What's wrong with this so-called paradox?

[gib65]

A friend and I are in a debate about relativity. He thinks he's found a paradox that overturns SR. I'm a little more skeptical. Here's the thought experiment:

You have a train at rest. You set it up with a stop-clock at the exact center of the train. You are standing there at the exact center of the train with the stop-clock. In your hands you hold two more stop-clocks. You synchronize them. You hand them to two assistants. They casually walk, at the same time and at the same rate, to each end of the train and place each stop-clock there, then leave the train. The stop-clocks at the ends of the train are set to emit a flash of light at a pre-determined time. That pre-determined time is the same for both. The flashes of light are aimed at the stop-clock in the middle. When the stop-clock in the middle receives two flashes of light at the same time, and only at the same time, it will stop.

That's the setup.

Now the train gets moving. It reaches a velocity close to the speed of light. At some point in the journey, the stop-clocks at the ends of the train emit their light flashes.

The question my friend and I are debating is: does the stop-clock in the middle stop or not?

You would think that from the point of view of you travelling on the train with the middle stop-clock, you would see the stop-clock stop. The two stop-clocks which you observed being synchronized should remain synchronized since they are in the same reference frame as you. So you would think they would emit their light flashes at the same time relative to you, and you should therefore see the middle stop-clock stop.

But for anyone not on the train (like your two assistants who also observed the stop-clocks being synchronized), the middle stop-clock should not stop because even if the flashes of light are emitted at the same time, the front flash will reach the middle stop-clock before the rear flash. After getting off the train once it stops, bringing the middle stop-clock with you, and rendezvousing with your assistants, what will they see?

If everyone has to observe the same events happening, then either a) everyone has to observe that the stop-clock indeed stopped, in which case your assistants would have to conclude that the flashes went off at different times (even though they observed the stop-clocks being synchronized), or b) everyone has to observe that the stop-clocks did not stop, in which case you would have to conclude that the flashes went off at different times (even though you observed the stop-clocks being synchronized).

Which will happen? And whatever the answer to this is, what does this imply about the timing of events that are initially synchronized?

[Strange]

This is a variant of the train/lightning example that Einstein used in his book to explain relativity of simultaneity.

 

The bit you are missing in your description is the meaning of "at the same time". The people external to the train will not see the flashes light emitted simultaneously. This must be true, because they will see the flash from the middle clock, they know the distances and speed of light are the same therefore they see the flash at the back occur first in order for it to reach the middle at the same time.

[gib65]

Yes, that's what I thought, but it still leave a question in the air:

 

What must the assistants conclude? They saw the stop-clocks being synchronized. Therefore, they must conclude that the one at the front slowed down relative to themselves more than the one at the rear. I know time dilation occurs as velocity increases, but is there an additional rule that says time dilates (towards the slower end) more for things at the front (or the end closer to the direction of travel) than for the rear?

[xyzt]

Therefore, they must conclude that the one at the front slowed down relative to themselves more than the one at the rear.

No, this is an incorrect conclusion.

[Robittybob1]

This is a variant of the train/lightning example that Einstein used in his book to explain relativity of simultaneity.

 

The bit you are missing in your description is the meaning of "at the same time". The people external to the train will not see the flashes light emitted simultaneously. This must be true, because they will see the flash from the middle clock, they know the distances and speed of light are the same therefore they see the flash at the back occur first in order for it to reach the middle at the same time.

 

No, this is an incorrect conclusion.

What is correct then?

[Strange]

Off the top of my head, I think the answer is that relativity of simultaneity means that from the external (stationary) perspective, the front and back of the train accelerate at different times and so one clock ends up lagging the other. (A variant of Bell's spaceship paradox.)

[md65536]

Strange is right.

 

Remember the twin paradox. When the accelerating twin changes inertial frame, the coordinate time of the distant twin that it accelerates toward jumps forward (only relative to a local clock). The coordinate time of a twin that it accelerates away from jumps backward. The "jump" is merely a change in relative simultaneity.

 

From the other thread... if a train is a certain length at rest on the tracks and all parts of it accelerate similarly and it is again the same length in its new rest frame, it must be that from the track's perspective, the back of the train accelerates first and the front last. But from the moving train perspective, the front accelerated first and the back last. The observer in the middle was in both frames... so how can that be?

 

Just before accelerating, the middle observer understands that she will accelerate first before the front of the train. Suppose with sync'd clocks the middle observer will accelerate at time t=1, and the front at time t=2. The middle will accelerate toward the stopped front of the train, so the coordinate time of the front of the train will "jump" ahead relative to the accelerating middle, let's just suppose it jumps ahead 2 units of time to t'=3. Now an instant after time t=1, the middle observer is now in the moving frame, and the front of the train has already started moving at its time t'=2, and continued to move up until its current time t'=3. The clocks are no longer in sync. The distance to the location on the tracks where the front of the train started from is now length contracted and is no longer half the length of the train... how can the train still be the original length? Well, the front of the train has been moving between its time t'=2 and t'=3, which occurred in an instant for the instantly accelerated middle, and the front is now ahead of that point, at the right distance.

 

To get the exact details you'd have to do the math. As always all the details work out consistently.

 

Meanwhile the observer at the front of the train observers local time passing normally, while other coordinate times jump (backward).

 

 

Since the clocks are no longer in sync after the train accelerated, they won't emit the flashes at the same time.

Edited January 29, 2015 by md65536

[xyzt]

Strange is right.

 

Remember the twin paradox. When the accelerating twin changes inertial speed, the coordinate time of the distant twin that it accelerates toward jumps forward (only relative to a local clock). The coordinate time of a twin that is accelerates away from jumps backward. The "jump" is merely a change in relative simultaneity.

This has nothing to do with acceleration or with the twin paradox. Please stop confusing people with your misinterpretations.

 

What is correct then?

[math]t'=\gamma(t-vx/c^2)[/math] tells you that the coordinate time is a function of BOTH time AND location. This doesn't mean that "the clock in the front slows down more than the clock in the rear".

[Phi for All]

This doesn't mean that "the clock in the front slows down more than the clock in the rear".

 

Sorry, who are you attributing this quote to?

[Robittybob1]

This has nothing to do with acceleration or with the twin paradox. Please stop confusing people with your misinterpretations.

[math]t'=\gamma(t-vx/c^2)[/math] tells you that the coordinate time is a function of BOTH time AND location. This doesn't mean that "the clock in the front slows down more than the clock in the rear".

Looking at the formula there is (t - vx/c^2) so is vx/c^2 a measure of time? Can you say that in words? "velocity * x / c squared" what is x? You call it location but how do you put a value on location?

[swansont]

what is x? You call it location but how do you put a value on location?

 

"One meter to the right of the origin", as an example, for x = +1m. Same as in physics 101.

[phyti]

The end clocks are synchronized to the center clock while the ship*is at rest. When the ship reaches high speed, the axis of simultaneity is different from the ground aos. The signals will not arrive at the center simultaneously.

If the end clocks are synchronized upon reaching the target speed, then they will emit signals at a preset time, and arrive simultaneously.

* A train moving at relativistic speed would be too risky!

[xyzt]

 

Sorry, who are you attributing this quote to?

It is a paraphrase to md65536's claim that :

 

 

 

Meanwhile the observer at the front of the train observers local time passing normally, while other coordinate times jump (backward).

Edited January 29, 2015 by xyzt

[pzkpfw]

I would expect (but am very happy to be shown wrong) that the end clocks would remain synchronised in the frame of the train, as they're both experiencing the same acceleration on that train. So they'd both fire at their pre arranged time, with the flashes reaching the centre clock at the same time, and that clock does stop as expected.

 

The outside observer, considering themselves as at rest and the train moving, would also see that centre clock stop (it can't both stop and not stop). From their point of view that clock was moving towards the light from the front clock and away from the light from the rear clock, so for the flashes to have reached that clock at the same time, the rear clock must have (considered in the frame of the outside observer) flashed before the front clock. Well, that's the relativity of simultaneity, as Einstein showed, observers in relative motion can't agree on whether events happen at "the same time".

 

The only paradox is if one demands that simultaneity must be the same for all observers, which creates a centre clock stops and doesn't stop situation. But that's getting it all upside down.

[gib65]

This has nothing to do with acceleration or with the twin paradox. Please stop confusing people with your misinterpretations.

[math]t'=\gamma(t-vx/c^2)[/math] tells you that the coordinate time is a function of BOTH time AND location. This doesn't mean that "the clock in the front slows down more than the clock in the rear".

 

 

 

"One meter to the right of the origin", as an example, for x = +1m. Same as in physics 101.

 

So what does this mean? That time dilates more the further away something is from the direction of travel compared to something close to the front?

[Phi for All]

It is a paraphrase to md65536's claim that :

 

Ah, OK. I'm not used to putting quotation marks around paraphrasing.

[Strange]

 

Ah, OK. I'm not used to putting quotation marks around paraphrasing.

 

Journalists do it all the time. (Not that that is any justification. )

[studiot]

I'm sorry I can't see the paradox here.

 

Why is there any suggestion given the original conditions, that the light flashes would reach the centre clock together?

 

Whenever the flashes are emitted they will take a finite time to travel to the centre, but in that time the centre will have moved, closer to emission point of one flash and further from the emission point of the other.

Edited January 29, 2015 by studiot

[xyzt]

 

 

 

So what does this mean? That time dilates more the further away something is from the direction of travel compared to something close to the front?

No, it doesn't.

[pzkpfw]

...

 

Why is there any suggestion given the original conditions, that the light flashes would reach the centre clock together?

 

Whenever the flashes are emitted they will take a finite time to travel to the centre, but in that time the centre will have moved, closer to emission point of one flash and further from the emission point of the other.

 

That's from the point of view of an outside observer, which is no better opinion than that of an observer on the train who (while it isn't accelerating) is entitled to consider themselves as at rest.

 

The relativity of simulatenity doesn't say one special observer can consider events to be simultaneous and others can't, it says events simultaneous for one observer won't be simultaneous for an observer moving in relation to that observer. (Or vice versa).

Edited January 29, 2015 by pzkpfw

[studiot]

Thank you pzkpfw,

 

But the original conditions state

 

 

You have a train at rest.

.............................................
Now the train gets moving.

 

Which is not a simple rest frame for the train.

[pzkpfw]

It's true the setup isn't very clear about whether the train is still accelerating or not at the time the flashes occur, but that's not really what I was commenting about in your post. You just seemed to be implying an observer outside the train as having a "special" or "absolute" rest frame.

 

Since they were at rest with the train and clocks at the start, I can see that's why you'd see them as "having" the frame within which the clocks are "still synchronised" (and I can't say I disagree (so my post #14 may be backward)); but that wasn't clear in your post.

Edited January 30, 2015 by pzkpfw

[swansont]

 

 

 

So what does this mean? That time dilates more the further away something is from the direction of travel compared to something close to the front?

 

It means that for something in a moving frame, the time depends on the position. Referring to time dilation is somewhat ambiguous here, because time dilation is the result of the rate changing (clocks run slow) but that's the same for the whole frame. That gives you [latex] t'=\gamma t [/latex] But the time in that frame also depends on where you are; simultaneity is also relative to what frame you are in.

 

"time dilates more" the longer you wait, which is why it's not a good descriptive phrase to use. Because gamma doesn't have to be bigger for that to happen (small gamma, large t), and gamma being large is another possible interpretation of "time dilates more"

I'm sorry I can't see the paradox here.

 

Why is there any suggestion given the original conditions, that the light flashes would reach the centre clock together?

 

Because only simultaneous arrival at a single position triggers an event.

[gib65]

 

No, it doesn't.

Thanks for clearing that up, xyzt, now I understand.

 

It means that for something in a moving frame, the time depends on the position. Referring to time dilation is somewhat ambiguous here, because time dilation is the result of the rate changing (clocks run slow) but that's the same for the whole frame. That gives you t'=\gamma t But the time in that frame also depends on where you are; simultaneity is also relative to what frame you are in.

Ok, it seems we have two explanations so far (according to how I'm interpreting it): 1) position does affect time (paraphrasing what you just said), and 2):

 

Just before accelerating, the middle observer understands that she will accelerate first before the front of the train. Suppose with sync'd clocks the middle observer will accelerate at time t=1, and the front at time t=2. The middle will accelerate toward the stopped front of the train, so the coordinate time of the front of the train will "jump" ahead relative to the accelerating middle, let's just suppose it jumps ahead 2 units of time to t'=3. Now an instant after time t=1, the middle observer is now in the moving frame, and the front of the train has already started moving at its time t'=2, and continued to move up until its current time t'=3. The clocks are no longer in sync. The distance to the location on the tracks where the front of the train started from is now length contracted and is no longer half the length of the train... how can the train still be the original length? Well, the front of the train has been moving between its time t'=2 and t'=3, which occurred in an instant for the instantly accelerated middle, and the front is now ahead of that point, at the right distance.

Assuming I've got these right, am I also right in assuming that 2) explains 1)? That is, the reason why objects closer to the front will be slightly "ahead" in terms of time is because of the non-simultaneous accelerations of the front and the back at the beginning of the train's journey?

[studiot]

Thank you pzkpfw and swansont for replying and thank you gib65 for not replying.

 

But you all missed my point. As a matter on interest this problem is different from Einstein's train because there the flashes occurred in the track system, not on the train.

 

I said that I don't see a paradox because of the given conditions set out in the OP and I highlighted the offending words.

 

'The train is moving'.

 

It doesn't matter if the train is accelerating or keeping constant speed, it is specified as moving.

 

So how does the observer on the train know he is moving?

And what is his (the whole train system) velocity?

It is a relative velocity, but relative to what?

 

By specifying that the train moves the OP has provided additional information ie an additional condition.

This condition is tantamount to specifying an absolute system of reference.

 

As I understand the properly posed analysis the external trackside observer does not see the train clocks as keeping the same time or travelling the same distance.

So to her the light from the end clocks travels different distances in different times.

These alterations to times and distances are just appropriate for the light to arrive at the 'centre' (she does not see this as the centre) clock together and trigger the stop mechanism.