GRED Answer: Fast train cars connected by string

[Henning Makholm]

My brain is raped. You're saying that two sleepers are longer apart on paper, but if I put my measuring rod agains them, they are shorter apart. Well, if this is the case, your paper does not describe my experience and is, as you said, "mathematical abstraction" - so why consider it? Btw, what's that bit about "time coordinates are crazy"?

Sorry about the brain. You really shouldn't let it go dressed that way. Just sayin'.

It's really quite simple, but perhaps I managed to complexify it in my explanation. I'll try again:

Relativity is all about events, that is, something that happens at a particular time and a particular place. A reference frame means a system for labeling events with time and space coordinates. The rotating frame we're talking about here is constructed in the following way:

We asssign each event space coordinates that tell where in the train it happened. All events with the same space coordinates will have happened at the same place in the train, measured for example as meters forward of the midpoint of some chosen car. (The coordinates of events in the next car over will have space coordinates that reflect how long the distance between cars look to the passengers. This is unproblematic as long as we're only interested in the situation after the train is already moving with constant speed). This choice of space coordinates is really unavoidable given that we want to construct a rotating frame that follows the train. But we still have to choose time coordinates for each event. Here we choose to label each event with the timestamp that somebody on the ground would assign to it. This choice is dictated by our wish to exhibit the symmetry of the circular track and train. We can multiply all time coordinates by a common factor in an attempt to account for the time dilation for clocks that travel with the train (let's imagine that we do this, though it won't have any implication for what follows), and we can add a common offset to all time coordinates (pointless but harmless). But if we use different factors or different offsets for any two places on the train, we will have broken the symmetry -- an observer on the platform will see some cars reach 12 o'clock (according to rotating-frame coordinates) before or after it happens in other cars, and this will destroy our assumption of symmetry.

So both time and space coordinates are more or less forced upon us by the idea "let's construct a rotating frame and do our analysis there". However, something horrible has happened: the time and space coordinates do not fit together right. To see this, consider one lightbulb flashing in each end of our railway car, such that the passengers in the car consider the flashes to have happened at the same time (i.e., the light from both flashes reaches the midpoint of the car simultaneously). There's a classic SR gedankenexperiment that concludes in precisely this circumstance that an observer on the platform will not consider the two flash events to be simultaneous. But this means that the two events have different time coordinates in the rotating frame, by definition (above) of said time coordinates. The passenger on the train will have seen two flash events and believe them to be simultaneous -- but they have different time coordinates according to the rotating frame. This is what I mean by the time coordinate being "crazy".

When I say that the distance between sleepers increase in the rotating frame, what I mean is neither more nor less than the following: Select an event on sleeper A. Select an event on sleeper B, such that the two events have identical time coordinate. Subtract the space coordinates of those two events. The difference is larger than the distance found by the ground observer.

More "practically", suppose that the ground observer places a lightbulb on each of the two sleepers, and arranges for them to flash at the same time in his frame. By design those two flashes have the same time coordinate in the rotating system. To find their space coordinates, we must intercut to the passenger. He sees (through a window in the car floor?) flash A happen right under the middle of his car. A bit later, according to his subjective time, he sees flash B happen about a meter to the rear. Therefore he can conclude that the sleeper distance in the rotating coordinate system is one meter. However, he won't himself believe this to be the true distance between sleepers. Why, while he was waiting for sleeper B to flash, sleeper A has been hurtling rearwards, and was perhaps only 30 cm from sleeper B "when" the latter flashed. So the measuring-stick distance he would prefer between the sleeper is 30 cm. (Meanwhile, the observer on the ground measures 60 cm between the sleepers).

In summary, the rotating coordinate system is one that does not match the experience of the passenger. And the reason for the whole mixup was that we wanted to have time coordinates that made the symmetry of the experiment manifest. All we need to do to alleviate the passenger's confusion is to let him use the coordinates of the intertial frame in which the car is momentarily at rest. In this frame the track contracts as it should do, the car in front of us reaches its top speed and turns off its engine a short time before ours does, in short everything behaves as it should.

The only thing we lose in the inertial frame is the symmetry of the experiment. For example, we cannot expect that the number of sleepers we count between our car and the next one at some instant in the inertial comoving frame is also the same as the number of sleepers between two cars at the other side of the circle at the same instant. They are not in a symmetric situation; they have a different velocity with respect to the frame where we do our sleeper-counting.

So, did Einstein err when he constructed the rotating frame in the first place? Not at all, because he was doing GR here, not SR. And in GR it is generally expected that coordinates by themselves are meaningless; you need to transform the coordinates into a locally inertial frame before you try doing SR-like physics with them. Our fallacy was to attempt to extract a "spatial geometry" from the metric by ignoring the time coordinate, and nevertheless expect the spatial geometry to give sensible results about things that involve time (namely: the motion of sleepers under the train).

[makc]

if my brain had an anus, it would be bleeding now I don't really see how your explanation above help. Same way I could leave my clock on the ground, spin around myself once, pick up the clock and say, "oh look, pluto just orbited around me with speed faster than light". Surely this is not the way to construct good reference frames, is it? what I want here is description of situation in terms of clocks and rods car passenger carries with him. co-moving inertial observer is not good for this, I mean... come on, he even thinks track is shortened to ellipse.

edit: ha ha, now that I read this post some hours later, I see that that's not really different if I don't put my clock on the ground, is it I need a fresh look at this, so I'm leaving this train at rest for a day.

[Henning Makholm]

if my brain had an anus, it would be bleeding now :( I don't really see how your explanation above help. Same way I could leave my clock on the ground, spin around myself once, pick up the clock and say, "oh look, pluto just orbited around me with speed faster than light". Surely this is not the way to construct good reference frames, is it?

No, that's kind of my point. The rotating frame is not a very good reference frame.

[wonderboy]

What about friction and drag? Would friction not cause the track to heat up at these speeds (depending on the size of the track (but I'm thinking that if the trains are spaced equally then the friction would be constant enough to heat the track up sufficiently enough to make it fundamentally useless)

Also, if the track is sufficiently big that for most of the time the trains are moving in an apparent straight line, would there not be a slip stream in which the second and third train would make use of thus speeding up more quickly than the train in front?


In this situation, the string will droop.

Paul.

[alter-ego]

[....]Our fallacy was to attempt to extract a "spatial geometry" from the metric by ignoring the time coordinate, and nevertheless expect the spatial geometry to give sensible results about things that involve time (namely: the motion of sleepers under the train).

I think I'm reading that the gist of what you're saying is clock synchronization is an issue in the rotating frame. It seems to me that a passenger trying to measure a circumfrence would find it a problem. I'm curious, do you still think the strings break?

[alter-ego]

What about friction and drag? Would friction not cause the track to heat up at these speeds (depending on the size of the track (but I'm thinking that if the trains are spaced equally then the friction would be constant enough to heat the track up sufficiently enough to make it fundamentally useless)

Also, if the track is sufficiently big that for most of the time the trains are moving in an apparent straight line, would there not be a slip stream in which the second and third train would make use of thus speeding up more quickly than the train in front?


In this situation, the string will droop.

Hi Paul,
I think your concern about heat, friction, slip streams, etc are not warrented. There are forces in a non-inertial reference frame that have a fundamental impact on relativistic analysis which you should be thinking about, but ignore the other practical phenomena. They are just distractions.

[Henning Makholm]

[....]Our fallacy was to attempt to extract a "spatial geometry" from the metric by ignoring the time coordinate, and nevertheless expect the spatial geometry to give sensible results about things that involve time (namely: the motion of sleepers under the train).

I think I'm reading that the gist of what you're saying is clock synchronization is an issue in the rotating frame. It seems to me that a passenger trying to measure a circumfrence would find it a problem.

That's about what I'm saying, yes.

I'm curious, do you still think the strings break?

Yes, of course. My own argument for the strings breaking works entirely in the ground frame.

The reason why I'm considering other frames at all is to convince myself (and such readers as I can convince along with myself) that I can find flaws in any competing argument that reaches a different conclusion.

[makc]

I actually still not convinced, but I don't mind if they break. I just want to "see" it. As my mental powers aren't quite good for this, I started writing a program yesterday, that would show me how someone moving in certain way would see the track So far I tested it on inertial motion and it generally matches special relativity picture, but I have some numeric errors that get worse near observer trajectory. Which is why I can't safely apply it to cars (yet), because their trajectory coincides with the track. I'm thinking to rewrite my code to higher precision or maybe rearrange operations, but that will have to wait another day...

[RJN]

I believe the best answer is "The strings break." This is a classic problem in relativity correctly identified elsewhere as the Ehrenfest Paradox. Much of the surrounding discussion that arrives at the "strings break" answer appears correct. Nevertheless, here my brief explanation.

In sum, the moving trains will actually physically contract via the famous Lorentz contraction. Since the track is not moving, however, it will not contract. Therefore, the shorter trains-and-strings can no longer remain connected around the track, and the strings will break.

***

Here are some interesting tidbits. First, people rotating with the trains ("train people") will have their rulers contract, too, as measured by observers at rest with the track ("track people"), so that these train people measure the circumference of the track to be longer than the track people. Specifically, if the track people measure the circumference to be "C", the train people will measure it to be C/sqrt(1-v2/c2).

Next, one reason I believe this is because Albert Einstein himself worked on the Ehrenfest Paradox and came to the conclusion of the previous two paragraphs. And he was REALLY good at solving these (and many other) sorts of things.

Still, although I believe this solution, I found it initially unsettling. First, in examples involving relative motion between inertial frames (no-acceleration frames), the Lorentz contraction was never really a physical contraction, but rather a perceptual contraction that observers in each frame could apply to the other frame. In other words, if two trains traveling in straight lines pass, and each train has cars attached by strings, each train sees the other train Lorentz contracted, but none of the strings break or droop because no real forces are being applied.

But here in the Ehrenfest Paradox, the strings actually break. Strange! What force breaks them? Why can't the train people see the track contracted and so cause the strings to droop? The answer to both of these questions is that the train people are accelerating and therefore not in an inertial frame, and therefore what they measure is more complicated. Apparently, the force that causes the acceleration can possibly be tapped to break the strings.

What makes this, for me, perhaps less unsettling is to picture the analogous situation involving gravity. There, one could consider trains again at rest on the track, still connected by stings, but instead of accelerating them around the circular track, this time put a heavy mass at the very center. For spectacular renditions in the Disney movie version, this can be a black hole. What happens to the strings in this case? I believe they break, then, too.

[Henning Makholm]

I actually still not convinced, but I don't mind if they break. I just want to "see" it. As my mental powers aren't quite good for this, I started writing a program yesterday, that would show me how someone moving in certain way would see the track

FWIW, here is what I get for a train of 12 cars that almost fill the track when stationary. (The cars are slightly curved, because I didn't bother to write code aligning straight cars with a chord). For comparison I have planted 12 trees (green dots) in a circle just inside the track. The trees don't move.

The train standing still

0.png (23.38 KiB) Viewed 4976 times

The train moving at 0.75c, from the ground

ground.png (29.37 KiB) Viewed 4976 times

The train moving at 0.75c, in the instantaneous frame of someone in the middle of the blue car.

train.png (22.56 KiB) Viewed 4976 times

I'm afraid this result does not look very intuitively enlightening. The oblique car ends are (as far as I can figure out) a real effect, coming about because the car are moving partly sideways with respect to the observer, and so are relativistically squished along a diagonal direction.

[Henning Makholm]

Still, although I believe this solution, I found it initially unsettling. First, in examples involving relative motion between inertial frames (no-acceleration frames), the Lorentz contraction was never really a physical contraction, but rather a perceptual contraction that observers in each frame could apply to the other frame. In other words, if two trains traveling in straight lines pass, and each train has cars attached by strings, each train sees the other train Lorentz contracted, but none of the strings break or droop because no real forces are being applied.

As MAReynolds pointed out early in the thread, essentially the same effect does occur on a straight line. Have a line of cars stand on a straight track. Let them all start accelerating at the same rate (as measured on each car's own accelerometer). The cars will separate, and strings, if any, between the cars will break. If a long train is to stay connected while accelerating to relativistic speeds, the rear cars need to accelerate faster.

Based on this, I'd insist that it is the acceleration along the track that is ultimately responsible.

Note that flatlanders living on a cylinder could make exactly the same thought experiment with a straight (to them) track going around their cylinder. They would not be able to lay the blame any specific effect of "rotation", so we shouldn't have to either.

What makes this, for me, perhaps less unsettling is to picture the analogous situation involving gravity. There, one could consider trains again at rest on the track, still connected by stings, but instead of accelerating them around the circular track, this time put a heavy mass at the very center. For spectacular renditions in the Disney movie version, this can be a black hole. What happens to the strings in this case? I believe they break, then, too.

I'm not sure about this analogy. The black hole draws things inward, whereas centrifugal force pushes things outward. That makes it difficult be sure whether the analogy would work in a direct or opposite way.

[makc]

As MAReynolds pointed out early in the thread, essentially the same effect does occur on a straight line. Have a line of cars stand on a straight track. Let them all start accelerating at the same rate (as measured on each car's own accelerometer). The cars will separate, and strings, if any, between the cars will break. If a long train is to stay connected while accelerating to relativistic speeds, the rear cars need to accelerate faster.

Straight track fits my feeble brain much better ok, so moving car on a straight track would see the car in front of itself where ground observer would see it at some point later, and the car behind it would be seen where it was at some point before, according to ground observer - this can be used to argue that strings will break But, in case of circular track, 1st and last cars are connected, and so such effect should be only true locally at circle track, if at all. Or else you would eventually see 1st and last cars overlapping, which is obviously not possible.

edit: actually this picture is nice. now I have less motivation to fix my program, thanks

[RJN]

essentially the same effect does occur on a straight line. Have a line of cars stand on a straight track. Let them all start accelerating at the same rate (as measured on each car's own accelerometer). The cars will separate, and strings, if any, between the cars will break. If a long train is to stay connected while accelerating to relativistic speeds, the rear cars need to accelerate faster.

Actually, I didn't mention acceleration, just velocity. So if I'm sitting here on my unaccelerated train connected by strings, it shouldn't matter that other trains pass me by with any velocity or acceleration -- my strings won't break. And the same goes for any constant velocity train that passes me by.

As for acceleration, I agree that there should be some way to accelerate the train cars to an inertial frame without breaking the strings. Exactly how that works might indeed be complicated.

I'm not sure about this analogy. The black hole draws things inward, whereas centrifugal force pushes things outward. That makes it difficult be sure whether the analogy would work in a direct or opposite way.

Yes, the acceleration directions (relative to the center) are in opposite directions. Still, although I frequently miss (other) minus signs, I believe the strings break in both cases.

- RJN

[makc]

Actually, I didn't mention acceleration, just velocity. So if I'm sitting here on my unaccelerated train connected by strings, it shouldn't matter that other trains pass me by with any velocity or acceleration -- my strings won't break. And the same goes for any constant velocity train that passes me by.

Same goes to circle train then - if we make the cars move at constant angular velocity first and then connect them, nothing breaks.

[alter-ego]

Actually, I didn't mention acceleration, just velocity. So if I'm sitting here on my unaccelerated train connected by strings, it shouldn't matter that other trains pass me by with any velocity or acceleration -- my strings won't break. And the same goes for any constant velocity train that passes me by.

Same goes to circle train then - if we make the cars move at constant angular velocity first and then connect them, nothing breaks.

That's true, and if you then slow down the train, the strings will droop. Although I didn't find a direct quote, Einstein supposedly said that to avoid the deformation stage during spin up, the cylinder or disk (as originally presented by Ehrenfest) should be done in a hot melt, and then solidified when accelaration is complete. Obviously, the atoms are "free" in a fluid stage, but they are still being redistributed while the melted disk is spun up.

[RJN]

Same goes to circle train then - if we make the cars move at constant angular velocity first and then connect them, nothing breaks.

Well, yes, but perhaps I should be more clear. In my opinion, there IS a way to linearly accelerate train cars connected in a line to a constant linear velocity without the strings that connect them breaking. However, there is NO way to accelerate train cars connected around a circle to a constant angular velocity without the strings breaking. Yes, this does carry the presumption that all of the train cars undergo the same acceleration as measured from the inertial frame at the center.