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

[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?"

 

[phyti]

This is the question everyone is asking, what are the initial conditions.
If it's what BT described, then the rod and the surface are moving at the approx. the same speed. Because the hole is shorter than the rod (as stated originally), the rod cannot translate sideways through the hole.

 

[Phrak]

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.

You've got it for the most part. But ignore gravity; no there's no 'falling' going on. Also, don't think of the force that's pushing it through the hole, as acting at it's center of mass. Whatever force is used is distributed in such a way that the rod, or broom stick is always parallel to the surface in the inertial frame of the hole.

It's best not to think of it as a rod, but a soft stick of taffy. The bead idea I gave earlier is actually very appropriate, but no one seems to take it up.

(Sigmoidal is the term used to describe the transfer function of an artificial neuro network neuron--soft saturating logic, idealized as the arctangent because the derivative is easy to deal with.)

 

[Fredrik]

The bead idea I gave earlier is actually very appropriate, but no one seems to take it up.

It's a good idea, but even better is to just imagine a rod that's been sliced up into thin slices that are perpendicular to the wall. That's how I think about it. The reason why this is better is that when you consider circular beads,you won't clearly see what the shape of the rod is going to be in the inertial frame that's co-moving with the rod before the push.

The shape of the rod's cross-section is a parallelogram that isn't a rectangle (assuming an instantaneous velocity change and that we're looking at the rod after every part of it has changed its velocity). Note that the slices are still perpendicular to the wall.

 

[Bible Thumper]

A gentle push is a gentle push, you guys. I understand that in reality, the increase in mass the rod takes up will mean far more than a gentle push will be needed for our rod. However, we have to presume the OP meant 'gentle push' when he said 'gentle push.'

And if he meant gentle push, I can't understand why we have to be regarding the rod as necessarily flexible. I mean, can't we make it rigid by forging it from neutron starstuff? Then it will be almost infinitely dense, thus infinitely rigid.

Please explain why it has to be flexible. Don't mention kinematics, either, because we already agreed the OP meant 'gentle push' when he said 'gentle push.'

 

[Fredrik]

I can't understand why we have to be regarding the rod as necessarily flexible. I mean, can't we make it rigid by forging it from neutron starstuff? Then it will be almost infinitely dense, thus infinitely rigid.

No. This was answered earlier in the thread. If it's rigid in one inertial frame, it isn't rigid in others. Suppose that it's rigid in a particular frame. When we give it a push, all the different parts will begin their acceleration at the same time. That means that in other frames, the different parts will begin their acceleration at different times. That makes it "not rigid" in those frames.

So it doesn't matter what it's made of. The properties of spacetime make it impossible for rigid bodies to exist.

The scenario we're discussing in this thread is keeping the rod parallel to the wall at all times in the wall's rest frame. This is only possible if the different parts are all pushed at the same time. We can't just push one part of the rod and let the force propagate through the rod, because the speed of that propagation would have to be infinite. (Otherwise the rod doesn't remain straight, and we have already said that it does).

 

[Bible Thumper]

No. This was answered earlier in the thread. If it's rigid in one inertial frame, it isn't rigid in others. Suppose that it's rigid in a particular frame. When we give it a push, all the different parts will begin their acceleration at the same time. That means that in other frames, the different parts will begin their acceleration at different times. That makes it "not rigid" in those frames.

So it doesn't matter what it's made of. The properties of spacetime make it impossible for rigid bodies to exist.

The scenario we're discussing in this thread is keeping the rod parallel to the wall at all times in the wall's rest frame. This is only possible if the different parts are all pushed at the same time. We can't just push one part of the rod and let the force propagate through the rod, because the speed of that propagation would have to be infinite. (Otherwise the rod doesn't remain straight, and we have already said that it does).

I wasn't regarding this kind of independence. I mean, I know things may appear nonrigid to one person but completely rigid to another.
Phrack mentioned kinematics, and that the application of force on the rod will turn the problem into a kinematic one. He supplements this with his bead-on-a-string thought experiment. The bead-on-a-string thinking can go both ways: with different observations or with the introduction of kinematic ideas (which would complicate things greatly and unnecessarily).

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... :)

 

[Fredrik]

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,...

Again, no. If a push applied to the center of the rod keeps the rod parallel to the wall in the wall's rest frame, then the rigidity is a property of the rod, not of the push. If it's a property of the rod, then it's rigid in all frames, and that contradicts special relativity.

At the very least, this problem becomes much more complicated when we consider a push applied to the center of the rod. Just think about the fact that the deformation of the rod will propagate at the speed of sound in the material the rod's made of, in the rod's rest frame. That speed, transformed to the wall's rest frame, must be high enough to reach the front of the rod before the front of the rod reaches the front of the hole. And you can't push the rod until at least half of it is over the hole. If the propagation speed in the wall's frame is greater than the rod's speed, then you have to wait even longer before you make the push.

There's a lot to think about here. If we do the math, we may find that even if we assume that the rod is infinitesimally thin and that its rest length is only infinitesimally greater than the hole's rest length, the speed of sound needs to be greater than the speed of light for this to work. I don't know if that's what we'll find. My point is that this is something we would have to check.

I'm not saying it's not an interesting problem, but you should try to understand the simple problem first.

 

[Phrak]

A gentle push is a gentle push, you guys. I understand that in reality, the increase in mass the rod takes up will mean far more than a gentle push will be needed for our rod. However, we have to presume the OP meant 'gentle push' when he said 'gentle push.'

Point taken. I was being flipant. Think of the rod as very thin and the surface as well--and then give it a good push.

 

[atyy]

A gentle push is a gentle push, you guys. I understand that in reality, the increase in mass the rod takes up will mean far more than a gentle push will be needed for our rod. However, we have to presume the OP meant 'gentle push' when he said 'gentle push.'

Point taken. I was being flipant. Think of the rod as very thin and the surface as well--and then give it a good push.

I don't understand why your bead idea isn't correct. In anyone (global) inertial frame, there is no rigid body if we push at only one point. So even in one inertial frame, we have to push at all points simultaneously in order to achieve rigidity. Since rigidity is achieved by simultaneous pushing and not the internal structure of the rod, it seems reasonable and equivalent to make the problem as simple as possible and consider the rod as independent beads.

 

[Bible Thumper]

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.
The effect of this pushing, however, is to see the rod enter the slit, front-end first, for the person on the rod.

 

[Fredrik]

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.
The effect of this pushing, however, is to see the rod enter the slit, front-end first, for the person on the rod.

Everything in this post is wrong.

* The rod's motion before the push defines an inertial frame. (If that's what you meant, you weren't wrong about that, but you stated your claim a bit carelessly).

* A force applied to a point doesn't propagate at infinite speed through the material. It propagates at the speed of sound, which can't be greater than the speed of light.

* Even if it did, so that the rod remains parallel to the wall in the rod's original rest frame, then this would contradict both what we have said before (parallel in the other frame) and what you said next.

* If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first.

 

[Bible Thumper]

Everything in this post is wrong.

* The rod's motion before the push defines an inertial frame. (If that's what you meant, you weren't wrong about that, but you stated your claim a bit carelessly).

* A force applied to a point doesn't propagate at infinite speed through the material. It propagates at the speed of sound, which can't be greater than the speed of light.

* Even if it did, so that the rod remains parallel to the wall in the rod's original rest frame, then this would contradict both what we have said before (parallel in the other frame) and what you said next.

* If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first.

What about atomospheric resistance on the rod? You forgot that. If the rod is going at nearly the speed of light, atmospheric resistance will heat the rod up to the point it will melt.

 

[Fredrik]

What about atomospheric resistance on the rod? You forgot that. If the rod is going at nearly the speed of light, atmospheric resistance will heat the rod up to the point it will melt.

Can you really not see the difference between pointing out that what you're saying is wrong and being too concerned with practical matters in a discussion about a thought experiment?

 

[Bible Thumper]

Can you really not see the difference between pointing out that what you're saying is wrong and being too concerned with practical matters in a discussion about a thought experiment?

Fredrik, this little guy: <--- he means, "sarcasm". Knowing that, realize I was being sarcastic when I brought up air resistance. Why was I being sarcastic? Because you were bringing up the rate at which force is distributed thru the rod, that's why. If you want to bring up the rate at which the force runs thru the rod, why not other matters, such as air resistance?

• A 'gentle push' can imply an accelerating force on the rod, at which consideration, the rate at which the force runs thru the rod will depend on the rate at which the 'gentle push' acceleration takes place.
• Since the rate of the external force (assuming an impact force) runs thru the rod will depend on factors, such as the material of which the rod is made from.

Thus I identify two scenarios that alone tell us they weren't meant to be taken into consideration in the initial question. In other words, factors such as the rate at which the force is transmitted thru the rod (from a 'gentle push' which means the force goes thru the rod very slowly--as slowly as the gentle push itself--to an impact force, which goes thru the rod much faster and is material-dependent) were never meant to be taken into consideration.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html" link helps show that the pole will remain always parallel to the surface relative to the slit, but goes into the slit nose-first relative to someone on the pole.

 

[atyy]

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html" link helps show that the pole will remain always parallel to the surface relative to the slit, but goes into the slit nose-first relative to someone on the pole.

There is no gentle push in the pole-barn paradox, because the pole is moving at constant velocity and does not accelerate.

 

[Bible Thumper]

There is no gentle push in the pole-barn paradox.

I thought his main concern was with contraction, as evidenced by this:

"If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first."

For this reason and only this reason I posted the barn-pole paradox link. Regard the barn-pole paradox after the 'gentle pushing' is done.

 

[atyy]

I thought his main concern was with contraction, as evidenced by this:

"If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first."

For this reason and only this reason I posted the barn-pole paradox link. Regard the barn-pole paradox after the 'gentle pushing' is done.

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

 

[Bible Thumper]

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

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

 

[Fredrik]

If you want to bring up the rate at which the force runs thru the rod, why not other matters, such as air resistance?

Because an infinite propagation speed contradicts the theory that we're trying use to solve this problem. That makes it absolutely necessary to avoid it, even in the most mathematically idealized thought experiment we can think of. Air resistance on the other hand is only a problem in an actual experiment performed by actual physicists.

The difference between the two is enormous. That's why I asked if you really don't see it, and apparently you don't. It's like the difference between being able to eat a trillion hamburgers and being able to eat oneself*. One of them is impossible in practice, one of them is impossible in principle. I wouldn't have said anything if the scenario you described involved a person who eats a trillion hamburgers.

*) Yes, I'm aware of how that can be interpreted.

the pole will remain always parallel to the surface relative to the slit, but goes into the slit nose-first relative to someone on the pole.

Yes, that's what I've been saying in every post in this thread, but it's not what you've been saying.

 

[Fredrik]

I thought his main concern was with contraction, as evidenced by this:

"If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first."

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.