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I've lately become interested in the paradoxes of relativity and there are a few things I can't understand. Can someone help me with this one?

Alice, Bob, and Charlie are all at a train platform. Bob takes off on a train heading north at 0.5c and Charlie takes off headed east at 0.5c. They're all carrying stopwatches that start at the moment of departure and stop after 1 minute of travel, relative to the trains.

When they stop, here's what I get from the calculations:

  • From Alice's POV, she's advanced 1.15 minutes while Bob and Charlie advanced 1 minute. (1)
  • From Bob's POV, he advanced 1 minute while Charlie advanced 0.6 minutes.
  • From Charlie's POV, he advanced 1 minutes while Bob advanced 0.6 minutes.

Bob's velocity relative to Charlie's is 0.8c (2) and therefore should see his clock advance at 1.67 times the speed of Charlie's (3). Said another way, he sees Charlie's watch advance at 60% of the speed of his own. The same holds true for Charlie if I understand correctly.

These statements don't seem to reconcile. When they all get together, what will their clocks show?


xeorem
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2 Answers2

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I will first give some specific remarks, and after that I will give some general statements, which are more important than the specific remarks. (I actually recommend immediately scrolling to the 'general statement'.)


In the scenario that you describe you have that each of the protagonists has a journey profile where their velocity is not uniform.

Initially all three are stationary with respect to each other, they then accelerate to a specified velocity, and then they decelerate to zero velocity relative to the starting state.

In such a scenario the relative velocity between Bob (heading North) and Charlie (heading East) is far from straightforward.

It would appear that for your evaluation of the relative velocity between Bob and Charlie you used the calculation that is valid for when both Bob and Charlie have a uniform velocity throughout.

Those are repairable issues: it is fairly straightforward to state the scenario in such a way that it matches the calculation.

Set up the scenario such that Bob and Charlie have a uniform velocity throughout. That means that initially they are both approaching the location where Alice is residing, in such a way that when the Alice clock is at $t=0$ Bob and Charlie fly right by that location.

More generally, any scenario can be simplified to a scenario with one or more flyby's with clocks being compared at the instant of flyby. While you need more protagonists that way the advantage is that any scenario can be set up with uniform velocities.



General statement

It's just not a good idea to say to yourself: let's see how I can come up with a variation of the journey scenario such that it is opaque to me.

It will always be possible to find a new way of bogging yourself down.


The only way to make progress is to transcend the individual scenarios. The choice is: generalize, or keep bogging yourself down forever.


In terms of special relativity the underlying structure is the Minkowski metric.

The Minkowski metric subsumes the Euclidean metric. (I mention that because in the scenario you present two of the protagonists are traveling along lines that are perpendicular to each other. That falls within the scope of the Minkowski metric, in the sense that Pythagoras' theorem falls within the scope of the Minkowski metric.)


About the concept of the metric of a space (and generalized: the metric of a spacetime)

Pythagoras' theorem expresses the metric of Euclidean space.

An example of a metric that is not Euclidean is the metric of how on a chess board the chess King is moving. From one corner to another corner along a row/column is 8 moves. Along a diagonal is 8 moves too! So, the metric of the chess board itself is Euclidean, of course, but for the way the King moves the distance relations are not Euclidan.

Minkowski metric

Instead of evaluating individual cases the endeavour should be to come to terms with the very concept of Minkowsk spacetime.

In terms of Newtonian mechanics there is also a statement about relation between spatial displacment and passage of time. There is the newtonian statement: 'An object in inertial motion will in equal intervals of time cover equal intervals of space'. In terms of newtonian mechanics, while space intervals and time intervals stand in a relation to each other, it's not a single metric.

The shift from newtonian mechanics to Special Relativity is the shift from not-a-single-metric to the Minkowski metric.

Cleonis
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What you are forgetting is that time dilation is the result of the fact that time in each of the three reference frames is not running at a different rate to time in the other frames but is out of synch with time in the other frames. That comes under the heading of the relativity of simultaneity, which is invariably overlooked when people struggle with seeming paradoxes in SR.

To take the relativity of simultaneity into account, you cannot just compare elapsed times on individual clocks. Let's suppose the journeys start at 12:00. You say that Alice stops her watch at 12:01:15 in the frame of the station- what time is that in Bob's frame? Ie, suppose Bob's train was infinitely long, and every carriage bore a clock that was synchronised in the frame of the train- on the carriage passing Alice at the time she stops her watch, what will the clock say? What you will find is that the time on that clock differs from the time according to Bob's watch. From Bob's point of view, Alice does not stop her watch at the same instant that he does. The same is true of Charlie. So the seeming paradox arises because the three people are not all stopping their watches at the same instant in every frame, because if a pair of events (eg Alice and Charlie stopping their watches) happen simultaneously in one of the frames, they will happen out of synch in the others. When you take that into account, you will find that the paradox disappears.

Marco Ocram
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