The Paradox of Relativity Length Contraction

[Sagittarius A-Star]

It is quite possible to have shear in this sense without shear forces being present.

I don’t see Rindler making any claims about what would actually happen to e.g. a glass rod. It is only Gron’s additional claims I disagree with.

As there are no shear forces, to my understanding Gron's glass should not break, if this is an SR scenario (pseudo-gravitation, without tidal effects).

 

[PAllen]

As there are no shear forces, to my understanding Gron's glass should not break, if this is an SR scenario (pseudo-gravitation, without tidal effects).

Wrong. There are no shear forces because there is no time for them to apply. Instead, each element becomes separated from others before any shear forces can apply. This constitutes catastrophic breakage - conversion of a coherent body to dust. It is exactly shear forces in a less extreme case, that would hold the body together with only a slight 'droop'.

 

[PeterDonis]

As there are no shear forces, to my understanding Gron's glass should not break

No shear forces does not mean no forces, period. Gravity (or "pseudo-gravity" if you want to call it that since we are only using SR to model this scenario) counts as a force here. And as @PAllen has said, for the rod to fall through the hole given that it is moving horizontally relative to the hole with relativistic speed, gravity must be extremely strong, so strong that it is much, much stronger than any internal forces inside the rod, and hence overwhelms them so they are negligible.

 

[Sagittarius A-Star]

gravity must be extremely strong, so strong that it is much, much stronger than any internal forces inside the rod, and hence overwhelms them so they are negligible.

I think, if the rest-lenght of the rod would be 1 LY and the Lorentz-contracted length of the hole 0.1 LY, then also a gravitational acceleration of $9.81 m/s^2$ would have enough time to break the rod. Reason: The shear forces are not only negligible, but zero in this scenario.

 

[alan123hk]

Let us consider the situation described by W. Rindler https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf

After thinking about it, I can finally draw a process diagram of the rod falling into the hole from the perspective of observer B, as shown in the figure below. This seems to explain why the rod falls into a hole smaller than it.

Now the remaining problem seems to be that I must believe in the theory of relativity. Observer B must see this descending trajectory calculated by Observer A through the Lorentz transformation. That is, the huge gravitational field has enough power to make the rod bend and fall into the hole, as described by the equation.

But what I find strange is that in observer A's frame of reference, the rod should fall freely and has nothing to do with the physical parameters of stiffness. But in the reference frame of observer B, it seems to be closely related to the physical parameters of the stiffness of the rod, unless it is compressed into small pieces by gravity. So does this affect its falling speed or its final collision position with the right wall of the hole, resulting in a value different from the value predicted by the equation derived from the free fall condition?

 

[cmb]

Might I ask if this is actually a 'quantum' question. I don't really understand the significance of the rod and rigidity so I am thinking more 'basketball'.

At a practical level, to fall through the hole means it has to accelerate in the direction of the hole to some degree, and in practice it's always going to bounce off the far edge, whatever the particulars of the object are. The question is really, how far down could it get (under some massive acceleration that pulls it down by more than a radius in the time to cross), not so much whether it will bounce off the other wall adversely.

So I was thinking what is similar to this scenario, if anything? Does a small particle traveling over a grating of comparable size tell us anything? Like, is this a quantum tunnelling problem, if it were a smaller scale (the question does not disqualify the hole and rod to unit Angstrom sizes) and in fact the depth of the hole is not insignificant?

 

[PAllen]

I think I can resolve the differing conclusions here. To do so, let's modify the set up a bit, to remove some features that are initially distracting. Imagine a table top, rather than a hole (i.e. no right side of the hole). Imagine that 'gravity' only exists to the right of the table top. We have a rod moving at near c to the right. We then can consider 3 force timing scenarios:

A) the rod starts experiencing force as soon as it passes the table edge (i.e. the most realistic scenario).
B) the rod (via a magic trap door, that can even act as 'force shield' until it is pulled away downards in the tabe frame) experiences force only after it has cleared the table top (in the table frame), with force application along the rod simultaneous in the table frame.
C) the rod experiences force only after it has cleared the table in the rod frame, applied simultaneously in the rod frame. This is the only case that satisfies Born rigidity. Imagine a magic trapdoor pulled down in the rod frame only after the table edge has reached the rod left end.

Once we are clear on this, we can re-introduce the right boundary of the hole, and the results should then be obvious. I will describe all three scenarios only from the rod frame, hereafter, as I think that is clearest for understanding the dynamics (and it cannot make a difference to the physics).

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:

A) and B) : the wall smashes the rod, 'vaporizing' it, with destruction moving to the left at near c (not greater than c), quickly overtaking the shear wave.

C) If the wall is too close to the table edge, the rod is mostly over the far side of the wall before force starts simultaneously. If we declare we want a true rigid motion, we insist that the left edge of the rod magically resists the force and nothing happens. If the wall is far enough away from the table edge (its length contracted distance from the table edge is greater than the rod length), then we get case C as described without the wall. Then, the wall hits the rod, vaporizing it right to left at near c. (Here, we obviously relax rigidity; if we keep rigidity, then, as the wall approaches, force is applied to the rod in a uniquely timed way such that it transitions to being at rest between the table and wall - and the constraints of hole size guarantee that it fits).

Hopefully, this clears up several confusions, including some of my own imprecisions earlier.

 

[Mister T]

Computer simulation of Rindler's length contraction paradox:

Grøn and Johannesen said:
Assume that our rod is a very thin rod made of glass
so that it breaks if one tries to bend it. Now look at
the figures. As observed in Σ the rod is straight and
unbroken all the time. But as observed in Σ' the rod is
severely bent, and should surely be broken-at least
according to conventional elasticity theory. The fact
that the rod does not break even if it is bent very
much, tells us that one needs a relativistic theory of
elasticity. Also this shows that it is not obvious what
we should call a 'physical effect' in the theory of
relativity. Demanding Lorentz invariance for the
existence of a physical effect, we must conclude that
while the breaking of a rod is indeed a physical effect,
the bending of a rod is not a physical effect.

Source (PDF):
https://ur.booksc.org/dl/46255893/29b9bd?openInBrowser

via:
https://ur.booksc.org/book/46255893/b0c94a

This reminds me of a thought experiment I came up with when I took my first class in relativity (I have since refined it as a result of my posting it here in this forum several years ago):

Shown in the figure is a pair of clocks joined by a stout metal rod through their centers. The clocks are synchronized in their rest frame so that the hands rotate in sync. A delicate glass rod connects the tips of the clock hands. If one of the clock hands advances more than the other, the glass rod would twist, stretch, and break. But as long as the clock hands move in sync the glass rod will not be stressed enough to break.

An observer at rest with respect to the clocks will see them in sync and so expect the glass rod to be intact and unbroken. An observer in motion along the line joining the clocks will conclude that the clocks are not synchronized. Should he expect the glass rod to be broken? Can the glass rod be broken according to one observer and not broken according to another? Explain your answers.

 

[PAllen]

This reminds me of a thought experiment I came up with when I took my first class in relativity (I have since refined it as a result of my posting it here in this forum several years ago):

Shown in the figure is a pair of clocks joined by a stout metal rod through their centers. The clocks are synchronized in their rest frame so that the hands rotate in sync. A delicate glass rod connects the tips of the clock hands. If one of the clock hands advances more than the other, the glass rod would twist, stretch, and break. But as long as the clock hands move in sync the glass rod will not be stressed enough to break.

View attachment 294379An observer at rest with respect to the clocks will see them in sync and so expect the glass rod to be intact and unbroken. An observer in motion along the line joining the clocks will conclude that the clocks are not synchronized. Should he expect the glass rod to be broken? Can the glass rod be broken according to one observer and not broken according to another? Explain your answers.

this is really trivial in the context of kinematic decomposition and Born rigidity. The glass will not break if the problem apparatus is set up in the rest frame of the two clocks. In any other frame, the coordinate representation of the glass will be 'deformed' but the kinematic decomposition will still show absence of shear, compression, or stretch.

 

[Ibix]

Might I ask if this is actually a 'quantum' question.

It isn't. It's just that in general whether an accelerating rod is straight or not is a frame dependent interpretation. This is because an accelerating object has a curved worldtube, and most frames' planes of simultaneity don't cross the worldtube perpendicular to the curvature.

 

[Sagittarius A-Star]

Let us consider the situation described by W. Rindler https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf
...
Now the remaining problem seems to be that I must believe in the theory of relativity. Observer B must see this descending trajectory calculated by Observer A through the Lorentz transformation. That is, the huge gravitational field has enough power to make the rod bend and fall into the hole, as described by the equation.

View attachment 294358​

In reference frame B:

The point, where the falling starts (by loosing support from the trap door), is moving with
$x'/t'_S = - c^2/v$ from the right-hand side to the left-hand-side of the rod.

On the right side of this (moving) point, the vertical slices of the rod fall freely with
$z' = - \frac{1}{2}a\gamma^2 (t'-t'_S)^2 = - \frac{1}{2}a\gamma^2 (t'+x'v/c^2)^2$.

This can be Lorentz-transformed into frame A. If the gravitation $a\gamma^2$ is "huge" or not does not matter.

But what I find strange is that in observer A's frame of reference, the rod should fall freely and has nothing to do with the physical parameters of stiffness. But in the reference frame of observer B, it seems to be closely related to the physical parameters of the stiffness of the rod, unless it is compressed into small pieces by gravity. So does this affect its falling speed or its final collision position with the right wall of the hole, resulting in a value different from the value predicted by the equation derived from the free fall condition?

Only because gravity cuts the rod (near the upward-force from the moving trap-door) into millions of vertical 1-atom-thick slices, those slices have a free fall condition, independently of their neighbors.

The reason, why the rod gets cut is, that the interval between the two events of start falling of two "horizontal" neighbor atoms is spacelike. So, a force resisting the shear could not be fast enough.

 

[PAllen]

I was finally able to look at the Gron article. I have no disagreement except the last paragraph of the discussion. Even the first two paragraphs of the discussion are consistent with points I've made in this thread. In the last paragraph he makes a statement about glass breakage which is unsupported by any analysis or discussion within the article. It is false. Note, that his version of hole is the 'thin right edge' case in my discussion above. Contrary to his claim, if the setup is done at lab scale, so a rod has to move at least centimeters through a modest size hole while crossing at near c, a fracture wave beginning on the right edge of the rod will start propagating to the left of the rod at speed of sound, and will have progressed a tiny bit before all the rod has passed into the hole in the rod frame, and it will continue propagating after this if nothing interferes. The result will be a delayed, pulverized rod.

 

[PeterDonis]

I think, if the rest-lenght of the rod would be 1 LY and the Lorentz-contracted length of the hole 0.1 LY, then also a gravitational acceleration of $9.81 m/s^2$ would have enough time to break the rod.

Yes, but no known type of object would have a gravitational acceleration of 1 g, a size much larger than 1 LY (so the region where the experiment is done could be taken to be locally flat), and a surface on which things can stand and move horizontally.

If you formulate the experiment inside an accelerating "rocket" in empty space, much more than 1 LY wide, that would do it.

For a rod that long, it would probably be much easier to intuitively see that it can't be treated as rigid.

 

[PeterDonis]

in the reference frame of observer B, it seems to be closely related to the physical parameters of the stiffness of the rod

No, it has nothing whatever to do with the rod's material properties, because, as noted in earlier posts, the force of gravity is many orders of magnitude stronger than the rod's internal forces so the latter can simply be ignored. The only reason the rod falls "freely" in frame A (i.e., all parts of the rod start accelerating downward in this frame at the same time) is that the scenario in Rindler's paper is explicitly set up that way; that's the function the trapdoor serves. It has nothing to do with the rod's material properties; gravity is many orders of magnitude stronger than the rod's internal forces in frame A just as well as in frame B, it's just that Rindler's specific scenario is carefully constructed so that in frame A the force of gravity is applied in just the right way to keep the rod straight in that frame.

 

[PeterDonis]

Might I ask if this is actually a 'quantum' question.

There is no quantum mechanics involved at all in the scenario being discussed in this thread.

So I was thinking what is similar to this scenario, if anything?

All of the questions you ask here are off topic in this thread, but the simple answer to all of them is "no". As above, QM is not involved at all in this thread.

 

[PeterDonis]

If I have time I'll try doing the computations.

I've done some initial computations, and from what I can see, the congruence of worldlines describing the rod will have nonzero expansion, shear, and vorticity once the rod is accelerating downwards. These computations are at least "I" level so if there is interest I will post them in a separate thread.

 

[PAllen]

I've done some initial computations, and from what I can see, the congruence of worldlines describing the rod will have nonzero expansion, shear, and vorticity once the rod is accelerating downwards. These computations are at least "I" level so if there is interest I will post them in a separate thread.

Sure I’d be interested. I once did some approximate computations that led me to believe the shear components were largest.

 

[PeterDonis]

I’d be interested.

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

 

[Dale]

Now the remaining problem seems to be that I must believe in the theory of relativity.

Why is that a problem?

 

[pervect]

I've done some initial computations, and from what I can see, the congruence of worldlines describing the rod will have nonzero expansion, shear, and vorticity once the rod is accelerating downwards. These computations are at least "I" level so if there is interest I will post them in a separate thread.

I was just about to post that I expected some vorticity. Consider a gyroscope mounted on a sliding block in an inertial frame. The gyroscope does not precess. Now, when the block hits the trap door, it starts to accelerate vertically due to the external force applied. (Rindler mentions that the vertical acceleration could be due to a magnetic force, or the equivalent of a hailstorm).

When the block starts to accelerate vertically, a gyroscope attached to the block will start to experience Thomas precession, due to the fact that $\vec{a} \times \vec{v}$ is nonzero, see https://en.wikipedia.org/wiki/Thomas_precession. Some of the old "sliding block" threads also talk about how a gyroscope attached to a vertically accelerating, horizontally sliding block precesses and why. [add] See also https://arxiv.org/abs/0708.2490v1, or the sliding block mega-thread https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/

I'm surprised to hear that there is also expansion and shear, but I have not done any calculations. I don't believe the motion of the block can be Born rigid, as it is basically changing it's state of rotation. Specifically, it is changing it's state of rotation relative to a gyroscope.

[add] To be more precise, I expect the results of the calculation to be that the expansion, shear, and vorticity of a sliding block in an inertial frame is zero, while the expansion and shear of a sliding block in an accelerated frame are also zero, while the vorticity is non-zero. The transition region from one to another would be very fiddly so I'm not considering it.

 

[PAllen]

).
.

[add] To be more precise, I expect the results of the calculation to be that the expansion, shear, and vorticity of a sliding block in an inertial frame is zero, while the expansion and shear of a sliding block in an accelerated frame are also zero, while the vorticity is non-zero. The transition region from one to another would be very fiddly so I'm not considering it.

But this case is not about sliding at all. Rindler’s construction, in the rod initial rest frame, has a boundary of proper acceleration moving right to left along the rod (in SR, with “gravity” being considered an ordinary force). At this boundary, there is obvious shear, while the distance between elements of the rod in the ‘falling‘ portion can be shown to be increasing with time from any element’s local rest frame (thus, expansion). Vorticity is also present.

 

[pervect]

Are we talking about https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf?

I was considering the variant where Rindler says:

Before proving our assertion, let's make the problem more concrete. The hole shall be filled with a trap door which shall be removed (downwards and with sufficient acceleration to allow the rod to fall freely)

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.

I was assuming there was a pressure on the top of the rod due to hail, counterbalanced by upwards pressure from the floor. Then, when the floor is removed, there is no counterbalancing pressure. Ignoring the vertical elastic response of the rod, this is basically the same as suddenly applying pressure uniformly to the top of the rod, modulo the issue of the vertical compression of the rod.

The rod starts to accelerate downwards due to the pressure from the hail when the trapdoor is removed. There is technically no floor, but the uniform pressure on the top of the rod in the frame of the rod is analogous to the same as the pressure that a sliding block experiences on Einstein's elevator, making the analysis similar. Since I've already thought about the sliding block case , it was naturally to leverage this work.

 

[Sagittarius A-Star]

Are we talking about https://home.agh.edu.pl/~mariuszp/wfiis_stw/stw_rindler_lcp.pdf?

I was considering the variant where Rindler says:

Before proving our assertion, let's make the problem more concrete. The hole shall be filled with a trap door which shall be removed (downwards and with sufficient acceleration to allow the rod to fall freely)

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's not ambiguous at all, because the sentence you cited partly, continues in the following way:

... by the observer A at the instant when to him the hind end of the rod passes into the hole.

The next two sentences make it also clear:

This precaution eliminates the tendency of the rod to topple over the edge. All points of the rod will then fall equally fast, and the rod will remain horizontal, in the frame of A.

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

 

[alan123hk]

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

I really don't understand. For example, let me show a calculation that may be wrong, the acceleration should be constant $~\frac {d^2z'}{dt'^2} = a'=a\gamma^2$ in the reference frame of observer B. If $\gamma=2$, then the effect of relativistic length contraction is already obvious, but the acceleration is only $a'=4a$, where $a$ is the acceleration measured by observer A. Let $a=9.8$, then gravitational field $a'=39.2 m/s^2$ only. So please allow me to ask a question that may be considered ignorance, how does this huge gravitational force that can crush everything actually arise ?

 

[Sagittarius A-Star]

If $\gamma=0.5$, then

Gamma ($=\frac{1}{\sqrt{1-v^2/c^2}})$ is always $\geq1$. But even that is not a "huge" factor in your example in the OP.

 

[alan123hk]

is always >=1. you need the 2nd derivative...

Sorry, corrected

 

[Sagittarius A-Star]

the acceleration should be constant $~\frac {dz'}{dt'} = a'=a\gamma^2$ in the reference frame of observer B.

For acceleration you need the 2nd derivative: $a' = \frac {d^2z'}{dt'^2}$.

This, together with time-dilation, is the reason for the square of the gamma-factor.

 

[PAllen]

I really don't understand. For example, let me show a calculation that may be wrong, the acceleration should be constant $~\frac {d^2z'}{dt'^2} = a'=a\gamma^2$ in the reference frame of observer B. If $\gamma=2$, then the effect of relativistic length contraction is already obvious, but the acceleration is only $a'=4a$, where $a$ is the acceleration measured by observer A. Let $a=9.8$, then gravitational field $a'=39.2 m/s^2$ only. So please allow me to ask a question that may be considered ignorance, how does this huge gravitational force that can crush everything actually arise ?

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.

If you use ordinary acceleration, the most that would happen (any frame) is a slight interference with the bottom surface layer of atoms by the right hole edge. This would likely be undetectable in the real world. That is, for ordinary gravity, the rod would just skim over the hole.

 

[Sagittarius A-Star]

If you use ordinary acceleration, the most that would happen (any frame) is a slight interference with the bottom surface layer of atoms by the right hole edge. This would likely be undetectable in the real world. That is, for ordinary gravity, the rod would just skim over the hole.

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

I think, if the rest-lenght of the rod would be 1 LY and the Lorentz-contracted length of the hole 0.1 LY, then also a gravitational acceleration of $9.81 m/s^2$ would have enough time to break the rod. Reason: The shear forces are not only negligible, but zero in this scenario.

See also:
https://www.physicsforums.com/threa...ivity-length-contraction.1010138/post-6576955

 

[alan123hk]

An element of the rod has to move downward by 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.

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.

 

[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.