B The Paradox of Relativity Length Contraction

[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.

 

[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.