B The Paradox of Relativity Length Contraction

[Sagittarius A-Star]

For thicknesses close to the atomic scale, there is still the strength of the gravitational field calculated in the figure.

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In the figure you wrote:

$\alpha = \sqrt{1-v^2/c^2}$
$a=2h[\frac{v}{l_0(1-\alpha)}]^2$
Let $h=0.4nm, l_0=1m, v=\frac{\sqrt{3}}{2}c$
Then $a=0.416 m/s^2$

Are you sure, that your numerical result for $a$ is correct?
According to my calculation $a=2h[\frac{v}{l_0(1-\alpha)}]^2 = 6hc^2/(1m)^2$.

 

[alan123hk]

Are you sure, that your numerical result for a is correct?

You are correct. I 'm sorry that my calculation result is an extremely outrageous error.

By the way, I think that based on these calculation results, I have reason to believe that there may not be any substance in this world that can withstand the pressure generated by this so-called length contraction fallacy without being crushed, then some of the main components mentioned in this imaginative experiment, including the table that carries the sliding rod and the trap door of the hole, are almost impossible to exist.

 

[PeterDonis]

there may not be any substance in this world that can withstand the pressure generated by this so-called length contraction fallacy without being crushed

The rod is not being subjected to compressive forces when it is over the hole, it's being subjected to shear forces. It's not being crushed, it's being torn. The relevant parameter is not its resistance to static pressure but the sound speed inside the material--how quickly it can propagate changes in its internal forces in response to being torn. By hypothesis, since the "shear events", the events at which each horizontal point on the rod is subjected to an unbalanced downward force, are spacelike separated, it is impossible for the sound speed in the rod to be fast enough to compensate--it would have to be faster than the speed of light.

some of the main components mentioned in this imaginative experiment, including the table that carries the sliding rod and the trap door of the hole, are almost impossible to exist.

The table is only being subjected to pressure, and while the rod does move over the table, so the pressure at a given horizontal point on the table does vary, the resistance to pressure is entirely vertical and nothing needs to propagate horizontally. There is no shear on the table, and no spacelike separated events are involved in the table's resistance to compression.

Whether the removal of the trap door fast enough to not impede the free fall of the rod is possible given the parameters Rindler uses might be worth further investigation.

 

[alan123hk]

the shear stress generated by gravity is enough to crush any object made of material.

The rod is not being subjected to compressive forces when it is over the hole, it's being subjected to shear forces. It's not being crushed, it's being torn. The relevant parameter is not its resistance to static pressure but the sound speed inside the material--how quickly it can propagate changes in its internal forces in response to being torn. By hypothesis, since the "shear events", the events at which each horizontal point on the rod is subjected to an unbalanced downward force, are spacelike separated, it is impossible for the sound speed in the rod to be fast enough to compensate--it would have to be faster than the speed of light.

Your narrative is of great reference value. I admit that I used the word "crush" improperly. But I know that when the rod is over the hole, the shear force acts on the rod.

The table is only being subjected to pressure, and while the rod does move over the table, so the pressure at a given horizontal point on the table does vary, the resistance to pressure is entirely vertical and nothing needs to propagate horizontally. There is no shear on the table, and no spacelike separated events are involved in the table's resistance to compression.

Whether the removal of the trap door fast enough to not impede the free fall of the rod is possible given the parameters Rindler uses might be worth further investigation.

Thank you for your very useful comment. Actually I'm thinking about something simpler. The gravitational field should be spread over the entire area including the table and should not just be concentrated in the space of the hole. I mean under such a huge gravity field, everything is very likely to be crushed by its own weight, including the rod, the table that supports the rod, and the ground that supports the table, and so on.

 

[PeterDonis]

The gravitational field should be spread over the entire area including the table and should not just be concentrated in the space of the hole.

Yes, that is true.

under such a huge gravity field, everything is very likely to be crushed by its own weight

Not necessarily. For example, everything could be made of neutronium. This is a thought experiment and we don't have to limit ourselves to materials we can make on Earth.

 

[PAllen]

Yes, that is true.Not necessarily. For example, everything could be made of neutronium. This is a thought experiment and we don't have to limit ourselves to materials we can make on Earth.

There is a problem with this idea. If you assume the material is rigid enough not to turn to nucleon soup under the proposed forces, then it also will behave Born rigid traversing the hole, and thus will be unable to get through the hole. The required rigidity will prevent the local deformation required in the hole frame analysis.

I prefer the magical assumption that the force is magically turned on at some time, simultaneously everywhere per some frame. Which frame’s simultaneity you use determines the nature of the result. And being some kind of pseudogravity, if the body is not obstructed, the force applies to all parts of the body at once (per chosen frame), producing only modest internal stresses.

 

[PeterDonis]

If you assume the material is rigid enough not to turn to nucleon soup under the proposed force

Under the compressive force exerted while the rod is sitting on the table.

As I said in post #153, by hypothesis, the events on the rod worldlines at which the force on each point of the rod becomes unbalanced (because the support is removed) and the downward acceleration starts are spacelike separated (they must be, because all of these events occur at the same coordinate time in the hole rest frame). That means it is impossible for any material, no matter how strong, to respond quickly enough to prevent shear and hence to prevent the rod from bending in at least some frames. The sound speed in the material would have to be faster than the speed of light, which is impossible.

 

[Sagittarius A-Star]

I think the whole non-inertial aspect of this thread's problem set up is a distraction from the basics of understanding how length, simultaneity, and orientation are all frame dependent.

(and no right side of hole at all, which is just a distraction for this discussion)

I am not a fan of simplifying the given problem, because then some interesting aspects get lost.

For example regarding the causality:

• In the hole's frame, the front-bottom corner "Q" of the rod hits the right edge of the hole, after the left-hand side of the rod started falling.
• In the rod's frame it is the opposite sequence: The front-bottom corner "Q" of the rod is hit by the right edge of the hole, before the left-hand side of the rod will start falling.

That means that both mentioned events cannot influence each other by internal forces.

 

[PAllen]

Under the compressive force exerted while the rod is sitting on the table.

As I said in post #153, by hypothesis, the events on the rod worldlines at which the force on each point of the rod becomes unbalanced (because the support is removed) and the downward acceleration starts are spacelike separated (they must be, because all of these events occur at the same coordinate time in the hole rest frame). That means it is impossible for any material, no matter how strong, to respond quickly enough to prevent shear and hence to prevent the rod from bending in at least some frames. The sound speed in the material would have to be faster than the speed of light, which is impossible.

Yes, but if one assumes this matter has a speed of sound near light speed, and rigidity such that it can retain its shape under compressive forece (that is, that its bonding forces are of the same order as the g forces), and assuming a lattice of e.g. elements separated by 1 angstrom, the the delay between an element staring to fall and its neighbors getting the news is such that the neighbors will have only moved a small fraction of an angstrom in the element local frame before a resistance force can be generated. The degree of assumed rigidity means that the such resistance force is able to overcome the gravity after this delay. The result is that, viewed from the rod frame, there is a snap back front trailing the force application front by a couple of angstrom, at the same superluminal speed of force application (because it is an independent local reponse to each force application event - it doesn’t have travel, e.g. from an end of the rod; it just has a local communication delay). The result is that by the time the rod has moved less than a micron overall, in the hole frame, the whole rod will have snapped into some rigid configuration, and will therefore not fit through the hole.

 

[PAllen]

I am not a fan of simplifying the given problem, because then some interesting aspects get lost.

For example regarding the causality:

• In the hole's frame, the front-bottom corner "Q" of the rod hits the right edge of the hole, after the left-hand side of the rod started falling.
• In the rod's frame it is the opposite sequence: The front-bottom corner "Q" of the rod is hit by the right edge of the hole, before the left-hand side of the rod will start falling.

That means that both mentioned events cannot influence each other by internal forces.

But that is irrelevant to whether the rod fits. Those issues can be addressed using a hole in some magic surface of zero thickness, as Gron does in his analysis. To me, the collision with a right wall is only a distraction - precisely because the spacelelike relation of events means it has no relation to how all
parts of the rod get through the hole.

Anyway, I don’t mean to suggest the right wall variant should be ignored, only that one had better first understand the simpler variant without the wall. Once that is understood, adding the wall is straightforward, and I don’t think will add much confusion for the neophyte. But only after they first understand the version without a wall

 

[PAllen]

You have made the relevant calculations, I trust them., But I think this wasn’t the original Rindler ‘s thought . He simply spoke of gravity.
Suppose that gravity is 1/100 of g: the fall of the rod happens the same, because its length is contracted wrt the hole, as per the observer stationary in the hole reference frame.Maybe the Rindler-Shaw paradox is better formulated?

But with ordinary gravity and a lab scale problem, the only answer is that the rod skips over the hole with no “knowledge“ of its existence. If it only falls a tiny fraction of an atomic radius over the hole, nothing will be noticed. Even at ordinary speed, a hockey puck can be hit over a hole bigger than the puck with no visible deflection hitting the far side of the hole.

 

[PeterDonis]

rigidity such that it can retain its shape under compressive forece

"Retain its shape" in the sense that, while supported, it retains a rectangular rod shape, yes, but it can still have significantly reduced thickness as compared to its unstressed thickness.

such resistance force is able to overcome the gravity after this delay.

The resistance force can't "overcome" gravity, because, by hypothesis, the rod is unsupported in the hole rest frame while all this is happening. There will be an induced vibration in the rod due to the internal forces as shear waves pass through the rod, but the net effect will still be for the rod as a whole to fall as long as it is not supported, and the downward acceleration of its center of mass will be the "acceleration due to gravity", even if individual particles of the rod have different accelerations due to internal forces. The rod can't magically suspend itself.

By hypothesis, the acceleration is large enough for the length contracted rod to fall through the hole in the hole rest frame. In that frame, the downward acceleration is applied at the same instant to all parts of the rod. That is the hypothesis I am working with. Under that hypothesis, while the rod might indeed "snap" to a configuration that looks rigid in some frame, it will still fall through the hole.

In the frame in which the rod is at rest prior to starting to fall, the support is removed first from the front end of the rod. Note that in this frame, the downward acceleration on unsupported parts of the rod is increased by a factor of $\gamma^2$. This means that it is quite possible in this frame for the shear on the rod to happen too fast for the rod's internal forces to compensate, so that the rod bends in this frame, even if in the hole rest frame the rod's motion looks rigid.

It is an independent local reponse to each force application event - it doesn’t have travel, e.g. from an end of the rod; it just has a local communication delay

In the hole rest frame, this is true, since there is no time delay horizontally across the rod in this frame. But in this frame, as above, the "local communication" can't possibly prevent the rod from falling, because all of it is unsupported at the same time.

In the rod rest frame (the one in which the rod is at rest prior to starting to fall), the support is being removed horizontally along the rod at almost the speed of light, so each part of the rod is losing its support at about the same time as it is receiving the information that it needs to pull back on the part of the rod just ahead of it. So "local communication" is not sufficient to keep the rod from falling (or more precisely in this frame, bending) in this frame either.

 

[Sagittarius A-Star]

One could also introduce a non-gravitational force to cause the rod to "fall" through the hole; that formulation has also been mentioned in this thread.

And this proposal is even mentioned in the paper of W. Rindler (in the 1st paragraph above the two pictures).

 

[PAllen]

"Retain its shape" in the sense that, while supported, it retains a rectangular rod shape, yes, but it can still have significantly reduced thickness as compared to its unstressed thickness.

Ok, I was assuming little or no shape change, which would mean extreme bond strength.

The resistance force can't "overcome" gravity, because, by hypothesis, the rod is unsupported in the hole rest frame while all this is happening. There will be an induced vibration in the rod due to the internal forces as shear waves pass through the rod, but the net effect will still be for the rod as a whole to fall as long as it is not supported, and the downward acceleration of its center of mass will be the "acceleration due to gravity", even if individual particles of the rod have different accelerations due to internal forces. The rod can't magically suspend itself.

Yes, I see this now. The 'atoms' only have a delayed 'awareness' of changed distance and orientation of nearby neighbors; they have no 'awareness' of common acceleration (assuming no tidal forces, as we are).

By hypothesis, the acceleration is large enough for the length contracted rod to fall through the hole in the hole rest frame. In that frame, the downward acceleration is applied at the same instant to all parts of the rod. That is the hypothesis I am working with. Under that hypothesis, while the rod might indeed "snap" to a configuration that looks rigid in some frame, it will still fall through the hole.

Agreed. It is actually easier to see how this works in the hole rest frame. There, you have simultaneous horizontal force applied, rod begins to move down horizontally while also moving horizontally. At a very small delay (by my calculations, under the model under discussion, the rod will travel less than a micron before local neighbors begin to react). In the hole frame, this slightly delayed reaction will occur simultaneously across the rod (just like the resistance to a horizontal cleaver hitting a cutting board occurs at a delay, but all across the cleaver simultaneously, in this frame). As I've described elsewhere, this looks like a reaction front following a small distance behind the force application front, superluminally, in the rod initial rest frame.

At this point, the rod will start reshaping itself into some shape corresponding to a rigid motion. I believe this actually means the rod will curve and start to bend (in the hole rest frame), concave up, with the left tilting more and more downward over time.

In the frame in which the rod is at rest prior to starting to fall, the support is removed first from the front end of the rod. Note that in this frame, the downward acceleration on unsupported parts of the rod is increased by a factor of $\gamma^2$. This means that it is quite possible in this frame for the shear on the rod to happen too fast for the rod's internal forces to compensate, so that the rod bends in this frame, even if in the hole rest frame the rod's motion looks rigid.

Yes, this happens initially. But distances between neighbors will have only shifted by order of 1% before the neighbors start to react. So with this assumed bond strength, there should be no problem snapping to a Born rigid configurations (with vibrations, etc. before it settles). With ordinary matter, bond strengths can't resist, and the interatomic distance just continue to grow until all bonds break. But interestingly, even for ordinary matter, this might not happen until the rod is way past the hole (in hole in surface formulation, rather than a hole with walls). That is, while it is invariant whether a the rod is experiencing stress, and this is definitely true for the formulation of horizontal applied downward acceleration in the hole frame - whereas horizontal downward acceleration in the rod frame need not have stress - it actually appears to me (I did some calculations for diamond), that the diamond rod will not have severe enough stress to break before being all the way under the surface well to the right of the hole. Thus, for a hole with a wall, the diamond will not have broken yet before hitting the wall - despite invariant internal stresses.

 

[PeterDonis]

At this point, the rod will start reshaping itself into some shape corresponding to a rigid motion. I believe this actually means the rod will curve and start to bend (in the hole rest frame), concave up, with the left tilting more and more downward over time.

Now that I think of it, the nonzero vorticity of the congruences we calculated earlier would indicate the same thing; the internal forces will tend to correct for the vorticity in the way you describe.

 

[PeroK]

I brought up the comparison of the required acceleration to the surface gravity acceleration of a neutron star. I assumed it was obvious we were not concerned with the tidal forces near a neutron star. I thought the statement that this was over 100 times the value of g for the surface a neutron star was instructive. It implies that chemical bonding forces effectively don’t exist, and even the nuclear forces are overwhelmed.

Precisely. Any good student should notice that at relativistic speeds the collison with the far side of the hole would be indistinguishable from the friction between the surfaces! When the rod is over the hole we need some gigantic force (it doesn't need to be gravity) that does something physically significant in the short time during which the rod passes over the hole. The Earth's surface gravity would not achieve this.

In fact, considering a mechanism other the gravity may help resolve the problem, as the student may have a blind spot about gravity not acting until the CoM is over the edge of the hole. As a hangover from too many classical mechanics problems!

If, instead, we have powerful daemons lurking in the hole, who grab the rod and pull it down to its fate, then the solution is more obvious. Even in the rod frame, where the hole is short, the first daemon grabs hold of the tip of the rod as soon as it appears over the edge!

That said, I don't know whether there is any published research on the compatibility of daemons and special relativity.

 

[PeterDonis]

Moderator's note: A number of off topic posts have been deleted.