The Paradox of Relativity Length Contraction

[Nugatory]

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

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

 

[ergospherical]

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

Even better:

 

[Dale]

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

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

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

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

 

[alan123hk]

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

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

 

[Dale]

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

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

 

[PeterDonis]

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

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

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

all the observers in the different inertial systems are equal

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

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

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

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

 

[PeterDonis]

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

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

 

[PeterDonis]

The rod is not rigid in any frame of reference

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

 

[Nugatory]

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

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

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

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

 

[Dale]

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

 

[PeroK]

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

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

 

[PeterDonis]

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

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

 

[PeroK]

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

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

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

 

[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.

 

[Mister T]

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).