Understanding the Twin Paradox: Simplifying with a Clear Experiment

[Phil42]

TL;DR Summary

Is there an answer to the simplest possible version of a twins travelling 'paradox'.

I have seen many attempts to rationalise the 'Twins Paradox', but none of the seem satisfactory. They usually use acceleration or asymmetric differences in inertial frame, or other aspects of special relativity that tend to obfuscate the problem/explanation.

So proposing an experiment that simplifies the problem as far as possible:

Take two distant points A and B, centre C. Take two travellers P and Q.

At a calculated time, P starts from beyond A and accelerates towards AB such that her speed (V) is constant by the time point A is reached and she passes C at point at time t.

Q starts at such time as he will also pass point C at time t, accelerating so that his constant speed is the same (but in the opposite direction) as P's, at point B. Because the clocks at their starting points cannot be assumed synchronised, they may need to do a trial run and agree that one will use an offset for the starting time.

So P and Q pass each other at point C and synchronise their clocks. As P passes B and Q passes A, they stop said clocks.

Now, each one sees the other as travelling at a constant speed, over a known distance, and calculates what they expect their opposite number's clock to read.

They then travel and meet each other with their stopped clocks, and place them side by side.

During the timed part of the experiment, no one has accelerated and (as follows), no one has changed course.

How do the readings on the two clocks relate to each other?

 

[Nugatory]

I have seen many attempts to rationalise the 'Twins Paradox', but none of the seem satisfactory. They usually use acceleration or asymmetric differences in inertial frame, or other aspects of special relativity that tend to obfuscate the problem/explanation.

If you haven’t already done so, go back and read the sticky at the of this subforum “When Discussing the Twin Paradox: Read This First”. The two analyses most used by real physicists (as opposed to science writers whose own grasp of the subject is a bit shaky) are the Doppler and spacetime interval ones; neither involve frames or acceleration.

So proposing an experiment that simplifies the problem as far as possible…

Both stopped clocks will read the same, it pretty much has to be that way because of the symmetry of the situation.
Comparing with the Doppler analysis of the twin paradox, if we place a strobe light at the central point both P and Q will have received the same number of flashes while on the way to their respective destinations.
Comparing with the spacetime interval analysis, we have three relevant events:
E0: P and Q both pass point C and start their now-synchronized clocks.
E1: P passes point B and stops their clock.
E2: Q passes point A and stops their clock.
The spacetime intervals between E0 and E1 and between E0 and E1 have the same lengths. (The length of a path through spacetime is the time measured by a clock moving on that path, the same way that the length of a path through space is the distance measured by the odometer of a car driving along that path).

So no frames are needed so far to explain the completely expected result that both clocks read the same. If we do choose to involve frames in our explanation we have to also allow for the relatively of simultaneity - overlooking it is one of the most common mistakes people make.
Using the frame in which points A, B, and C are at rest, both clocks are ticking at the same rate, events E1 and E2 are simultaneous, both clocks are stopped at the same time so they read the same.
Using the frame in which P is at rest, Q’s clock is running slow but event E2 happens after event E1 so Q’s slower clock is ticking for longer and ends up reading the same when it does stop.
Using the frame in which Q is at rest, P’s clock is running slow but event E1 happens after event E2 so P’s slower clock is ticking for longer and ends up reading the same when it does stop.

 

[PeroK]

TL;DR Summary: Is there an answer to the simplest possible version of a twins travelling 'paradox'.

A simpler version is to have two clocks travel on the same circular path at the same speed, but in opposite directions. There is continuous, symmetric time dilation, yet the clocks show the same elapsed time every time they pass.

This highlights that time dilation itself is not really a "thing". Differential ageing is the thing. By "thing" I mean an invariant quantity - that all reference frames agree on. In the above example, there is time dilation but no differential ageing.

PS although, of course, this is not the twin paradox. Perhaps it's the twin paradox paradox?

 

[Orodruin]

What you describe here is not a twin paradox situation. It is a time dilation situation.

 

[Dale]

TL;DR Summary: Is there an answer to the simplest possible version of a twins travelling 'paradox'.

Your scenario isn't a twin's paradox, and it can be further simplified.

So proposing an experiment that simplifies the problem as far as possible:

Take two distant points A and B, centre C. Take two travellers P and Q.

You should specify that points A, B, and C are locations defined in a given inertial frame.

At a calculated time, P starts from beyond A and accelerates towards AB such that her speed (V) is constant by the time point A is reached and she passes C at point at time t.

Q starts at such time as he will also pass point C at time t, accelerating so that his constant speed is the same (but in the opposite direction) as P's, at point B. Because the clocks at their starting points cannot be assumed synchronised, they may need to do a trial run and agree that one will use an offset for the starting time.

So P and Q pass each other at point C and synchronise their clocks.

None of this prep is necessary, and serves only to confuse and distract. Especially the stuff about accelerating. Replace this whole section with:

In the given inertial frame, P moves inertially with velocity V and Q moves inertially with velocity -V. They set their clocks to zero when they pass each other at point C.

As P passes B and Q passes A, they stop said clocks.

Now, each one sees the other as travelling at a constant speed, over a known distance, and calculates what they expect their opposite number's clock to read.

They then travel and meet each other with their stopped clocks, and place them side by side.

During the timed part of the experiment, no one has accelerated and (as follows), no one has changed course.

How do the readings on the two clocks relate to each other?

The readings are equal and the calculations expect them to be equal.

 

[Nugatory]

So proposing an experiment that simplifies the problem as far as possible:

A bit of a footnote here, your thought experiment isn’t a twin paradox situation at all. In a twin paradox we have two points in spacetime (the departure event when clocks are started and the reunion event where they are stopped and compared) and we’re comparing the proper time along two different paths between these events. Because we only have two events there are no simultaneity confusions and we see a real physical consequence: the twins are standing side-by-side and one is physically older than the other.

In your setup we have three events (the E0, E1, E2 from my post above) and we’re comparing the E0 to E1 distance with the E0 to E2 distance. It’s sort of like comparing various driving routes between Warsaw and Berlin with routes between Warsaw and Moscow: sure, we can find routes that happen to be the same length, but so what?

 

[Phil42]

Thanks for the replies.

I apologise for referring to the 'twin paradox'. This occurred because, so many times, I have seen someone explaining special relativity, and then producing the classic twin paradox as an example. As these examples never came with an explanation, and as they were always preceded by a discussion of special relativity, I assumed that the paradox (at least in the minds of the writers) related to special, rather than general, relativity.

As to the answers given, they all appear to agree with what I assumed to be the case: The clocks show the same time.

The reason this seems a paradox to me is that you have two observers, P and Q, both of whom see the other moving, with each view symmetrical with the other, so, surely, both P and Q would expect the other's clock to be indicating a time earlier than their own?

Why are they not puzzled by the fact that they are the same?

To put it flippantly: who turned off special relativity for the experiment?

 

[Ibix]

As these examples never came with an explanation, and as they were always preceded by a discussion of special relativity, I assumed that the paradox (at least in the minds of the writers) related to special, rather than general, relativity.

It is purely a special relativistic example. Simple proof: Einstein stated and resolved it in his 1905 paper "On the Electrodynamics of Moving Bodies" which set out what became known as special relativity, ten years before GR was thought up.

Why are they not puzzled by the fact that they are the same?

Because they know about the relativity of simultaneity. Whether two things happen at the same time or not is frame dependant, so they simply say the other guy's slower-ticking clock stopped after their own normally ticking clock.

Einstein's train is the usual scenario used to demonstrate the relativity of simultaneity.

 

[PeroK]

Thanks for the replies.

Why are they not puzzled by the fact that they are the same?

To put it flippantly: who turned off special relativity for the experiment?

As I said above, time dilation is not the reason for one twin being older when they meet again. The difference in ages is due to the different paths the twins take through spacetime. If the twins take symmetric paths, then their ages and clocks will show the same elapsed time when they meet again.

Special relativity is not synonymous with time dilation.

 

[Dale]

To put it flippantly: who turned off special relativity for the experiment?

The problem is that you “turned off” the relativity of simultaneity in your analysis. That is an essential part of special relativity.

you have two observers, P and Q, both of whom see the other moving, with each view symmetrical with the other, so, surely, both P and Q would expect the other's clock to be indicating a time earlier than their own?

In each frame the resting clock finds that the other’s clock runs slower, but each also finds that the other clock was stopped after their own clock. Thus they show the same time.

 

[PeterDonis]

As to the answers given, they all appear to agree with what I assumed to be the case: The clocks show the same time.

Yes.

The reason this seems a paradox to me is that you have two observers, P and Q, both of whom see the other moving, with each view symmetrical with the other, so, surely, both P and Q would expect the other's clock to be indicating a time earlier than their own?

No, each would expect the other's clock to read the same as his own. Why? Because you made the situation exactly symmetric between the two of them, just as you say!

When we say that ordinary SR time dilation is symmetric, we are only looking at one piece of the puzzle. You also, as @Dale has said, have to look at relativity of simultaneity, which is also symmetric in your scenario (because you set it up that way), and which exactly cancels the effect of time dilation so that both stopped clocks end up showing the same elapsed time.

Another way to look at it is this: in a scenario where we say that a moving clock "runs slow", we are implicitly assuming only one standard of simultaneity, our own. But here you can't do that with either P's or Q's standard of simultaneity, because you're not using either one to determine when their clocks stop. You're using the standard of simultaneity of the frame in which A and B are at rest, since P and Q stop their clocks when they reach those points. And in that frame, both P and Q are moving, and both of their clocks run slow, by the same time dilation factor, because they are both moving at the same speed in that frame. So by the only standard of simultaneity that you can correctly apply to the entire experiment, you would expect P's and Q's stopped clocks to read the same elapsed time.

 

[Orodruin]

People have an unhealthy propensity to misunderstand or simply not internalise the concept of relativity of simultaneity when they first encounter special relativity. Like most popular texts bring up Einstein’s train-bank-lightning example, but somehow the idea just doesn’t manage to make the internal sense of that there is an absolute meaning to simultaneity disappear.

If only someone wrote something about it in their signature so we could stop writing volumes about it …

 

[Sagittarius A-Star]

Why are they not puzzled by the fact that they are the same?

Because they used the Lorentz transformation.

 

[Histspec]

As to the answers given, they all appear to agree with what I assumed to be the case: The clocks show the same time.

The reason this seems a paradox to me is that you have two observers, P and Q, both of whom see the other moving, with each view symmetrical with the other, so, surely, both P and Q would expect the other's clock to be indicating a time earlier than their own?

Why are they not puzzled by the fact that they are the same?

In the frame in which points A, B and C are constantly at rest, let's say Q moves with -0.75c from C to A, and P moves with +0.75c from C to B. If both clocks P and Q are initially indicating 0 when they depart from C, it is clear that they both will indicate the same time, let's say 5 seconds.

How is this compatible with mutual time dilation? Well, we have to consider relativity of simultaneity:

From Q's perspective, the thin red lines of simultaneity tell us, that 1 second Q time corresponds to ~0.3 seconds P time, and 2 seconds Q time corresponds to ~0.5 seconds P time, ... and finally 5 seconds Q time corresponds to ~1.4 seconds P time. So P is retarded with respect to Q.

From P's perspective, the thin blue lines of simultaneity tell us, that 1 second P time corresponds to ~0.3 seconds Q time, and 2 seconds P time corresponds to ~0.5 seconds Q time, ... and finally 5 seconds P time corresponds to ~1.4 seconds Q time. So Q is retarded with respect to P.

 

[Ibix]

Just to emphasize what @Histspec's diagram is showing, the horizontal axis shows position in space according to a person using the frame where A, B and C are at rest and P and Q are moving with equal and opposite velocity. The vertical axis shows time according to the same frame. If you've come across displacement-time graphs in physics at school, that's pretty much all it is, except that time is shown vertically and displacement horizontally.

So if you draw a horizontal line across that you see space (or, at least, the contents of the line ACB) at one instant of time. If you draw the line higher up, you see the blue and red lines representing P and Q cross the line further apart, as they are further apart at a later time.

Where that is really interesting in relativistic physics is that it models everything as spacetime - so that graph suddenly gets promoted from "just a graph" to "a map of spacetime" (or at least, a map of time and one spatial dimension - the whole thing would be a 4d object). And, just like there are no grid lines on a map except the ones you draw, there are no grid lines on this except the ones Histspec drew. And two people don't have to draw the same gridlines. Relativity tells you how observers in motion with respect to each other would usually choose to draw them, and those are the fine red and blue lines. They aren't the same as each other, or the horizontal ones. So they have a different ideas of what "all of space at one instant of time" means. And that's why the time dilation being symmetric isn't a contradiction - A, B and C may say that P and Q arrived at their destinations simultaneously and their clocks stopped simultaneously, but P and Q do not say that. They both say that the other reached their destination later, so are not surprised that the other's slower-ticking clock stopped showing the same time as their own.