B The Paradox of Relativity Length Contraction

[PAllen]

Do you agree, that in the following example, the rod becomes dust at ordinary acceleration?

For your one light year rod, there would still be kinematic shear and stretch because of timing of force application, followed by shear force readjustment. Some amount of small vibration settling down into a slightly drooped shape would happen. This would be sort of analogous to a monoatomic rod at ordinary scales. There would be no breakage precisely because forces would be mild. (Assuming a thin right edge for the hole, rather than a wall).

 

[PAllen]

At my level, it seems that I can't immediately understand it, but anyway, I would like to thank you for providing precious reference information.

Not sure what you don't understand. You want to get the whole rod through the hole. If the rod is a centimeter thick and e.g. a foot long, and the hole is a bit smaller, then, for the contracted rod to fall through, it needs to move at least a centimeter down in the time it takes to traverse the hole. This traversal time is of the order of a nanosecond (1 foot per nanosecond is approximately the speed of light). So the acceleration required is of order a centimer per nanosecond squared.

 

[PAllen]

...

A) The rod experiences force starting from its right edge, propagating left at near c to the left. Each element of the rod initially responds to the force without resistance, immediately producing kinematic shear. Starting from the right edge of the rod, moving left at the speed of sound in the material, shear forces kick in, operating to restore the rod to equilibrium shape. It is during this phase that one can distinguish dust from various other material models. Obviously, the table edge will have moved a vast distance to the left (traveling at near c) before equilibrium is established. Note, that if the acceleration was great enough (depending on material) - producing excessive kinematic shear displacement - the shear propagation will be a fracture wave rather than a restoring force.

B) The rod experiences force starting from the right edge, propagating left at a speed greater than c. Every other aspect of the analysis in A remains the same.

C) The rod experience force simultaneously (in the rod frame, as for all of these descriptions). There is never any kinematic shear, or restoring shear forces, or change of shape. (For true rigid motion , force application has to be timed in a unique way along the vertical thickness of the rod, but we can ignore this detail for the purposes of this analysis).

So now we can reintroduce the right side. If we introduce it as a thin edge, and the right side of the rod has had time to get below this edge, then nothing changes in the above analysis. If we introduce it as a wall, that is magically resistant, traveling to the left at near c:
...

I realize a few small correction are in order here. In A) and B) cases, the left 'traveling' response to the initial kinematic deformation is not limited to the speed of light. Instead, because each force application is considered independent (we are not requiring a deformation at one end to traverse the rod), the response simply has a delay time related to what might be called a molecular relaxation delay - the time for a given molecule to 'communicate' with its nearest neighbors, with either severed bonds settling into new 'dust' state or some vibrational response initiated (depending on the problem specifics and the material). I would guess this is of the order time it takes sound in the material to cross one inter-molecule distance. Thus, a front of molecular response would trail the force application boundary at the same speed as this boundary, with this delay. Thus for A, the trailing response front would travel left at near c, and for B, at a superluminal rate.

For C), it is actually quite easy to state the vertical requirement for Born rigidity. It would be the force application has to be slightly greater at the top surface of the rod compared to the bottom surface, consistent with the Rindler congruence.

 

[Sagittarius A-Star]

There would be no breakage precisely because forces would be mild. (Assuming a thin right edge for the hole, rather than a wall).

Then, according to the LT, the rod must become distorted in the reference frame of the hole, when falling.

 

[PAllen]

Then, according to the LT, the rod must become distorted in the reference frame of the hole, when falling.

No, I think it would be horizontal in the hole frame, by construction, and slightly distorted and not quite horizontal in the rod frame (assuming we remain talking about the Rindler example from post #2, with a thin right hole edge, rather than a wall). Given the tiny angle of deformation required over a light year, even the most rigid material imaginable would easily accommodate the deformation non-destructively.

 

[Sagittarius A-Star]

No, I think it would be horizontal in the hole frame, by construction, and slightly distorted and not quite horizontal in the rod frame (assuming we remain talking about the Rindler example from post #2, with a thin right hole edge, rather than a wall). Given the tiny angle of deformation required over a light year, even the most rigid material imaginable would easily accommodate the deformation non-destructively.

The rod can only stay horizontal in the hole frame, if in the rod frame, according to LT, the right-hand side of the rod falls forever with a higher velocity than the left-hand side of the rod, because the acceleration has started there earlier. That would not be compatible with "non-destructively".

 

[PAllen]

The rod can only stay horizontal in the hole frame, if in the rod frame, according to LT, the right-hand side of the rod falls forever with a higher velocity than the left-hand side of the rod, because the acceleration has started there earlier. That would not be compatible with "non-destructively".

True, but I think this would reach a breaking point on a time scale longer than it takes the rod to get through the hole. It's hard to realize how much give even e.g. diamond would have over a 1 ly long rod of e.g. 1 cm thickness. But, yes, eventually the rod must break if the conditions are maintained long enough.

 

[PAllen]

True, but I think this would reach a breaking point on a time scale longer than it takes the rod to get through the hole. It's hard to realize how much give even e.g. diamond would have over a 1 ly long rod of e.g. 1 cm thickness. But, yes, eventually the rod must break if the conditions are maintained long enough.

Or, alternatively, there would be time for the rod to be pulled horizontal in the rod frame by molecular forces before it broke. I would have to calculate this, at least to order of magnitude, to know which is a more plausible model.

 

[Dale]

This is a bit ambiguous due to the relativity of simultaneity, I was assuming the trap door was removed "at the same time" as seen in the rod's frame, which seems simplest to analyze. But it's not entirely clear as which simultaneity convention is implied when Rindler states that the floor is removed.

It is removed simultaneously in the ground frame. This leads to the rod being straight in the ground frame, as shown in the diagrams that he draws.

 

[Dale]

You can work around this by e.g. having a rod whose thickness would be less than an atom, if you want to consider that acceptable.

And it should be obvious that such a rod will not be rigid.

 

[PeroK]

At my level, it seems that I can't immediately understand it, but anyway, I would like to thank you for providing precious reference information.

I think I mentioned in a previous thread that I believe it's a bad idea to try to learn SR using these so-called paradoxes. Instead, you should learn SR properly first and then tackle these paradoxes.

If after 100 posts on this paradox you still don't understand the resolution, then that confirms my point.

In any case, I suggest you consider whether you are trying to learn SR in an efficient manner.

 

[alan123hk]

In any case, I suggest you consider whether you are trying to learn SR in an efficient manner.

Your suggestion might be a good idea. The problem is that I am reading a book on basic special relativity, and I read each chapter in order. These famous "paradoxes" have been mentioned in the previous chapters of this book. So far, I understand almost everything except for incomplete understanding or not completely satisfied with the explanation of the "paradox" in the book, so I can't restrain myself from trying to find an explanation that I think is satisfactory. I will actively consider your suggestions, but by the way, I think only a small part of these 100 posts are directly related to the basic issues I want to discuss.

 

[Nugatory]

I am reading a book on basic special relativity,

Which one? We may have found the problem, as many basic ones are fairly awful (failure to explain what is meant by a reference frame, failure to introduce relativity of simultaneity early and often, confusing time dilation and the twin paradox, overemphasis on length contraction and time dilation are among the more common failings).

 

[alan123hk]

Which one? We may have found the problem, as many basic ones are fairly awful (failure to explain what is meant by a reference frame, failure to introduce relativity of simultaneity early and often, confusing time dilation and the twin paradox, overemphasis on length contraction and time dilation are among the more common failings).

Thank you for your reminder, I don't know if the explanation of this book is not detailed and clear enough, but this very old book is written in Chinese, my native language, so when I read this book, it feels more fluent and easier to understand from a language perspective.

 

[Sagittarius A-Star]

No, I think it would be horizontal in the hole frame, by construction

In Rindlers scenario, the right wall of the hole hits the right-hand side of the rod in the rod frame, before the rod is completely in free fall (see figure b). So, Rindlers "by construction" statement is only valid until this event.

The following table shows in the rod-frame the z'-coordinate of 8 atoms on the bottom of the rod (columns = x'-position, rows = instance in time, increasing downwards.

The shown time granularity is $D * v / c^2$, with D= distance between the atoms.

0	0	0	0	0	0	0	-1
0	0	0	0	0	0	-1	-2
0	0	0	0	0	-1	-2	-4
0	0	0	0	-1	-2	-4	-8

Because the point, where the falling starts (due to missing support of the trap-door), is moving to the left with more than c, for example the atom in the 3rd table row and z'-position "-2" thinks, that its originally right neighbor is also at z'=-2 (instead of z'=-4). So, the right neighbor can freely fall and the rod stays horizontally in the hole-frame.

The news can propagate no faster than the speed of light and cannot reach the adjacent atom before a time interval D/c has passed. Until then the shear forces cannot act.

Each segment of the rod is in free fall and never "knows" that the adjacent segments are at different heights. The shape of the rod in frame B is determined solely by how long each segment has been falling and has nothing to do with the elastic properties of the material of which it is composed. A rubber rod and a steel rod would have precisely the same shape.

Source:
"Understanding Relativity, A Simplified Approach to Einstein's Theories", LEO SARTORI, UNIVERSITY OF NEBRASKA-LINCOLN

I found also a link, but I don't post it because I am not sure regarding copyright.

 

[PAllen]

In Rindlers scenario, the right wall of the hole hits the right-hand side of the rod in the rod frame, before the rod is completely in free fall (see figure b). So, Rindlers "by construction" statement is only valid until this event.

The following table shows in the rod-frame the z'-coordinate of 8 atoms on the bottom of the rod (columns = x'-position, rows = instance in time, increasing downwards.

The shown time granularity is $D * v / c^2$, with D= distance between the atoms.

0	0	0	0	0	0	0	-1
0	0	0	0	0	0	-1	-2
0	0	0	0	0	-1	-2	-4
0	0	0	0	-1	-2	-4	-8

Because the point, where the falling starts (due to missing support of the trap-door), is moving to the left with more than c, for example the atom in the 3rd table row and z'-position "-2" thinks, that its originally right neighbor is also at z'=-2 (instead of z'=-4). So, the right neighbor can freely fall and the rod stays horizontally in the hole-frame.

The news can propagate no faster than the speed of light and cannot reach the adjacent atom before a time interval D/c has passed. Until then the shear forces cannot act.

You missed that I said "with a thin right hole edge, rather than a wall", i.e. the rod just continues to fall and move to the right in the hole frame (or the hole continues to move to the left in the rod frame). This is right in the post your replied to. The rest of your post is generally consistent with what I have been saying. Except that I proposed that nearby atoms will apply a counter to free fall only at a delay of molecular distance divided by speed of sound. I think this means that for your light year long rod and ordinary g, there is plenty of time for back reaction before breakage; while for any lab scale case, the only response is for molecules to stabilize to an independent, unbound state. Ultimately, everything depends on the distance between a molecule and its neighbors by the time it can "communicate" with them. If this is too large, "dust" is the result. If not, attractive forces will pull the molecules back together, thus straightening the rod in the rod frame (and curving it in the hole frame).

 

[Sagittarius A-Star]

Except that I proposed that nearby atoms will apply a counter to free fall only at a delay of molecular distance divided by speed of sound.

But after that time, the nearby atom has already itself a new z'-coordinate and sees then no need to react to the (same) old z'-coordinate of it's neighbor.

I think this means that for your light year long rod and ordinary g, there is plenty of time for back reaction before breakage;

I think, also in the "LY-long rod" case the back reaction does not happen for the reason, I described above.

 

[PAllen]

But after that time, the nearby atom has already itself a new z'-coordinate and sees then not need to react to the (same) old z'-coordinate of it's neighbor.

I disagree. When 'news' of displacement of neighbor reaches a given molecule, it responds to that. It does not respond to a yet later position of the molecule until later yet. Note, irrespective of the faster than c propagation of where force is applied, the relative speed of molecules is quite small in the light year rod case. So there is plenty of time for restoring forces. In the lab scale model, the only problem is that displacement becomes too large before a response can occur.

 

[Sagittarius A-Star]

I disagree. When 'news' of displacement of neighbor reaches a given molecule, it responds to that. It does not respond to a yet later position of the molecule until later yet.

I don't understand this argument. I think it is always 'news' of an apparent non-displacement of the neighbor (same z'-coordinate). The horizontal "signal" from the neighbor at z'=-2 reaches the atom, when it has arrived in the meantime itself at z'=-2.

 

[PAllen]

I don't understand this argument. I think it is always 'news' of an apparent non-displacement of the neighbor (same z'-coordinate). The horizontal "signal" from the neighbor at z'=-2 reaches the atom, when it has arrived in the meantime itself at z'=-2.

Horizontal versus vertical are irrelevant. For simplicity, imagine a molecule connected to its neighbors by springs. Neighbor undergoes displacement, wave propagates on spring, given molecule responds to wave produced by initial displacement, and only later finds out about any further change of position by neighbor. Note that if there were anything really mysterious about FTL initial displacement, then a rod sliding off a table at 100 mph, with a trapdoor pulled at 2g at the right moment (and no right side of hole at all, which is just a distraction for this discussion) would have all the features of this case. The 1 g force would propagate FTL right to the left in the rod frame, but not quite simultaneous. Do you really think any strange breakage would occur in this case?? On the other hand, there would certainly be molecular restoration forces, and some minute flexing in the rod before equilibrium is reached.

 

[Sagittarius A-Star]

Do you really think any strange breakage would occur in this case??

In this 100 mph example obviously no breakage occurs, according to our experience. That would also be valid for glass. But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with $z'=z$, without being slowed down by internal forces. Maybe, Gron is therefore correct with the glass?

 

[PAllen]

I don't understand this argument. I think it is always 'news' of an apparent non-displacement of the neighbor (same z'-coordinate). The horizontal "signal" from the neighbor at z'=-2 reaches the atom, when it has arrived in the meantime itself at z'=-2.

There is another problem with this. I claim the news travels at the speed of sound, not light, for mechanical purposes. Then there is definitely news of a different z.

 

[PAllen]

In this 100 mph example obviously no breakage occurs, according to our experience. That would also be valid for glass. But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with $z'=z$, without being slowed down by internal forces. Maybe, Gron is therefore correct with the glass?

No, he is wrong on glass at lab scale example where accelerations are centimeter per nanosecond squared. Gron was talking about a case where a glass rod went through a small hole at lab dimensions.

 

[Sagittarius A-Star]

And what about my argument "But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with $z'=z$ , without being slowed down by internal forces."?

 

[PAllen]

And what about my argument "But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with $z'=z$ , without being slowed down by internal forces."?

I completely disagree. The separation between molecules in their local rest frames would increase without bound. Either there are restoring forces, or dust is the result. In cases where the growth of displacement is less than the speed of sound for the material, restoring forces would produce a modified equilibrium shape in the rod frame, such that further free fall involves no further displacement from each element’s rest frame.

 

[PAllen]

In this 100 mph example obviously no breakage occurs, according to our experience. That would also be valid for glass. But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with $z'=z$, without being slowed down by internal forces. Maybe, Gron is therefore correct with the glass?

Let me be clear about something. If the trap door were pulled in rod frame, then rod would simply free fall with no stresses. If moving trap door is pulled in its own frame, then there would be small stesses and restoration forces leading to delayed equlibrium for the rod in its own frame (for ordinary g forces; for extreme cases, there would transition to dust state) There cannot be absence of stress in both cases.

 

[Sagittarius A-Star]

then there would be small stesses and restoration forces leading to delayed equlibrium for the rod in its own frame.

After this delay, the falling rod must be then distorted in the hole's frame, according to LT.

 

[PAllen]

After this delay, the falling rod must be then distorted in the hole's frame, according to LT.

Of course, so what?

 

[PAllen]

The following table shows in the rod-frame the z'-coordinate of 8 atoms on the bottom of the rod (columns = x'-position, rows = instance in time, increasing downwards.

The shown time granularity is $D * v / c^2$, with D= distance between the atoms.

0	0	0	0	0	0	0	-1
0	0	0	0	0	0	-1	-2
0	0	0	0	0	-1	-2	-4
0	0	0	0	-1	-2	-4	-8

Because the point, where the falling starts (due to missing support of the trap-door), is moving to the left with more than c, for example the atom in the 3rd table row and z'-position "-2" thinks, that its originally right neighbor is also at z'=-2 (instead of z'=-4). So, the right neighbor can freely fall and the rod stays horizontally in the hole-frame.

The news can propagate no faster than the speed of light and cannot reach the adjacent atom before a time interval D/c has passed. Until then the shear forces cannot act.

Each segment of the rod is in free fall and never "knows" that the adjacent segments are at different heights. The shape of the rod in frame B is determined solely by how long each segment has been falling and has nothing to do with the elastic properties of the material of which it is composed. A rubber rod and a steel rod would have precisely the same shape.

I thought of another fundamental problem with this table. Interactions are two way communications. Even in this table, if you model each element doing a round trip communication with its neighbor, the round trip time increases without bound. Meanwhile, the rod free falling in its own rest frame horizontally, has constant round trip times. It is the round trip time that matters - interactions are two way exchanges. And this is why the frame invariant kinematic decomposition is what really determines whether stresses are present to respond to (at whatever delay). And this clearly shows the hole frame horizontal fall has ever growing stress, while the rod frame horizontal fall has none. And thus, if the acceleration in the hole frame scenario for initial state is not too extreme, an equilibrium will be reached that is Born rigid in the rod frame and bent in the hole frame.

And I repeat how amazing it is to me that none of these sources even consider the kinematic decomposition, when this was first published in 1909!

 

[PAllen]

I thought of another fundamental problem with this table. Interactions are two way communications. Even in this table, if you model each element doing a round trip communication with its neighbor, the round trip time increases without bound. Meanwhile, the rod free falling in its own rest frame horizontally, has constant round trip times. It is the round trip time that matters - interactions are two way exchanges. And this is why the frame invariant kinematic decomposition is what really determines whether stresses are present to respond to (at whatever delay). And this clearly shows the hole frame horizontal fall has ever growing stress, while the rod frame horizontal fall has none. And thus, if the acceleration in the hole frame scenario for initial state is not too extreme, an equilibrium will be reached that is Born rigid in the rod frame and bent in the hole frame.

And I repeat how amazing it is to me that none of these sources even consider the kinematic decomposition, when this was first published in 1909!

Oh, and yet another thing about that table: if you consider a signal from one element to its right neighbor (rather than to its left), the right neighbor has the information on the displacement as soon as it gets this signal. But really, changing timings in a continuing sequence of two way probes is a better model of how molecular bonds would detect and respond to changes in their neighbors; and exactly this is equivalent to the kinematic decomposition.

 

[PeterDonis]

Ok, when I get a chance I'll start a new thread and post a link to it here.

The new thread with the kinematic decomposition calculations is here:

https://www.physicsforums.com/threa...-for-rod-and-hole-relativity-paradox.1010444/

 

[RW137]

In this kind of problem, you will not come to a conclusion with length contraction alone. You have to apply space-time physics: you have to consider events.

Call the end points of the rod, A and B (from left to right) and those of the hole, C and D. The rod moves to the right in the reference frame of the hole, whereas the hole moves to the left in the reference frame of the rod.

Let ‘B at C’ be the reference event. From there, B moves from C to D and C moves from B to A. In space-time physics, the question is not about the rod matching the hole but about the simultaneity of the events ‘B at D’ and ‘C at A’.

Compute the (x, t) coordinates of these events, in each frame. Then choose one of the frames (for instance the rod frame) and compute the space interval as well as the time interval of the events. You will find out, that the space interval dominates: the events are space-like. This means that there exists a third reference frame in which the events are simultaneous. In this reference frame both lengths L0 are contracted to the proper distance of the events. Now, you can say, that in this reference frame the rod matches the hole. However, in all other reference frames the two events will not happen simultaneously.

Note, that length contraction is not a compression of some rigid structure. You cannot establish a connection between space-like events.

 

[PAllen]

In this kind of problem, you will not come to a conclusion with length contraction alone. You have to apply space-time physics: you have to consider events.

Call the end points of the rod, A and B (from left to right) and those of the hole, C and D. The rod moves to the right in the reference frame of the hole, whereas the hole moves to the left in the reference frame of the rod.

Let ‘B at C’ be the reference event. From there, B moves from C to D and C moves from B to A. In space-time physics, the question is not about the rod matching the hole but about the simultaneity of the events ‘B at D’ and ‘C at A’.

Compute the (x, t) coordinates of these events, in each frame. Then choose one of the frames (for instance the rod frame) and compute the space interval as well as the time interval of the events. You will find out, that the space interval dominates: the events are space-like. This means that there exists a third reference frame in which the events are simultaneous. In this reference frame both lengths L0 are contracted to the proper distance of the events. Now, you can say, that in this reference frame the rod matches the hole. However, in all other reference frames the two events will not happen simultaneously.

Note, that length contraction is not a compression of some rigid structure. You cannot establish a connection between space-like events.

Note that if you require the two rod ends to enter the hole simultaneously in the rod frame, a hole whose contracted width is greater than the rod length is required. If, instead, you require the the two ends to enter the hole simultaneously in the hole frame, then a big (proper length) rod can get through a small hole. Both of these statements are true, and they are not in conflict because they are different physical setups..

 

[PAllen]

I think the whole non-inertial aspect of this thread's problem set up is a distraction from the basics of understanding how length, simultaneity, and orientation are all frame dependent. It is simply a fact that irrespective of any holes, if you want to change a rod moving horizontally, with horizontal orientation, in some frame, to a state of motion that remains horizontal (per this frame) but later has a vertical component to the velocity, an actual, physical deformation of the material is required. This is actually closely related to the Ehrenfest paradox.

To sidestep this largely separate feature of SR (that the Herglotz-Noether theorem places very tight constraints on what motions can be stress free in SR), and focus on understanding how length, simultaneity, and orientation are all frame dependent, we should simply posit a pure inertial version of this problem.

Consider a rod with a horizontal orientation in some frame, moving mostly to the right at high speed, but with a modest downward component as well. Then it will easily pass through a horizontal hole shorter than its rest length, horizontally.

Then, instead of looking at the rod frame, we look at a frame in which the rod has no horizontal motion (because this is much simpler than looking at the rod rest frame) in which the surface with hole is moving horizontally (to the left let's say). Now, we find that:

1) The surface is still horizontal in this frame.
2) The rod is not horizontal. It tilts down on its right side, while moving downward at modest speed.
3) As a result, it has no trouble sliding through the left moving gap due to its tilt along with its downward motion.

 

[Sagittarius A-Star]

To sidestep this largely separate feature of SR (that the Herglotz-Noether theorem places very tight constraints on what motions can be stress free in SR), and focus on understanding how length, simultaneity, and orientation are all frame dependent, we should simply posit a pure inertial version of this problem.

Here is a related paper with summaries of other publications and the pure inertial version of this problem:

Differing perceptions on the landing of the rod into the slot
...
In the conventional rod and slot paradox, the rod, if it falls, was expected to fall into the
slot due to gravity.
...
Marx [6] observed that as the velocity of the rod increased, the observation from the co-moving
frame of the slot would be rod rotated and contracted but the line of motion of the end points of
the rod would remain the same. This indicated that the rod’s passing or not passing through the
slot was not affected by the magnitude of the approach velocity.
...
In this paper we present a variant of the rod and slot paradox with motion in two directions but
only with constant velocity. The pedagogical contribution of our paper shows how the noninvariance
of angles and proper lengths comes into play.

Source:
https://arxiv.org/abs/0809.1740

 

[RW137]

Note that if you require the two rod ends to enter the hole simultaneously in the rod frame, a hole whose contracted width is greater than the rod length is required. If, instead, you require the the two ends to enter the hole simultaneously in the hole frame, then a big (proper length) rod can get through a small hole. Both of these statements are true, and they are not in conflict because they are different physical setups..

 

[RW137]

The rod as well as the hole have the same length L0. I said, that there exists a third reference frame (different from the rod and hole frames) in which the two events are simultaneous. If you do the math, you will find out, that in this reference frame, the rod and the hole still have the same length, but are contracted by the factor 0.816.

 

[PeroK]

The rod as well as the hole have the same length L0. I said, that there exists a third reference frame (different from the rod and hole frames) in which the two events are simultaneous. If you do the math, you will find out, that in this reference frame, the rod and the hole still have the same length, but are contracted by the factor 0.816.

Independent of $v$?

The third frame is simply the one where both rod and hole have the same speed.

 

[alan123hk]

I found the following article about the rod and hole paradox. I think this is an ingeniously and subtly shifting the focus to the moment before the entire rod is completely bent and falling. In this very short moment, the stress must propagate within the rod at a finite speed. Based on this consideration, the same result can be independently obtained from the perspective of the stationary frame (table) or the moving frame (rod). I don't know if everyone is satisfied with this approach or agree that this is a complete analysis of the problem, but it is indeed the way I have always expected to deal with it.

https://www.researchgate.net/publication/37422707_The_rod_and_hole_paradox_re-examined

 

[PeterDonis]

I found the following article about the rod and hole paradox.

It's actually not the same.

The first part of the paper discusses a "car and hole paradox" which is quite different, since it assumes that the car is only supported by its wheels, not continuously over its length, and puts a spoiler at the front of the car that extends well beyond the front wheel, and is assumed not to start falling until not only the front wheel starts falling, but until the information that the front wheel has started falling propagates at finite speed to the front of the spoiler. This is obviously a different scenario from the standard rod and hole paradox and cannot be used to draw useful conclusions about it.

The second part of the paper discusses what purports to be the standard rod and hole paradox, but makes a crucial change: it lowers the far end of the hole. It also ignores the crucial points already raised in this thread, that to realize this scenario with a rod of ordinary length, the "force of gravity" (or whatever force causes the rod to fall once its support is withdrawn) must be many, many orders of magnitude larger than 1 g, and also many, many orders of magnitude larger than the internal forces in the rod. So the claim in the paper that the rod's "proper stiffness" is unchanged does not justify the paper's conclusion that the rod will not free-fall once its support is removed.

The paper does make one valid point: that if we assume the downward force is gravity, the rod will be compressed under its own weight while it is supported, so the initial motion of the rod once the support is removed will be for it to uncompress itself. This means that the top of the rod will initially not be in free fall when the support is removed (I am describing things here in the hole's rest frame); it will still be experiencing an upward force from the layer of the rod below it, but this force will rapidly decrease to zero and the top of the rod will start accelerating downward, just initially with less than the full "acceleration due to gravity".

However, the only way the paper can get this valid point to lead to a conclusion that the rod will not fall into the hole is, as mentioned above, to lower the far end of the hole. The drawings in the paper make it appear that the lowering is slight; but again, if you actually run numbers for a rod of ordinary size with the $\gamma$ values in question, you will find that the lowering has to be huge.

Also, given that the bottom of the rod does immediately start free-falling when the support is removed, if the speed of sound in the rod is slow, that just means the rod will end up stretching vertically, not just "uncompressing" to its unstressed height. The paper does not take this possibility into account at all.

Finally, the paper assumes "exactly horizontal stress propagation", without ever realizing that this can only be true in one frame. This issue has also been brought out in previous posts in this thread.

Unfortunately, this is a good illustration of why the "researchgate" site is not a trustworthy source (as we have often found in previous threads here on PF).

 

[alan123hk]

Unfortunately, this is a good illustration of why the "researchgate" site is not a trustworthy source (as we have often found in previous threads here on PF).

I can understand and agree with many of the points you mentioned in your reply.
For a junior-level person like me, I shouldn't and don't have the ability to evaluate the technical views of this article. But what I admire most is that I think it really tries to apply the same laws of physics from the perspective of different observers. I appreciate this way of dealing with problems, because when different observers come to the same conclusion, the conclusion becomes very convincing.

 

[PeroK]

I found the following article about the rod and hole paradox ...

I must admit I don't get the problem. What happens when water flows over a cliff? The water starts to fall as soon as it over the edge. In the rod and hole paradox, we must imagine an extraordinary gravitational force that would treat any object like water and either bend it like a waterfall and/or rip it apart.

This does not seem to require any knowledge of SR or the Lorentz Transformation. Eventually, no matter how strong your material is, it will bend! Everything is held together by molecular forces and these are not infinitely strong. I honestly don't get what is difficult about the concept of all materials bending or breaking under sufficient force?

 

[alan123hk]

I must admit I don't get the problem. What happens when water flows over a cliff? The water starts to fall as soon as it over the edge. In the rod and hole paradox, we must imagine an extraordinary gravitational force that would treat any object like water and either bend it like a waterfall and/or rip it apart.

This does not seem to require any knowledge of SR or the Lorentz Transformation. Eventually, no matter how strong your material is, it will bend! Everything is held together by molecular forces and these are not infinitely strong. I honestly don't get what is difficult about the concept of all materials bending or breaking under sufficient force?

I understand what you mean. To be honest, although I still have some things that I don't understand, after careful consideration, I feel that even the so-called contradictions I mentioned before have not completely disappeared, but the severity seems to be far less serious than I thought before.

Perhaps some people have blind spots in their thinking, and they are particularly difficult to understand for certain things

 

[PAllen]

I understand what you mean. To be honest, although I still have some things that I don't understand, after careful consideration, I feel that even the so-called contradictions I mentioned before have not completely disappeared, but the severity seems to be far less serious than I thought before.

Perhaps some people have blind spots in their thinking, and they are particularly difficult to understand for certain things

Would please carefully study my posts #77 and #103 (which makes a few minor corrections in #77). These consider the three different problem setups, with all physical description described in the rod rest frame rather than the hole rest frame. I remain convinced that this summary is actually more conceptually complete and accurate than any of the referenced sources. Please ask further questions in relation to these posts so hopefully any remaining doubts can be resolved.

 

[robphy]

Regarding
https://www.researchgate.net/publication/37422707_The_rod_and_hole_paradox_re-examined
…

Unfortunately, this is a good illustration of why the "researchgate" site is not a trustworthy source (as we have often found in previous threads here on PF).

I haven’t been following this thread in depth. And I haven’t looked at the content of any of the papers referenced.

But I wanted to point something out.
Although the trustworthiness of researchgate is questionable,
this particular reference was peer-reviewed and published in the European Journal of Physics.

From the reference given there, it reads
January 2005
European Journal of Physics 26(1)
DOI:10.1088/0143-0807/26/1/003

http://dx.doi.org/10.1088/0143-0807/26/1/003
Harald van Lintel and Christian Gruber 2004 Eur. J. Phys. 26 19

So, in this case, researchgate is a being used as repository by one of its authors to the content of his peer-reviewed article.

 

[PeterDonis]

what I admire most is that I think it really tries to apply the same laws of physics from the perspective of different observers.

Every solution to the rod and hole paradox discussed in this thread does the same thing. In fact, they do it better than the paper you cited, because they don't leave out the key items I mentioned.

 

[PeterDonis]

in this case, researchgate is a being used as repository by one of its authors to the content of his peer-reviewed article.

While it's good to point this out, I still see the same issues with the paper itself that I described earlier.

 

[alan123hk]

gravity is many orders of magnitude stronger than the rod's internal forces in frame A just as well as in frame B,

An element of the rod has to move downward by at least its thickness before the right hole edge reaches it. At lab scales, this would be e.g. a centimeter per nanosecond squared for this to happen. This works out to be much larger than the surface gravity of a neutron star. You can work around this by e.g. having a rod whose thickness would be less than an atom, if you want to consider that acceptable.

I must admit I don't get the problem. What happens when water flows over a cliff? The water starts to fall as soon as it over the edge. In the rod and hole paradox, we must imagine an extraordinary gravitational force that would treat any object like water and either bend it like a waterfall and/or rip it apart.

Thanks to PeterDonis, PAllen and PeroK for the tips. I just had time to do actual calculations (I admit I was careless and a bit lazy before), as shown below.

For thicknesses close to the atomic scale, there is still the strength of the gravitational field calculated in the figure. I finally began to believe that in this case, the shear stress generated by gravity is enough to crush any object made of material.

For another example, in the moving frame, a copper rod with a diameter of 1 cm (h=1cm) and a length of 100 cm has a mass of about 0.7 kg, the relative speed of rod and hole is $~\frac{\sqrt{3}}{2}c~$, so according to the calculation formula, if it needs to fall completely into the hole, its weight is $~1.5 \cdot 10^{16}$ Newton, and this huge weight may cause it to be crushed to dust on the table before reaching the hole.

​

Edit: I think the following situation is impossible, so delete it.
However, I feel that for very hard objects, such as objects that are much harder than diamonds, I am not sure whether there will be exceptions, and I don’t know how to perform more in-depth calculations. In any case, if this violates the laws of physics or is illogical, then please ignore it.

 

[Sagittarius A-Star]

In "Spacetime Physics" of Taylor and Wheeler, the "length contraction paradox" of W. Rindler is called
"paradox of the skateboard and the grid".

Length Contraction Paradox
...
The answer hinges on the relativity of rigidity.

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

L-12 paradox of the skateboard and the grid
...
The answer hinges on the relativity of rigidity.

Source:
https://www.eftaylor.com/spacetimephysics/04a_special_topic.pdf

Unfortunately, the paradox of the skateboard and the grid has no solution.

 

[alan123hk]

We can also make the same calculation for the moving frame (rod), as show below.

So this result is the same as the result of the stationary frame (table), that is to say, in the moving frame (rod), when the rod hits the wall on the right side of the hole, the falling distance is exactly h.