I Meter stick slides over a meter wide hole at a high speed

[Hiero]

Consider a meter stick sliding along its length towards a hole which is a meter wide in the direction of motion. Suppose the meter stick moves with speed so that the gamma factor is 10.

Then in the stick’s frame, the hole is 10cm and should be easy to cross. But in the hole’s frame, the stick is only 10cm and so should easily fall in.

Does the stick fall or not? Why?

I just can’t seem to wrap my head around this scenario. (Seems similar to the ”ladder paradox” but that is just an example of the relativity of simultaneity; I don’t see how that will save us in the stick-hole version.)

 

[Orodruin]

Instead of looking at a stick that is sliding along, which would need acceleration to fall down - leading to necessarily looking at the acceleration process and understanding that the stick is necessarily deformed in at least one of the frames, consider a situation where there is a one meter stick oriented horizontally and falling vertically towards a plate that is moving with a gamma factor of 10. The plate has a 1 m hole in its rest frame. This hole in the original frame will be 10 cm due to length contraction and the stick will not be able to fall through.

If you Lorentz transform this to the plate rest frame, then the x-distance between the stick end-points will indeed be 10 cm, but the stick will no longer be horizontally oriented, it will be tilted. Due to this and to the movement of the stick in the horizontal direction, the stick will not be able to pass through the hole in this frame either - the same conclusion as for the original frame.

 

[Hiero]

@Orodruin, I understand that scenario. To be quantitative, let’s say the stick moves vertically at speed v_y in the first frame (where it’s horizontal). In the plate’s frame the meter stick will have Δy = v_yv_x/c^2 meters but also a vertical speed of v_y/ϒ_x and so the time between when the top and bottom ends reach the plate is ϒ_xv_x/c^2 in which time it moves horizontally by v_x times that. So altogether we need the the gap to be 1/ϒ_x+ϒ_xv_x^2/c^2 = ϒ meters long which is more than one, hence they still collide.I don’t quite understand in what ways this is equivalent to the original setup. Are you saying the stick will cross the hole? I don’t see the connection.

 

[Hiero]

Actually, I think I see part of your point. The motion of falling down the hole tends to rotate the rod in the right direction for the forward-end to stay above the forward edge.

So then in the stick’s frame, it will bend into the hole a little, despite being a “perfectly rigid” stick? Is there then subtler bending in the hole’s frame as well?

If I wanted to analyze this within the realm of special relativity, is it possible? At least to some good approximation?

 

[Orodruin]

So then in the stick’s frame, it will bend into the hole a little, despite being a “perfectly rigid” stick?

A perfectly rigid stick simply does not exist. The reason to look at the other scenario is to avoid the entire problem of specifying exactly how the rod is allowed to accelerate.

If I wanted to analyze this within the realm of special relativity, is it possible? At least to some good approximation?

Yes, but you will have to accept that a perfectly rigid object does not exist and that the answer will depend on what you specify for how the rod accelerates

 

[pervect]

I'd strongly suggest thinking about a simpler paradox first. When the OP understands the simpler paradox, they can maybe consider coming back to this one. They're very similar, the simpler paradox only involves inertial motion, not accelerated frames of reference and/or gravity. The simpler paradox is "The Barn and Pole Pardox", and there's a lot written about it, for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html

 

[Hiero]

A perfectly rigid stick simply does not exist.

Can’t we say something like it’s the most ideal rigidity? In other words it’s the least amount of bending for the scenario? I’m just wondering because this bending has nothing to do with any stiffness constants, right?@pervect Of course I understand that problem; I even mentioned it in the OP (under a different name, “ladder paradox,” but I linked the details).
That’s a trivial example of the frame dependence of simultaneity.

Not to be rude, but how do you think I could correctly analyze Orodruin’s situation in post 2 but not understand the barn thing...Anyway I am not seeing how the stick will make it over the hole... it won’t right?

 

[Ibix]

If you imagine doing this at non-relativistic speeds then the rod will tip as it moves over the gap and start to tumble. Handling objects that start to spin in relativity is complex because they cannot behave rigidly and you have to specify a material model. That goes double for a rod sliding off a table, where part of it is supported and part of it isn't, so it must flex. That's why we keep ducking the problem.

I can offer another slightly modified version. This happens explicitly in zero g. The rod is very thin compared to its length, as is the tabletop, and it is traveling very close to the table. As it passes over the gap in the table it is struck simultaneously (in the table frame) along its length by an anvil which is moving perpendicular to the table in the table frame, knocking the rod through the hole in the arbitrarily thin tabletop in arbitrarily short time. What does that look like from the rod frame?

 

[Hiero]

What does that look like from the rod frame?

Well the anvil will be angled downwards and the impulse will no longer be simultaneous, the front of the rod gets hit down first. I believe the front end (or any point) gets hit when it’s the same fraction of the way along the gap as in the first frame.
Am I missing any key details?

 

[Ibix]

Am I missing any key details?

In the table frame the whole rod is hit simultaneously. In the rod frame, as you say, the rod is hit at different times along its length. You might like to think about whether or not it'll suffer any bending stresses.

 

[Hiero]

@Ibix indeed the front end will be want to bend down. So going back to the original problem, is the result that in either frame the stick does not cross the gap? In the stick’s proper frame it falls in by this bending effect. Is this right?

To be honest I thought the answer was supposed to be that the stick will cross. I’ve been so confused trying to figure how it will come back up in the hole’s frame. If it’s actually the case that the rod falls in the hole, then I am much more content with neglecting the quantitative details.

I am curious though, is there any concept of ideal rigidity? Of course “perfect” rigidity is out the window, but is there some concept of “most rigid”? (Any elementary references on rigidity in SR would be cool.)

 

[jartsa]

Consider a meter stick sliding along its length towards a hole which is a meter wide in the direction of motion. Suppose the meter stick moves with speed so that the gamma factor is 10.

Then in the stick’s frame, the hole is 10cm and should be easy to cross. But in the hole’s frame, the stick is only 10cm and so should easily fall in.

Does the stick fall or not? Why?

I just can’t seem to wrap my head around this scenario. (Seems similar to the ”ladder paradox” but that is just an example of the relativity of simultaneity; I don’t see how that will save us in the stick-hole version.)

The stick bends in both frames. The frames disagree about the bending force.

So how about if we say that on the stick there is a mechanism that makes the first 10 cm of the stick 'weightless'. Like for example a small physicist standing on the stick on the 15 cm mark and pulling on a rope attached on the front of the stick.

In the frame where the hole is ten times wider the downwards force is ten times smaller. That might explain something. I am referring to the special relativistic transformation of transverse forces: F`= F/gamma.

 

[pervect]

<<snip original confusing stuff>>

So, let us proceed, knowing that you understand the simpler case of the barn and pole. , I will assume that you understand that it's the relativity of simultaneity that explains the apparent paradox. (If you don't, we may have to revisit this earlier problem, as it will then be obvious that we do not share the same understanding of it).

What's not obvious, though, is how relativity of simultaneity applies to the new problem. The short answer is that the relativity of simultaneity drastically changes our understanding of rigid bodies, rigid body mechanics, and rigid body motions in special relativity.

In Newtonian physics, rigid bodies have six degrees of freedom - three translational degrees of freedom, and three rotational degrees of freedom. In special relativity, there is no such thing as a rigid body, which is Orodruin's point in #5. We do have something called a Born-rigid body in special relativity, but it's not the same as a rigid body.

The reason a Born rigid body is not the same as a rigid body is that Born rigid bodies cannot change their state of rotation without violating the rigidity conditions.

So the short answer to your problem is that you are (apparently) incorrectly trying to apply techniques that used to work in Newtonian physics, techniques built upon rigid bodies, to a relativistic situation where rigid bodies no longer exist in exactly the same sense.

There are two routes one could take from this point. The easier route is to study what happens to something like a steel bar, which is not perfectly rigid. Because it's not perfectly rigid, steel bars can rotate just fine in special relativity - it's just that when they change their state of rotation, they must change their shape. They must deform. When we allow the bar to deform, there's no problem.

The point of Orodruin's #5 is that we can understand how the steel bar deforms when it goes over the hole in the floor with a bit of work. With just a litlte more work, we can come to the understanding that the speed of sound in the bar serves as a quantitative measure of it's rigidity, and have a nice neat answer that because the speed of sound in steel is about 5km/second, our notions of the steel bar remaining "rigid" as it goes over a hole in the floor start to become very poor approximations well before we reach relativistic velocities. If we study how a steel bar moving at 10 times the speed of sound in steel, 50km/secm, responds to the situation of going over a hole in the floor, using purely Newtonian methods, we will see that it's better to treat the bar as if it had no rigidity. With a bit of imagination, we can look for things with low speeds of sound. The low-speed-of-sound realm will give us some insight into the relativistic behavior as well - some of the details will change, but studying how a slinky drops



will give us a lot of insight into the low-speed-of-sound realm. Imagining a collection of slinky's going over the hole, each separate slinky representing one section on the bar, will give us better insight into it's behavior at 50km/sec than thinking of it as rigid.

The other route one could take is to study Born rigidity in more depth. This can be rather interesting, but it may not answer your original question, because in the end we'll come to the conclusion that Born rigid objects don't change their state of rotation.

 

[Hiero]

Easily, I'm afraid

By this you mean the rod makes it easily over the hole?

It's clear, on a qualitative level, how bending could cause the rod to fall in the hole, in the rod's frame.

I cannot understand, on a qualitative level, how any bending could save the rod from falling in the hole, in the hole's frame.

So if the answer is that the rod passes the hole, please say that explicitly, because I cannot fathom it.

Anyway thanks for your response; I will be sure to look into "born rigidity," even if it doesn't help with this problem.

 

[Orodruin]

So if the answer is that the rod passes the hole, please say that explicitly, because I cannot fathom it.

Again, the problem of your setup is acceleration. Without specifying when the rod starts accelerating and how different parts of the rod accelerate, there is no definite answer to your question.

 

[PAllen]

Well, let's ask how far a hockey puck falls under 1 g crossing a 1 meter hole, moving at .999 c. The answer is about 200,000 times less than the radius of one hydrogen atom. Normal equilibrium interactions between a solid body and air are much larger than this. Thus the puck would not have any 'awareness' of the hole at all. This underscores @Orodruin 's point that the features of this problem are all dependent on what acceleration profile you specify.

 

[pervect]

By this you mean the rod makes it easily over the hole?

Nope, that's not what I meant at all. I'll just snip that and focus on the rest.

It's clear, on a qualitative level, how bending could cause the rod to fall in the hole, in the rod's frame.

I cannot understand, on a qualitative level, how any bending could save the rod from falling in the hole, in the hole's frame.

In the hole frame, the rod bends, and the tip of the rod hits the edge of the hole. I usually stop the analysis there, because I start to imagine the consequences of the tip of the rod hitting the edge of the hole at relativistic velocities.

In the rod frame, we seem to agree that the tip of the rod hits the edge of the hole.

So, in either frame, the tip of the rod hits the edge of the hole.

 

[Hiero]

Well, let's ask how far a hockey puck falls under 1 g crossing a 1 meter hole, moving at .999 c. The answer is about 200,000 times less than the radius of one hydrogen atom.

That’s a good point. Perhaps I should have made it a kilometer stick! That should be a million times what you said so like 5 hydrogen atoms? Light sure is fast huh.

Again, the problem of your setup is acceleration. Without specifying when the rod starts accelerating and how different parts of the rod accelerate, there is no definite answer to your question.

Well to avoid the complexities of rotation Ibix mentioned, let us think of a false floor over the hole, which will be pulled away the moment the back end goes over the edge, in the hole’s frame.

The entire stick will begin accelerating at once in that frame, similar to Ibix’s distributed impulse example.
With that, do we need to specify anything else?

As a side question - I’m honestly not sure how to handle gravity. The thing I’m unsure of is that a downwards acceleration in the stick’s frame should be 1/ϒ^2 times smaller in the hole’s frame. So one frame is g the other is g/ϒ^2 ? I don’t feel like this is right, but I’ve never thought of gravity beyond Newton.

 

[sweet springs]

Though I am afraid it is just a duplicate of above discussions:

In Hole system
Top end of Rod touches Hole bottom and Bottom end of Rod enter Hole top are simultaneous.

In Rod system
When Top end of Rod touches Hole bottom, Bottom end of Rod does not enter Hole top yet.
When Bottom end of Rod enter Hole top, Top end of Rod smashes hit by Hole bottome or penetrate Hole.

So Rod and Hole cannot keep sound in the process. Bottom end of Rod cannot stop entering Hole. Elastic signal from Rod top "Don't push me anymore!" cannot reach Rod end to prevent it. The two or at least one must be broken.

 

[Orodruin]

Well to avoid the complexities of rotation Ibix mentioned, let us think of a false floor over the hole, which will be pulled away the moment the back end goes over the edge, in the hole’s frame.

What makes you think that this avoids rotation in the rod frame? Also, you are probably thinking of using gravity to accelerate the rod through. Don’t. It will just add another layer of complication and the closest you will get in SR is a non-inertial reference frame with a constant proper acceleration in the vertical direction. It would be much better for your own sanity not to care about the acceleration mechanism and instead just specify a scenario fully in one frame. You can then Lorentz transform to the other to find out what the situation looks like there.

I don’t feel like this is right, but I’ve never thought of gravity beyond Newton.

The point is that you shouldn’t think of gravity in SR. How gravity is described in relativity is given by GR.

 

[pervect]

Well to avoid the complexities of rotation Ibix mentioned, let us think of a false floor over the hole, which will be pulled away the moment the back end goes over the edge, in the hole’s frame.

I'm not sure of why you are doing this, and I also want to point out the hidden frame dependence. The event "when the back end goes over the edge in the hole's frame" is well defined, but you say that the floor is pulled away at "the same instant of time". The idea of the 'same instant of time' over an extended spatial region is frame-dependent because of the relativity of simultaneity. "The same instant of time" in one frame of reference, plotted on a space-time diagram, is different than "the same instant of time" in another frame.

This is key point in understanding the pole-in-barn paradox, something that you mentioned earlier that you were familiar with when I asked. It's also something that a lot of people get wrong. Thus it's hard for me to tell if you actually are familiar with the relativity of simultaneity, or whether you only think you are familiar with it.

As a side question - I’m honestly not sure how to handle gravity. The thing I’m unsure of is that a downwards acceleration in the stick’s frame should be 1/ϒ^2 times smaller in the hole’s frame. So one frame is g the other is g/ϒ^2 ? I don’t feel like this is right, but I’ve never thought of gravity beyond Newton.

You'll be missing out on at least two other effect. If we have a rocketship accelerating to simulate gravity, and we have a block or train sliding across it's floor, a gyroscope attached to the sliding block will precess. This is called Thomas precession.

See for instance https://arxiv.org/abs/0708.2490v1

As an application where both the reference frame and the gyroscope accelerates we will consider a gyroscope on a train that moves along an upward accelerating platform as shown in Fig. 3.

Figure 3: A train at two consecutive times moving with velocity v relative to a platform that accelerates upward. A gyroscope with a torque free suspension on the train will precess clockwise for v >0.

A second effect, which I shall loosely call 'curvature of the track', is described in the above paper in figure 11.

A graduate level treatment that systematically reveals all the effects of a frame of reference associated with a train moving across the floor of an accelerating spaceship involves tensors and Christoffel symbols. If you are familiar with these graduate level techniques, I can point to a few long threads where the topic is discussed in some mathematical detail, and a few metrics are derived. (Different people have different approaches). If you don't have that background, working through https://arxiv.org/abs/0708.2490v1 is probably more viable attempt to gain some understanding of the two effects I mentioned above.

 

[Ibix]

@Ibix indeed the front end will be want to bend down. So going back to the original problem, is the result that in either frame the stick does not cross the gap? In the stick’s proper frame it falls in by this bending effect. Is this right?

To be honest I thought the answer was supposed to be that the stick will cross. I’ve been so confused trying to figure how it will come back up in the hole’s frame. If it’s actually the case that the rod falls in the hole, then I am much more content with neglecting the quantitative details.

I am curious though, is there any concept of ideal rigidity? Of course “perfect” rigidity is out the window, but is there some concept of “most rigid”? (Any elementary references on rigidity in SR would be cool.)

Here is a 2+1d Minkowski diagram of the example I gave - the rod (blue) is just in front of the table (green, with a gap) at the beginning of the experiment, moving in the +x direction at 0.6c. It is struck by a force in the y-direction that pushes it below the table as it passes the gap. I haven't illustrated the anvil, so all you see is the effect of the impact - the "kink" in the blue worldsheet.

You were wondering what that looked like in the frame where the rod is stationary. Here's the answer:

Now you can see that the kink, which was parallel to the x axis, is not parallel to the x' axis - it's not a simultaneous strike in this frame. In fact, the rod kind of ripples through the hole. The ripple is faster than light, though (not a problem - there's no direct causal dependency of one part of the impact on any other part), so there's never any bending stress on the rod - each adjacent element of rod is already in motion by the earliest time it could become aware that the next-door element started moving.

This is basically why you can't have a general model of rigidity in relativity - something that is a rigid motion in one frame isn't necessarily rigid in any other. You can define things like Born rigidity, but they can't apply to things with changing proper acceleration.

Does that help?

 

[PAllen]

Just a fine pint clarification of @Ibix lasr post: Born rigid acceleration is possible for any acceleration profile at all including arbitrary changes in 4-acceleration (including direction). However, once the profile is specified for one point of a born rigid object, the motion of the whole object is determined. Any change in some other part of the body different from this, which would be needed to account for rotation, is not possible.

 

[Herman Trivilino]

I am curious though, is there any concept of ideal rigidity?

Think of the rod as being made of delicate glass. It breaks whenever it's bent, and you are free to choose the amount of bending it can sustain before breaking. The smaller your choice, the closer we get to your ideal. However small your choice, the rod makes it through the hole unbroken.

 

[PAllen]

Think of the rod as being made of delicate glass. It breaks whenever it's bent, and you are free to choose the amount of bending it can sustain before breaking. The smaller your choice, the closer we get to your ideal. However small your choice, the rod makes it through the hole unbroken.

Can you clarify? Seems to me that most variants of the scenario have the rod breaking in all frames. For example, any scenario where the rod tilts as it goes through the hole must involve breakage unless the you allow a lot of deformation. If you take the anvil suggestion instead, the impact along different parts of the rod is not simulaneous in the rod frame, so again it must break, unless you allow substantial deformation. If you plug in numbers you need something more like ideal chewing gum rather than ideal glass to avoid breakage.

 

[PAllen]

Now you can see that the kink, which was parallel to the x axis, is not parallel to the x' axis - it's not a simultaneous strike in this frame. In fact, the rod kind of ripples through the hole. The ripple is faster than light, though (not a problem - there's no direct causal dependency of one part of the impact on any other part), so there's never any bending stress on the rod - each adjacent element of rod is already in motion by the earliest time it could become aware that the next-door element started moving.

I must disagree with part of this. By any standard definition (e.g. shear tensor) as well as physical common sense, there is enormous shear stress in your scenario, and this feature is invariant. The FTL motion of the kink simply means that the shear, instead of propagating (limited to speed of sound in material), is causally independently induced at each point along the rod, not that there is no shear. Note that in the MCIF of a given rod element, you will have adjacent elements have acceleration of trillions g, and relative speed of large fraction of c.

Any attempt to do this with real materials would result in a rod being converted to a compressed plasma moving at high speed downward (in the original rod rest frame) once the process is complete. Note, this downward moving plasma line will be rotated relative to the original rod (per the original rod rest frame).

Note, that the illustration provided does not account for the fact that after the anvil accelerates a rod element to near c relative to the original rod, there is no proposed counter anvil to decelerate an element, so every rod element ends up continuing to move at high speed downward relative to the original rod. That is, retraction of the anvil does nothing to stop the downward motion induced on each rod element.

 

[Orodruin]

To be honest, I think it is rather meaningless to discuss this issue with ”real” materials for the reasons already mentioned (also, stress energy tensor calculations are typically beyond introductory courses). The interesting stuff is how a series of events in one frame appears in another. Questions such as ”does it break” should preferably be postponed for later.

 

[Herman Trivilino]

Can you clarify? Seems to me that most variants of the scenario have the rod breaking in all frames.

I was thinking that in the variant posed by the OP the rod makes it through the hole unbroken.

My point is that if the rod doesn't bend in its own rest frame, then it doesn't break. That's invariant regardless of how delicate the rod.

 

[PAllen]

I was thinking that in the variant posed by the OP the rod makes it through the hole unbroken.

My point is that if the rod doesn't bend in its own rest frame, then it doesn't break. That's invariant regardless of how delicate the rod.

For the rod not to bend in its own rest frame, the hole must be bigger than the rod per the rod rest frame. That means the hole must be much bigger than the rod in the hole rest frame. The OP actually describes a case where the hole is too small in the rod's initial rest frame, in which case it cannot get through the hole without bending in the rod rest frame, and this establishes that it can't get through the hole without shear in any frame (despite appearances in the hole rest frame).

 

[Ibix]

I must disagree with part of this. By any standard definition (e.g. shear tensor) as well as physical common sense, there is enormous shear stress in your scenario, and this feature is invariant. The FTL motion of the kink simply means that the shear, instead of propagating (limited to speed of sound in material), is causally independently induced at each point along the rod, not that there is no shear.

Hm. The problem as I stated it was one given to us as undergrads (you are, of course, correct that there should be another anvil to stop the y-direction motion). I think it was just presented as a variant on the rod and barn problem, but we certainly ended up arguing about whether or not the rod broke. We concluded that it didn't on the grounds I stated before, that each subsequent atom of the rod is in motion before it can communicate with its neighbours.

Unfortunately I've been sketching out the stress-energy tensor for the rod in my head and it does seem likely that there will be space-space off diagonal elements in it, which implies shear stresses. So you may be right. I need pen and paper.

As an amusing aside, I've been googling stress and strain tensors and now I see adverts offering me counselling services.

 

[Orodruin]

Unfortunately I've been sketching out the stress-energy tensor for the rod in my head and it does seem likely that there will be space-space off diagonal elements in it, which implies shear stresses. So you may be right. I need pen and paper.

Not to mention the space-diagonal entries that would result from the stretching of the rest-length.

Also, the atom by atom is only true horizontally.

I think it would be safe to say that any rod of this sort hit in this way would be utterly pulverised.

 

[jartsa]

Hm. The problem as I stated it was one given to us as undergrads (you are, of course, correct that there should be another anvil to stop the y-direction motion). I think it was just presented as a variant on the rod and barn problem, but we certainly ended up arguing about whether or not the rod broke. We concluded that it didn't on the grounds I stated before, that each subsequent atom of the rod is in motion before it can communicate with its neighbours.

Unfortunately I've been sketching out the stress-energy tensor for the rod in my head and it does seem likely that there will be space-space off diagonal elements in it, which implies shear stresses. So you may be right. I need pen and paper.

As an amusing aside, I've been googling stress and strain tensors and now I see adverts offering me counselling services.

A tilted anvil hits a thin horizontal rod?

How about if we first consider a horizontal anvil hitting a horizontal thin rod:

First every molecule of the rod is at its preferred position, as the rod is not under stress, then every molecule is displaced from that position. It seems like the rod is given potential energy, because some force is pulling every molecule towards its original position.

I would say the rod disintegrates, then fuses back to its original form, releasing the potential energy as radiation.

(The thickness of the rod is just one atom-layer)

 

[PAllen]

Hm. The problem as I stated it was one given to us as undergrads (you are, of course, correct that there should be another anvil to stop the y-direction motion). I think it was just presented as a variant on the rod and barn problem, but we certainly ended up arguing about whether or not the rod broke. We concluded that it didn't on the grounds I stated before, that each subsequent atom of the rod is in motion before it can communicate with its neighbours.

Unfortunately I've been sketching out the stress-energy tensor for the rod in my head and it does seem likely that there will be space-space off diagonal elements in it, which implies shear stresses. So you may be right. I need pen and paper.

As an amusing aside, I've been googling stress and strain tensors and now I see adverts offering me counselling services.

I have been debating writing up this little essay, but given continued interest, I will.

A key to this is use Born rigidity, not as a realizable notion, but as a frame invariant definition of motion without shear, compression, or twist. Note that this does effectively chose a preferred local frame at every event in the world tube of a body - the local MCIF of the body element at that point (mathematically, at that event along a world line of a congruence). It is relative to this local MCIF that closest congruence world lines must be at rest. For a body that starts out inertial, this means it is easiest to derive Born rigid acceleration in the starting rest frame of the body. Further, despite its idealized character, we can say of two motions that if one is Born rigid and another severely deviates from it, that only the former approximates a possible motion that won't break a reasonably rigid body.

A key fact about Born rigid motion is that the world line of one element of the congruence ('atom of the body') uiniquely determines the motion of all others if the world line has proper acceleration.

So, now the question to ask is what is Born rigid motion of a rod at rest that starts to move downward orthogonal to its length? By symmetry, the motion must start simultaneously across the length of the rod in this frame - there is nothing that can prefer either end. On the other hand, it is also true that the top surface of the rod must begin moving slightly before the bottom surface for Born rigidity to be maintained, however, for our purposes, this plays minimal role because we can make the rod as thin as we want. Then, given a chosen motion of the left edge of the rod, we know that the Lorentz transform of this motion is the only possible rigid motion in any frame.

What does this motion look like in a frame in which the rod is initially moving inertially to the right at high speed? The left edge starts to move down, and the start of downward motion propagates FTL to the right along the rod. But note, it is in this frame's coordinates that the propagation is FTL - compared to the left edge of the rod, which is moving near c, the propagation of downward motion initiation is only moving a little faster than the left edge of the rod. The faster the initial motion of the rod, the longer it takes the motion point to reach half way along rod. Thus, the faster the motion of the rod, the larger a hole would have to be for the whole rod to have time to move through without hitting at either end. This is the description in this frame of the fact that in the rod rest frame, the faster a hole moves, the larger its rest size must be for the rod to get through it laterally unbroken.

Since the above Born rigid motion is uniquely determined by e.g. specifying the motion of the rod left edge, then a motion different from the (e.g. one where the downward motion is simultaneous in the hole frame) is severely non-rigid, and this fact is frame invariant. @Ibix has already provided the description of this case in the rod rest frame. This description is correct, the only issue is claiming it won't break the rod - it must because only one possible motion is Born rigid, and it is not this one. As a final nuance, note that an explanation of why the motion boundary (kink in @Ibix description) is FTL is that you have the front of the rod going through the hole close to the right side of the hole, while the left edge goes through the hole near the left side of the hole, while the hole is moving to the left near c. Thus, the start of motion must proceed FTL right to left for this to be possible.

 

[Ibix]

I have been debating writing up this little essay, but given continued interest, I will.

Thanks. I need to digest that a bit, I think.

 

[rrogers]

Can’t we say something like it’s the most ideal rigidity? In other words it’s the least amount of bending for the scenario? I’m just wondering because this bending has nothing to do with any stiffness constants, right?@pervect Of course I understand that problem; I even mentioned it in the OP (under a different name, “ladder paradox,” but I linked the details).
That’s a trivial example of the frame dependence of simultaneity.

Not to be rude, but how do you think I could correctly analyze Orodruin’s situation in post 2 but not understand the barn thing...Anyway I am not seeing how the stick will make it over the hole... it won’t right?

Try this. Plot the spacetime diagram; the stationary "reference" frame has a spacelike gap whose edges move forward in time. Project them forward. Now do the same thing for the moving hole and you can see how they can intersect. But in the "reference" you must make sure the "simultaneous" image of the stick is correct; it's shorter but that comes out as being extended in time; i.e. tilted. I have always found that space-time diagrams make everything clearer than talk and equations. You do have to think but some of us think better with pictures.

 

[James Demers]

I think most commenters have lost the point of the OP's original question, which had to do with the apparent paradox of relative lengths (rod and hole) being different in the two frames of reference. Let's do away with gravity, and acceleration, and rotation and bending. How about the surface with the hole being in motion, on a path 90 degrees to the path of the rod? When the rod reaches the hole, can it pass through cleanly or not? The surface is arbitrarily thin and moving plenty fast; the question is how to resolve the conflicting answers in the two frames of reference: Riding along on the surface, one should see a short rod that easily passes through the hole, but riding on the rod, one would see an approaching hole that's too small to get through.
(I think the answer turns out to be that the rod ends up coming in at an angle, which - within limits - allows it to pass through even if it's "too long".)

 

[PAllen]

I think most commenters have lost the point of the OP's original question, which had to do with the apparent paradox of relative lengths (rod and hole) being different in the two frames of reference. Let's do away with gravity, and acceleration, and rotation and bending. How about the surface with the hole being in motion, on a path 90 degrees to the path of the rod? When the rod reaches the hole, can it pass through cleanly or not? The surface is arbitrarily thin and moving plenty fast; the question is how to resolve the conflicting answers in the two frames of reference: Riding along on the surface, one should see a short rod that easily passes through the hole, but riding on the rod, one would see an approaching hole that's too small to get through.
(I think the answer turns out to be that the rod ends up coming in at an angle, which - within limits - allows it to pass through even if it's "too long".)

That’s a related but different problem. It is worth discussing. To be more precise, consider two problem statements:

1) In a frame with a hole at rest, and a rod moving inertially, oriented parallel to the hole surface in this frame, but with rod trajectory slight skewed from horizontal, a rod with rest length longer than the hole should fit through due to length contraction.

2) In a frame with a rod at rest, and a hole in a surface parallel to the rod in this frame, with said surface and hole moving slightly skewed from horizontal (but surface remaining parallel to rod per this frame), the hole should hit rod in its central region, even though its rest diameter is larger than the rod.

The answer is that these are simply different scenarios, each of which will occur as stated, but each will look different in the rest frame of the other entity.

What distinguishes the rod moving exactly parallel to the hole surface, this being true in both frames, then starting to accelerate through the hole, is that the constraint that the rod not shear, gives a physical preference to the rod initial rest frame. All the rod elements share a common initial simultaneity, and, while starting to change motion, each must not separate from or approach its neighbors.

 

[Orodruin]

Let's do away with gravity, and acceleration, and rotation and bending.

You can get away from gravity (never a good idea to look at in SR), acceleration, and bending (the bending was an artefact of acceleration). What you cannot get rid of is rotation. Your scenario is similar to the one I described in my first post. If the plane is horizontal and moving up in ”hole” frame, it will not be horizontal in the rod frame.

 

[pervect]

I think most commenters have lost the point of the OP's original question, which had to do with the apparent paradox of relative lengths (rod and hole) being different in the two frames of reference. Let's do away with gravity, and acceleration, and rotation and bending. How about the surface with the hole being in motion, on a path 90 degrees to the path of the rod?

In what frame if the path of the hole at 90 degrees to the path of the rod?

If we let $\vec{v}_h$ be the 3-velocity of the hole, and $\vec{v}_r$ be the three velocity of the rod, then the condition that the two velocities be at 90 degrees is that $\vec{v}_h \cdot \vec{v}_r = 0$

In Newtonian mechanics $\vec{v}_h \cdot \vec{v}_r$ is an invariant. So if $\vec{v}_h \cdot \vec{v}_r = 0$ in one frame, it's zero in all frames. But this is not true in special relativity. The relationship in SR is that $u_h \cdot u_r$ is invaraint, where $u_h$ and $u_r$ are 4-vectors, not 3-vectors.

For those not familiar with 4-vector notation, we can re-write the invariant 4-vector dot product in 3-vector notation as follows, using my favorite sign convention:

$$u_h \cdot u_r = \vec{v}_h \cdot \vec{v}_r -\frac{c}{\sqrt{1- \frac{\vec{v}_h \cdot \vec{v}_h}{c^2}}} \frac{c}{\sqrt{1- \frac{\vec{v}_r \cdot \vec{v}_r}{c^2}}} $$

As a consequence, we can see that the condition that $\vec{v}_h$ and $\vec{v}_r$ be orthogonal, i.e. that $\vec{v}_h \cdot \vec{v}_r = 0$, depends on one's choice of frame.

 

[Lord Crc]

So how about a 15m rod on an inertial tracectory passing two optical through-beam sensors spaced say 10m away from each other, tripping the sensors. The sensors are stationary with respect to each other and they are connected with equally long cables to an ideal AND gate, which in turn powers a signal light.

If the rod moves slowly the signal light will turn on.

What if the rod moves fast enough to appear 5m in the rest frame of the sensors, will the signal light turn on?

 

[Sorcerer]

So how about a 15m rod on an inertial tracectory passing two optical through-beam sensors spaced say 10m away from each other, tripping the sensors.

In which rest frame? The rest frame of the sensors? That's what I think you mean here. Is that right?

The sensors are stationary with respect to each other and they are connected with equally long cables to an ideal AND gate, which in turn powers a signal light.

If the rod moves slowly the signal light will turn on.

What if the rod moves fast enough to appear 5m in the rest frame of the sensors, will the signal light turn on?

All I can say is if it the light turns on in one frame, it turns on in the other.

There's also this to consider: if the rod appears 5m in the rest frame of the sensors, how far apart are the sensors in the rest frame of the rod?
Also, what do you mean by "if the rod moves slowly the light will turn on?" Do you mean if the rod moves at a non-relativistic speed with respect to the sensors? If that's the case, then why would it appear to be 5m with respect to the sensors? There wouldn't be any noticeable length contraction if the rod is moving slowly with respect to the sensors.

 

[Herman Trivilino]

The sensors are stationary with respect to each other and they are connected with equally long cables to an ideal AND gate, which in turn powers a signal light.

This means they turn on the signal light when the sensors send signals simultaneously in the rest frame of the sensors. The signal light can still turn on if the signal-sending is not simultaneous in other frames.

Simultaneity is not absolute, but the turning on of the signal light is absolute. If it turns on in one frame it turns on in all frames.

 

[pervect]

So how about a 15m rod on an inertial tracectory passing two optical through-beam sensors spaced say 10m away from each other, tripping the sensors. The sensors are stationary with respect to each other and they are connected with equally long cables to an ideal AND gate, which in turn powers a signal light.

If the rod moves slowly the signal light will turn on.

What if the rod moves fast enough to appear 5m in the rest frame of the sensors, will the signal light turn on?

Is this different (and if so how), from the well-known Barn and Pole paradox, about which a lot has been written, some of it in this thread. See also for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html

Making the "doors" of the barn electronic doesn't make any difference to the problem statement, except for making it a bit more practical.

The resolution is the usual one - the "doors" to the barn are both closed at the same time in the barn frame, but not in the pole frame, due to the relativity of simultaneity.
See the references and other threads for more details.

 

[Ibix]

What if the rod moves fast enough to appear 5m in the rest frame of the sensors, will the signal light turn on?

This appears, as others have said, to be a restatement of the rod and barn paradox. The only addition is transmission lines and an AND gate. As others have noted, if it turns on in one frame it turns on in all. The reason is that the transmission delays are different in a frame where the equipment is moving. In general, if the leading edge of the "sensor activated" pulses moves at speed ±v in the equipment rest frame then it does $(u\mp v)/(1\mp uv/c^2)$ in the moving frame. This will conspire with the changed travel distances to light up the indicator.

 

[Sorcerer]

This is a bit of an off topic rant, but I can’t help it...

Special relativity is entirely self consistent. Brilliant minds have been devising new thought experiment tests for it for more than 100 years, and yet it remains self-consistent. Mathematicians have formulated it into a subset of mathematical structures (I believe the term is “manifold”) that show exactly how it is logically self-consistent.

I understand people get confused. I get confused about some aspects of it, too. But what grinds my gears is when someone thinks they’ve discovered a smoking gun that shows it is not self-consistent, as if the mathematicians have not already generalized it and derived every theorem, lemma, and variant imaginable ad naseum.

The only time special relativity is “inconsistent” is the time when the person proposing the thought experiment starts off with an assumption that already violates the well established laws and postulates that define special relativity. In other words, it happens when they create straw man arguments and proceed to tear them down, completely whiffing on their strange desire to prove that special relativity is not self-consistent within its scope of applicability.

As I said twice already, the mathematicians have gotten ahold of it, nearly from the beginning. You can rest assured that they have covered every conceivable contingency and have written thousands of tedious proofs from the most specialized aspects of SR to the most general./end rant.

 

[James Demers]

This is a bit of an off topic rant, but I can’t help it...

what grinds my gears is when someone thinks they’ve discovered a smoking gun that shows it is not self-consistent

/end rant.

I don't see these thought experiments as attempts to disprove Special Relativity - they're logical puzzles that challenge our ability to think in S.R. terms.

 

[Ibix]

But what grinds my gears is when someone thinks they’ve discovered a smoking gun that shows it is not self-consistent

I don't see these thought experiments as attempts to disprove Special Relativity - they're logical puzzles that challenge our ability to think in S.R. terms.

You see both. Generally I think people have just missed something. For example, SR is rather less generous in letting you hand-wave over details like the transformation of signal speeds than Newton, which is what I think has caught Lord Crc here.

When someone simply isn't accepting that there is an oversight in their scenario, though, that's what the "report" button is for.

 

[Orodruin]

You see both. Generally I think people have just missed something. For example, SR is rather less generous in letting you hand-wave over details like the transformation of signal speeds than Newton, which is what I think has caught Lord Crc here.

The biggest hurdle is typically the relativity of simultaneity. For some reason people seem perfectly happy to think about time dilation and length contraction and try to find ”inconsistencies” in those while ignoring the issue of simultaneity, which is crucial for understanding what is going on.

 

[Sorcerer]

The biggest hurdle is typically the relativity of simultaneity. For some reason people seem perfectly happy to think about time dilation and length contraction and try to find ”inconsistencies” in those while ignoring the issue of simultaneity, which is crucial for understanding what is going on.

Every. Time. Or so it seems when it comes to crackpots I’ve encountered, that is.

But as was said above, not everyone is doing that. A lot of people just want to understand. I just had a moment lol. Too many cranks in one week.