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

[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.