1. Apr 11, 2013

### Ookke

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.

2. Apr 11, 2013

### A.T.

See Bell's[/PLAIN] [Broken] 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.

Last edited by a moderator: May 6, 2017
3. Apr 12, 2013

### pervect

Staff Emeritus
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.

4. Apr 12, 2013

### Fredrik

Staff Emeritus
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.

5. Apr 12, 2013

### Ookke

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.

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".

Last edited by a moderator: May 6, 2017
6. Apr 12, 2013

### PAllen

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...

7. Apr 12, 2013

### 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)

8. Apr 12, 2013

### Staff: Mentor

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.

9. Apr 12, 2013

### Fredrik

Staff Emeritus
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.

10. Apr 12, 2013

### 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.

11. Apr 12, 2013

### PAllen

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.

12. Apr 12, 2013

### jartsa

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?

Last edited: Apr 12, 2013
13. Apr 13, 2013

### PAllen

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.]

Last edited: Apr 13, 2013
14. Apr 13, 2013

### jartsa

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)

15. Apr 13, 2013

### PAllen

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'.

16. Apr 14, 2013

### Staff: Mentor

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.

17. Apr 14, 2013

### jartsa

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.

18. Apr 14, 2013

### Fredrik

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

19. Apr 14, 2013

### jartsa

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.

20. Apr 14, 2013

### PAllen

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.