I understand the ladder paradox but am confused by a variation of it where an action is taken to trap an object that is Lorentz-contracted.

The same question was asked here, but to the best of my understanding they don't address the actual question asked:

In the ladder and barn paradox what happens if we leave the doors closed?

The original scenario is confusing to me becuase it invokes limitations of object rigidity to explain why the door can't be closed to trap the lorentz-contracted object in the room. Also the answers in that post don't really seem to answer the question.

To avoid running into the same issue with this question I'll construct a new scenario that could be run in an actual laboratory (albeit an extremely sophisticated one), as opposed to requiring The Flash, barns and extremely rigid ladders.

Here is the scenario / experiment:

A rod travels in the direction of its lengthwise axis at relativistic speed. We set up a crystal and project a beam of light through it to create a spectrum of colours, such that the rod will pass through the red light first, and the violet last. Both of the rod and the wall eclipsed by the rod as it passes by measure the light landing on it. The beam is projected in such a way that all colors of the spectrum reach the path of the rod at the same time from the perspective of a stationary observer in the room.

The light is completely orthogonal to the path of the rod.

We project a pulse of light, such that, from the perspective of the room, the light hits the rod and some red and violet light lands on the sensors behind the rod, as it was shortened by Lorentz contraction due to its speed.

The orange, yellow, green and blue light landed on the rod.

From the rod's perspective, the experiment and spectrum were contracted. From the rod's perspective the light hit the rod at some point in time, but the red and violet light couldn't have passed through in front of and behind the rod, since they were too close. If the spectrum lands centered on the rod, then it will have eclipsed the full spectrum of colours.

What will we see when we retrieve the rod and inspect the sensors of both the rod and the sensors in the room? We can't both see the full spectrum landing on the rod and still have light landing on the sensors in the room...

• Not sure how to actually answer, but think about this: In the vanilla version of the paradox where the ladder is not trapped, nothing is accelerated; not the barn, not the ladder, not any observer. In the "doors locked" version it's exactly the same up until the moment when the leading edge of the ladder touches the locked door. After that moment, something must bend or break. Relativity does not allow any object to be truly rigid, because if any object was truly rigid, then the speed of sound in the material of which it was made would be infinite Commented Jan 25, 2021 at 14:30
• I think that the punchline is, when the front of the ladder hits the closed door, and if the door holds, then it is not possible for the "news" that the ladder is stopping to reach the tail end of the ladder before the tail end of the now, very squashed, ladder is inside the barn. I don't know how to prove that that's always the case though. Commented Jan 25, 2021 at 14:39
• @quant You have to remember that the ladder is not rigid (in relativity, no object can be completely rigid). As the Flash decelerates the ladder contracts. Commented Jan 25, 2021 at 14:53
• @gandalf61, If The Flash holds the ladder at its midpoint, then the front half will stretch when he stops, and the back half will be compressed. But, in a simpler version of the paradox, he doesn't even need to be there. The ladder simply is hurtling through space at constant speed. (Maybe it got all of that momentum when The Flash threw it like a javelin.) What stops it is the patented, absolutely indestructable exit door failing to open when the front end of the ladder hits. Meanwhile, the entrance door still can close after the back end of the rapidly collapsing ladder passes through. Commented Jan 25, 2021 at 23:20
• Please choose only one scenario and delete the other. You should ask only one question per post
– Dale
Commented Jan 26, 2021 at 12:47

I think this answer will disappoint you since it doesn't provide much insight into the other question, but with respect to the rest frame of the rod, the light arrives at an oblique angle due to aberration. The cross-sectional width of the rod along the path of the light is narrower than the light beam, so the red and violet light will make it past. It will then proceed toward the wall, which is rapidly receding behind the rod. (The fact that the light must hit this moving target is one way of seeing that its path must be angled in this frame.)

• Both answers helped me to understand the scenario but I'll mark this one since it provided a link to the specific concept that I was misunderstanding (aberration). Thanks! Commented Jan 28, 2021 at 22:43
• It's fascinating to me (as a layperson I suppose) that these seemingly unrelated phenomena conspire to make these paradoxes evaporate. I guess if I could visualise 4 dimensions then all of this would seem like an obvious consequence of geometry... Commented Jan 28, 2021 at 22:45
• @quant You don't need to visualize 4 dimensions for understanding the given problem but only two-dimension (one for time and one for space). You can draw a spacetime diagram and see the solution as a consequence of geometry. Commented Jan 30, 2021 at 8:34

"The beam is projected in such a way that all colours of the spectrum reach the path of the rod at the same time from the perspective of a stationary observer in the room."

But that doesn't mean that all colours will reach at the same time from the perspective of the rod's frame. The Violet will reach the path of the rod before the rod even reaches that point and the red will reach the path of the rod when the rod has already passed that point.

I have here ignored the fact that the colours won't appear the same to the rod's observer as to the observer in the room.