B The Paradox of Relativity Length Contraction

[PeterDonis]

Post #2 is a different problem from that posted by the OP in post #1.

Yes, but neither one is the same as the barn door paradox (although of course there are some similarities).

That might explain a lot of the confusion.

I think the key general statements that have been made in the thread are applicable to both versions, the post #1 version and the post #2 version. But I agree it is good to be clear about which version is being discussed if any statements are being made that are particular to a given version.

 

[Herman Trivilino]

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

To avoid the issues of rigidity altogether, you can consider the train-tunnel paradox. Both the paradox discussed in this thread and the train-tunnel paradox are designed to teach the same lesson.

Do a google search for relativity train tunnel paradox.

 

[PeterDonis]

To avoid the issues of rigidity altogether, you can consider the train-tunnel paradox.

This is also known as the "barn door paradox" or "barn and pole paradox". It has been mentioned briefly in this thread.

Both the paradox discussed in this thread and the train-tunnel paradox are designed to teach the same lesson.

To the extent that the lesson is length contration and the relativity of simultaneity, yes, that's true.

However, the fact that there is no such thing as a perfectly rigid body in SR is also a valid lesson to be taught, and the paradox discussed in this thread can be (and has been) used to teach that lesson as well.

 

[PeroK]

To avoid the issues of rigidity altogether, you can consider the train-tunnel paradox. Both the paradox discussed in this thread and the train-tunnel paradox are designed to teach the same lesson.

Do a google search for relativity train tunnel paradox.

They are both illustrated meticulously in the Khutoryansky video I linked to.

 

[alan123hk]

@PeterDonis Thank you for your more specific and substantive explanation of the issues I really care about in the two replies (#41 and #44). Your explanation is indeed convincing and logical. Although I think there are some other questions that still confuse me, it does not mean that I think there is any problem with your statement.

 

[PeterDonis]

I think there are some other questions that still confuse me,

If there are, feel free to ask about them.

 

[Sagittarius A-Star]

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

Please consider Rindlers scenario with a trap-door, which makes the scenario easier to analyze.

In frame B, the stiffness is lost for a very short time, because:

1. The left side of the rod is prevented from falling by the upward force from the (moving) tabletop and, in Rindlers scenario, for an additional distance from a trap door.
2. The right side of the rod is falling by gravitation.
3. The point, where the falling starts (by loosing support from the trap door), is moving with more than the speed of light to the left (* proof see below). That is faster than the speed, at which material-internal forces could react.
4. The movement and length-contraction of the hole stops, as soon, as the right edge of the hole hits the right side of the rod.

The left edge of the hole is moving with velocity $v$ to the left.

* The trap-door is removed in frame A at time $t=0$. That means with the Lorentz transformation:
$0=t= \gamma (t'+vx'/c^2)$
$=>$
$x'/t' = -c^2/v$
$=>$
The point, where the falling starts (by loosing support from the trap door), is moving with more than the speed of light to the left.

 

[pervect]

Who was it that said that there are not rigid bodies in special relativity, just rigid motions? I thought that this distinction, along with the supporting text as to why the distinction is made, might be helpful to the original poster, but I couldn't recall the source. Additionally, I've found it much better to go to the extra effort to check exactly what the source says than to rely on my memory, especially nowadays.

 

[Sagittarius A-Star]

Computer simulation of Rindler's length contraction paradox:

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

 

[PAllen]

Computer simulation of Rindler's length contraction paradox:
Source (PDF):
https://ur.booksc.org/dl/46255893/29b9bd?openInBrowser

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

Neither of those links works for me, but I must disagree with the analysis. The kinematic decomposition of a congruence allows an Invariant definition of shear and expansion/contraction. Whatever frame you compute it in, Rindler’s scenario has shear. Meanwhile, the case of horizontal fall through a sufficiently large hole in the rod rest frame (with horizontal defined in this frame, of course) shows zero expansion or shear (while being bent in the hole rest frame). Thus, I agree you can have frame variant bending that has no physical effect. I disagree that that there is some missing theory of elasticity in SR, and I disagree which case has the actual physical shear.

 

[PeterDonis]

Whatever frame you compute it in, Rindler’s scenario has shear.

Does it? As I understand the scenario in Rindler's paper, in the hole rest frame (the one in which the rod is length contracted), the trap door is withdrawn in such a way that all parts of the rod start accelerating downward with the same coordinate acceleration at the same time. This should result in a Born rigid motion.

 

[Sagittarius A-Star]

Neither of those links works for me

Now, the links also don't work for me. A few minutes ago, they did. Strange!

Here is another link, but not with free access:
https://iopscience.iop.org/article/10.1088/0143-0807/14/3/001/pdf

 

[PAllen]

Does it? As I understand the scenario in Rindler's paper, in the hole rest frame (the one in which the rod is length contracted), the trap door is withdrawn in such a way that all parts of the rod start accelerating downward with the same coordinate acceleration at the same time. This should result in a Born rigid motion.

No, because the ‘the same time’ here has nothing to do with the same time in local frame of a congruence line. The congruence lines are moving at high speed relative to the hole, so motion does not begin simultaneously in the local frame for nearby elements. In the corresponding rod rest frame case, coordinate simultaneity matches local rest frame simultaneity in the horizontal direction. The vertical motion has to begin with just the right timing over the vertical dimension of the rod, but as long as this is done, the whole motion is Born rigid.

 

[PeterDonis]

No, because the ‘the same time’ here has nothing to do with the same time in local frame of a congruence line.

Ah, I see. If I have time I'll try doing the computations.

 

[Sagittarius A-Star]

Whatever frame you compute it in, Rindler’s scenario has shear.

Does this mean, that in frame B, in which the hole is moving to the left, the right-hand side of the rod is not accelerating downwards with exactly $\gamma^2 a$ in the hole, because the gravitational force is counteracted by material-internal forces? Then the Rindler solution of the problem would be wrong.

 

[PAllen]

Does this mean, that in frame B, in which the hole is moving to the left, the right-hand side of the rod is not accelerating downwards with exactly $\gamma^2 a$ in the hole, because the gravitational force is counteracted by material-internal forces? Then the Rindler solution of the problem would be wrong.

No, each element of the rod is moving with exactly that acceleration once it starts accelerating. The issue is the timing. If an element of the rod has a nearby element starting to move before it does, in its local rest frame, there must be shear, expansion, or contraction. In this case, the major component is shear.

I don’t see anything wrong with Rindler’s overall presentation. He claimed that what looks rigid on one frame, won’t look rigid in another. I see no claim as to Born rigidity or any invariant notion of elasticity. He also claims that if you set the problem up as he described, then his solution for both frames is correct, which I also agree with. 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.

In particular, in an idealized problem, I claim the rod breaks in Rindler’s set up, and this true in both frames, despite appearances. In a realistic case, the rod breaks in any conceivable version of the scenario. Simply, the rod sliding over a frictionless surface with an implied gravity consistent with the rod falling through the hole despite its speed, implies that a glass rod would collapse into neutron star material in a nanosecond. One is talking about the order of $10^{18}$ Newtons of force. [edit: this is actually several orders of magnitude greater than the surface gravity of a neutron star, so who knows what would ‘really’ happen]

 

[Sagittarius A-Star]

If an element of the rod has a nearby element starting to move before it does, in its local rest frame, there must be shear, expansion, or contraction. In this case, the major component is shear.

But Sartori writes about Rindlers paradox:

And what is wrong with the argument that predicts that the moving narrow hole should simply pass under the stationary block, as in figure 6.17?

To answer these questions, let us picture the block as being composed
of many small vertical segments, each attached to one thread. Before any
threads have been cut (fig. 6.17b), every segment is in equilibrium. For
each segment that is over the hole, the force of gravity is balanced by the
upward pull of a thread.

Figure 6.17c shows the thread that holds up the end segment Q being
cut. That segment is no longer in equilibrium because there is nothing to
balance the downward pull of gravity. Therefore figure 6.17c cannot be
right. The end segment must begin to fall, initially with the acceleration of
gravity. The rest of the block is still suspended and momentarily remains
horizontal.

If no additional threads were cut, the fall of the end segment would be quickly arrested by upward forces, called shear forces, exerted by the adjacent segment. The block would resemble a cantilevered beam set into a wall. Each part of such a beam is held up by shear forces exerted by the adjoining parts; the beam does not remain strictly horizontal but "droops" a little.

In the present problem, however, the second segment itself begins to fall shortly after the first. Moreover, the information that the first thread has been cut and the end segment is falling is not communicated instantaneously to the rest of the block. The news can propagate no faster than the speed of light and cannot reach the adjacent segment before a time interval D/c has passed, where D is the distance between the two segments. Until then the shear forces cannot act. For at least a time interval D/c, therefore, the end segment is in free fall. Before the interval D/c has passed, the next thread has been cut and the next segment of the block has begun to fall. This must be true because the interval between the cutting of the two threads is spacelike. (Remember the two events are simultaneous in frame S.) The process continues until all the threads have been cut and the entire block is falling freely. Under these circumstances, the shear forces never get a chance to act. Each segment of the block is in free fall and never "knows" that the adjacent segments are at different heights. The shape of the block in frame S' 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 block and a steel block would have precisely the same shape.

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

 

[PeterDonis]

But Sartori writes about Rindlers paradox

By "shear", @PAllen does not mean "shear forces", he means part of the kinematic decomposition:

https://en.wikipedia.org/wiki/Congr...atical_decomposition_of_a_timelike_congruence

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

 

[PAllen]

But Sartori writes about Rindlers paradox:Source:
"Understanding Relativity, A Simplified Approach to Einstein's Theories", LEO SARTORI, UNIVERSITY OF NEBRASKA-LINCOLN

with quote:

"The process continues until all the threads have been cut and the entire block is falling freely. Under these circumstances, the shear forces never get a chance to act. Each segment of the block is in free fall and never "knows" that the adjacent segments are at different heights"

This is wholly irrelevant to the notion of Born rigidity. There exists an alternative motion that is Born rigid motion where such a thing never happens at all - that is its very definition. Further, the argument actually shows that under such extreme circumstances, the body is acting as if it is dust, not as if it is a coherent material body (dust is, in fact, the best description of any matter under extreme relativistic regimes). The argument shows that bonds between elements effectively don't exist under these circumstances. The argument that the adjacent element doesn't have a chance to exert counter forces does not mean they that the change in distance is not real, instead it means that steel and diamond cannot be distinguished from dust in these circumstances - and thus they 'become dust' - disassociated particles. What would you call each element being independently in free fall, with changing distance from its neighbors (and no interaction with its neighbors), if not dust?

I honestly don't know why so few books on relativity deal with congruences and their kinematic decomposition. They make so many questions like this invariant with respect to any coordinate choice. I also find it surprising that so few books cover the implications of Herglotz-Noether theorem on rigid motion (both of these were well covered by Pauli's book in 1921, but seem to have been poorly covered in first courses ever since).

 

[PAllen]

But Sartori writes about Rindlers paradox:Source:
"Understanding Relativity, A Simplified Approach to Einstein's Theories", LEO SARTORI, UNIVERSITY OF NEBRASKA-LINCOLN

"And what is wrong with the argument that predicts that the moving narrow hole should simply pass under the stationary block, as in figure 6.17?"

As to this, it should be added that this is what would indeed happen if the rod were required to undergo only rigid motion (in the Born sense). This gets at the feature that there is no such thing as rigidity in the real world. But for accelerations like ordinary gravity, the motion would behave rigidly, and the rod would simply pass over the hole - it would 'fall' far less than diameter of one atom, which would have no consequence. It is precisely the extreme force needed to make the rod fall through the hole when the rod is moving near c relative to the hole that requires the material to act like dust. (Or, in the case of @Ibix piston, the piston immediately renders the body into dust, given the force applied).

 

[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.

 

[Herman Trivilino]

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.