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Ok, let's say we are standing on a huge inertial frame, a flat 2-dimensional space (just like people living on this planet used to think of the ground they were standing on). Off in the distance ahead of us are 2 train tracks, parallel to each other and separated from each other by a few meters. They are both going off to the east and the west from us, the viewer (using east & west directions assumes we are looking north). And our line-of-sight to them is perpendicular to the straight-lines of these tracks, which go off to the right and left for as far as we want. Let's say if we go way up above this scene, the main points of interest are at the vertices of a huge isosceles triangle, but with the base of the triangle at the top and the 2 equal sides going down from there to join at a 90° angle. There is a train on one end of the tracks to the left, sitting there ready to go towards the right. There is a train on the right side of the tracks at that vertex of the triangle, pointed to the left ready to go. There are clocks at all three vertices of this triangle, as well as people: A clock at the train on the left (call this train A), a clock at the train on the right (call this train B). And we the viewer at the 3rd vertex of the triangle also have a clock. All these clocks are connected to this vast inertial frame we a standing on, and they are all still with respect to the inertial frame and with respect with each other. So they are all synchronized (am I allowed to do that, to synchronize clocks at various points of an inertial frame?). They all show the same time and they are all ticking at the same rate.

Now, people at train A get on their train and start going at a predetermined time (say 10 o'clock AM). Likewise, people at train B get on their train (train B) and start their train engine moving towards train A, also at 10 o'clock. They both are going to increase their speed from a dead stop at exactly the same rate. They will go faster and faster. They will both move at a faster and faster rate, which will start to approach the speed of light. But they will both be moving with respect to their inertial frame at the same rate, only they will be going in opposite directions and on separate tracks. Finally they have their speed up to something like 2/3 of the speed of light (with respect to the inertial frame this is taking place on) and that will be their cruising speed.

Now, according to the special theory of relativity, people in train A will see the clock in train B as going at a slower pace then their own clock, because train B is going at a high speed in relation to train A. Like wise, the people in train B will see the clock in train A as ticking at a slower pace than their own clock. For example, when the clock in train A gets to show 11 AM, they might see the clock in train B as showing 10:50 AM. So they are both barreling along towards each other. Finally they are both approaching the midpoint between where they started from (this will be directly between the left vertex and right vertex of this isosceles triangle, and it will be directly "above" the bottom vertex where us the viewers are watching this from). They put on their brakes at the same time and slow down. Slower and slower and finally come to a stop right next to each other. The people in train A will have been watching the clock in train B tick along slower than their own clock, so once train A has stopped at let's say 12 o'clock their time, they expect the clock on train B to be showing something like 11:30 AM. Likewise, the people on train B have been watching the clock on train A, seeing it ticking at a slower rate than their own clock, so when they stop and get off train B at a point when their own clock says 12 o'clock, they expect the clock in train A to show something like 11:30 AM. Now from the vantage point of us the viewers of these 2 trains from the lower vertex of this huge isosceles triangle, we who are sitting on this huge inertial frame that everything is taking place on, we would have seen both the clock on train A and the clock on train B as ticking at the exact same rate, although they would be ticking slightly slower than our own clock, since they are both moving with respect to us and even some significant fraction of the speed of light. But now everyone is stopped and standing on this one huge inertial frame of reference again. Now, how will the clocks on train A and on train B compare? To us the viewers of these trains, they should both show the same time. To train A occupants, train B's clock should be retarded in relation to their own. To the train B occupants, train A's clock should be retarded in relation to their own. But only one reality will manifest itself once the occupants of train A and train B meet at the middle, correct? What will they see when they view each other's clocks?

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    $\begingroup$ Can you reduce the amount of text? Summarize it more effectively? $\endgroup$ – Yashas Mar 4 '17 at 18:29
  • $\begingroup$ I think your inertial frame needs more than two dimensions, otherwise you really are just standing on it, not in it. $\endgroup$ – John Mar 4 '17 at 21:06
  • $\begingroup$ Sorry for verbosity. Basically 2 trains with initially synchronized clocks travel directly towards each other approaching speed of light for some period. They finally decelerate and meet in the middle, having both moved exactly as mirror images of each other. Special Relativity claims the clock of mirror-image-train of each train will be slowed down relative to original train's clock. This leads to contradiction when the 2 trains meet and compare clocks. $\endgroup$ – physicitpeks Mar 5 '17 at 11:55
  • $\begingroup$ Trains A and B both start moving at 10:00 according to the observer on the ground, and also according to the passengers on both trains an instant before they start moving. An instant after they start moving, the passengers on Train A say that Train B has already been moving for hour, and that its clock runs slow. When the trains pass (an hour later according to the observers on either train, both clocks read 11:00. The A-people say: You guys on B have been traveling 2 hours, but your clock has only advanced 1 hour. It's slow. The B-people say the same about the A-people. $\endgroup$ – WillO Apr 8 '17 at 22:58
  • $\begingroup$ After they decelerate, they'll have different stories to tell (they'll now both say that both their clocks ran slow the whole time), but you don't need to think about that to resolve your paradox. It's simpler to think about the case where they just pass each other. $\endgroup$ – WillO Apr 8 '17 at 22:58
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First, this is simpler (and nothing is lost) if we assume the trains jump instantaneously to their cruising speeds. Also, for the moment, let's suppose the trains don't stop when they pass, but just keep right on going --- which does not affect your apparent paradox.

The observer on the ground says: Both trains started with their clocks set to 10:00, both clocks ran slow, and when the trains passed two hours later, both clocks said 11:30.

The observer on the $A$ train sees this series of events:

9:15AM: Train $B$ starts moving, with its clock set to 10:00.

10:00: Train $A$ starts moving, with its clock set to 10:00.

11:30: The trains meet, with both clocks saying 11:30.

Therefore Train $B$ traveled for 2.25 hours, during which time its clock advanced by 1.5 hours --- so its clock ran slow.

The observer on the $B$ train says exactly the same, with, of course, the roles of $A$ and $B$ reversed. So they are in perfect agreement (as they must be) that both clocks say 11:30 when they meet, but still disagree about which clock ran slow.

Of course an observer on the ground, at the point where the trains meet, says that the two clocks are synchronized throughout and that both run equally slow. Your story has the trains coming to a stop, after which all the passengers agree with that ground observer. But your apparent "paradox" has nothing to do with that part.

The basic error is to think that if the clocks are initially synchronized in one frame, they are initially synchronized in all. In fact, the opposite is true: If the clocks are spatially separated and synchronized in one frame, then they cannot be synchronized in any other frame. In particular, if they are initially synchronized in the ground frame, they can't be initially synchronized in their own frames.

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  • $\begingroup$ As pointed out in my comment, if the clocks are initially synchronized in one frame, they are initially synchronized in all frames. Consider a frame of reference S in which two clocks are synchronized and at rest, possibly at different spatial locations. Now these two clocks from any observer's (say the observer is in frame S') perspective may be obtained by a Lorentz transformation. This is standard special relativity. The time measured in S relative to that in S' is given by the standard time dilation formula. It does not distinguish between the two clocks. See for yourself. $\endgroup$ – John Apr 9 '17 at 17:10
  • $\begingroup$ @John: "If the clocks are initially synchronized in one frame they are initially synchronized in all frames." This isn't true. Two points determine a line. If two clocks at separate events are synchronized in Frame $A$, then Frame $A$ is fully determined by the condition that the line connecting the two events is a line of simultaneity. There can be no other frame in which the clocks are synchronized. (This asumes one-dimensional space, of course. In higher dimensions, one has to be a little more careful with the language, but the conclusion remains true.) $\endgroup$ – WillO Apr 9 '17 at 17:33
  • $\begingroup$ Thanks, I see that you are right now. However, I am not convinced you have interpreted the OP's confusion correctly. He is not confused about why A and B see different times as they pass. He's confused by why when they stop at the station and compare their clocks they should have the same time despite that wasn't the case when they were on the train. This is why you can't make them not decelerate. It just changes the problem entirely. $\endgroup$ – John Apr 9 '17 at 18:03
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It seems that you have come up with a symmetrical thought experiment and you seem to have confused yourself by arriving at a contradiction. The apparent contradiction is resolved if you consider an observer standing at the axis of symmetry of the problem (actually anywhere along the platform would do, but I don't want to break the symmetry of your thought experiment).

Clearly, from this observer's point of view, both trains are doing exactly the same thing from the start to the end. Hence, this observer sees that the clocks on both trains run at exactly the same rate at any instant of time. There is no paradox, as expected.

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  • $\begingroup$ So you say that both the clocks from train A and train B show the same time when the trains come to a stop in the middle. But relative to train A, train B's time will be slowed down when the speed of train B relative to train A approaches speed of light, which I am assuming happens for some period during the trip. That is what Special Relativity claims, does it not? Thus there IS a contradiction. When trains stop in middle, train A occupant should see that the clock from train B shows earlier time. What am I missing? Thank you for your response. $\endgroup$ – physicitpeks Mar 5 '17 at 12:10
  • $\begingroup$ That is not what special relativity claims. If both trains are not accelerating, then special relativity does indeed tell us that observers in either train would see the other train as having a slower-running clock. However, in order for them to actually come together to compare their clocks, at least one of them has to decelerate/accelerate; in your case, both would have to decelerate. When they decelerate, they will see the opposite. $\endgroup$ – John Mar 5 '17 at 15:11
  • $\begingroup$ So you're saying during deceleration, train B's clock runs at faster rate than A's? That could explain dilemma I'm having with the clocks when 2 trains meet. But I've never heard of other clock as going anything but slower than observer's clock. Do you have reference for that? I found web page that seems to confirm what I had thought was correct: SR If what this web page states about the rate of clocks in accelerated frames is true, then problem I have with the clocks at conclusion of my thought experiment still remains. $\endgroup$ – physicitpeks Mar 6 '17 at 14:55
  • $\begingroup$ I guess an intuitive answer to that is that when you decelerate you necessarily feel a force; consequently, you will realise: oh I am moving. It is no surprise that your clock will run slower then. This is the key idea in resolution of the twin paradox too, where, when the twin that is sent off to space decelerates and turns around before heading back to the Earth, this deceleration phase will more than make up for the extra time that he observes to have passed for himself compared to his twin. $\endgroup$ – John Mar 6 '17 at 16:23
  • $\begingroup$ Deceleration has nothing to do with this, because the OP's question does not require the trains to decelerate. They can pass each other in the middle without changing speed, and it doesn't change the question. $\endgroup$ – WillO Apr 8 '17 at 22:52

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