Triplets Paradox: Is There a Solution?

[villy]

One person (A) stays on Earth, while another (B) goes on a long journey and returns later.
We can calculate the time interval for A and the proper time interval for B. Let's say we get Δt=100y and Δτ'=50y.

We then consider a second traveler (C) going on a different journey, departing and returning at the same moments.
We perform the same calculations between A and C and get Δt=100y and Δτ"=70y.

How can we have a solution for the B and C pair, Δτ'=50y and Δτ"=70y, without knowing their relative motion?
If we assume they're both traveling in straight lines , the angle between those lines can be anything from 0 to 180 and it wouldn't impact any of the previous calculations.

The only way I can see to resolve this is if all the values were equal. Am I missing something?

 

[Ibix]

I don't understand clearly what you are asking. Is it "can B predict C's age at the final meetup"? If so the answer is obviously yes - the easiest way to do it is to work in A's inertial rest frame, as you did.

If they want to work in their own rest frames they can do so, but it is significantly more mathematically complex since they are non-inertial. Radar coordinates are probably simplest, but even they are non-trivial.

 

[DaveC426913]

One person (A) stays on Earth, while another (B) goes on a long journey and returns later.
We can calculate the time interval for A and the proper time interval for B. Let's say we get Δt=100y and Δτ'=50y.

We then consider a second traveler (C) going on a different journey, departing and returning at the same moments.
We perform the same calculations between A and C and get Δt=100y and Δτ"=70y.

How can we have a solution for the B and C pair, Δτ'=50y and Δτ"=70y, without knowing their relative motion?
If we assume they're both traveling in straight lines , the angle between those lines can be anything from 0 to 180 and it wouldn't impact any of the previous calculations.

The only way I can see to resolve this is if all the values were equal. Am I missing something?

Sorry, can you elaborate? How did we perform the calculation for B and C in para 2, if we didn't know their relative motion? This feels like a cart-before-horse-in-a-barn-with-a-javelin thing, but that might just be me.

 

[PAllen]

Note that your argument (assuming B and C are out and back along a straight line per A) elegantly proves that the angle between their paths doesn't matter. Working this out in other coordinates will be more complex, but your argument establishes that it can't matter.

 

[DaveC426913]

Note that your argument (assuming B and C are out and back along a straight line per A) elegantly proves that the angle between their paths doesn't matter. Working this out in other coordinates will be more complex, but your argument establishes that it can't matter.

Is the OP not saying that C's path can be at any angle up to 180 degrees relative to B?
And therefore, the dilation factor between B and C will be angle-dependent, yes?

Oh, you mean when they arrive back home at compare their clocks... right. The direction they traveled is irrelevant.

I was thinking the OP meant in-situ.

 

[PAllen]

Sorry, can you elaborate? How did we perform the calculation for B and C in para 2, if we didn't know their relative motion? This feels like a cart-before-horse-in-a-barn-with-a-javelin thing, but that might just be me.

The OP is correct, we don't need to consider the relative motion of B and C for the problem as posed. Each of B and C can go out and back along some straight line paths, at arbitrary angle relative to each other. Knowing that they both depart at the same time and return at the same time, but with different speeds relative to A, it is trivial to arrange the proper times as per the OP. Then this result is invariant, not coordinate dependent, and is also independent of the angle between B and C paths.

 

[DaveC426913]

Right. They're all "...departing and returning at the same moments".

 

[Dale]

How can we have a solution for the B and C pair, Δτ'=50y and Δτ"=70y, without knowing their relative motion?

Proper time is invariant. So whatever reference frame you use $\Delta \tau’=50$ and $\Delta \tau’’= 70$. That is the nice thing about invariants. You can calculate them in any convenient frame and then you know the result in any other frame.

 

[Ibix]

And therefore, the dilation factor between B and C will be angle-dependent, yes?

Yes, as will the details of the messing around to account for the turnaround, even once you've selected a general policy for how to do that messing around. So if the OP wants B and C to have a minute-by-minute account of the other's age then the angle matters, as does the choice of coordinate system (the "general policy" I referred to above). But all such schemes must give the same final ages (although it's easy to make mistakes) because those are invariants. The maths involved is somewhat beyond special relativity 101, though.

 

[villy]

I'm unsure what to make of this.
Let me propose a change in the setup:

We have an object A. Another object, B passes by it at a constant velocity. Let's say we calculate γ'=2.
Another object, C, also passes by A at the same time, with the same constant velocity, We again calculate γ''=2.
So, from the perspective of A, the proper times of B and C are in sync.

Depending on the angle between those trajectories, we have a varying relative velocity between B and C.
Ignoring A, we calculate γ=1.5 between B and C, so from their perspectives, they're not in sync.

So, even if from our perspective (A) 2 other clocks (B and C) appear in sync, they might actually not be?
Is this correct?

 

[Ibix]

So, even if from our perspective (A) 2 other clocks (B and C) appear in sync, they might actually not be?

There is no "actually" in sync. Different frames will have different opinions on whether or not any pair of clocks is synchronised, but none is "righter" than another.

If they turn around and meet up then their ages at the meeting are invariants that all (correct) calculation methodologies must agree on. This must be so because they can compare their ages directly and unambiguously when they meet.

You may be struggling with applying time dilation formulae to observers who accelerate. That's because the standard formulae assume that the one doing the measurement is inertial at all times, and the formulae do not work correctly for observers who accelerate.

 

[Nugatory]

So, from the perspective of A, the proper times of B and C are in sync.

Careful - those aren’t proper times, they are coordinate times. $\gamma$ is used to transform coordinate times in one frame to coordinate times in another frame (that’s why it appears in the Lorentz transformation equations) but doesn’t enter into calculations of proper time unless we’re going out of our way to complicate the problem.

To calculate the proper time between two events (for example, traveling twin turns around and reunion on earth) we calculate the spacetime interval between those two points and that calculation is most easily carried out using coordinates from a single frame.

(A corollary, often missed in less modern treatments of relativity, is that the Twin Paradox is not a time dilation problem, and introducing time dilation just gets in the way of understanding it).

 

[villy]

So, knowing only that from our perspective, 2 other clocks appear in sync, we don't have the information to say anything about how they appear to each other.

I still can't really wrap my head around it.

In my initial formulation, with the return trips, we could say what the relationship between B and C was (based on the relationships between AB and AC were).
We could compare Δτ'=50y against Δτ"=70y because of the common event, the return (Δt=100y).

But can't we do the same calculations for the second setup?
We do them for the same Δt=100y and, since we set up equal speeds, we'll get equal results: Δτ'=Δt/γ'=50y and Δτ"=Δt/γ"=50y.
Yet if we do a calculation between B and C, for Δt=50y, we'll get Δτ=Δt/γ=33y.
I realize this isn't technically a common event, but we can't have that without acceleration.
Isn't it correct to say then when A measures 100y have passed, B will measure 50y (and by extension, the same for C)?
But the following calculation says then when B measures 50y, C should measure 33y.

How would the correct synchronization between all 3 be made?

Sorry if I don't address more advanced math, but much of it is still beyond me.
What I'd like to understand is how the results fit together.

 

[Vanadium 50]

I still can't really wrap my head around it.

The solution is usually not to add more complications, like going from twins to triplets to octuplets....

A good start is to label every event (A launches, B turns around etc), assign coordinates to those events and transform as needed. Don't just multplly by gammas.

 

[villy]

Sorry, I'm not trying to complicate the the situations, just attempting to highlight the point I'm confused about.
In the first situation we were able to calculate the difference in elapsed time between BC, based on the differences between AB and AC.
In the second situation, can't we do the same?
As far as I understand, the common event (the return) only serves as confirmation. Mathematically, it only amounts to Δt=100y for the calculations between both AB and AC. Can't we do this in the second situation?
How then are the 100/50/50 and 100/50/33 results reconciled?

Edit: Is it correct to say that there's no such thing as synchronizing 3 clocks, we have to look at them in pairs?

 

[Dale]

Let me propose a change in the setup

This is a different scenario and so the answers will be different. Be aware that you will now be getting answers that seem contradictory even if they are simply applying the same physics to different scenarios

@ all respondents, please identify which scenario you are responding to

 

[Dale]

In my initial formulation, with the return trips, we could say what the relationship between B and C was (based on the relationships between AB and AC were).
We could compare Δτ'=50y against Δτ"=70y because of the common event, the return (Δt=100y).

Yes.

But can't we do the same calculations for the second setup?

No.

The proper time is a sort of “length” in spacetime. It is calculated as $ds^2=-c^2 dt^2+ dx^2 + dy^2 + dz^2$. This one formula contains all of special relativity in it. So special relativity is, at its heart, a new kind of geometry.

In your first scenario you have some paths and you have specified a clear beginning event, a clear ending event, and the spacetime path “lengths”. The length of a line segment is well defined. Transforming your coordinates does not change any of that.

In your modified scenario, the paths have a definite beginning, but no definite end. With no definite end there is no way to calculate a definite “length”. You can use coordinates to define the ends, but when you do so then obviously the “lengths” will depend on the coordinates you choose.

 

[Ibix]

As far as I understand, the common event (the return) only serves as confirmation.

No. Unless you are at the same place there is not a unique notion of "at the same time", so there is no unique answer to the question "what time does B's clock read at the same time as C's reads 20 years". That is why the meeting up is crucial to the twin paradox experiment, because it removes all ambiguity about what we mean by "at the same time".

Your variant with no meet up, therefore, does not have a unique end to the experiment. You can say it stops "when" A is 50 years old (or whatever), but since there are multiple ways for anyone not at the same place as A to define "when", that is not clearly defined.

The time dilation formulae assume a process for defining "when" called the Einstein synchronisation covention. Unfortunately, applying this process gives different definitions of "when" for your three different observers. Naively applying time dilation formulae without understanding this will confuse you a lot.

Remember that time is a dimension in relativity. You would not be surprised if A and B disagreed which direction was "forward" and hence which direction is "beside me" - they're just facing different directions. Relativity merely extends this to disagreement about which direction in spacetime is "forward in time" and hence which direction is "at the same time as me".

 

[villy]

Isn't it possible to define the end of the experiment in the second scenario as when the spacetime distance between A and B is a given value? The spacetime distance between A and C will be equal, right?
Since B and C are moving at the same velocity with respect to A, wouldn't the lengths of the paths that B and C travel through time (from the start to end point) be equal?

 

[PeterDonis]

Isn't it possible to define the end of the experiment in the second scenario as when the spacetime distance between A and B is a given value?

There is no such thing as "the spacetime distance between A and B". You have to pick specific events on A's and B's worldlines to define a "spacetime distance" between them. But picking those events requires you to already know what defines the end of the experiment, since that's what picks out particular events on A's and B's worldlines as being "the events when the experiment ends".

 

[villy]

Right, I get your point, no such thing as the distance between 2 intersecting straight lines.
I guess what I'm looking for is the distance between a point on the worldline of object A (Δt=100y) and the worldline of B. Is that something that makes sense under this framework?

 

[PeterDonis]

the distance between a point on the worldline of object A (Δt=100y) and the worldline of B. Is that something that makes sense under this framework?

No, since "the worldline of B" is not a point.

 

[Vanadium 50]

label every event (A launches, B turns around etc), assign coordinates to those events and transform as needed.

I know you think it's simpler if you avoid doing this. It really isn't. For example, it will clarify that your two scenarios are different.

 

[Dale]

Isn't it possible to define the end of the experiment in the second scenario as when the spacetime distance between A and B is a given value?

No. In addition to what Peter Donis said:

You have to pick specific events on A's and B's worldlines to define a "spacetime distance" between them.

Not only specific events, but a specific path between them. In special relativity we can fudge that a bit since the straight path between two distinct events is unique, but I think it may be helpful to be clear for the OP.

@villy I would recommend actually drawing the spacetime diagrams for the scenarios you are imagining. This may help you understand the issues.

 

[villy]

No, since "the worldline of B" is not a point.

No, I meant geometrically, there is a method to define the distance between a point and a line. Isn't there a operation we can define in this framework? An operation that always points to the same point on the worldline of B, given a point on the worldline of A.Let me phrase it like this:
Excuse my poor naming scheme, I'm not sure what would be appropriate, and simple to type without special formatting.
Let's call the starting points A0, B0 and C0
We choose a point on the worldline of A, when A measures the passage of 100 years, Aa1.
From the perspective of A, we calculated a time dilation factor of 2, so we have two corresponding points on the worldlines of B and C, Ba1 and Ca1. Δt(B0,Ba1)=Δt(C0,Ca1)=50 years.
I realize that from the perspective of B, Ba1 and Ca1 are not simultaneous.
So, from the perspective of B, the point Ba1 is simultaneous with a point Cb1 on the worldline of C. Symmetrically we have Ca1 and Bc1. Given the time dilation of 1.5 i set, Is it correct that Δt(B0,Bc1)=Δt(C0,Cb1)=33 years ?

 

[PeterDonis]

geometrically, there is a method to define the distance between a point and a line

That involves picking a particular point on the line and taking the distance between that point and the given point. But it only works if there is a unique method of picking a particular point on the line.

 

[PeterDonis]

Let's call the starting points A0, B0 and C0

Aren't they all starting from the same point?

We choose a point on the worldline of A, when A measures the passage of 100 years

I assume you mean 100 years from the starting point?

From the perspective of A, we calculated a time dilation factor of 2

Which scenario are you talking about? The one in your OP, or the one in your post #10? I assume it's the latter?

we have two corresponding points on the worldlines of B and C

"Corresponding" to what? And "corresponding" how?

This is the point at which @Dale's suggestion to draw the spacetime diagram is a very, very, very, very good one.

 

[villy]

"Corresponding" to what? And "corresponding" how?

I mean the point on the wordline of B that appears to be simultaneous with Aa1, from the perspective of A.
Otherwise yes to all your assumptions.
As for the diagram, I am picturing it, it's just that it requires 2 dimensions for space, so a 2D ilustration would only serve to increase confusion. It is all straight lines though, right?

 

[Dale]

it's just that it requires 2 dimensions for space, so a 2D ilustration would only serve to increase confusion

Maybe you should figure out 1 dimension of space + 1 dimension of time before moving on to 2D space + 1D time. I think you are trying to run before you can walk.

 

[villy]

True, I do have plenty I need to understand better.
I understand that everything can be worked out properly for 2 objects.
It just doesn't seem to fit together with 3 objects for me.
I do appreciate all the answers.

Edit: I'm not trying to learn how to run, or expecting anyone to be able to teach me in a couple of posts. I'm trying to get the general gist of it.

 

[PeterDonis]

I mean the point on the wordline of B that appears to be simultaneous with Aa1, from the perspective of A.

Ok.

As for the diagram, I am picturing it, it's just that it requires 2 dimensions for space

This is a huge increase in complication, because the two Lorentz boosts in question--from A's rest frame to B's rest frame, and from A's rest frame to C's rest frame--are in different directions. That means the transformation from B's rest frame to C's rest frame, or vice versa, is not a pure Lorentz boost; it's a boost plus a spatial rotation. I doubt that anyone's powers of visualization are up to handling that by intuition. Cases like this need to be worked by grinding through the math. But as @Dale has pointed out, before even trying that, you need to be sure you have a firm grasp of the simple 2D (1D space + time) case.

You can at least get a start on the 3D (2D space + time) case by looking at the two obvious edge cases, though: the case where B and C are moving in the same direction (in which case they are at rest relative to each other--easy), and the case where B and C are moving in opposite directions, in which case the relative speed between B and C can be obtained from the simple relativistic velocity addition formula.

It is all straight lines though, right?

As I understand things, you are specifying that the objects always move inertially, correct? If that is the case, then yes, all of the lines of interest will be straight lines.

 

[PeterDonis]

Depending on the angle between those trajectories, we have a varying relative velocity between B and C.
Ignoring A, we calculate γ=1.5 between B and C

Where are you getting 1.5 from? Or are you just specifying it arbitrarily?

 

[villy]

Just an arbitrary number to account for a difference in velocity.

 

[PeterDonis]

Just an arbitrary number to account for a difference in velocity.

How do you know that you can even achieve $\gamma = 1.5$ between B and C, given that $\gamma = 2$ between A and B, and between A and C?

 

[villy]

How do you know that you can even achieve $\gamma = 1.5$ between B and C, given that $\gamma = 2$ between A and B, and between A and C?

As far as I can tell it could be anything varying from 1, for an angle of 0 to a value that has to be larger than 2 for an angle of 180.

You are correct though, this has gotten quite further away from where I'm comfortable with the math.
When I said I'm visualizing it, I only meant the simple diagram from the perspective of A, with two symmetrical straight lines going out. I wasn't claiming I'm doing any transformations with it.
Just imagining the relative positions of the points based on the values (33<50).