I Can an Altered Thought Experiment Reconcile the Special Relativity Conundrum?

[hatflyer]

I can't reconcile how a slightly altered thought experiment can be explained.

It's the 1 where a train is going past, with an observer on the ground. Say instead of a train it's a cable car, held up by a cable in the front and 1 in the back to an overhead line.

The observer on the ground shoots 2 lasers simultaneously (to him) so that each laser destroys the support cables, the 1 in the front and the 1 in the back. To that observer, they get destroyed simultaneously, and so the cable car falls in a horizontal orientation.

But, to an observer in the middle of the cable car, the front cable snaps before the rear 1. So he sees the front of the cable car fall first, then the rear cable snaps and it falls a bit later. Thus the cable car would seem to fall with the front pointed downward, not horizontally as seen by the ground observer.

Those opposing views can't be, as the cable car falls in only 1 orientation.

Which is it?

Thanks.

 

[Ibix]

You can't ignore the speed at which the cable relaxes. Basically the car won't fall until the information about the break propagates from the break to where the car is suspended. That happens at the speed of sound in the cable, which is much less than the speed of light and is frame dependent - and not the same speed in both directions in frames where the cable is moving. This will conspire with the relativity of simultaneity so that if the information about the cable breakages arrives simultaneously in one frame it will do so in all.

 

[Dale]

Those opposing views can't be, as the cable car falls in only 1 orientation.

The orientation is not invariant. There is nothing wrong with two reference frames disagreeing about the orientation.

 

[phinds]

You can't ignore the speed at which the cable relaxes. Basically the car won't fall until the information about the break propagates from the break to where the car is suspended. That happens at the speed of sound in the cable, which is much less than the speed of light and is frame dependent - and not the same speed in both directions in frames where the cable is moving. This will conspire with the relativity of simultaneity so that if the information about the cable breakages arrives simultaneously in one frame it will do so in all.

I don't think the introduction of the speed of sound in the cable is really relevant to the OP's conundrum and his question could be reformulated to avoid any need for consideration of that. His problem, I think, is that he is ignoring (as you point out) the relativity of simultaneity but in the following way: he thinks that he can fire the lasers simultaneously AND have the return pulses hit his eyes at the same time AND have both of those event happen along with the pulses hitting the cables at the same time in the TRAIN's FOR so that it drops horizontally. The geometry of all that just doesn't work.

 

[phinds]

The orientation is not invariant. There is nothing wrong with two reference frames disagreeing about the orientation.

NUTS ! I see that I overthought the problem.

 

[Herman Trivilino]

If each end of the car hits the ground at the same time in one frame, then they can't hit at the same time in the other frame.

 

[hatflyer]

I'm not quite getting it. The front cable is seen by the rider as snapping first. Doesn't he thus see the front cable itself collapse downward before he sees the rear cable collapse? Then I thought time frames in a direction perpendicular to the time frame in the direction of motion were the same to both observers, so that at least on the end of the cable car (not towards the observer) it would start to fall before the rear does.

Obviously that's wrong, but confused if we are talking about the ends of the car they would at least start at separate times.

 

[hatflyer]

PS. So the observer in the car sees the cable in the front snap before he sees the rear cable snap (i.e. he sees the moment the laser hits the cables), yet he sees himself falling horizontally?

 

[PeterDonis]

I thought time frames in a direction perpendicular to the time frame in the direction of motion were the same to both observers

There are not different "time frames" in different directions. The spatial coordinates in directions perpendicular to the direction of motion are the same in both frames; but anything involving time is affected regardless of what direction it's in.

 

[hatflyer]

There are not different "time frames" in different directions. The spatial coordinates in directions perpendicular to the direction of motion are the same in both frames; but anything involving time is affected regardless of what direction it's in.

Sorry, I meant velocities.

 

[phinds]

Sorry, I meant velocities.

velocities involve time, thus the relevance of Peter's comment

 

[hatflyer]

So if the observer in the car shoots at gun outside the side window, don't both observers measure the speed of the bullet as the same speed?

 

[FactChecker]

The orientation can only be defined by the positions of the front and end of the car at the same time. So coordinate systems that do not agree on what is simultaneous will not agree on the orientation.

 

[jartsa]

A military helicopter has a propeller at the rear and at the front, enemy fire destroys first the front propeller, then after a short while the rear propeller is destroyed. Now we have helicopter that spins around, right?

That war story was told in the frame of the helicopter, I mean it was the point of view of the helicopter crew.

So the cable car also spins around, right?

 

[hatflyer]

The orientation can only be defined by the positions of the front and end of the car at the same time. So coordinate systems that do not agree on what is simultaneous will not agree on the orientation.

That doesn't seem possible. What if there was a marble at the center of the cable car. Both observers must agree on whether the marble stays at the center or rolls to the front of the car. So surely they agree on orientation of the car, right?

 

[PeterDonis]

So if the observer in the car shoots at gun outside the side window, don't both observers measure the speed of the bullet as the same speed?

No. See here:

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[Dale]

What if there was a marble at the center of the cable car. Both observers must agree on whether the marble stays at the center or rolls to the front of the car.

Yes. But the direction the marble rolls does not depend only on the orientation.

 

[PeterDonis]

Both observers must agree on whether the marble stays at the center or rolls to the front of the car.

Yes.

So surely they agree on orientation of the car, right?

No. This scenario includes proper acceleration--the car is being accelerated upward as long as the cables hold--so you can't analyze it the way you would analyze purely inertial motion. In the frame in which the car is moving, because of the upward proper acceleration, its surface is not flat--it is concave upward. So even though the front end falls first in this frame, the concavity prevents the marble from moving forward.

 

[jartsa]

That doesn't seem possible. What if there was a marble at the center of the cable car. Both observers must agree on whether the marble stays at the center or rolls to the front of the car. So surely they agree on orientation of the car, right?

In the ground frame the marble starts moving, because sound waves from the front reach the marble first, because of anisotropic speed of sound in the car structure.

And in the car frame the marble starts moving, because sound waves from the front reach the marble first, because of the non-simultaneity of the laser shots.

 

[Herman Trivilino]

You don't need the cables to illustrate your paradox. Suppose the car is in free fall and hits Earth's surface. If the ends hit simultaneously in the car's rest frame, they don't hit simultaneously in Earth's rest frame. A marble on the floor of the car stays put in the car's rest frame because the car is level in that frame. It therefore stays put relative to the car's floor in Earth's rest frame, not because the car is level, but because both the car and the marble are in motion in Earth's rest frame.

 

[hatflyer]

In the ground frame the marble starts moving, because sound waves from the front reach the marble first, because of anisotropic speed of sound in the car structure.

And in the car frame the marble starts moving, because sound waves from the front reach the marble first, because of the non-simultaneity of the laser shots.

In the ground frame, doesn't the force translation from the sudden loss of upward force from the front and back reach the center simultaneously? You add the speed of the car to the speed of travel of the force effects atom by atom from the back, and subtract it from the front?

Anyway, I still can't see how an observer on the ground, seeing both cables snap simultaneously, would have any reason to believe the car would tilt forward. There is no introduction of rotation (i.e. the front falling more than the rear).

Likewise, if I'm in the car and see the car as being at rest (same thing), if I see my front cable snap first, I'm expecting at least the tip of the car to drop first.

Sorry if an above explanation set this straight for me, but it hasn't yet.

 

[jartsa]

In the ground frame, doesn't the force translation from the sudden loss of upward force from the front and back reach the center simultaneously? You add the speed of the car to the speed of travel of the force effects atom by atom from the back, and subtract it from the front?

Anyway, I still can't see how an observer on the ground, seeing both cables snap simultaneously, would have any reason to believe the car would tilt forward. There is no introduction of rotation (i.e. the front falling more than the rear).

Likewise, if I'm in the car and see the car as being at rest (same thing), if I see my front cable snap first, I'm expecting at least the tip of the car to drop first.

Sorry if an above explanation set this straight for me, but it hasn't yet.

Well I can understand the intuition that says that the car stays level in the ground frame. It's a good intuition. But there's this other intuition that says that the car starts spinning in the ground frame. See post #14. Now I just happen to be quite sure that the latter intuition is right, even though the first intuition and the other posters disagree, and even though there's no easy explanation for the spinning in the ground frame.But how about a fighter jet that gets hit by a grenade on the front, and an identical grenade on the rear, simultaneously in the ground frame, and non-simultaneously in the jet frame. The rear and the front get the same upwards speed, so the jet does not spin, but moves upwards, front first in the jet frame, front and rear level in the ground frame.

Oh yes, there's the question what does a marble aboard the jet do, and why? I guess it does not roll anywhere, because the plane stays level in the ground frame, and the direction of gravity changes in the plane's frame. When the plane accelerates, the direction of gravity changes, right? I mean the direction depends on velocity.

 

[bahamagreen]

The observer on the ground shoots 2 lasers simultaneously (to him) so that each laser destroys the support cables, the 1 in the front and the 1 in the back. To that observer, they get destroyed simultaneously...

I think the problem is that "shoots 2 lasers simultaneously" and "the support cables...destroyed simultaneously" can not be the platform observer's experience.

The observer in the car will assume she is at rest and that the observer on the platform is in relative motion.

For the car observer, she observes that the concentric rings of leading edges of radio signals emitted by the platform observer will be closer together on her side than the far side.

If you draw this picture you will see that her exterior to interior interception of a circumference must always be with the front of her car first.

If you convert this to laser line distances, the platform observer must fire on the back cable attachment before firing on the front one in order to observe both release simultaneously; that is, if he fires the lasers simultaneously he will observe the front release being hit first.

 

[phinds]

I think the problem is that "shoots 2 lasers simultaneously" and "the support cables...destroyed simultaneously" can not be the platform observer's experience.

Yes. As I pointed out in post #4. I think the problem is that the OP is trying to use human "intuition" in a situation where it not only doesn't help, it actively hurts.

 

[hatflyer]

I think the problem is that "shoots 2 lasers simultaneously" and "the support cables...destroyed simultaneously" can not be the platform observer's experience.

If you convert this to laser line distances, the platform observer must fire on the back cable attachment before firing on the front one in order to observe both release simultaneously; that is, if he fires the lasers simultaneously he will observe the front release being hit first.

.

I don't agree with that at all. Use bullets if somehow lasers and radio signals add complexity. The ground observer will indeed see both cables snap at the same time. It's simply 2 signals of light bouncing off the cables straight back to the observer at the simultaneous moment in his frame. Why would anything be different? You shoot at 2 things at the same time, they reflect right back that they just started to snap at the same time..It's not intuition. It's just following the events in time.

The moving observer will view the front and rear cables snap at different times. That's when things get confusing.

 

[Dale]

In the ground frame, doesn't the force translation from the sudden loss of upward force from the front and back reach the center simultaneously?

No, they do not. You need to use the relativistic velocity addition formula.

You add the speed of the car to the speed of travel of the force effects atom by atom from the back, and subtract it from the front?

You need to use relativistic velocity addition

 

[PeterDonis]

I think the problem is that "shoots 2 lasers simultaneously" and "the support cables...destroyed simultaneously" can not be the platform observer's experience.

I don't see why not. The 2 lasers can be shot from the same point in space--i.e., the shots are "simultaneous" in the sense that they happen at the same event (point) in spacetime. (I believe this was the OP's intent.) All that then needs to be arranged is that that single point in spacetime is chosen such that the lightlike paths from it in opposite directions reach the two support cables at events that are simultaneous (in the usual SR sense) in the platform's frame. This is the same setup that is used in Einstein's classic simultaneity thought experiments, so I don't see why it wouldn't be possible.

 

[bahamagreen]

I don't see why not. The 2 lasers can be shot from the same point in space--i.e., the shots are "simultaneous" in the sense that they happen at the same event (point) in spacetime. (I believe this was the OP's intent.) All that then needs to be arranged is that that single point in spacetime is chosen such that the lightlike paths from it in opposite directions reach the two support cables at events that are simultaneous (in the usual SR sense) in the platform's frame. This is the same setup that is used in Einstein's classic simultaneity thought experiments, so I don't see why it wouldn't be possible.

Do you mean using one laser to hit a midpoint between the two supports from which a pair of lasers emit in opposite directions to the supports (all this from the platform observer's position)? I think he would observe the back support being cut first; to observe both support cuts simultaneously the "midpoint" would need to be offset toward the front of the car...?

 

[hatflyer]

No, they do not. You need to use the relativistic velocity addition formula.

You need to use relativistic velocity addition

So it's not symmetrical, in the sense that, using that formula, in 1 case the velocities of the car and the translation speed of the force are additive (from the front end), and in 1 case in a subtractive way (from the back end)? A brief look at the formula seems to show that it is not.

So can you conclude that the cable car is seen on the ground as having both ends start to fall simultanously, but in time the front falls further than the middle and the rear? And so the car hits the ground front first, not flat? Not sure why the rear and front would be different though.

 

[hatflyer]

Do you mean using one laser to hit a midpoint between the two supports from which a pair of lasers emit in opposite directions to the supports (all this from the platform observer's position)? I think he would observe the back support being cut first; to observe both support cuts simultaneously the "midpoint" would need to be offset toward the front of the car...?

No. 2 separate lasers on the ground, 1 aimed at the front end, 1 at the back end.

 

[FactChecker]

The problem of the ground person firing lasers toward two positions and cutting cables simultaneously is simple. It has nothing to do with the motion of the car.

 

[Orodruin]

Dale gave the correct explanation in #3. Why are we still debating this?

Coincidentally, I gave this related problem to my students on the exam last year.

During the French revolution, guillotines with a slanted blade were used to decapitate nobility. The guillotine blade at rest has the dimensions shown in the figure. Eager to save the nobility, the Scarlet Pimpernel rides by on his horse at velocity $v$. How fast does he have to ride in order for the guillotine blade to be horizontal in his inertial frame $S'$ if it falls at velocity $u$ in the guillotine rest frame?

It shows that what is slanted in one inertial frame is not necessarily slanted in another.

Spoiler

The problem is most easily solved by considering the two events when the ends of the blade reaches the $y' = y = 0$ plane. Letting the tip of the guillotine cutting the plane be the origin $x = x' = t = t' = 0$, the other end of the guillotine cutting the plane will have the coordinates
$$
t = \frac{\ell}{u} = \frac{\ell_0}{u\gamma_u}, \quad x = L,
$$
where we have taken into account that the blade is length contracted in the $y$-direction in $S$ with a factor $\gamma_u = 1/\sqrt{1-u^2}$. Transforming this to the $S'$ frame, we find that
$$
t' = \gamma_v (t - vx) = \gamma_v \left( \frac{\ell_0}{u\gamma_u} - vL\right).
$$
If the blade edge is horizontal in $S'$, then all points on the edge will cut the $y' = 0$ plane at the same time and since $t' = 0$ for the point cutting it, we find that
$$
\boxed{\frac{\ell_0}{u\gamma_u} - vL = 0 \quad \Longrightarrow \quad v = \frac{\ell_0}{uL\gamma_u}}.
$$

 

[hatflyer]

Dale gave the correct explanation in #3. Why are we still debating this?

Because the answer was too short. ;) What does each observer observe? Does the ground observer observe a horizontal car falling flat to the ground? Or do both see a front tilt down? But the front and the rear are both subject to the same force at the same time. So it's still not clear.

 

[Orodruin]

Assuming an idealised situation where the car falls down in a horizontal position in the ground rest frame, the car will not be horizontal in its rest frame. In the inertial frame that was originally the car's rest frame, the car will be slanted - just as the guillotine blade in #32.

 

[Dale]

Coincidentally, I gave this related problem to my students on the exam last year.

Excellent problem! Definitely a little on the macabre side

 

[hatflyer]

Assuming an idealised situation where the car falls down in a horizontal position in the ground rest frame, the car will not be horizontal in its rest frame. In the inertial frame that was originally the car's rest frame, the car will be slanted - just as the guillotine blade in #32.

Thanks. So the rider needs to tilt back to stay upright? How would that look to the observer on the ground?

 

[FactChecker]

There was a puzzling consequence pointed out by the OP in post #15 regarding whether a marble would roll toward the low end or not. I think that was answered by @PeterDonis in post #18, but the explanation is not trivial.

 

[Orodruin]

Thanks. So the rider needs to tilt back to stay upright? How would that look to the observer on the ground?

This depends on the inertial system in which you observe it.

 

[Dale]

So it's not symmetrical, in the sense that, ...

Yes, that is correct.

And so the car hits the ground front first, not flat?

I would have to work out the numbers to be sure what happens in any given frame, but if there is some frame where the ends hit simultaneously then there will be other frames where it hits front first and still other frames where it hits rear first.

 

[hatflyer]

This depends on the inertial system in which you observe it.

Not sure what you mean.
Before the cables snap, both observers see the rider standing vertically, 90 degrees from the car's floor and motion. After the breaks, if the ground observer sees the car still horizontal all the way down, he would see the rider vertical. But the rider sees a tilt, so he needs to tilt backwards, more than 90 degrees from the car's floor. How does that square?

I think it's like my marble question, will the marble roll forward or not? Seems the answer is it will not. Is that right?

[oops - now I'm getting opposing answers from you and Dale. ;) Does this car hit the ground horizontally or no in the ground observer's reference frame?]

 

[Dale]

Seems the answer is it will not. Is that right?

I don't know what the answer is. I would have to work it out, and the way it is posed is pretty complicated to work out and I would have to make a lot of clarifying assumptions. My experience is that working such a problem usually is a waste of time since it takes a lot of effort and then the requester disagrees with one of the clarifying assumptions or otherwise does not trust the results.

That is why I focused on the general principle. Direction is frame variant, so if something is horizontal in one frame it does not need to be horizontal in other frames. That is the key principle that you need to know here.

 

[hatflyer]

I don't know what the answer is. I would have to work it out, and the way it is posed is pretty complicated to work out and I would have to make a lot of clarifying assumptions. My experience is that working such a problem usually is a waste of time since it takes a lot of effort and then the requester disagrees with one of the clarifying assumptions or otherwise does not trust the results.

That is why I focused on the general principle. Direction is frame variant, so if something is horizontal in one frame it does not need to be horizontal in other frames. That is the key principle that you need to know here.

My problem is applying those principles to a problem in a thought experiment. Seems the answers have been all over the place.

The assumptions I think are pretty basic. Cut the cords on a moving car and watch it fall. In 1 reference they are cut simultaneously, so in the other not. Does it tilt in either/both frames of reference in relation to its original direction? I should think it should not require any computations. It's just a concept I've been mulling over and hoped I could get an answer to it.

 

[Dale]

I should think it should not require any computations.

Well, maybe someone with more background could do it without computations, but I can't.

The assumptions I think are pretty basic. Cut the cords on a moving car and watch it fall. In 1 reference they are cut simultaneously, so in the other not. Does it tilt in either/both frames of reference in relation to its original direction?

How long are the cords? Where are they cut? How long is the car? Where are the cables attached to the car? What is the speed of sound in the cords? What is the speed of sound in the material of the car? What is the stiffness of the car? Are we neglecting air? Is gravity considered uniform? Which frame is the ground spatially flat? Etc...?

My recommendation: get rid of gravity from the scenario and simplify it as much as possible.

 

[PeterDonis]

2 separate lasers on the ground, 1 aimed at the front end, 1 at the back end.

Are both lasers at the same point on the ground? Or are they separated?

 

[PeterDonis]

In the inertial frame that was originally the car's rest frame

If the car falls downward once the support cables are broken, then it did not start out at rest in an inertial frame. It started out at rest in a non-inertial frame that is accelerating upward.

Most of the discussion in this thread has ignored the proper acceleration, but without the proper acceleration, the car doesn't fall, because the support cables aren't supporting anything.

 

[PeterDonis]

I think it's like my marble question

Not quite, because the rider has a vertical length that the marble does not. That makes a difference, because now we have to think about the motion of a whole vertical rod (the rider), instead of just a point (the marble).

 

[Janus]

One of the things you have been assuming that the car can be treated as if it were infinitely rigid

So it's not symmetrical, in the sense that, using that formula, in 1 case the velocities of the car and the translation speed of the force are additive (from the front end), and in 1 case in a subtractive way (from the back end)? A brief look at the formula seems to show that it is not.

So can you conclude that the cable car is seen on the ground as having both ends start to fall simultanously, but in time the front falls further than the middle and the rear? And so the car hits the ground front first, not flat? Not sure why the rear and front would be different though.

Okay. one thing that needs to be considered here is that the car itself can not be treated as if it is perfectly rigid (there are no perfectly rigid objects in Relativity.)
So even just hanging there, the car will sag a bit in the middle and be under stress. If you cut the cables simultaneously in any given frame, the car will try to restore to its normal stress-free state. This release of stress will propagate from the ends toward the middle at the speed of sound for the car material.
If the cables are cut simultaneously in the Ground frame, then the propagation speeds relative to the car as measured from the ground frame will be different coming from both ends, meaning they won't meet at the center of the car, but closer to the rear of the car And it is not until they meet at this point that this point will begin its fall.
They will then reflect off of each other and head back to the ends. But now the reflection heading towards the front will be traveling faster relative to the car then the one traveling back towards the rear. By the time either has returned to the ends, the ends will have over-shot the stress-free point. they will reach the maximum stress point in the opposite direction and then start moving upwards relative to the car. Basically, the ends of the car will "vibrate" back and forth around an equilibrium point. Since the equilibrium point id closer to one end than the other, the period of oscillation will differ for the two ends. So which end hits the ground first will depend on where in the oscillation each end is when the ground is reached.
As far as the car observer goes, the two ends are released at different times, and the forces from the ends propagate at the same speed relative to the car. Thus they meet closer to the rear than the front, and since this equilibrium point is closer to one end than the other, the ends oscillate with different periods. (just as noted from the ground frame.) Working out the exact oscillations in each frame would be quite a chore, but be assured both frame will end up agreeing as to which end, if either, hits first. As to what an observer at the midpoint would experience, he too would be subject to propagation speed effects, so it is a bit of a complex problem to attack.

 

[MikeLizzi]

Following up on Dale’s post #43 suggesting that you simplify the problem as much as possible. Here is a much simpler problem I worked on in the past. I believe it contains most of the geometric consequences that are your concern. There are 4 pictures.

Picture 1:
The initial conditions as observed within a train car. Train tracks are moving at constant velocity horizontally beneath. A barbell in the car has just been released from a fixture in the ceiling. To avoid the complexity of gravity/acceleration, the tracks, car and barbell are in a gravity free environment and the barbell is given some constant downward velocity by some mechanism not shown.

Picture 2:
The barbell has fallen vertically and landed on the floor of the car. No bouncing allowed.

Picture 3:
The initial conditions as observed from the train tracks. The car is moving horizontally and is width contracted. The barbell is moving diagonally and is tilted. (There are other smaller geometry changes to the barbell too).

Picture 4:
The lead end of the barbell hits the floor first. This is not a paradox. It’s just a glaring example of relativity of simultaneity.

The purpose of the exercise is to think about what forces are transmitted through the bar. The bar itself is one of those mythical weightless, frictionless, perfectly rigid things.

According to the observer in the car, when it hits the floor, equal vertical forces are applied to stop the ends of the bar. The bar itself, being weightless, stops when the ends stop. Note that there is no way to generate a horizontal force so no forces are transmitted axially through the bar.

But the observer on the tracks might think that there will be some axial forces transmitted from the lead end to the far end of the bar. That would be a paradox. One inertial observer concludes that forces are present in a body and another inertial observer does not. Fortunately, a careful examination will show that is not the case.

 

[FactChecker]

Picture 4:
The lead end of the barbell hits the floor first.

You mean trailing end, right?
Your example is very good. It shows that acceleration is not essential to the issue. I am afraid that I can't see how there is no axial force and it seems as though the bar would have to "flop down" on the floor like a rubber hose. I'll have to think about that, but it may be beyond me.

EDIT: I guess that at the point of contact of the bar with the floor, there is acceleration. So that might account for an angle of the bar at that point.

 

[hatflyer]

Following up on Dale’s post #43 suggesting that you simplify the problem as much as possible. Here is a much simpler problem I worked on in the past. I believe it contains most of the geometric consequences that are your concern. There are 4 pictures.

Picture 1:
The initial conditions as observed within a train car. Train tracks are moving at constant velocity horizontally beneath. A barbell in the car has just been released from a fixture in the ceiling. To avoid the complexity of gravity/acceleration, the tracks, car and barbell are in a gravity free environment and the barbell is given some constant downward velocity by some mechanism not shown.
View attachment 204304

Picture 2:
The barbell has fallen vertically and landed on the floor of the car. No bouncing allowed.
View attachment 204305

Picture 3:
The initial conditions as observed from the train tracks. The car is moving horizontally and is width contracted. The barbell is moving diagonally and is tilted. (There are other smaller geometry changes to the barbell too).
View attachment 204306

Picture 4:
The lead end of the barbell hits the floor first. This is not a paradox. It’s just a glaring example of relativity of simultaneity.
View attachment 204307The purpose of the exercise is to think about what forces are transmitted through the bar. The bar itself is one of those mythical weightless, frictionless, perfectly rigid things.

According to the observer in the car, when it hits the floor, equal vertical forces are applied to stop the ends of the bar. The bar itself, being weightless, stops when the ends stop. Note that there is no way to generate a horizontal force so no forces are transmitted axially through the bar.

But the observer on the tracks might think that there will be some axial forces transmitted from the lead end to the far end of the bar. That would be a paradox. One inertial observer concludes that forces are present in a body and another inertial observer does not. Fortunately, a careful examination will show that is not the case.

Thanks much.
Ok, so if there is a ball at the midpoint of the barbell (and is subject to the same initial force downward), what does the ball do as seen by the tracks observer? For the car at rest perspective, the ball stays put, right in the middle. It never touches the ends. I assume for the tracks view, the ball must also stay in the center, even if the bar goes from tilted to finally horizontal in the end.

So what if you flip your experiment, so that, to the car at rest perspective, the bar receives the downward force at the front before the force is applied at the back in such a way that there is no rotation, he would see the bar fall diagonally. To the ground observer, if he sees the force to the front and rear of the bar applied at the same time, can it be so that he could see the bar falling horizontally? What happens to the ball in the middle here? To the ground observer, it would not move. To the car observer, the bar being tilted, somehow he also concludes the ball doesn't move? If he sees a tilted bar, why wouldn't he see the ball move?

Thanks.