I Relativity paradox: rocket landing in a cylinder

[PeterDonis]

I think this is actually possible if one assumes the whole rocket, by whatever mechanism, is decelerating together

There is the limiting case of Born rigid deceleration case that was mentioned earlier. This, as has been said, requires every single piece of the rocket to have its own rocket engine that decelerates it in just the right way, and the whole deceleration has to be pre-programmed since it cannot be coordinated in real time (that would require faster than light communication between the parts of the rocket, which is impossible).

In this (unrealizable in practical terms) limiting case, the events at which the parts of the rocket stop will still be spacelike separated, so their time ordering will be frame-dependent, and they cannot causally affect each other. But in the cylinder rest frame (which is inertial throughout since the cylinder remains in free fall the whole time), these events will all be simultaneous; in this frame every part of the rocket comes to rest in just the right position relative to the cylinder, at the same instant of time, such that the bottom of the rocket just touches the bottom of the cylinder as it comes to a stop and the head of the rocket just touches the top of the cylinder as it comes to a stop.

assuming the rocket is in a gravity field, then so is the cylinder: each is in an accelerated frame, not an inertial one.

"In an accelerated frame" is the wrong way to put it. All objects are "in" all frames; they aren't "in" one rather than another.

The rocket's state of motion is accelerated for at least some portion of the scenario, while the cylinder's is not--as stated above, it is in free fall the whole time. If one adopts a non-inertial frame for the rocket, one can (sort of) view a "gravitational field" as existing in this frame, pointed towards the bottom of the rocket, which will cause the cylinder, which was initially flying upward very fast, to decelerate, just as a stone thrown upward will decelerate. However, viewing the problem this way this adds nothing useful to the analysis.

 

[PeterDonis]

as when a rock thrown upwards slows down under gravity with all its parts slowing together.

This is not a valid analogy. The thrown rock is in free fall; its "deceleration" is only coordinate deceleration, not proper deceleration. The rocket's deceleration is proper deceleration. They're not the same thing.

 

[OscarCP]

This is not a valid analogy. The thrown rock is in free fall; its "deceleration" is only coordinate deceleration, not proper deceleration. The rocket's deceleration is proper deceleration. They're not the same thing.

I don't recall the cylinder being in free fall mentioned in the original statement of this thread, the one with the question.
As to the equivalence between the end of the complete series of events as observed in each frame, I have encountered a difficulty with this: let's assume that if in its own rest fame the rocket's bottom touches first the relativistically shortened cylinder's bottom if and only if before the top does, the rocket explodes. It cannot be that if, in the cylinder's frame is the top that touches first, the rocket does not explode, because the rocket cannot end up both exploded and not exploded: this is not quantum physics. So what gives?

 

[PeterDonis]

I don't recall the cylinder being in free fall mentioned in the original statement of this thread

Hm, yes, I see it says the cylinder is "on Earth", which would mean it is indeed not in free fall. I suspect that's not what the OP intended, or what most of the previous posts in this thread (including mine) have implicitly assumed.

Having the cylinder sitting at rest on the Earth's surface does change things a bit, but actually not much. The cylinder's proper acceleration is now not zero (it's 9.8 meters per second squared upward), but it's still constant (i.e., not adjustable with rocket thrust or anything like that). So the deceleration profile of the rocket would be somewhat different in detail, but qualitatively everything that's been said in this thread would still hold. And the (unrealizable in practice) requirements for the Born rigid deceleration limiting case would be pretty much the same--the only difference is that the end state is now not free fall but upward acceleration at 9.8 meters per second squared to match the cylinder.

let's assume that if in its own rest fame the rocket's bottom touches first the relativistically shortened cylinder's bottom if and only if before the top does, the rocket explodes.

You can't just assume that; you have to describe exactly how the explosion would be triggered by actual physical signals, whose speed is limited to the speed of light, from actual physical sensors located at particular points. When you do that, you will find that the explosion trigger condition is invariant, not frame-dependent; it might look weird in different frames in terms of the relative timing of the signals, but the fact of the explosion either occurring or not occurring will not depend on which frame you choose.

 

[OscarCP]

Hm, yes, I see it says the cylinder is "on Earth", which would mean it is indeed not in free fall. I suspect that's not what the OP intended, or what most of the previous posts in this thread (including mine) have implicitly assumed.

Having the cylinder sitting at rest on the Earth's surface does change things a bit, but actually not much. The cylinder's proper acceleration is now not zero (it's 9.8 meters per second squared upward), but it's still constant (i.e., not adjustable with rocket thrust or anything like that). So the deceleration profile of the rocket would be somewhat different in detail, but qualitatively everything that's been said in this thread would still hold. And the (unrealizable in practice) requirements for the Born rigid deceleration limiting case would be pretty much the same--the only difference is that the end state is now not free fall but upward acceleration at 9.8 meters per second squared to match the cylinder.You can't just assume that; you have to describe exactly how the explosion would be triggered by actual physical signals, whose speed is limited to the speed of light, from actual physical sensors located at particular points. When you do that, you will find that the explosion trigger condition is invariant, not frame-dependent; it might look weird in different frames in terms of the relative timing of the signals, but the fact of the explosion either occurring or not occurring will not depend on which frame you choose.

Good point. That makes sense. Thanks.

 

[Ibix]

let's assume that if in its own rest fame the rocket's bottom touches first the relativistically shortened cylinder's bottom if and only if before the top does, the rocket explodes.

You're implying the existence of a "simultaneity detector" of some kind. You can build such things, but they detect simultaneity in one frame only, and detect some specific time separation when analysed in other frames because simultaneity is frame dependent. So the explosion or lack thereof is invariant, but frames will differ on the "because A happened before B" part - some will say things like "because A happened no less than a nanosecond after B".

 

[OscarCP]

I think I understand now: I am better at this when I can visualize the events.

In the cylinder frame the rocket top touches first, the signal that says "top just touched and rocket has started to stop" propagates at a finite speed and, as it does, the portion of the rocket above the wave front in the rocket body stops and stretches to its rest length in the frame of the cylinder. When the wave reaches the rocket's bottom, this is now at is full rest length from top to bottom in the cylinder's frame, and its bottom is now touching the cylinder's bottom, but the signal just received is still that the top just touched. So, as far as the bottom of the rocket is concerned, it and the top both have just touched at the same time: no explosion takes place.
A similar reasoning can be made in the rocket's frame with the same conclusion. In this frame it is the cylinder that stretches above the wave traveling downwards along the rocket's body.

I think that in the case of the original question, a similar reasoning will lead to the conclusion that both the top and bottom of the rocket end up damaged, as observed in both frames, only with a different order of the damaging events as seen from each.

Isn't this also related to the classical question of simultaneity of events for one observer not being necessarily so for another when they are moving relative to each other, and it is all because signals propagate at a finite speed not faster than that of light in vacuum, a limit speed that is a constant no matter from which system light is observed?

 

[PeroK]

@OscarCP take a look at how a slinky falls under gravity:

 

[OscarCP]

Thanks, PeroK. Veritasium is a great site. What is shown in the video with the slinky looks pretty much like what I reasoned myself to arrive ay my previous response.
The one about the tennis racket and the golf club is explained by the fact that it takes time for the shock of the contact of the implement with the ball to propagate along it and then the arm and then as electric impulses along its nerves so it can finally be felt in the brain.

 

[PeroK]

My point is that even a seemingly rigid object will not fall perfectly rigidly under gravity when dropped. The slinky is simply an extreme example.

 

[OscarCP]

PeroK: Quite right: the rocket cannot be perfectly rigid as, for example, one part, the lower one, will still be falling while the upper part is already stationary relative to the cylinder. There are also different reference frames for different parts of the rocket: the one already stopped and the one still moving. And there is a stress between these two parts that can only be physically meaningful in a non-rigid, deformable body. In the frames of the rocket, that part that stops first is at the bottom, and the stopping wave propagates towards the head that is still falling. But the bottom does not "know" that until the downward-traveling wave saying "the top just stopped" reaches it and so from the bottom point of view it will be as if the top had stopped after the bottom did.
In the first case there is tension, in the second, compression. In both cases not only there is no explosion, but both the top and the bottom are damaged, although in a different order depending on whether all this is seen from the rocket or the cylinder.
Which I think is also the correct answer to the original commentator's question.

 

[PeterDonis]

even a seemingly rigid object will not fall perfectly rigidly under gravity when dropped.

Actually, an object in free fall that has internal forces that give it a constant shape in free fall will fall rigidly under gravity to the extent that tidal gravity is negligible across its length. The slinky falls the way it does because it is not rigid at all--its internal forces do not maintain a constant shape even in free fall.

 

[PeterDonis]

one part, the lower one, will still be falling while the upper part is already stationary relative to the cylinder

In the limiting case of Born rigid deceleration, there will be a frame (the cylinder rest frame) in which this is not the case--all parts of the rocket stop simultaneously. (If the cylinder has nonzero proper acceleration, just pick its momentarily comoving inertial frame at the appropriate instant.)

There are also different reference frames for different parts of the rocket

As I think I've already commented, you need to keep the concept of "reference frame" distinct from the concept of "state of motion". Different parts of the rocket have different states of motion (except in certain limiting cases)--they are moving relative to each other. But that does not mean they are "in different frames". There is no such thing as an object being "in" one frame but not another: everything is "in" all frames. Its coordinate motion might be different in different frames, but it is "in" all of them.

 

[OscarCP]

Actually, an object in free fall that has internal forces that give it a constant shape in free fall will fall rigidly under gravity to the extent that tidal gravity is negligible across its length. The slinky falls the way it does because it is not rigid at all--its internal forces do not maintain a constant shape even in free fall.

I would say that this is an approximation. I think that PeroK has been discussing an exact Special Reativity solution to the problem. Not even in non-relativistic mechanics is a perfectly rigid body possible, except as an approximation: a body where the largest dimension is less than the distance a sound wave travels through the body in some very short time interval, so sound can be considered, as an approximation, to be traveling inside this body at an infinite speed. Only in this approximate sense a body can be seen as perfectly rigid and all its parts to move together all the time a force is exerted somewhere on it.

 

[OscarCP]

In the limiting case of Born rigid deceleration, there will be a frame (the cylinder rest frame) in which this is not the case--all parts of the rocket stop simultaneously. (If the cylinder has nonzero proper acceleration, just pick its momentarily comoving inertial frame at the appropriate instant.)As I think I've already commented, you need to keep the concept of "reference frame" distinct from the concept of "state of motion". Different parts of the rocket have different states of motion (except in certain limiting cases)--they are moving relative to each other. But that does not mean they are "in different frames". There is no such thing as an object being "in" one frame but not another: everything is "in" all frames. Its coordinate motion might be different in different frames, but it is "in" all of them.

I think that is precisely what I meant to say. Apologies if I did not expressed myself as rigorously as you have.

 

[Nugatory]

I would say that this is an approximation.

And you would be wrong. The limiting case of Born rigid acceleration that @PeterDonis describes is an exact solution (although suitable only for a thought experiment because there is no practical way of applying exactly the right amount of force to each part of the rocket).

Quite a few posts back I suggested that you Google for “Born rigid motion”. The concept is essential if you want to understand the relativistic behavior of accelerating bodies at more than a handwaving level.

 

[PeterDonis]

I would say that this is an approximation. I think that PeroK has been discussing an exact Special Reativity solution to the problem.

In special relativity (flat spacetime, no gravity), an object that is in free fall does move perfectly rigidly, in the sense that all of its parts remain at rest relative to each other for all time. (This is an example of Born rigid motion, in fact--the simplest possible such example.)

In the presence of spacetime curvature, such an object, falling freely, will only have non-rigid motion (parts not remaining at rest relative to each other) to the extent that tidal gravity is not negligible over the length of the object, as I said.

 

[PeterDonis]

Not even in non-relativistic mechanics is a perfectly rigid body possible when the body is subjected to forces, except as an approximation

See the bolded qualifier I added above. It makes a big difference. A freely falling body is not subjected to any forces (since in relativity gravity is not a force).

Also, as I have mentioned, the case of Born rigid acceleration is a limiting case (unrealizable in practical terms) in which the motion of the body is rigid. The motion of actual bodies subjected to forces, like rockets, can approach Born rigid motion very closely in some cases.

 

[PeroK]

Actually, an object in free fall that has internal forces that give it a constant shape in free fall will fall rigidly under gravity to the extent that tidal gravity is negligible across its length. The slinky falls the way it does because it is not rigid at all--its internal forces do not maintain a constant shape even in free fall.

"When dropped" was the operative phrase. If we assume it is held from above and that external force is removed, then there must be a finite delay between a change in the internal forces near the top and bottom of the object. There must be a transition between its extended length while suspended and its natural length in free fall.

 

[PeterDonis]

"When dropped" was the operative phrase. If we assume it is held from above and that external force is removed

Ah, I see. I had missed that part.

 

[OscarCP]

And you would be wrong. The limiting case of Born rigid acceleration that @PeterDonis describes is an exact solution (although suitable only for a thought experiment because there is no practical way of applying exactly the right amount of force to each part of the rocket).

Quite a few posts back I suggested that you Google for “Born rigid motion”. The concept is essential if you want to understand the relativistic behavior of accelerating bodies at more than a handwaving level.

I wrote that because I thought we were no longer considering physically impossible or practically unrealizable bodies: rigid, Born.