Can an Altered Thought Experiment Reconcile the Special Relativity Conundrum?

[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.

 

[Herman Trivilino]

Imagine the device known as a carpenter's level in free fall towards a horizontal surface. The device is horizontal as it falls, as verified by the fact the bubble is in the center. Both ends of the device hit the surface at the same time. All of this is in a frame of reference in which the device falls vertically downward. The vertical direction being of course perpendicular to that horizontal surface.

When all of this is observed from a passing train the two ends of the device do not hit the surface at the same time, the bubble is in the middle, and the direction of the fall is not perpendicular to the surface.

The trailing end of the device will strike the surface $\frac{Lv}{c^2}$ before the leading end, where $L$ is the proper length of the device, $v$ is the speed of the train relative to the surface, and $c$ is the speed of light.

As @Dale stated in Post #3, what's horizontal in one frame isn't necessarily horizontal in another.

 

[FactChecker]

@Mister T , What would the bar look like after the trailing end hits and the leading end is still falling? Is the trailing end flat on the ground for some length and then the bar makes an angle straight to the leading end still in the air? I guess then one observer would see the bar as straight all the time and the other would see it as bent as it hit the ground. That seems strange, but I don't see any other possibility.

 

[PeterDonis]

Here is a much simpler problem

Your problem differs from the OP in that in your case, the barbell's ends hitting the floor are simultaneous in the train frame instead of the ground frame. I think it's instructive to consider both cases, though.

 

[Orodruin]

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.

Indeed, I wilfully and explicitly ignored the physics of how things would actually happen as described and instead considered a case where the entire cart is a horizontal line that starts accelerating down at the same time as the supports are removed (in the ground rest frame). This is of course a simplifying assumption and something I was pretty sure the OP has not considered but just assumed. The problem is that this

The assumptions I think are pretty basic. Cut the cords on a moving car and watch it fall.

is not as basic as the OP thinks it is. Obviously, if cutting the cord is the reason things start falling, then it will also take some time before the structure in the middle no longer supports the wagon and the OP is clearly assuming that this occurs instantaneously.

While I think it is fine to ask what happens to the idealised line, the OP runs into much deeper waters than he realizes when he asks about the marble because it is a case he tries to cover with "intuition". The marble has several problems, one of them being that the OP assumes that it is gravity that accelerates it towards the direction which the wagon is sloped. I am also not very fond of SR questions involving gravity for obvious reasons. Even classically, if the wagon falls in a sloped position at the same acceleration as the gravitational acceleration, the marble will not roll anywhere because it will be falling with exactly the same acceleration and there will be no normal force from the floor on the marble. I think it is fair to say that the OP has taken water over his head in thinking that this is a "pretty basic" question without many further qualifying assumptions.

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.

This problem is exactly equivalent to my guillotine problem in post #32.

@Mister T , What would the bar look like after the trailing end hits and the leading end is still falling? Is the trailing end flat on the ground for some length and then the bar makes an angle straight to the leading end still in the air? I guess then one observer would see the bar as straight all the time and the other would see it as bent as it hit the ground. That seems strange, but I don't see any other possibility.

If the bar stops as it hits the floor in the train rest frame, then it will be bent at a single point in the ground frame.

 

[MikeLizzi]

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.

The Trailing end hits the floor first. Yes, I need to correct that. Also, as you indicated, once one end hits the floor, acceleration is unavoidable and one must abandon the weightless, frictionless, rigid assumption.

 

[MikeLizzi]

@Mister T , What would the bar look like after the trailing end hits and the leading end is still falling? Is the trailing end flat on the ground for some length and then the bar makes an angle straight to the leading end still in the air? I guess then one observer would see the bar as straight all the time and the other would see it as bent as it hit the ground. That seems strange, but I don't see any other possibility.

If the bar undergoes any (Edit: permanent bending) bending with respect to the train track observer, that would constitute a paradox since there can be no bending with respect to the train car observer.
So, once one end hits the floor, one has to abandon the weightless frictionless, rigid assumption. I did a crude finite element analysis for this. It gets messy. The key is to recognize that each element in the tilted falling bar has a somewhat diamond shape (vertical sides). Each element experiences a shear force from the stopped end and passes it on. The shear force reduces the vertical component of the velocity for each element and each element changes shape to a contracted rectangle in sequence. The result is the bar pivots and shrinks till the other end hits the floor.

Edit: Now that I think about it, the rubber hose analogy from FactChecker has value.

 

[MikeLizzi]

This problem is exactly equivalent to my guillotine problem in post #32.If the bar stops as it hits the floor in the train rest frame, then it will be bent at a single point in the ground frame.

Sorry, I was occupied writing my reply and did not see your post #32 before I posted.
With respect to the bending, I tried to address that in my post above #56. The bending conclusion was the original paradox I had to deal with.

 

[MikeLizzi]

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.

With respect to placing a ball at the center of the barbell. As has been mentioned by other posters, since the ball and bar experience the same accelerations, the relationship (relative position/relative velocity) between them cannot change even in a Newtonian world.

With respect to your follow up question involving flipping the experiment, your suggestion is that a downward force is to be applied to one end before a similar force is applied to the other end but no rotation (angular velocity) is to occur. That can't happen. (Edit: I spoke too soon. The angular velocity will be canceled after the second force is applied) Regardless, your focus seems to be on the ball now. Hold a bar vertical with a ball beside it. Then let go. Do they not fall together as if glued?

 

[Dale]

This is of course a simplifying assumption and something I was pretty sure the OP has not considered but just assumed

This is similar to my experience in the past. The questioner has a whole bunch of unstated simplifying assumptions in mind, and hasn't the background to consider which are compatible with relativity and which are not. The fact that some of the unstated assumptions are incompatible with relativity is a big problem because nobody can work the problem they have in mind.

I am also not very fond of SR questions involving gravity for obvious reasons.

Me too. I won't even attempt such problems as an educational exercise for SR. Those should be reserved for courses in GR.

 

[Dale]

I guess then one observer would see the bar as straight all the time and the other would see it as bent as it hit the ground.

Yes, that is correct. This is one place where the "no rigid objects" thing is glaring and the described motion is not Born rigid.