What does a rod passing through a hole look like in the inertial frame?

[Fredrik]

I see. I thought Phrak was interested in the acceleration. I'm a bit confused what the latest incarnation of this problem is.

There is no new incarnation of the problem. Just about everything he (Bible Thumper) said contradicts SR or something else he said.

 

[Bible Thumper]

Contraction? No, not at all. What I was concerned about there is a contradiction.

You said that the rod is parallel to the wall in the rod's rest frame. That's the same thing as saying that every part of it does everything simultaneously in the rod's frame. Immediately after saying that, you said that the front went in first in the rod's frame. That's a contradiction.

Lulz...
This was my first proposal. I posted it yesterday:

"Since an external force ("gentle push") acts on our rod; and must necessarily on the center of the rod if we wish to keep the rod parallel to the plane it's moving upon, then I propose the rod will fall directly into the hole, but front-end first, relative to the person on the rod.
The man beside the slit OTOH will see the event taking place with the rod always parallel to the slit.
The rod, according to the person riding it OTOH, will seem to enter nose-first, then the remaining of the rod in a parabolic fashion, until the end of the rod falls in.
The person on the rod will be scared at first, as the slit will look radically shortened--way too short for his rod--right before it enters the slit. But as his rod appears to bend into it, he is relieved.
The person standing beside the slit sees a very short rod (much shorter in appearance than when he took his measurements with the rod at rest) entering effortlessly into the slit without incident... :) "​

So as you can see, I made no such assertion that the rod appears parallel to the surface while the rod is moving; in fact, I made it clear that the rod sees its front-end enter into the slit first thing.
The man standing on the surface, however, sees the rod moving into the slit while remaining always parallel to thwe surface the man is standing on.
I placed, in bold, the observed reference frames.
Period.

 

[Bible Thumper]

There is no new incarnation of the problem. Just about everything he (Bible Thumper) said contradicts SR or something else he said.

Lulz (x2)...
Just because I read and thump the Bible and just because skeptics think the Bible (G-d's holy Word) is full of contradictions, doesn't mean everything I type or say has to contradict necessarily... :tongue:

 

[Phrak]

I was under the impression the only thing he cared about was what the slit looks like from the pole's perspective.

It doen't matter what I care about; the problem should stand on it's own, right?

As a reminder, I should say I didn't pose the problem as well as could be, and the improved problem statement is in post #7. Maybe Frederik could clean it up, so the misleading stuff, like "a gentle push" is gone.

 

[Fredrik]

So as you can see, I made no such assertion that the rod appears parallel to the surface while the rod is moving

I didn't say that you made that assertion in every post. I said that you made it in #60, and you did. What you said there implies that the rod is parallel to the wall in the rod's original rest frame, even if you don't understand that:

The rod is its own intertial reference frame. For this reason the gentle push affects the entire rod at the same time, relative to anyone on the rod.

But OK, we don't have to debate what you said for ever. We apparently agree that this thread is about a rod that remains parallel to the wall in the wall's frame.

 

[Fredrik]

Maybe Frederik could clean it up, so the misleading stuff, like "a gentle push" is gone.

What we're interested in a scenario where we push the rod through the hole while keeping it parallel to the wall in the wall's rest frame. As long as we all agree on that, and that "the contracted length of the rod" < "the rest length of the hole" < "the rest length of the rod", the other details don't really matter.

I prefer to leave the acceleration unspecified because it's interesting to see how the shape of the rod in the rod's original rest frame depends on the details of the velocity change.

The fact that the front of the rod goes through the hole before the rear of the rod in the rod's original rest frame is however independent of the details of the velocity change.

 

[DrGreg]

It doen't matter what I care about; the problem should stand on it's own, right?

As a reminder, I should say I didn't pose the problem as well as could be, and the improved problem statement is in post #7. Maybe Frederik could clean it up, so the misleading stuff, like "a gentle push" is gone.

This is my precise interpretion of the problem. Phrak, is my interpretation correct?

There is a flat two-dimensional surface S and a one-dimensional rod R.

Initially R is straight and parallel to S and R moves inertially relative to S in the direction of its own length.

Let I be the initial inertial rest frame of the rod. The rest-length of the rod, measured in I, is L.

There is a linear slot cut in the surface S, parallel to R, wide enough for R to fit through, but its rest-length (measured in the S-frame) is less than L. However its rest-length is greater than $L/\gamma$, where $\gamma$ is the relative Lorentz factor between S and I. Thus, due to length contraction of the rod relative to S, the rod is short enough, in the S-frame, to parallel-slide through the hole, but in the I-frame it is not.

The rod now moves towards the hole in such a way that it it remains straight and parallel to the hole according to the S-frame, and then passes through the hole.

Aside: The actual practicalities of how this is achieved are not relevant to the problem. You'd need to pre-arrange for a continuum of forces to be applied along the whole length of the rod, all pre-programmed to exert a specific force at a specific time to achieve the desired effect. It would be difficult to engineer in practice, but, as a thought experiment, no laws of physics need be broken to achieve this.​

The question is then how does this appear in the I-frame, in view of the fact that the rod is longer than the hole is this frame?

The answer which has already been thrashed out in this thread is that the rod does not remain straight in this frame: it curves and passes through the hole at an angle.

(Note that rod curvature --relative to I -- is inevitable during acceleration, but afterwards it will straighten and so may well be straight by the time the rod meets the hole. But it will slide through at an angle, head first, so it doesn't matter that it's too long to fit. See diagrams below.)

 

[Bible Thumper]

W00t! W00t! Finally, someone who acquiesces!
He said:

This is my precise interpretion of the problem.
There is a linear slot cut in the surface S, parallel to R, wide enough for R to fit through, but its rest-length (measured in the S-frame) is less than L.
However its rest-length is greater than $L/\gamma$, where $\gamma$ is the relative Lorentz factor between S and I. Thus, due to length contraction of the rod relative to S, the rod is short enough, in the S-frame, to parallel-slide through the hole, but in the I-frame it is not.
The rod now moves towards the hole in such a way that it it remains straight and parallel to the hole according to the S-frame, and then passes through the hole.
The answer which has already been thrashed out in this thread is that the rod does not remain straight in this frame: it curves and passes through the hole at an angle.
(Note that rod curvature --relative to I -- is inevitable during acceleration, but afterwards it will straighten and so may well be straight by the time the rod meets the hole. But it will slide through at an angle, head first, so it doesn't matter that it's too long to fit.

And I said:

Originally posted by Bible Thumper:
I propose the rod will fall directly into the hole, but front-end first, relative to the person on the rod.
The man beside the slit OTOH will see the event taking place with the rod always parallel to the slit.
The rod, according to the person riding it OTOH, will seem to enter nose-first, then the remaining of the rod in a parabolic fashion, until the end of the rod falls in.
The person on the rod will be scared at first, as the slit will look radically shortened--way too short for his rod--right before it enters the slit. But as his rod appears to bend into it, he is relieved.
The person standing beside the slit sees a very short rod (much shorter in appearance than when he took his measurements with the rod at rest) entering effortlessly into the slit without incident... :)

 

[Phrak]

Greg. How do you do that? I want to draw pictures. It would make things so much clearer.

I'd draw things a little different, with the forces applied to the rod only when the rod is directly over the hole in the hole's frame.

Your last frame has an error--sort of. Any portion of the rod sticking-out, north of the hole, is parallel to the surface. Recall that when there are no forces applied, the rod is just Lorentz contracted.

If you wished, you could argue that the gentle push has boosted the rod to have a perpendicular component of velocity as large as its velocity along the surface, which is sort of what your picture respresents, but it simpler if you consider that the added velocity in the perpendicular direction is v_p << c and v_p << v_l.

I didn't forsee a force adding as much y velocity as x velocity.

I have a drawing that makes the Bell Inequality visually obvious, but I don't know how to post it. Can you give me a hint on how to make and send pics?

 

[Dale]

Your last frame has an error--sort of. Any portion of the rod sticking-out, north of the hole, is parallel to the surface.

No, that is incorrect. There is only one bend in the worldline. If you wanted two bends in the worldline then you would have to push it twice.

 

[Phrak]

No, that is incorrect. There is only one bend in the worldline. If you wanted two bends in the worldline then you would have to push it twice.

I'll be ding-donged. I am sure you're right.

 

[phyti]

The drawing is not accurate. Try drawing it as a Minkowski space-time diagram, and you will see the difference. The scenario requires 2 dimensions, one for the direction of motion and a small offset for the separation of the rod and hole.

 

[Fredrik]

Greg knows what the spacetime diagram would look like. See his post earlier in this thread. A 2+1-dimensional spacetime diagram is precisely what you need to use to figure out what the shape of the rod is going to be at different times in different frames, and I'm pretty sure that's what he did. I don't see anything wrong with the drawing.

 

[DrGreg]

The drawing is not accurate. Try drawing it as a Minkowski space-time diagram, and you will see the difference. The scenario requires 2 dimensions, one for the direction of motion and a small offset for the separation of the rod and hole.

The diagrams I drew were not spacetime diagrams, they were space-only diagrams. Each picture is a 2D spacelike slice through 3D (2+1) spacetime, i.e. a plane of simultaneity relative to the particular frame.

It's hard to draw unambiguous 3D spacetime diagrams on a 2D computer screen. I described such a diagram in words in post #48 of this thread.

I still stand by the accuracy of my pictures (for my interpretation of what the problem is).

 

[DrGreg]

Greg. How do you do that? I want to draw pictures. It would make things so much clearer.

I'd draw things a little different, with the forces applied to the rod only when the rod is directly over the hole in the hole's frame.

Your last frame has an error--sort of. Any portion of the rod sticking-out, north of the hole, is parallel to the surface. Recall that when there are no forces applied, the rod is just Lorentz contracted.

If you wished, you could argue that the gentle push has boosted the rod to have a perpendicular component of velocity as large as its velocity along the surface, which is sort of what your picture respresents, but it simpler if you consider that the added velocity in the perpendicular direction is v_p << c and v_p << v_l.

I didn't forsee a force adding as much y velocity as x velocity.

I have a drawing that makes the Bell Inequality visually obvious, but I don't know how to post it. Can you give me a hint on how to make and send pics?

For the avoidance of any doubt, note the arrows are not forces, they are velocity-directions (relative to the frame).

The "push" has to be applied before the rod gets to the hole, if you wait until it's next to it, it would be too late. As I've drawn it, the force is applied only for a very short period of time, when the rod's direction changes and when it appears bent in the I-frame. For the rest of the time the rod moves inertially.

In the S-frame, the forces are applied at all points of the rod simultaneously. (In my diagram, just before the 3rd blue snapshot only.) In the I-frame, the forces begin at the front and ripple their way through to the back. The rod is bent at the points where there are forces. (In my diagram, between the 2nd and 4th pink snapshots, moving from top to bottom of R.)

The relative x- and y-components of velocity depend on how early you push and how hard and how long -- an early push can be gentle, a late push will have to be much harder.

 

[DrGreg]

How to upload pictures

Greg. How do you do that? I want to draw pictures. It would make things so much clearer.

I have a drawing that makes the Bell Inequality visually obvious, but I don't know how to post it. Can you give me a hint on how to make and send pics?

There are two halves to this.

1. Do you know how to draw diagrams on your computer?

2. Do you know how to upload a picture?

1. There are lots of computer applications that can create diagrams. E.g. Microsoft Powerpoint or Microsoft Word and lots of others. Failing that, you could even draw on a piece of paper and scan it in, if you have access to a scanner.

2A. You need to save the picture in the right format. For computer-generated line diagrams the best formats are PNG or GIF. For a full list of supported types see step 2C below.

2B. If your application cannot save in your preferred format, you need to make a screen dump as follows:
(This assumes you use Microsoft Windows. Other operating systems vary.)
- arrange your picture so it is completely visible in the front window, at the right zoom level
- hold down ALT and press the "Print Screen" (or "Prt Sc" etc) key on your keyboard. This copies a picture of the front window to the clipboard.
- open Microsoft Photo Editor (or any image editor)
- select "Paste as New Image" from the Edit menu
- use the "select tool" to select the area you want to upload (exclude all the window borders).
- Use Image > Crop > OK
- File > Save As... to save as a PNG file.

2C. When you are composing your message using the forum's "advanced edit" option, press the "Manage Attachments" button. This let's you upload a file and includes a list of supported file types and limits on how big files can be. Avoid file types that others may not be able to read.

 

[Phrak]

Thanks, DrGreg. That's quit a post. Nicely written. I can use autoCad for perspective, but my skills at it are no to great. I think MS Word will convert drawings to jpg of gifs when inserted.