About man falling into grate paradox

[Ookke]

http://en.wikipedia.org/wiki/Ladder_paradox#Man_falling_into_grate_variation

In the resolution it says: "It should be stressed that, since this bending occurs in the rod's rest frame, it is a true physical distortion of the rod which will cause stresses to occur in the rod."

I wonder if this is generally accepted. As we assume uniform acceleration in outer frame (the downwards pushing force acts on all parts of rod simultaneously), the different parts of rod are never moving respect to each other, so no stress should occur. Also the stress on rod sounds like an absolute event that all frames should agree. The stress could even break the rod, which certainly all frames must agree whether or not it happens.

It seems that in rod's own frame there is some bending caused by relativity of simultaneity (different parts of rod seem to get acceleration at different times), but I think this kind of apparent bending doesn't need to have physical significance i.e. stress or breaking.

My guess would be that there is no physical bending in either frame. Any ideas? Thanks.

 

[A.T.]

As we assume uniform acceleration in outer frame (the downwards pushing force acts on all parts of rod simultaneously), the different parts of rod are never moving respect to each other, so no stress should occur.

See Bell's[/PLAIN] spaceship paradox. In the initial rest frame the string is not stretched, yet it breaks. Similary here, in the the grate's rest frame the rod is not bent, yet it breaks.

 

[pervect]

Rigid rods don't actually exist.

To model the behavior of an actual rod, you need a different model. A reasonably good mental image is a wet noodle.

You can then see that the wet noodle sags, and the tip catches the edge of the grate. Which is about as good a picture you'll get without throwing a lot of math at the problem.

 

[Fredrik]

I wonder if this is generally accepted. As we assume uniform acceleration in outer frame (the downwards pushing force acts on all parts of rod simultaneously), the different parts of rod are never moving respect to each other, so no stress should occur. Also the stress on rod sounds like an absolute event that all frames should agree. The stress could even break the rod, which certainly all frames must agree whether or not it happens.

It seems that in rod's own frame there is some bending caused by relativity of simultaneity (different parts of rod seem to get acceleration at different times), but I think this kind of apparent bending doesn't need to have physical significance i.e. stress or breaking.

My guess would be that there is no physical bending in either frame.

If you push all the parts of the rod at the same time in one of the frames, you will be pushing them at different times in the other frame. You seem to have realized that already. Why would you consider pushing different parts of the rod at different times in the rod's rest frame to be only "apparent" bending? The rod is already in an equilibrium shape in the sense that if you deform it slightly, internal forces will restore the shape, and you're talking about radically changing the shape of the rod in its own rest frame, into a non-equilibrium shape. This seems to me like an obvious case of "actual" bending.

 

[Ookke]

See Bell's[/PLAIN] spaceship paradox. In the initial rest frame the string is not stretched, yet it breaks. Similary here, in the the grate's rest frame the rod is not bent, yet it breaks.

I'm somewhat familiar with Bell's paradox and accept the resolution that the string breaks in both frames. The only difference is in the physical interpretation of what causes the string to break: in rockets' initial rest frame, the string breaks because of length contraction of the rope, but in their final rest frame, the string breaks because rockets accelerate at different times and the distance between them increases.

In this rod paradox, I fail to see what could be the physical cause for rod breaking in grate's frame.

Why would you consider pushing different parts of the rod at different times in the rod's rest frame to be only "apparent" bending? The rod is already in an equilibrium shape in the sense that if you deform it slightly, internal forces will restore the shape, and you're talking about radically changing the shape of the rod in its own rest frame, into a non-equilibrium shape. This seems to me like an obvious case of "actual" bending.

I was thinking something like this: Let's name the rod ends A and B, A gets pushed first in rod's own frame. Let's have an observer at B. If the push shock wave (or whatever, the information anyway) reaches B before B gets pushed, then I would consider it "actual" bending. But if B gets pushed before the information from A reaches it, I'm not convinced that it's actual bending in that case, hence "apparent".

 

[PAllen]

I was thinking something like this: Let's name the rod ends A and B, A gets pushed first in rod's own frame. Let's have an observer at B. If the push shock wave (or whatever, the information anyway) reaches B before B gets pushed, then I would consider it "actual" bending. But if B gets pushed before the information from A reaches it, I'm not convinced that it's actual bending in that case, hence "apparent".

Interesting definition??! So if I have a very long rod, and bend one end into an elbow, it is not really bent until information reaches the other end (say, a light year away)? Think about this...

 

[jartsa]

Maybe the rod thinks it is turned, but not bent?

If we suddenly remove the floor under a moving rod, is it really possible that the rod does not fall normally?
By normally I mean all potential energy turning into kinetic energy. (No potential energy used for bending or other such thing)

 

[Nugatory]

If we suddenly remove the floor under a moving rod, is it really possible that the rod does not fall normally?

You've just tripped over the relativity of simultaneity again (it's easy to do).
If we suddenly remove all of the floor underneath the rod in one frame... Its not so sudden in another.

 

[Fredrik]

Maybe the rod thinks it is turned, but not bent?

If we suddenly remove the floor under a moving rod, is it really possible that the rod does not fall normally?

There's no gravity in SR, so we would have to simulate the effects of gravity by having the rod slide on a floor that's doing constant proper acceleration in the direction perpendicular to the line of motion. This acceleration will make the problem much more complicated.

If we instead imagine giving every part of the rod a push at the same time, then...see nugatory's post above, and my previous post.

 

[jartsa]

Well I think the non-simultaneous pushes in the rod's frame just make the rod more vertical, in its own frame, then the rod fits into the grate, in its own frame.

 

[PAllen]

Well I think the non-simultaneous pushes in the rod's frame just make the rod more vertical, in its own frame, then the rod fits into the grate, in its own frame.

If there is no force on one end of the rod, and there is on the other, and the transmission of forces that normally make the rod appear rigid are limited to the speed of sound (much less than light), what you propose is logically impossible.

Frederik's wet noodle is a really good image of how so called rigid bodies behave in situations where relativistic effects are significant.

 

[jartsa]

If there is no force on one end of the rod, and there is on the other, and the transmission of forces that normally make the rod appear rigid are limited to the speed of sound (much less than light), what you propose is logically impossible.

Frederik's wet noodle is a really good image of how so called rigid bodies behave in situations where relativistic effects are significant.

Well then I'll say the rod does not feel any bending force, although it sees itself bending.

It may happen, by chance, that when a sound wave reaches a point on the rod, at that same time the pushing force starts to push the point, and no shear stress appears at that point.

If the speed of sound is slow then that is approximately what happens, isn't it?

 

[PAllen]

Well then I'll say the rod does not feel any bending force, although it sees itself bending.

It may happen, by chance, that when a sound wave reaches a point on the rod, at that same time the pushing force starts to push the point, and no shear stress appears at that point.

If the speed of sound is slow then that is approximately what happens, isn't it?

How do you have bending without shear stress? You know sound in a solid is propagating stress or compression. Do you think you can just make up whatever physics you want, with no logic or verifiability?

What is true is that in the rest frame of the rod (assuming the hole is moving to the left in this frame) the left most point of force application moves many time faster than the speed of sound. Thus stress is independently produced along the rod from right to left, followed by propagation of stress.

Let's get an idea of numbers here. Suppose the grate has a 1 meter opening. Suppose in the grate frame, a 2 meter rest length rod is moving such that it is 1/2 meter in the grate frame. Suppose a piston above the hole in the grate frame is timed to push the rod 1 millimeter down as it travels across the hole. Suppose the rod is 1 kg. Then, ballpark, the force applied from right to left in the rod frame (with force application boundary propagating to left at nearly c) is 10^15 Newtons. You think sudden application at one point of 10^15 Newtons produces no stress??

[edit: In the above figures, I should really talk about force/meter to describe the moving boundary of force application: 10^15 Newtons/meter. This doesn't change the substance of the argument.]

 

[jartsa]

How do you have bending without shear stress? You know sound in a solid is propagating stress or compression. Do you think you can just make up whatever physics you want, with no logic or verifiability?

Well now I'm again sceptical if there's any visual bending. Because if many antennas are sticking out from the rod and a still standing observer says some antennas do not touch, then the rod must agree.

Let's say forces push the upper side of the rod, non-simulteneously in the rod's frame. First thing that comes to mind is that a stretching force is felt on the upper side of the rod.

But the rod says the direction of the force is down and backwards. So there is a compression component too. Probably compression and streching cancel each other.

A still standing observer says the direction of force is downwards, but the acceleration of the rod is downwards and backwards. (The observer may consider the bacwards component of the acceleration an increasing time dilation of the forwards motion)

 

[PAllen]

Well now I'm again sceptical if there's any visual bending. Because if many antennas are sticking out from the rod and a still standing observer says some antennas do not touch, then the rod must agree.

Let's say forces push the upper side of the rod, non-simulteneously in the rod's frame. First thing that comes to mind is that a stretching force is felt on the upper side of the rod.

But the rod says the direction of the force is down and backwards. So there is a compression component too. Probably compression and streching cancel each other.

A still standing observer says the direction of force is downwards, but the acceleration of the rod is downwards and backwards. (The observer may consider the bacwards component of the acceleration an increasing time dilation of the forwards motion)

Your position is ridiculous. In the rod's frame, part of the rod has moved under force while part is not even feeling any force (certainly not even being reached by shear stress). Only in fantasy land is this not bending.

It is not really possible to discuss statements reject fact and logic by 'I say it is so'.

 

[Nugatory]

Let's say forces push the upper side of the rod, non-simulteneously in the rod's frame. First thing that comes to mind is that a stretching force is felt on the upper side of the rod.

But the rod says the direction of the force is down and backwards. So there is a compression component too. Probably compression and streching cancel each other.

What do you would think would happen if the rod weren't moving at all, we've just pushed it gently forward so that it's sticking out over the edge of the opening by a few centimeters and is just sitting there?

Before you answer, consider Pervect's wet noodle in #3 and PAllen's 10^15 Newtons in post #13.

The force pulling the end of the rod down into the hole is enormous. It has to be; we've been assuming that it is strong enough to change the path of the fast-moving length-contracted rod enough that it would fall into the hole.

 

[jartsa]

What do you would think would happen if the rod weren't moving at all, we've just pushed it gently forward so that it's sticking out over the edge of the opening by a few centimeters and is just sitting there?

Before you answer, consider Pervect's wet noodle in #3 and PAllen's 10^15 Newtons in post #13.

The force pulling the end of the rod down into the hole is enormous. It has to be; we've been assuming that it is strong enough to change the path of the fast-moving length-contracted rod enough that it would fall into the hole.

A steel rod is slowly pushed over an edge, and this happens on an extremely strong gravity field?

Let's see ... oh yes the edge cuts the rod like scissors. I know because once my job was cutting steel rods with pneumatic scissors.

 

[Fredrik]

A steel rod is slowly pushed over an edge, and this happens on an extremely strong gravity field?

Let's see ... oh yes the edge cuts the rod like scissors. I know because once my job was cutting steel rods with pneumatic scissors.

That force would probably completely annihilate the rod before it even reaches the hole.

 

[jartsa]

Your position is ridiculous. In the rod's frame, part of the rod has moved under force while part is not even feeling any force (certainly not even being reached by shear stress). Only in fantasy land is this not bending.

Maybe, but when the rod is in the hole, and the pushing has stopped, then every part of the rod has received the same impulse, so the rod says: "I'm not bent, but I fit in this hole"

"My front parts received the impulses first, that's why my front is lower in he hole than my rear, which helps me to fit in the hole", the rod continues.

 

[PAllen]

Maybe, but when the rod is in the hole, and the pushing has stopped, then every part of the rod has received the same impulse, so the rod says: "I'm not bent, but I fit in this hole"

"My front parts received the impulses first, that's why my front is lower in he hole than my rear, which helps me to fit in the hole", the rod continues.

Wrong. In the rest frame of the rod, the sequence of events is: force bends front rod into moving hole; back of hole smashes the front of the rod; all of this before most of the rod has experienced any force or stress. The process of bending into moving hole and smashing back proceeds until finally the whole rod has been bent and smashed into the moving hole. In the rod's rest frame there is never a time when it fits nicely in the hole. Also, it is never oriented nearly vertical.

You can't just make up whatever you feel like, which is what you insist on doing.

 

[Ookke]

So if I have a very long rod, and bend one end into an elbow, it is not really bent until information reaches the other end (say, a light year away)?

In a sense, yes. Because how do you know that the rod is bent, if the information has not reached you?

Let's say that we ride along the rod at the other end, the one that is pushed last. Because this is just a thought experiment, we may assume that the bending effect i.e. shock wave transfers at speed of light along the rod.

If the effect arrived to our position before we are pushed downwards, the rod would be undisputably bent: we would have direct observation that some parts of the rod are lower than our part. That's pretty much what bending is all about.

However, since all parts of the rod are pushed simultaneously in grate's frame, we can conclude that our part of the rod must get pushed before the effect along the rod arrives. This is because if events A and B are simultaneous in any frame (as pushing of different parts of rod are in grate's frame), they cannot be causally connected or notify each other beforehand.

So we get pushed downwards without any warning beforehand. And when the effect arrives along the rod, we note that the other parts of the rod are already at our level i.e. no bending, stress or breaking occur at our position at the other end of the rod. The same applies at any point of the rod.

 

[PAllen]

In a sense, yes. Because how do you know that the rod is bent, if the information has not reached you?

It is bent (or broken) because someone looking at that end of the rod can immediately detect it is bent (or broken).

Let's say that we ride along the rod at the other end, the one that is pushed last. Because this is just a thought experiment, we may assume that the bending effect i.e. shock wave transfers at speed of light along the rod.

It is valid to observe that what is normally referred to as tilting a rod leaning over an edge by pushing on one end is really bending it a little, that bend propagating at the speed of sound to the other causing the other end to raise; repeat this microscopic process until you have tilted rod. If there is significant movement of one part before the rest has moved you have breaking or shattering. That is really what would happen here as Frederik noted a few posts back. You would have the rod being pulverized all at once in the grate frame, while being pulverized and pushed sequentially in the other. Specifically, in the case of a grate (not a hole with walls), in the rod rest frame, you would have the rod pulverized and pushed below the small grate hole, a little at a time. During this process you would have pulverized rod below the grate, unaffected rod to the left of the grate hole, proceeding until the pulverized remainder of the rod is all below and right of the small grate hole. There is never a time when the rod is tilted or fits into the hole in the rod frame.

If the effect arrived to our position before we are pushed downwards, the rod would be undisputably bent: we would have direct observation that some parts of the rod are lower than our part. That's pretty much what bending is all about.

However, since all parts of the rod are pushed simultaneously in grate's frame, we can conclude that our part of the rod must get pushed before the effect along the rod arrives. This is because if events A and B are simultaneous in any frame (as pushing of different parts of rod are in grate's frame), they cannot be causally connected or notify each other beforehand.

So we get pushed downwards without any warning beforehand. And when the effect arrives along the rod, we note that the other parts of the rod are already at our level i.e. no bending, stress or breaking occur at our position at the other end of the rod. The same applies at any point of the rod.

All of the rest of this is counter-factual nonsense.

 

[Nugatory]

A steel rod is slowly pushed over an edge, and this happens on an extremely strong gravity field?

Let's see ... oh yes the edge cuts the rod like scissors. I know because once my job was cutting steel rods with pneumatic scissors.

That is indeed one conceivable outcome - the rod totally disintegrates as whatever portion of it moves over the edge is sheared away, sort of like slowly pushing a stick into a wood chipper.

Another possible outcome is that you push the rod forward quickly enough that its elastic limit is not exceeded as the front end of the rod is pulled into the hole, and then it bends (I'm imagining a giant English wheel or sheet metal brake - the analogy isn't quite right but if you've used metal fabrication tools you'll know what I mean) as it goes over the edge, ends up in a sort of 'C' shape.

But the important point here is that the forces involved are so great that there is no way that the rod will respond like a rigid object, rotating when force is applied to one end or the other. The applied force will move the end to which it is applied, and the other end will not move, and that will bend or break the rod.

 

[pervect]

But the important point here is that the forces involved are so great that there is no way that the rod will respond like a rigid object, rotating when force is applied to one end or the other.

It's not really so much a question of the forces being so great. It's more of a case of rigid bodies not existing in relativity. It's well known that rigid bodies and relativity don't mix. I'm not quite sure what fascination the topic or rigid bodies has has, except insofar as it probably demonstrates a general failure to understand the idea that rigid bodies don't make sense in relativity in the usual sense, one needs to go to notions such as Born rigidity.

Given that the speed of sound is glacial compared to the speed of light in any known material, the best intuitive approximation to body is to assume it's totally non-rigid, possessing the strength of a ball of water, or (as per my previous post) a limp, wet, noodle. It's usually easy to figure out what happens with this approach, and it'll pretty accurately describe the actual physics.

The reason it accurately describes the actual physics is that the contributions from the forces that usually hold a body together are basically totally negligible. The amount of stored energy in deformations of the material is much less than mass*c^2, and the energy of the body in motion is a good fraction of mass*c^2 (if it wasn't , the motion wouldn't be relativistic). So you can pretty much ignore the "restoring forces", their contribution is not significant.

If you do want to take into account restoring forces anyway, even though they are very slight, you can't assume they act instantaneously for reasons which should be obvious, but apparently aren't to a large number of posters :-(.

 

[PAllen]

It is also worth noting that all deformations (compression, stretching, shear) boil down to relative motion between elements of a body. From that point of view, the fact that one element of the rod is accelerating down at 10^15 m/sec^2 orthogonally to the sequence of elements to its left means you have incomprehensibly extreme shear deformation. That this happens can be invariantly expressed (e.g. expansion and vorticity tensors); using these one could compute in the grate frame, that despite appearance of contracted rod seeming to smoothly and horizontally drop through the hole, there is still extreme shear deformation occurring. This then becomes a variation of the same issues as Bell Spaceship 'paradox'.

 

[Nugatory]

It's not really so much a question of the forces being so great. It's more of a case of rigid bodies not existing in relativity.

Those are two different ways of saying the same thing, I think. Saying that rigid bodies don't exist is equivalent to saying that there is no body rigid enough to withstand the stresses involved; all objects become noodle-like. We can say that this is because the objects are weak or the forces are strong, but it comes down to the same thing - the object deforms or fails under stress.

In any case, we agree about keeping idealized rigid bodies out of the thought experiments, I think.

 

[PAllen]

Looking back at the wiki link in the OP, I see that the OP was led to think stresses exist in one frame but not in another (I think). The wiki does not say this, but its wording is subject to this misunderstanding. A fundamental point is that the stresses are present in in all frames; they are an invariant feature of the rod's motion. It's just much harder to 'see' them in the grate frame. The simplest approach, I think, is what I suggested: you have to compute the relative motion adjacent elements of the rod. This is invariant and thus computed in any frame, and for the set up I described in #13, will lead to extreme shear deformation.

 

[jartsa]

Wrong. In the rest frame of the rod, the sequence of events is: force bends front rod into moving hole; back of hole smashes the front of the rod; all of this before most of the rod has experienced any force or stress. The process of bending into moving hole and smashing back proceeds until finally the whole rod has been bent and smashed into the moving hole. In the rod's rest frame there is never a time when it fits nicely in the hole. Also, it is never oriented nearly vertical.

That sounds right.

Now let's consider a 2 meter deep hole into the bottom of which the front end of a 1 meter long rod collides, when pushed by a force large enough.

If velocities are close to c, then it's impossible for the front of the rod to reach the bottom of the hole before the rear of the rod has reached the edge of the hole.

So in this scenario a nearly vertical rod will be completely inside the hole.(it is impossible for the rod to move at nearly c to the right and at the same time at nearly c downwards, but it's possible to make the rod to move at nearly c downwards by pushing with a force, as we have been discussing, so the velocity to the right must decrease when the force is applied, so there's no such problem as rod hitting the side wall instead of the bottom.)

 

[PAllen]

That sounds right.

Now let's consider a 2 meter deep hole into the bottom of which the front end of a 1 meter long rod collides, when pushed by a force large enough.

If velocities are close to c, then it's impossible for the front of the rod to reach the bottom of the hole before the rear of the rod has reached the edge of the hole.

So in this scenario a nearly vertical rod will be completely inside the hole.


(it is impossible for the rod to move at nearly c to the right and at the same time at nearly c downwards, but it's possible to make the rod to move at nearly c downwards by pushing with a force, as we have been discussing, so the velocity to the right must decrease when the force is applied, so there's no such problem as rod hitting the side wall instead of the bottom.)

I specifically specified pushing the rod down only one millimieter so as not to interfere with the ability to have 2 meter rod length contracted to fit readily inside a 1 meter hole. Note how extreme this still is: you need to get from 0 to .001c in well under 10^-8 seconds. Zero to 676,000 mph (> 1 million kph) in under 10^-8 seconds. This is necessary to move down just one millimeter in the required time.

If you are going to talk about a rod with minimal length contraction entering a hole nearly vertically, that is a totally separate problem that I don't have an interest in pursuing (it is just boring non-relativistic mechanics). It seems to me you are completely changing the problem because you cannot admit that you are wrong to try to deny stress and deformation in the original problem: a two meter rod length contracted to 1/2 meter, pushed a little into a one meter hole, remaining horizontal in the grate frame. And the surprising but true fact about this is that the rod is undergoes severe shear deformation, and this fact is invariant though not readily visible in the grate frame.

 

[jartsa]

I specifically specified pushing the rod down only one millimieter so as not to interfere with the ability to have 2 meter rod length contracted to fit readily inside a 1 meter hole. Note how extreme this still is: you need to get from 0 to .001c in well under 10^-8 seconds. Zero to 676,000 mph (> 1 million kph) in under 10^-8 seconds. This is necessary to move down just one millimeter in the required time.

If you are going to talk about a rod with minimal length contraction entering a hole nearly vertically, that is a totally separate problem that I don't have an interest in pursuing (it is just boring non-relativistic mechanics). It seems to me you are completely changing the problem because you cannot admit that you are wrong to try to deny stress and deformation in the original problem: a two meter rod length contracted to 1/2 meter, pushed a little into a one meter hole, remaining horizontal in the grate frame. And the surprising but true fact about this is that the rod is undergoes severe shear deformation, and this fact is invariant though not readily visible in the grate frame.

Well I had a problem with the energy of the supposed stress, where does it come from, particularly in the still standing observers frame.

But maybe that problem can be solved.

Let's see ... two rapidly spinning discs are pushed so that they slam together and fuse together, the result is a slowly spinning object. The spinning energy of the two disc system turned into stress energy of the discs, which stress energy turned into heat when the discs collided.

Is the above correct?

The discs are identical, they have identical angular momentum, and they collide with the flat parts first. And all velocities are relativistic.

 

[Nugatory]

Well I had a problem with the energy of the supposed stress, where does it come from, particularly in the still standing observers frame.

In all frames, there will be some period of time when the rod is sticking out over the edge, so that one end is supported by the ground underneath and the other end is not. This is just an ordinary cantilever, no different in principle from a balcony on the side of a building. The rod doesn't even have to moving relative to the hole for there to be stress in the rod: gravity is pulling the projecting end of the rod down into the hole, and is being resisted by stress forces within the rod.

 

[PAllen]

Well I had a problem with the energy of the supposed stress, where does it come from, particularly in the still standing observers frame.

I don't know what you mean by the still standing observer's frame.

The energy for the stress comes from work done on the rod - a great force (10^15 Newtons) does work on it, moving it distance (1 millimiter). The energy (work) applied to the rod, stressing it, is on the order of a terajoule (about 1% of the energy released by the Nagasiki atomic bomb). In rod frame (the only relevant frame for analyzing stress), this forice is not applied simultaneously, but from right to left (assuming the rod is moving left to right).

Following earlier suggestions, I proposed a piston pushing the rod down the hole. This has many advantages over gravity. Dealing with gravity (from a source that would have to specified) on an object moving near c relative to it, with gravity needing to be extremely strong field, would be a complex problem in GR. Also, the piston ensures that the rod is seen to remain horizontal in the grate frame.

But maybe that problem can be solved.

Let's see ... two rapidly spinning discs are pushed so that they slam together and fuse together, the result is a slowly spinning object. The spinning energy of the two disc system turned into stress energy of the discs, which stress energy turned into heat when the discs collided.

Is the above correct?

The discs are identical, they have identical angular momentum, and they collide with the flat parts first. And all velocities are relativistic.

The above is correct, but not very relevant the rod pushed through a grate hole.

 

[jartsa]

In all frames, there will be some period of time when the rod is sticking out over the edge, so that one end is supported by the ground underneath and the other end is not. This is just an ordinary cantilever, no different in principle from a balcony on the side of a building. The rod doesn't even have to moving relative to the hole for there to be stress in the rod: gravity is pulling the projecting end of the rod down into the hole, and is being resisted by stress forces within the rod.

No the ground never pushes the rod. In the man falling into grate paradox "It is assumed that the rod is entirely over the grate in the grate frame of reference before the downward acceleration begins simultaneously and equally applied to each point in the rod"

Quoted part is from the wikipedia.

 

[PAllen]

No the ground never pushes the rod. In the man falling into grate paradox "It is assumed that the rod is entirely over the grate in the grate frame of reference before the downward acceleration begins simultaneously and equally applied to each point in the rod"

Quoted part is from the wikipedia.

Of course, the key point being simultaneously in the grate frame = not simultaneously in the rod frame. Thus, for a given element of the rod, the element to its right (for example) accelerates down and to the left before any force or acceleration is applied to the given element.

 

[Nugatory]

No the ground never pushes the rod. In the man falling into grate paradox "It is assumed that the rod is entirely over the grate in the grate frame of reference before the downward acceleration begins simultaneously and equally applied to each point in the rod"

Quoted part is from the wikipedia.

Ah - right - yes, as I've seen the paradox described in other sources that part isn't always specified.

It's still something of a red herring though, because if you choose to use the grate frame you're just choosing to use a frame in which the problem is harder to analyze, hence leading to the apparent paradox. If you're going to insist on starting with the grate frame, you might try drawing the world lines of three points (each end and the middle of the rod) in such a way that the acceleration will be simultaneous in the grate frame. You'll see that these are also the world lines of those three points if the rod is being bent by the progressive application of force from one end to the other it its rest frame.