B The Paradox of Relativity Length Contraction

[alan123hk]

A rigid rod with length $l_0$ slides on a smooth and flat tabletop along the length at speed of $~\frac {\sqrt{3}}{2}c~$, there is a hole of width $~l_0~$on the table.

The observer who is stationary relative to the desktop thinks that the length of the rigid rod $~ l=l_o \sqrt{1-\beta^2}=\frac {l_0}{2}~$, which is only one-half of the hole width. Therefore, when the rigid rod passes through the hole, it will fall into the hole due to gravity.

However, for the observer who is stationary relative to the rigid rod, since the tabletop moves relative to him at the speed of $~\frac {\sqrt{3}}{2}c~$ , the width of the hole he measured is $~ l'=l_o \sqrt{1-\beta^2}=\frac {l_0}{2}~$, so he thinks that the the rigid rod with length of $~l_0~$ will not fall into the hole with width only $~\frac {l_0}{2}~$.

Since the length contraction of the theory of relativity is not an illusion but a real effect, there is only one answer to whether the rigid rod falls into the hole, so how do we use the simplest way to explain this contradiction and find the answer?

 

[Sagittarius A-Star]

Here is the solution of W. Rindler:

Which is correct? The answer hinges on the relativity of rigidity.

Source:
https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf

 

[alan123hk]

In the twin time paradox, the two frames of reference are not completely symmetrical, because one feels the force of acceleration, and the other does not feel the force of acceleration.

However, in this contradiction of length contraction, it is assumed that both observers move in a straight line at a uniform speed, so they appear to be symmetrical.

I think that in the solution W. Rindler, the assumption is to start from the perspective of the observer who is stationary relative to the desktop. But if we infer what happens from the perspective of the observer who is stationary relative to the moving rigid rod, will the answer be different?

Because I am a beginner, maybe I am very ignorant of this problem.

 

[BvU]

Do you know about the Lorentz transform ?
What do the two observers each say about the two events (one end of rod reaches gap, other end reaches gap) ?

$\$

 

[alan123hk]

Do you know about the Lorentz transform ?

I know the Lorentz transformation, but I believe I may only use it correctly in very simple situations. When the situation is a little more complicated, I am already very confused.

 

[Dale]

However, in this contradiction of length contraction, it is assumed that both observers move in a straight line at a uniform speed, so they appear to be symmetrical.

Yes, they are indeed symmetrical. That isn’t the issue.

The issue is that “rigid” objects are not compatible with relativity. So the problem setup is incorrect in any frame. You must allow for non-rigidity in all frames.

 

[PeroK]

In the twin time paradox, the two frames of reference are not completely symmetrical, because one feels the force of acceleration, and the other does not feel the force of acceleration.

This is not generally true. The geometry of Minkowski space is not dependent on something accelerating. The twin paradox may be described completely without acceleration or gravity. It's simply the geometry of spacetime.

The length contraction paradoxes are presented here, for example:



Note: pay attention to the times on the clocks in the first half of the video.

 

[alan123hk]

Yes, they are indeed symmetrical. That isn’t the issue.
The issue is that “rigid” objects are not compatible with relativity. So the problem setup is incorrect in any frame. You must allow for non-rigidity in all frames.

I also found this sample problem in the book
You mean my interpretation is wrong, and I should not use the name rigid rod. If I put the word rigid in quotation marks, or indicate that it may not actually be rigid, then there is no problem with the setting of this question, am I right ?

Because the link below should have given the correct answer, I believe you are not saying that there is really a problem with the setting of the problem itself in the book, right?

https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf

 

[PeroK]

You mean my interpretation is wrong, and I should not use the name rigid rod.

If you have a perfectly rigid object, then when you move one end, the other end (no matter how far away) must move simultaneously. This requires information (in the form of a sound wave through the rod) to travel at infinite speed. And, you could use this perfect rigidity to send a message at infinite speed.

The resolution to this is to realize that the movement of any object, however rigid, depends on the speed with which each particle in the object is able to transmit a force to the next particle. This is the speed of sound in the object.

This paradox is slightly disengenuous as you would need a fantastically high force of gravity for an object to fall significantly in a few nanoseconds. In any case, as soon as you introduce a finite speed of sound in the rod, the paradox vanishes.

 

[PeroK]

https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf

Try the Eugene Khutoryansky video.

 

[Dale]

If I put the word rigid in quotation marks, or indicate that it may not actually be rigid, then there is no problem with the setting of this question, am I right ?

Yes. The rod cannot be rigid and so if you remove the word rigid or otherwise indicate that it is not actually rigid then the question is fine.

the link below should have given the correct answer

It did give the correct answer. Did you not read it? It shows clearly how even if the rod is undistorted in one frame it must be distorted in the other frame.

 

[Ibix]

Since the length contraction of the theory of relativity is not an illusion but a real effect, there is only one answer to whether the rigid rod falls into the hole, so how do we use the simplest way to explain this contradiction and find the answer?

I know the Lorentz transformation, but I believe I may only use it correctly in very simple situations. When the situation is a little more complicated, I am already very confused.

An easier computation is if the rod has a length-contracted length that is exactly the same as the hole's rest length. Then you write down the coordinates of the front and back of the rod at the instant it falls (it makes more sense to think of a gravity free environment and a piston that slams the rod downwards instantaneously in the frame of the table - that way there's no questions about tipping). Then you transform those coordinates into the frame where the rod is at rest. Notice that the time coordinates are not the same. What does this tell you about how the "instantaneous" slam looks like in the rod's rest frame? How fast does the "slam" travel along the rod?

 

[PAllen]

Another aspect not emphasized enough for such false paradoxes is that where simultaneity is involved, which frame is used for problem specification determines the actual problem. Furthermore the problem specified in two different frames (i.e. different simultaneity) are both correct, because they are different physical problem specifications.

In this case, if you require that the rod ends enter the hole simultaneously in the rod initial rest frame, then, indeed, it takes a hole with large size (per its own rest frame) for the rod to get through. In contrast, if you require the rod ends to get through simultaneously per the hole rest frame, then a much smaller hole will suffice for the same rod. Thus, both 'sides' of the 'paradox' are valid because they represent completely different physical situations.

As a slightly above B level response, there is a notion of Born rigidity. This can be defined in a frame independent way, and cannot be achieved in practice (it requires pre-programmed motion of all the elements of a body to preserve constancy of local distance from adjacent elements). If one requires the rod to be Born rigid, then it turns out the description in the rod rest frame is required, and a large hole is needed for the rod to get through with Born rigid motion.

 

[PAllen]

A quick follow up to my prior post is that once a problem has been specified in one frame, it has a description in any other frame which may seem very different in terms of coordinate dependent quantities.

 

[pervect]

As others have mentioned, by introducing the notion of "rigidity" in special relativity, you've made the problem much more complicated. At some point you might benefit from visiting the notion of rigidity in special relativity, such as the notion of Born rigidity - but I don't think that point is now. I believe you'd be much better off to first understand the problem of length contraction in a formulation that does not require rigidity, such as the standard pole in the barn formulation of the paradox. Such a formulation typically involves a pole that is too long to fit into the barn classically, and a pair of electronic doors at each end. The question becomes then whether the entire pole can be trapped inside the barn with both doors closed "at the same time". We can replace the doors with optical sensors that are blocked by the pole's motion, making the pole and barn very thin in the transverse direction works around the issue that the light beams necessarily take some time to propagate. Making the pole and barn thin allows one to ignore this issue, and come up with a way to tell when the end of the pole is co-located with the electronic "door" marked out by the sensor.

The resolution of this standard variant is that "at the same time" is frame dependent in special relativity, this is called "the relativity of simultaneity". One observer will see the doors closed "at the same time", the other observer will not. This can be confirmed by utilizing the Lorentz transform that various posters have mentioned, which is typically done in standard treatments of the problem.

It is very likely that you are either not familiar with the relativity of simultaneity at all, or struggling with understanding it properly. How familiar are you with the idea? Is this the first you've heard of it, or are you somewhat familiar with it but not grasping its relevance to the problem you're interested in?

If your understanding of physics is entirely based on rigid objects, you may find it alarming that the notion of rigidity becomes somewhat problematical in relativity. The good news is that physics can and does deal with non-rigid objects routinely. The bad news is that it takes some significant effort to learn even the basics of the needed math, such as partial differential equations as opposed to ordinary differential equations.

 

[alan123hk]

Did you not read it? It shows clearly how even if the rod is undistorted in one frame it must be distorted in the other frame.

I understand that the reasoning process in that article first assumes that the observer who is stationary relative to the hole is absolutely correct, and then writes the equation $~z=\frac{1}{2}at^2 ~$ for the fall of the "rigid" rod from his perspective, and then the Lorentz transformation is used to derive the equation of the parabola falling trajectory of the "rigid" rod seen by the observer who is stationary relative to the "rigid" rod. The proof is complete and the problem has been resolved.

But I have another idea, since the two observers are symmetrical and equal to each other, why can't we look at this problem from the perspective of the observer who is stationary relative to the rigid rod. He thinks that the "rigid" rod is stationary and the hole approaches him in the opposite direction close to the speed of light . Since the width of the hole he measured was much smaller than the width of the "rigid" rod, he is certain that It won't fall into the hole, so he also write his own equation $\left[~~z'=0~, ~\infty<t'<\infty\right]~$, and his conclusion is that after applying Lorentz transform, the observer who is stationary relative to the hole must also agree that $z$ does not change and the "rigid" rod will not fall into the hole.

Of course I understand that the above statement will be confirmed by experts here to be wrong, but I don't know where the problem is. Please advise where is the fallacy of the above reasoning process?

 

[PeterDonis]

Since the width of the hole he measured was much smaller than his own, he is certain that he would not fall into the hole

No, he is not. In the "B" frame that has been discussed so far, which is related to the "A" frame by a Lorentz transformation, this is because the rod cannot be assumed to be rigid. But we can also consider a (non-inertial) frame in which B is at rest the whole time, so that B falls with the rod when the rod starts to fall. In this (non-inertial) frame, the two ends of the hole are not always at the same height at the same time; the far end of the hole starts moving upward well before the forward end of the rod reaches it, so the forward end of the rod does not go over the hole but gets stopped because the hole is tilted at an angle so its far end is higher. (In A's frame, this looks like the rod falling into the hole; by relativity of simultaneity, the two ends of the hole are always at the same height at the same time in this frame.)

 

[PeterDonis]

since the two observers are symmetrical and equal to each other

No, they aren't, because the statement of the problem specifies that the rod falls "rigidly" in A's frame. More precisely, the rod falls in A's frame such that all parts of the rod are always at the same height at the same time (and the two sides of the hole are also always at the same height at the same time). Because of relativity of simultaneity, if these conditions are true in A's frame, they cannot be true in any other frame moving relative to A's, including the "B" frame. This makes the two frames asymmetrical. (This is the kind of thing @PAllen was referring to in post #13.)

 

[alan123hk]

@PeterDonis Thank you for your detailed explanation. Due to my level limitation, it seems that I may need a little more time to think before I can gradually understand it.

 

[Dale]

But I have another idea, since the two observers are symmetrical and equal to each other, why can't we look at this problem from the perspective of the observer who is stationary relative to the rigid rod

You certainly can. If you describe the situation completely in one frame then you can use relativity to transform it to another frame.

But I have another idea, since the two observers are symmetrical and equal to each other, why can't we look at this problem from the perspective of the observer who is stationary relative to the rigid rod. He thinks that the "rigid" rod is stationary and the hole approaches him in the opposite direction close to the speed of light . Since the width of the hole he measured was much smaller than the width of the "rigid" rod, he is certain that It won't fall into the hole, so he also write his own equation [ z′=0 , ∞<t′<∞] , and his conclusion is that after applying Lorentz transform, the observer who is stationary relative to the hole must also agree that z does not change and the "rigid" rod will not fall into the hole.

This is correct.

However, let’s be a little more specific what we mean by “rigid” here. What we mean is that in this frame it keeps its shape and never bends. Now, since it is only “rigid” and not rigid, the way to do that is to apply a force to the rod that will keep it from deforming, it will not maintain its shape due to its own internal structure. The force needs to exactly match the force provided by the floor, otherwise it will not remain “rigid”. Such a force can most easily be obtained by placing a floor covering the hole. So, essentially this version of the problem gets rid of the hole. With no hole the rod remains straight in both frames as you deduced.

Please advise where is the fallacy of the above reasoning process?

No fallacy. It is a correct analysis of a rather uninteresting scenario

 

[Sagittarius A-Star]

why can't we look at this problem from the perspective of the observer who is stationary relative to the rigid rod.

The rod is not rigid. In the text of W. Rindler, a "trap door" is described, which ensured, that in frame A the rod starts the falling in a horizontal orientation.

W. Rindler describes an inertial (not internal) frame B, fixed (only in x-direction) to the hind end of the rod (see figure 1b).

I found another error. It is in equation (3). In the second condition for x', a minus sign in missing. It should be:
$\text {when } x' \geq - c^2t'/v$.

In equation (3) you can see, that for example the hind end of the rod (at x'=0) is freely falling with a gravitation acceleration, which is in this frame $a \gamma^2$. The free fall cannot be counter-acted by internal forces, because the vertex of the parabola is moving faster than the speed of light (with velocity $c^2/v$).

In frame B, the bending of the rod makes it fall into the hole.

 

[alan123hk]

After referring to many valuable and professional replies here, a simple and direct logical explanation finally emerged in my mind.

Since the "rigid" object is incompatible with the theory of relativity, and this incompatibility has been confirmed by the result of the Lorenz transformation from observer A to observer B, then even if observer B measures that the width of the hole moving toward him close to the speed of light is much narrower than the "rigid" rod, he still cannot conclude that the "rigid" rod will not fall into the hole.

In this way, my previous inference becomes invalid. Although I may not be able to appreciate or understand more professional and rigorous proof method for the time being, this simple explanation may allow me to temporarily let go of this contradictory entanglement.

 

[PAllen]

Since the "rigid" object is incompatible with the theory of relativity, and this incompatibility has been confirmed by the result of the Lorenz transformation from observer A to observer B, then even if observer B measures that the width of the hole moving toward him close to the speed of light is much narrower than the "rigid" rod, he still cannot conclude that the rigid rod will not fall into the hole.

--because it's not really rigid. It may look that way in one frame's description, but rigidity simply does not exist for high speed interactions.

 

[Ibix]

In this way, my previous inference becomes invalid. Although I may not be able to appreciate or understand more professional and rigorous proof method for the time being, this simple explanation may allow me to temporarily let go of this contradictory entanglement.

The calculation I suggested in #12 is worth doing - it's paraphrased from my own undergraduate studies. It will show you that the rod bends through the hole in all frames except the table rest frame.

 

[PeroK]

Since the "rigid" object is incompatible with the theory of relativity, and this incompatibility has been confirmed by the result of the Lorenz transformation from observer A to observer B, then even if observer B measures that the width of the hole moving toward him close to the speed of light is much narrower than the "rigid" rod, he still cannot conclude that the "rigid" rod will not fall into the hole.

The point is that even without SR, it's clear that perfect rigidity is impossible. SR simply identifies the speed of light as an ultimate constraint on rigidity. The people who built bridges in the 19th Century would have know all about the lack of rigidity of materials they were using.

The speed of sound and shock waves are studied in classical physics. It's no mystery that waves must propagate through a medium and cannot travel a finite distance instantaneously. This has nothing to do with SR or the Lorentz Transformation, per se.

 

[PeterDonis]

even without SR, it's clear that perfect rigidity is impossible.

I'm not sure that is true. It was clear prior to SR that all of the materials people actually knew about were not perfectly rigid; but pre-relativity physics does not put any constraint in principle on how rigid a material can be. So it was possible prior to SR to believe that, if we could just find the right material, it would be perfectly rigid.

 

[PAllen]

The calculation I suggested in #12 is worth doing - it's paraphrased from my own undergraduate studies. It will show you that the rod bends through the hole in all frames except the table rest frame.

And conversely, if you specify the problem in the rod initial rest frame using a hole such that its length contracted width is big enough for the rod to fit through horizontally in this frame, you find that the rod bends in the hole rest frame, even though it has all the room in the world. Of these two different problem specifications, only one is Born rigid (this is unique because a Herglotz-Noether corollary specifies that the trajectory of one point of Born rigid body moving noninertially uniquely specifies the complete motion of the body). The one that is Born rigid is the one in the rod rest frame. This is the one where no locally comoving frame of any rod element ever sees nearby points change distance.

 

[alan123hk]

I still cannot give up thinking about this issue, because I believe that contradictions still exist. As the reply #20 has confirmed that the idea I described in #16 is basically not logically fallacy.

Since all the observers in the different inertial systems are equal, the observer A from his point of view after the Lorentz transformation thinks that the rod of observer B will lose its rigidity and bend into the hole, but observer B has the absolute right to think that material properties, such as the stiffness of a rod at rest in his frame of reference, will not change due to its speed relative to other frames of reference. In addition, the problem setting itself assumes that the rod can pass through a hole smaller than it from above without falling into the hole.

There is only one answer whether the rod falls into the hole. This answer must be agreed by Observer A and Observer B, so there should be some things that are not clearly specified in the problem setting or people have not considered it carefully. I think the answer must exist. It may be related to the rigidity of the object and the limited transmission speed of force in the object.

 

[PeroK]

I still cannot give up thinking about this issue, because I believe that contradictions still exist.

Did you not watch the Khutoryansky video?

thinks that the rod of observer B will lose its rigidity and bend into the hole,

You have to stop saying "thinks". These are physical measurements. There's no "thinks" about it.

The rod is not rigid. An infinitely rigid rod is an absurdity; that's the contradiction.

absolute right to think that material properties, such as the stiffness of a rod at rest in his frame of reference, will not change due to its speed relative to other frames of reference.

The rod is not rigid in any frame of reference; it bends as soon as part of it over the hole. Why do you think the rod is infinitely rigid in its rest frame? What would happen to an aircraft wing under extreme forces? It's going to bend and eventually break regardless of its state of motion.

I think the answer must exist. It may be related to the rigidity of the object and the limited transmission speed of force in the object.

That's precisely what the solution is. This is what paectically every post in this thread has said (and that is precisely what is shown in that video).

I really can't understand this hang-up about the rod being infinitely rigid.

 

[alan123hk]

Did you not watch the Khutoryansky video?

I have watched it, but I feel that it is not the only way to explain this problem, and it has not completely cleared the doubt in my mind.

You have to stop saying "thinks". These are physical measurements. There's no "thinks" about it.

OK

The rod is not rigid. An infinitely rigid rod is an absurdity; that's the contradiction.

I really can't understand this hang-up about the rod being infinitely rigid.

I certainly agree with this. But I think the degree of deformation of the rod will depend on the force measured by observer B and the stiffness of the rod itself, which may not be exactly the same as that calculated by observer A. Perhaps the degree of deformation is much smaller than the result of observer A's calculation.

The rod is not rigid in any frame of reference; it bends as soon as part of it over the hole. Why do you think the rod is infinitely rigid in its rest frame? What would happen to an aircraft wing under extreme forces? It's going to bend and eventually break regardless of its state of motion.

Yes, but if we imagine that the wall on the right side of the hole is a little lower than the left side, this extra clearance space is enough to accommodate the downward deformation of the rod, so that the rod does not fall into the hole, then the situation may be different?

 

[PeroK]

Yes, but if we imagine that the wall on the right side of the hole is a little lower than the left side, this extra clearance space is enough to accommodate the downward deformation of the rod, so that the rod does not fall into the hole, then the situation may be different?

In all frames of reference, the rod starts to bend as soon as it overhangs the edge.

In practice, something moving at this speed will travel a long way before it falls any distance under the usual gravity. This problem requires:

An infeasibly thin rod; an infeasibly thin surface; and/or, a massive gravitational force. It's very much a thought experiment from that point of view.

If the RHS of the gap is lower, then you'd need an even greater gravitational force to deflect the rod into the hole.

In any case, forces and stiffnesses of materials must be frame-dependent. This is a direct consequence of length contraction in the direction of motion. Analysing forces and accelerations in relativistic mechanics is the next step.

Remember that "stiffness" isn't some absolute quantity but something that is measured. An object may have a proper stiffness, as it has a proper length, but length and stiffness must be frame dependent. A metre stick traveling at very close to $c$ may have negligible length and be modeled more like a point particle.

 

[PeroK]

One of the things you cannot do if you want to learn SR is to hang on to invariant classical concepts at all costs. You have to accept that if space and time are frame dependent, then all else must be put under proper scrutiny. SR brings with it new definitions of kinetic energy, momentum and four-dimensional vectors. It's not just a veneer painted on classical physics. Quite the reverse. Anything you think is true from classical physics must be shown to be a special case of SR.

You can't hang on to classical absoluteness at every turn and hope to learn SR. Classical Mechanics is the special case.

 

[Nugatory]

But I think the degree of deformation of the rod will depend on the force measured by observer B and the stiffness of the rod itself, which may not be exactly the same as that calculated by observer A. Perhaps the degree of deformation is much smaller than the result of observer A's calculation.

When you speak of the deformation of an object, you are making a statement about where each part of that object is at the same time.

In a setup where the rod falls through the hole, for every point along the length of the rod there will be some event “this point on the rod passes below the level of the surface”. When we assert that the rod is not distorted as it passes through the hole, we’re asserting that all of these events happen at the same time; conversely if some parts of the rod drop below the level of the surface at more or less different times we find the rod more or less distorted.

Relativity of simultaneity means that “at the same time” is different in different frames, so the distortion will be different in different frames. No differences in forces or stiffness are required to explain the difference in shape.

 

[Nugatory]

Yes, but if we imagine that the wall on the right side of the hole is a little lower than the left side, this extra clearance space is enough to accommodate the downward deformation of the rod, so that the rod does not fall into the hole, then the situation may be different?

You could save yourself a ton of grief and unnecessary exploration of increasingly complicated scenarios by thinking of the rod as a length of string instead. No matter what material it is made of, “string” will be a better approximation of its behavior than “rigid rod”.

 

[PeroK]

No differences in forces or stiffness are required to explain the difference in shape.

Perhaps not, but if we look at a moving compressed spring, then length contraction implies that either the forces and/or the spring constant must be different in the moving frame.

 

[Nugatory]

Perhaps not, but if we look at a moving compressed spring, then length contraction implies that either the forces and/or the spring constant must be different in the moving frame.

or that the Hooke’s law is properly written with a gamma in it somewhere, but we usually don’t think about this because we’re usually working with a frame in which $\gamma=1$.

 

[ergospherical]

No matter what material it is made of, “string” will be a better approximation of its behavior than “rigid rod”.

Even better:

 

[Dale]

still cannot give up thinking about this issue, because I believe that contradictions still exist. As the reply #20 has confirmed that the idea I described in #16 is basically not logically fallacy

That same post #20 also showed that your idea in #16 doesn’t lead to any contradictions. So please don’t use my post as a justification for believing that there is a contradiction. There isn’t.

There is only one answer whether the rod falls into the hole. This answer must be agreed by Observer A and Observer B, so there should be some things that are not clearly specified in the problem setting or people have not considered it carefully. I think the answer must exist. It may be related to the rigidity of the object and the limited transmission speed of force in the object.

You can make many different scenarios. Each scenario can have a different answer, but for each scenario the two frames will be consistent. For scenarios related to this the non-rigidity is key.

 

[alan123hk]

That same post #20 also showed that your idea in #16 doesn’t lead to any contradictions. So please don’t use my post as a justification for believing that there is a contradiction. There isn’t.

I did not say that you agree that there is a contradiction. In fact, you never said it, and I never meant it.
I understand that from the beginning to the present in this thread, I should be the only one who still doubts, is not completely satisfied or cannot fully understand the explanations mentioned so far.

 

[Dale]

I understand that from the beginning to the present in this thread, I should be the only one who still doubts, is not completely satisfied or cannot fully understand the explanations mentioned so far.

That is fine, although if you remain unsatisfied with the previous explanation it would be better to quote specifically the unsatisfactory part and explain why that is confusing. Also, since multiple scenarios have been discussed in this thread it is important to be clear which you are discussing.

 

[PeterDonis]

I still cannot give up thinking about this issue, because I believe that contradictions still exist.

That's the wrong way to approach this. If you think there are contradictions somewhere in SR, you are wrong. Period. SR is a theory with massive experimental confirmation of its predictions.

If you approach this with the idea that obviously there can't be a contradiction in SR, but you are unable to understand how the views of what is happening in different frames fit together, that's one thing. But that's not how you are approaching it. And for that reason, you are having trouble understanding the correct answers you have been given. So you need to change your approach.

all the observers in the different inertial systems are equal

That's not what the principle of relativity says. The principle of relativity says that the laws of physics are the same in all inertial frames. (Note that this is the SR version of the principle, which is all we need to discuss here. The GR version says the laws of physics are the same in all frames, even non-inertial ones. But that's beyond the scope of this thread.) But that does not mean that every scenario must look exactly the same in every frame. It only means that every scenario must be a valid realization of the laws of physics in every frame.

One of the laws of physics in SR is that anything that has to propagate can only propagate at a finite speed, the speed of light. "Anything that has to propagate" includes forces being applied; if a force is applied to one part of an object, that force can only propagate through the object at the speed of light, not infinitely fast. That is what prevents objects from being perfectly rigid in SR.

In this particular scenario, the scenario is set up so that the force that causes the rod to fall is applied to all parts of the rod at the same time in frame A. So the force doesn't have to propagate through the rod in frame A; it is already set up in that frame to apply the same force at the same time to all parts of the rod. So the rod does not bend in frame A. But by relativity of simultaneity, that means that force is not applied to all parts of the rod at the same time in frame B. And that means the rod must bend in frame B, because a force is being applied to only part of it at a given time, and that force can only propagate through it at the speed of light.

In other words, the scenario as it is set up is not completely symmetrical between the frames. And that is perfectly fine in SR! SR does not say that all scenarios must be completely symmetrical between all frames. It only says the laws of physics are the same in all frames.

 

[PeterDonis]

Each scenario can have a different answer, but for each scenario the two frames will be consistent.

Note that "the two frames will be consistent" does not mean "the scenario will look exactly the same from the two frames" or "the two frames must be symmetrical in all respects". It only means the laws of physics will be the same in the two frames.

 

[PeterDonis]

The rod is not rigid in any frame of reference

True in the sense that forces can only propagate through it at a finite speed.

it bends as soon as part of it over the hole.

However, this is false in frame A because the force that is making the rod fall is applied to all parts of the rod at the same time, so it falls rigidly even though it is not a rigid object (since there are no perfectly rigid objects in SR). In other words, the applied force is specially set up so that the rod's motion is rigid in frame A and the rod does not bend in that frame. (It does, of course, bend in frame B.)

 

[PeterDonis]

It may be related to the rigidity of the object and the limited transmission speed of force in the object.

Most definitely. As has already been stated earlier in this thread. I have stated it again, perhaps with some more helpful detail, in post #41.

I think the degree of deformation of the rod will depend on the force measured by observer B

Not "the" force, since a force has to be applied at each point of the rod, and, as I noted in an earlier post just now, in frame B the forces are applied to different parts of the rod at different times. That plus the finite propagation speed of forces through the object is what causes the rod to bend in frame B.

and the stiffness of the rod itself, which may not be exactly the same as that calculated by observer A.

In principle I suppose the stiffness of the rod could be different in frame A and frame B (although I have not seen this discussed in any literature I'm aware of), but I don't think this will be a significant factor in practice. The key is the different times of application of the force to different parts of the rod in frame B.

if we imagine that the wall on the right side of the hole is a little lower than the left side

Then we are imagining a different scenario, which we would need to analyze separately. I would strongly advise you not to start bringing in different scenarios until you understand the original one.

 

[Nugatory]

it bends as soon as part of it over the hole.

However, this is false in frame A because the force that is making the rod fall is applied to all parts of the rod at the same time

I think PeroK is right when we're considering the most common description of the problem: the downwards force on the rod is its weight, resisted by the normal force from the surface except where the rod hangs over the edge of the hole.

PeterDonis is right when we're considering something like what @Ibix proposes in #12, where the downwards force is switched on across the length of the rod, simultaneous in one frame but no other.

 

[Dale]

Just to be clear it may be worthwhile to identify and describe the major versions.

The standard version is one where there is just a hole that the rod moves over. In that one the rod begins to bend as soon as the front end overhangs the lip of the hole. This bending occurs in both frames, so the rod is not straight in either frame. In both frames it crashes into the far wall.

Rindler’s version is one where there is a trap door that opens when the rear of the rod crosses the threshold. It is designed to open simultaneously across the hole in the ground’s frame. In this case the rod is straight in the ground frame. It falls after removing the trap door and therefore crashes into the far wall. In the rod’s frame the trapdoor does not open simultaneously, so the front of the rod starts falling before the rear does. The rod bends and crashes into the far wall. Rindler’s version is equivalent to the @Ibix version, just with the trap door being replaced with a piston.

The uninteresting version is the one you proposed above where the rod remains straight in the rod’s frame. This only can happen by covering the hole with something. In that case it is straight in both frames and does not crash into the wall in either frame.

The clearance version is the new one where the ground is not level, but is lower on the far side of the hole. This one depends entirely on how far from level it is. If it is sufficiently unlevel then there will be no crash in either frame. If it is just barely unlevel then there will be a crash in both frames. Either way both will agree.

I would recommend only discussing the original version or the Rindler version, and always being clear which you are discussing.

 

[PeterDonis]

I think PeroK is right when we're considering the most common description of the problem: the downwards force on the rod is its weight, resisted by the normal force from the surface except where the rod hangs over the edge of the hole.

In that case, the rod will bend in any frame, since part of it will hang over the edge of the hole while part does not in any frame. But in the Rindler paper referenced in post #2, Rindler specifies that a trap door is set up in the hole that is only removed once the entire rod is over the hole, so that, in frame A, the downward net force (gravity, no longer resisted by something holding the rod up) is applied at the same time in this frame to all points of the rod. That is the version that I have understood us to be discussing in this thread.

PeterDonis is right when we're considering something like what @Ibix proposes in #12, where the downwards force is switched on across the length of the rod, simultaneous in one frame but no other.

Yes, another way of realizing the frame A condition I described would be to set up the experiment in free fall and do something like what @Ibix described in post #12 to apply the force at the same time in frame A to all parts of the rod.

Edit: I see @Dale posted much the same thing.

 

[PeroK]

However, this is false in frame A because the force that is making the rod fall is applied to all parts of the rod at the same time, so it falls rigidly even though it is not a rigid object (since there are no perfectly rigid objects in SR).

I didn't read the Rindler version, but that's seems like a variation of the barn door paradox.

 

[PeterDonis]

I didn't read the Rindler version, but that's seems like a variation of the barn door paradox.

No, it's a version of the scenario we're discussing in this thread. (Rindler's paper is referenced in post #2.) The barn door paradox has two doors; Rindler's version of this thread's scenario has just one trap door, in the one hole.

 

[PeroK]

No, it's a version of the scenario we're discussing in this thread. (Rindler's paper is referenced in post #2.) The barn door paradox has two doors; Rindler's version of this thread's scenario has just one trap door, in the one hole.

Post #2 is a different problem from that posted by the OP in post #1. That might explain a lot of the confusion.

I've been responding to the problem in post #1, as posted by the OP.