's frame, we presented a qualitative argument in part. So let's try to determine the gap in time between the two occurrences in Anna's perspective using our Lorentz transformation.

Determine the time lag between these two events.

Substitute 0 for $∆t$and $L_0\cos \left({\theta }_0\right)$for $∆x^{\text{'}}$into equation (1).

$∆t^{\text{'}}=-\frac{u{L}_0\cos \left({\theta }_0\right)}{c^2}$

The negative sign here tells us that the top of the plank will pass through the wall first, then the bottom of the plank came later.

As a matter of fact, Anna, the top of the board will go through the wall before the bottom. The time lag between these two events is$∆t^{\text{'}}=-\frac{u{L}_0\cos \left({\theta }_0\right)}{c^2}$.