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

[Phrak]

A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the rod. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

What does this look like in the inertial frame of the rod?

 

[granpa]

the hole would appear to be smaller than the rod but the rod wouldn't be parallel to the surface as it went through the hole due to loss of simultaneity.

 

[bernhard.rothenstein]

A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the rod. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

What does this look like in the inertial frame of the rod?

Am. J. Phys. Vol 74 (2006) pp998-1001
A look at the paper quoted above brings you information and references about the subject you are interested in which is known as Rindler Effect. You can find the paper going on google and giving it to look for barn hole paradox

 

[Phrak]

The Rindler effect is about nonuniformly accellerating observers.

 

[Dale]

A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the rod. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

What does this look like in the inertial frame of the rod?

It sounds like you are making the rod perfectly rigid. Anyway, the answer is easy, and granpa already mentioned the essence. Simply Lorentz transform the worldline of the rod ends and the edges of the hole for the details.

 

[MikeLizzi]

Phrak, the way you present the problem, you are setting yourself up for a long winded dialog about acceleration. If you don't want that, there is a strictly inertial version of the problem.

Consider this reference frame: The rod is moving to the right, constant velocity. The surface (parallel to the rod) is moving up, constant velocity. The hole in the surface is on an intersection course with the rod. Now there is no need to include any acceleration. If the proper length of the rod is slightly larger than the proper length of the hole, and velocities are high enough, relativistic physics will predict that the rod will get swept into the hole as the surface rises.

Now your question, describe the action from the ref frame of the rod is more straightforward.

 

[Phrak]

Phrak, the way you present the problem, you are setting yourself up for a long winded dialog about acceleration.

You're right. I'm guilty of posing an ill-posed question, but the acceleration of the rod is kinda central. I'll try again:


A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the length of the rod measured at rest. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

What does this look like in the initial inertial frame of the rod?

I'm curious to see if anyone can come up with the correct answer.

 

[Ich]

As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

Simply examine what this statement means in both reference frames. It still is an ill-posed question.

 

[Doc Al]

where at all times it remains parallel to the surface.

Those gentle pushes may occur at the same time from the hole frame, but from the rod's frame the front of the rod gets pushed down first.

 

[Phrak]

Those gentle pushes may occur at the same time from the hole frame, but from the rod's frame the front of the rod gets pushed down first.

What does the rod look like?

 

[Doc Al]

What does the rod look like?

Like a limp piece of spaghetti as it is bent into the hole.

 

[phyti]

You're right. I'm guilty of posing an ill-posed question, but the acceleration of the rod is kinda central. I'll try again:

A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the length of the rod measured at rest. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

What does this look like in the initial inertial frame of the rod?

I'm curious to see if anyone can come up with the correct answer.

If you are saying the hole is shorter than the rod when they are together and no relative motion, what you are describing will never happen.

 

[Doc Al]

If you are saying the hole is shorter than the rod when they are together and no relative motion, what you are describing will never happen.

Why is that? If the rod is moving fast enough, it will be shorter than the hole diameter in the hole frame.

 

[MikeLizzi]

Phrak, I think I know what you are getting at.

From an observer on the surface, the rod and surface remain parallel and the contracted rod passes thru the hole without incident. From an observer on the rod, the rod and surface are not parallel and this fact allows the uncontracted rod to again pass thru the hole without incident.

If this is also your opinion, I don't think you will get much support for it in this forum.

 

[Phrak]

Phrak, I think I know what you are getting at.

From an observer on the surface, the rod and surface remain parallel and the contracted rod passes thru the hole without incident. From an observer on the rod, the rod and surface are not parallel and this fact allows the uncontracted rod to again pass thru the hole without incident.

The observer is not on the rod. But could you explain what you mean?

If this is also your opinion, I don't think you will get much support for it in this forum.

Support?? I'm not running for office, and I'm not selling anything, I swear. But if you're in the market I can get help you get some prime Wet Lands in Florida. But no---that wouldn't be right. Watch yourself, it's a concrete Rain Forest out there.

 

[Phrak]

the hole would appear to be smaller than the rod but the rod wouldn't be parallel to the surface as it went through the hole due to loss of simultaneity.

What does the rod look like?

 

[Dale]

I thought granpa already described how it would look quite well. Perhaps you could be more specific on the form of the answer you are seeking. Otherwise, from what I can tell it has already been provided.

 

[Phrak]

I thought granpa already described how it would look quite well. Perhaps you could be more specific on the form of the answer you are seeking. Otherwise, from what I can tell it has already been provided.

"...wouldn't be parallel" can mean a lot of things.

But Doc Al had something to say about spaghetti.

 

[Dale]

"...wouldn't be parallel" can mean a lot of things.

Then do you want the angle that it goes through the hole at? You haven't given enough information to specify that, so I think granpa's statement is the best you can do. If you want any more detail you will have to exactly specify everything in one frame and then simply transform it to any other frame you like.

 

[Phrak]

Dale, as far as I can tell granpa could, and probably was talking about a stright rod.

In the inerial frame of the observer the rod remains parallel to the hole. Then, necessarily, the rod cannot be ridgid (Doc Al's oblique reference to spaghetti).

In the inerital frame, in the initial trajectory of the rod, the rod is mechanically deformed as it is pushed through the hole. The leading and trailing ends remain parallel to the plane, above and below it.

The rod has an lazy sigmiodal bend over the hole, as the force pushing it through the hole propagates from front to rear.

 

[phyti]

Dale, as far as I can tell granpa could, and probably was talking about a stright rod.

In the inerial frame of the observer the rod remains parallel to the hole. Then, necessarily, the rod cannot be ridgid (Doc Al's oblique reference to spaghetti).

In the inerital frame, in the initial trajectory of the rod, the rod is mechanically deformed as it is pushed through the hole. The leading and trailing ends remain parallel to the plane, above and below it.

The rod has an lazy sigmiodal bend over the hole, as the force pushing it through the hole propagates from front to rear.

If parts of the rod are moving differently from other parts, all parts are not in the same frame, and not in an inertial frame, and not all parallel to the surface.

You need more specific details.

 

[Dale]

The rod has an lazy sigmiodal bend over the hole, as the force pushing it through the hole propagates from front to rear.

Is it even possible for the rod to be straight in one frame and sigmoidal in another? The Lorentz transform is a linear transform. Since you had mentioned it was parallel at all times in the hole's frame I understood that to mean straight at all times in the hole's frame. So I took "gentle" to mean "Born-rigid" acceleration.

You really need to describe it completely in one frame. Once you have done that it is easy to describe it in any other frame, but until then it is just a game of "Read Phrak's mind".

 

[Jonathan Scott]

Is it even possible for the rod to be straight in one frame and sigmoidal in another? The Lorentz transform is a linear transform. Since you had mentioned it was parallel at all times in the hole's frame I understood that to mean straight at all times in the hole's frame. So I took "gentle" to mean "Born-rigid" acceleration.

You really need to describe it completely in one frame. Once you have done that it is easy to describe it in any other frame, but until then it is just a game of "Read Phrak's mind".

For the rod to move it has to accelerate. That acceleration cannot be simultaneous in different frames, and cannot be constant, as it must start and stop. Hence the rod must bend from some frame.

 

[Fredrik]

From an observer on the surface, the rod and surface remain parallel and the contracted rod passes thru the hole without incident. From an observer on the rod, the rod and surface are not parallel and this fact allows the uncontracted rod to again pass thru the hole without incident.

If this is also your opinion, I don't think you will get much support for it in this forum.

Why not? It's the correct answer.

If the rod remains parallel to the surface in the surface's rest frame, then every part of the rod was given a push simultaneously in that frame. That means that the different parts of the rod didn't get pushed simultaneously in the frame that's co-moving with the rod before the push. Since they didn't get pushed simultaneously in that frame, the rod can't possibly remain parallel to the surface in that frame.

The rod doesn't get rotated or bent in the co-moving frame. This is what happens: Its cross-section will change from a rectangle to another kind of parallelogram. The front and back are still perpendicular to the surface, but the long sides aren't making 90 degree angles with the front and back anymore. They aren't parallel to the surface anymore, but they are still straight. The linearity of the Lorentz transformation is sufficient to guarantee that.

There's no need to consider forces propagating in the rod because of the requirement that every part of it gets an identical velocity boost towards the hole simultaneously in the hole's rest frame. (I assume that this is what the OP had in mind, or at least would have had in mind if he had been careful enough to specify the details completely). That's a constraint that completely determines the shape of the rod.

 

[Phrak]

If parts of the rod are moving differently from other parts, all parts are not in the same frame, and not in an inertial frame, and not all parallel to the surface.

You need more specific details.

See post #7, after MikeLizzi told me the problem statement needed improving.

 

[Phrak]

Is it even possible for the rod to be straight in one frame and sigmoidal in another? The Lorentz transform is a linear transform. Since you had mentioned it was parallel at all times in the hole's frame I understood that to mean straight at all times in the hole's frame. So I took "gentle" to mean "Born-rigid" acceleration.

You really need to describe it completely in one frame. Once you have done that it is easy to describe it in any other frame, but until then it is just a game of "Read Phrak's mind".

I think that everything you need to know is there. Give it a couple of days to sink in, and I think you can convince yourself that ridgid objects are an exception to the rule in special relativity.

 

[Phrak]

There's no need to consider forces propagating in the rod because of the requirement that every part of it gets an identical velocity boost towards the hole simultaneously in the hole's rest frame. (I assume that this is what the OP had in mind, or at least would have had in mind if he had been careful enough to specify the details completely). That's a constraint that completely determines the shape of the rod.

I'm sure the OP doesn't know what you're talking about. The details of forces and crosssections are made immaterial by ensuring that the rod be parallel to the surface in the inerital frame of the hole.

 

[Fredrik]

I'm sure the OP doesn't know what you're talking about. The details of forces and crosssections are made immaterial by ensuring that the rod be parallel to the surface in the inerital frame of the hole.

That's what I said.

Edit: OK, there's a small difference. I specified the velocity change to be instantaneous, and you left the acceleration unspecified (and suggested that it isn't instantaneous). That part isn't really relevant though.

 

[Bible Thumper]

I'm guessing the rod is rigid during slow velocities, but turns spaghetti-like under near-light speeds?

 

[Bible Thumper]

So the only way the rigid rod can turn spaghetti-like is:

1. The rod is traveling at near-light velocities. And,
2. There is an external force acting on the rod at that speed (an external force may be a stationary object relative to the hole, such as a meatball sitting right beside the hole).

 

[HallsofIvy]

I'm guessing the rod is rigid during slow velocities, but turns spaghetti-like under near-light speeds?

No, there is no such thing as a "rigid" rod at any velocities. "Rigidity" would be a property of the rod itself, not "relative" to some external coordinate system so it wouldn't make sense to talk of it dependent of "slow" or "near light" speeds which must be relative to an external coordinate system.

 

[Fredrik]

If you understand simultaneity it's very easy to understand why no objects can be absolutely rigid in SR. Just think about what happens if all the different parts of an object begin to accelerate simultaneously in one frame. (The object can obviously not stay the same shape if the different parts don't begin their acceleration simultaneously). If it happens simultaneously in one frame, it doesn't happen simultaneously in other frames, so the object will change its shape in those other frames.

 

[Dale]

The rod has an lazy sigmiodal bend over the hole, as the force pushing it through the hole propagates from front to rear.

Sorry Phrak, I still don't buy it. How can a linear transform turn a straight line into a sigmoidal curve? You should really do this rigorously.

 

[Jonathan Scott]

Sorry Phrak, I still don't buy it. How can a linear transform turn a straight line into a sigmoidal curve? You should really do this rigorously.

The transform is not linear for the whole object, because such transforms are only possible for constant velocity. Another form of transform assuming a form of rigidity is possible for an object with constant acceleration, but this doesn't apply here either.

To calculate this rigorously, you'd need a specific acceleration profile to describe what happens.

However, to get an idea, consider each part of the rod to be guided in parallel by separate tracks dipping through the hole, which are static in the frame of the hole, where the spacing between the tracks is such that when the rod arrives at full speed all parts switch from above to below at the same time. In the frame of the rod, the front parts switch first, so half way through the front will have completed the switch, the middle will be switching and the back has not yet started to switch, so the rod has a slight S-curve in the middle. The detailed shape of the curve depends on the acceleration profile at the switch, which is not specified.

 

[Fredrik]

You don't need to know the acceleration as a function of time, since it has been specified that the rod remains parallel to the wall in the wall's rest frame at all times. When the acceleration is over, it's moving with a constant velocity and is still parallel to the wall, in the wall's frame. There's no need to consider anything but inertial frames and Lorentz transformations at this point, and the rod remains straight in all inertial frames because of the linearity of the Lorentz transformation.

Edit: I didn't read everything in Johnathan's post before I wrote the above. I agree of course that in an inertial frame where the different parts of the rod don't begin their acceleration at the same time, the rod will have a strange shape when its parts are accelerating.

 

[Phrak]

That's what I said.

My mistake, Fredrik. I'd thought you were pointing out an insufficiency in problem statement.

 

[Phrak]

Dale, forget that last post to you; sorry about that. I don't give you enough credit for what you know. I went back and read some of your previous posts. This is not as easy a problem as it sounds, and I stated it a bit flipantly. And I've managed to PO a lot of folks who thought this should be easy!

 

[Phrak]

I've been thinking about how the solution to this problem is so easily deceptive. I heard it a long time ago before studying relativity where it was accompanyed by an incorrect answer, so it's become a bit ingrained, I guess.

First, the word "rod" invokes the concept of ridgidity. Second, it's far from obvious that the requirement that the rod be always parallel in one inertial frame implies that forces can be ignored It's becomes a purely kinematic problem. Only the trajectories of the elements of the rod have to be transformed to new coordinates.

Since it's purely kinematics you can think of the rod as a string of disconnected beads. They are given a downward impulse as they pass over the hole. If the train of beads is fast enough, only one or two are over the hole at once. So each bead gets an impulse in sucession at different times. Make sense?

I could get some valid criticism, because the rod gets stretched out of shape getting pushed down the hole. I think it gets stretched longer by a 1/cosine factor due to it's oblique velocity on leaving the hole.

 

[Ich]

Since it's purely kinematics you can think of the rod as a string of disconnected beads. They are given a downward impulse as they pass over the hole. If the train of beads is fast enough, only one or two are over the hole at once. So each bead gets an impulse in sucession at different times. Make sense?

No. The chain of events where the beads are being pushed is simultaneous in the wall's frame, which means it propagates always with FTL speed along the rod in the rod's initial frame, not at v/sqrt(1-v²), as you suggest.

 

[Dale]

The transform is not linear for the whole object, because such transforms are only possible for constant velocity.

The Lorentz transform is linear, which is why it can be expressed in matrix form.

I can see how you could get a simple bend in the object (making it a V shape), because the "bend in time" that exists in one frame can get transformed into a "bend in space" in another frame. But I just don't see how it gets curved. A linear transformation always keeps all straight segments straight.

Dale, forget that last post to you; sorry about that. I don't give you enough credit for what you know. I went back and read some of your previous posts. This is not as easy a problem as it sounds, and I stated it a bit flipantly. And I've managed to PO a lot of folks who thought this should be easy!

No problem!

 

[Ich]

I can see how you could get a simple bend in the object (making it a V shape), because the "bend in time" that exists in one frame can get transformed into a "bend in space" in another frame. But I just don't see how it gets curved. A linear transformation always keeps all straight segments straight.

Finite acceleration gives a "curve in time", which transforms to a curve in space.

 

[Fredrik]

I can see how you could get a simple bend in the object (making it a V shape), because the "bend in time" that exists in one frame can get transformed into a "bend in space" in another frame. But I just don't see how it gets curved. A linear transformation always keeps all straight segments straight.

It seems to me that it will have two straight parts only if the velocity change is instantaneous. If it takes a while, the shape should have one curved part and one straight part (the part that hasn't started accelerating yet.

Think e.g. about how every part will reach half of their maximum speed towards the hole at the same time in the wall's rest frame. That means that they won't in the frame that's co-moving with the rod before the push. So when one point in the middle has reached that speed, some parts will be moving faster and some slower. This should give the rod a weird shape.

Another way of looking at it is to imagine a spacetime diagram with two spatial dimensions drawn, showing the events in the wall's rest frame. The world sheet of the rod will be a curved surface (when the acceleration is smooth) and the shape we're talking about is a diagonal slice of that curved surface.

 

[Fredrik]

No. The chain of events where the beads are being pushed is simultaneous in the wall's frame, which means it propagates always with FTL speed along the rod in the rod's initial frame, not at v/sqrt(1-v²), as you suggest.

What he said sounds good to me, assuming that he's describing things using the frame that's co-moving with the rod before the push.

I don't think of it as a "chain of events" or as something "propagating". I imagine a large number of really tiny things pushing different parts of the rod at times that have been scheduled in advance.

 

[Ich]

I don't think of it as a "chain of events" or as something "propagating". I imagine a large number of really tiny things pushing different parts of the rod at times that have been scheduled in advance.

Ok, wrong wording. What I wanted to say is that these events are necessarily spacelike separated, which means that they have to be scheduled in advance and can not be triggered at the same place (but different times) in the wall frame.

 

[Dale]

Finite acceleration gives a "curve in time", which transforms to a curve in space.

But then it isn't parallel at all times in any frame.

It seems to me that it will have two straight parts only if the velocity change is instantaneous.

Yes, that is exactly what I was thinking. I think that is implied by the "always parallel" requirement. Of course, if that is not implied then the problem is ill-posed (even the second version) since there would be more than one "always parallel" worldline (worldsheet).

 

[Jonathan Scott]

But then it isn't parallel at all times in any frame.

It doesn't make any difference whether the change of velocity is a step (which of course implies infinite acceleration, but we'll ignore that) or gradual. If the change is simultaneous in one frame, it propagates along the rod in some other frame, so there could be points which have reached the new alignment, points which are still switching over, and points which have not yet started to switch. The parts of the rod which have started and finished switching will be parallel, but the ones which are still switching will be somewhere in between.

If the switching over is simultaneous in one frame, the separation between the events giving the same stage of switching at different points on the rod must remain spacelike in all other frames, so the switching process at different points propagates faster than c and it cannot of course be observed directly, only in retrospect.

 

[Dale]

OK, I can accept that explanation. Then the problem is again incompletely specified.

 

[DrGreg]

This whole thing is an exercise in spacetime geometry, no forces need be considered.

In fact, as there are just 2 relevant space dimensions to consider, it's an exercise in visualising 3D (2+1) spacetime geometry.

A 1-dimensional rod is represented as a 2-dimensional surface ("worldsheet") in spacetime. If the rod always moved inertially the sheet would be flat, but as we know it accelerates for some of the time (in a direction other than along its own length), the worldsheet is curved or bent. If there were an instantaneous change of velocity, there would be a sharp crease in the worldsheet. For a gradual change, a smooth bend.

For the purpose of this visualisation in 2 space dimensions, we can consider the wall to be a one-dimensional straight line with a gap (hole) in the middle. Thus the wall's worldsheet is a 2-dimensional flat plane (with a strip missing cutting it into two pieces).

Now let us take spacetime axes relative to the wall frame, with wall-time vertically up and 2D-wall-space in a horizontal plane. (So the wall's worldsheet is a vertical plane with a vertical slot missing.) We are told the rod is always parallel to the wall in this frame which means that every horizontal cross-section of the rod's worldsheet is a horizontal straight line (parallel to the wall's horizontal cross-section).

However, the planes of simultaneity for the initial rest-frame of the rod are not horizontal. Therefore, in some of these planes (i.e. at the appropriate times), the cross-section of the rod's curved worldsheet will be a curved (or bent) line.

The shape of the rod in this frame will look like a space-time graph of the rod in the wall-frame (in the space-direction perpendicular to the wall).

 

[Bible Thumper]

The rod has an lazy sigmiodal bend over the hole.

Excellent use of www.thesaurus.com, Phrak!
Most people would have been content with 'curvilinear', or even 'sinusoidal'. But 'sigmiodal? Good going! :)

 

[Bible Thumper]

A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the rod, when both are measured at rest. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.
What does this look like in the inertial frame of the rod?

Phrak, is this how the question goes? There's still the obvious confusion with what's happening here. Is this what is happening?:

So what we have is a rod not unlike a broomstick. This rod is horizontal and moving at near-light speeds. There is also a horizontal surface the rod is quickly moving and levitating over that is parallel to the rod. There's a hole in the horizontal surface that's roughly the same shape as the rod (thus the hole is more like a slit rather than a hole) but a bit shorter than the rod when the length measurements are taken at rest.
This rod then approaches the slit. As it does so, an external force acts on the rod in such a way that the rod may fall into this slit. This means that the external force acts on the center of the rod, causing it to fall toward the slit while remaining always parallel to the horizontal surface.

"What does this look like in the inertial frame of the rod?"