Train traveling near light speed

[uniqueland]

TL;DR

how many times did a near light speed train circle the earth if you are a passenger vs an observer at the train station?

A train is traveling around the Earth at just under light speed. Light would circle the Earth around 7 times per second so let's say this train cricles the Earth 6 times per second. There is a physical ticker on the track of the train that records revolutions. Each time the train makes one revolution the ticker adds one revolution to the counter total. Disregarding the impossible g forces of impossible acceleration and deceleration and say you reached your max speed pretty much instantly, the passenger on the train might experience a trip of say 10,000 seconds whereas the observer at the train station experience 10 years went by. But the physical COUNTER on the track recorded the number of physical revolutions that the train actually made during that time that the train started and ended. The physical counter's display was shown on both a display at the train station and on a display on the train itself too. If the counter would not add another revolution unless the physical train actually passed by the counter, thereby causing that physical counter to add another revolution to the total, how can the counter on the train be different than the counter at the train station?

 

[Ibix]

let's say this train cricles the Earth 6 times per second.

As measured by who? I'll assume the Earth frame measures this. At this speed, the time dilation factor is about 1.94.

how can the counter on the train be different than the counter at the train station?

It wouldn't be. It would show six times the number of seconds measured by the Earth. The train observer would measure it ticking upwards 6×1.94=11.5 times per second, but a journey duration of 1/1.94 what the Earth measures, so the same total count.

Note that a rotating frame is non-inertial and naive time dilation and length contraction calculations (which assume an inertial frame) won't necessarily work to get a description of the train's measurements. In particular, time dilation is not symmetric in this case.

 

[Nugatory]

how can the counter on the train be different than the counter at the train station?

It cannot, of course. They both count the same number of orbits, but with a different elapsed time per orbit.

 

[mfb]

From an instantaneous view from the train, Earth is a very oblate spheroid: It is much shorter in travel direction, allowing the train to travel more distance over the ground per train second than the train can travel per Earth second.

If the train is fast enough it can circle Earth thousands of times per second according to the clock on the train. Train and Earth will agree on the number of revolutions, but not on the time that passed.

 

[Ibix]

Incidentally, what didn't you understand about the answers the last time you asked this question?

 

[Quasimodo]

How is this different from a train moving on a straight line?

Just draw your lines on a spacetime diagram on an infinite cylinder radius 6,375 km!

 

[Ibix]

How is this different from a train moving on a straight line?

It's not inertial. That makes naive calculations go wrong. Certainly you can unwrap the cylinder to draw the relevant part of spacetime in the Earth frame, but drawing a spacetime diagram in "the frame of the train" goes wrong because a naive attempt at such a thing produces lines of simultaneity that don't close.

 

[Quasimodo]

It's not inertial. That makes naive calculations go wrong. Certainly you can unwrap the cylinder to draw the relevant part of spacetime in the Earth frame, but drawing a spacetime diagram in "the frame of the train" goes wrong because a naive attempt at such a thing produces lines of simultaneity that don't close.

We agree on the inertial part. But the lines of simultaneity "in the frame of the train" have all points each at different times. That they have to be closed is a naive notion inhereted from their closed condition at v=0.

 

[Ibix]

But the lines of simultaneity "in the frame of the train" have all points each at different times.

...I can't parse this sentence.

That they have to be closed is a naive notion inhereted from their closed condition at v=0.

No. That maps between the coordinate space and the actual manifold need to be bijections in their domain of validity is a fundamental requirement of defining coordinates. You can get away with it in this case by excluding a line up the cylinder, but then you can't talk about the time for a complete orbit in this system.

 

[Quasimodo]

but then you can't talk about the time for a complete orbit in this system

I certainly can!

 

[Dale]

How is this different from a train moving on a straight line?

As you mentioned in the description the g forces are very high in this scenario while they are zero for the straight line. This indicates that the straight line is inertial and this is not.

 

[Quasimodo]

As you mentioned in the description the g forces are very high in this scenario while they are zero for the straight line. This indicates that the straight line is inertial and this is not.

Yes, but OP disregards g-forces etc., so an infinite cylinder radius Earth is the rigtht space-time to use, if we want him to understand his scenario.

 

[Dale]

Yes, but OP disregards g-forces etc., so an infinite cylinder radius Earth is the rigtht space-time to use, if we want him to understand his scenario.

I don’t see how a cylinder vs a sphere changes anything. You would still have g forces etc. Gravity would be weird on a cylindrical planet, but I don’t think we were considering gravitational effects anyway.

 

[Ibix]

I don’t see how a cylinder vs a sphere changes anything.

I think @Quasimodo is talking about this kind of thing: https://www.physicsforums.com/threads/a-clocks-double-life.974236/post-6202880 (edit: see also posts #30 and #33 in that thread). The problem is that it doesn't really help as an intuitive explanation of this scenario (one of my purposes when I drew those diagrams was to illustrate the break in synchronisation). However you slit the cylinder to unwrap it either the Earth or train clocks have to leave the chart and re-enter, and apart from the Earth frame this results in a clock correction. And it was pointed out (in the thread I linked above, I think) that a chain of locally-Einstein-synchronised clocks on the train aren't globally Einstein-synchronised anyway except in the Earth frame - so you bump up against the complexities of rotating frames anyway.

 

[Herman Trivilino]

Think about the train making only one revolution. This is the twin paradox. The train passenger's clock will show less elapsed time than the station clock.

This is different from the train moving in a straight line because the train will never return to the station. The train clock and the station clock will never again share the same location so that their readings can be compared.

Note that this experiment is done every day in circular particle accelerators. Particles that decay live longer because they are moving in the circular beam tube.

 

[Quasimodo]

I think @Quasimodo is talking about this kind of thing: https://www.physicsforums.com/threads/a-clocks-double-life.974236/post-6202880 (edit: see also posts #30 and #33 in that thread). The problem is that it doesn't really help as an intuitive explanation of this scenario (one of my purposes when I drew those diagrams was to illustrate the break in synchronisation).

Correct. If the passenger on the train tried to synchronize the cars of the train with light beams, there would be a break in synchronization for a stationary observer (his watch and every point on the train would read multiple times), but if he ignores this and just reads the passing stations clocks outside his window, there is nothing wrong with this model.

 

[Quasimodo]

This is different from the train moving in a straight line because the train will never return to the station.

Of course it returns! Same place, at a different time along the line. This is one revolution.

 

[NoahsArk]

If I am understanding this correctly, the train driving over the ticker is considered an event, just like an explosion happening on the train every X number of miles. The number of events will be the same in both the train's and the Earth's perspective. What will be different is the time elapsed between the events. So, at a speed of 6/7 the speed of light, someone on Earth would observe the train's clocks at moving at almost half (1.94 as Ibix said) times slower than its own clocks. So for each revolution (or "tick"), Earth's clocks measure 1/6 of a second, while the train's clocks would measure about 1/12 of a second. For all six revolutions one second would pass on earth, and about half a second would pass on the train.

Someone please correct me if I've answered this question incorrectly. In the past I've only asked questions here, and this is my first attempt at answering one.

 

[Ibix]

Correct. If the passenger on the train tried to synchronize the cars of the train with light beams, there would be a break in synchronization for a stationary observer (his watch and every point on the train would read multiple times), but if he ignores this and just reads the passing stations clocks outside his window, there is nothing wrong with this model.

Right. So you are just picking a rather complicated way of saying "the train observer is free to use the Earth's rest frame for analysis"?

Of course it returns! Same place, at a different time along the line. This is one revolution.

But a train traveling in a straight line won't, which I believe was what @Mister T was talking about in the paragraph you quoted. That's part of why the train-round-the-Earth scenario is distinct from the straight track.