I Twin Paradox: Who Is Right, A or B?

[Ibix]

But if they are essentially stationary relative to each other there is an objective answer to the question of which was born first according to either of the clocks they carry with them (because these clocks run at the same speed and can be synchronised at any time in a unique, consistent way).

Using Einstein synchronisation is a matter of choice. Using Einstein synchronisation in a particular frame is a matter of choice. And any other choice amounts merely to picking a different coordinate system which, apart from making the maths more complicated, has no consequences.

It is an objective fact that if you use Einstein's synchronisation procedure and if you take the twins to be at rest then you will say that they are twins. But the twins may have made different choices - so they may not agree with you. There are rather a lot of subjective caveats to your "objective answer".

Of course we do this all the time on our own planet. If you ask someone "how old are you?", they rarely answer "with respect to a clock moving at which speed?".

And "I was only doing 30mph relative to the car I crashed into" is not a defence against a charge of doing 60 in a 30 limit, although I have yet to see a law that specifies the reference frame in which speed limits are set. We adopt standard conventions all the time, but they should not be elevated to the status of "objective".

 

[Dale]

Hey, maybe them twins ain't twins!

It must have been a painful delivery if they are

 

[Dale]

But if they are essentially stationary relative to each other there is an objective answer to the question of which was born first according to either of the clocks they carry with them (because these clocks run at the same speed and can be synchronised at any time in a unique, consistent way).

That doesn’t change the fact that their starting ages are frame dependent.

Also, it isn’t terribly informative since (regardless of the motion of the twins or the clocks) there is always an objective answer of which was born first according to any given pair of synchronized clocks at the births. The problem is that different pairs of synchronized clocks at the births disagree on their respective objective answers.

 

[Elroch]

I don't have any problem agreeing with your unambiguous statements.

1. "their starting ages are frame dependent"
2. "there is always of which was born first according to any given pair of synchronized clocks at the events"
3. " different pairs of synchronized clocks at the births disagree on their respective objective answers"

I think the phrases "terribly informative", the implication implied by "since" and the nature of the "problem" referred to at the end need a little clarification. Let me illustrate with an analogous statement (in italics to indicate it is not my view).

The time the Earth has existed is frame dependent. It isn't terribly informative since there is always an objective answer of how long the Earth has existed given a choice of frame. The problem is that different choices of frame disagree on their respective objective answers.

I emphasise that I am not disagreeing with the physics that has been described. I would suggest it is correct to say that picking the stationary frame is no more informative than using any other (just simpler). i.e. nothing more can be inferred about observations in the real world (which is all that really matters, right?)

As no-one else has done so, perhaps a sketch of a couple of freehand (sorry) space time diagrams will throw light on the central point to someone. They always do to me.

The first is the story in the rest frame of the two planets. Those familiar with the special relativistic pseudometric will immediately see the blue path has shorter proper time associated with it. The second diagram is in a frame chosen to coincide with the velocity of the traveling "twin". Here the blue path has visibly larger proper time. This illustrates the traveling twin has only "aged less" in some frames.

 

[Ibix]

The first is the story in the rest frame of the two planets. Those familiar with the special relativistic pseudometric will immediately see the blue path has shorter proper time associated with it. The second diagram is in a frame chosen to coincide with the velocity of the traveling "twin". Here the blue path has visibly larger proper time. This illustrates the traveling twin has only "aged less" in some frames.

Worth noting that those two spacetime diagrams don't show the same thing. In both frames you have the red and blue lines starting simultaneously at $t=0$ and $t'=0$, so you've cut them off at different events. This is misleading, since the whole point is that the twins are only born at the same time in at most one frame. It is this difference in birth times, coupled with the paths of different "lengths", that resolves the apparent paradox.

I would also say that your last sentence is confusing at best. If you mean that the traveling twin has aged less than the other during the journey in some frames, then I agree. However, if you mean that the traveling twin has aged less when they meet up in some frames, then I disagree - that's true in all frames.

 

[Elroch]

Worth noting that those two spacetime diagrams don't show the same thing.

Yes, while that was key, it is worth emphasising. It is of course only possible for the proper times to be different in the two diagrams because they start at different points in the lives of the twins (or they are different pairs of twins).

In both frames you have the red and blue lines starting simultaneously at $t=0$ and $t'=0$, so you've cut them off at different events. This is misleading, since the whole point is that the twins are only born at the same time in at most one frame. It is this difference in birth times, coupled with the paths of different "lengths", that resolves the apparent paradox.

You rightly draw attention to something that I should have clarified. The two different horizontal axes cannot possibly both correspond to both births for the same pair of twins. They can represent two different sets of twins being born at the same time according to a stationary observer in one case and a moving one in the other. Alternatively, they can correspond to two pairs of points in the lives of the "twins" which are respectively considered simultaneous to the two different observers.

I would also say that your last sentence is confusing at best. If you mean that the traveling twin has aged less than the other during the journey in some frames, then I agree. However, if you mean that the traveling twin has aged less when they meet up in some frames, then I disagree - that's true in all frames.

Suppose the second diagram does show the birth of the twins at time zero in the moving observer's frame (or just a chosen time in his frame from which he wishes to see how A and B age). Then he infers that A (the traveling twin) ages more from there to when they meet up, and he is right.

 

[Dale]

The time the Earth has existed is frame dependent. It isn't terribly informative since there is always an objective answer of how long the Earth has existed given a choice of frame. The problem is that different choices of frame disagree on their respective objective answers.

I am not sure I get your point. It sounded at first like you were trying to make an objectionable analogous statement. But then your analogy was not at all objectionable to me.

 

[Elroch]

I am glad it was not inadvertently objectionable! You achieve a very good tone in your contributions.

 

[vanhees71]

Yes, the age at reunion is frame independent. What is frame dependent is the initial age and the rate of aging.

I don't understand the latter statement. The proper time of each twin is a scalar and thus frame independent, right?

If you want to compare clocks of two observers, this must be done at one event, i.e., when they meet. Then you can compare the clock readings of the observers.

Take as an example the most simple practical setup for a test of the twin paradox: Some unstable particle/nucleus in a storage ring. It's lifetime is defined (sic!) as the mean proper time of this particle it takes for the particle to decay measured from a time where the particle's existence has been established with certainty. This lifetime can be compared to a clock reading at rest wrt. the laboratory. The measured mean lifetime in the lab is longer by a Lorentz $\gamma$ factor.

 

[Ibix]

If you want to compare clocks of two observers, this must be done at one event, i.e., when they meet. Then you can compare the clock readings of the observers.

Unfortunately, the setup in this version of the thought experiment doesn't allow that. The twins don't start in the same place - that's the whole problem. So there's an invariant answer to how old the traveller was when he left his planet and an invariant answer to how old both were when they meet. But "how old was the inertial twin when the traveller started travelling" depends on your simultaneity criterion, as does "how fast does each one age".

 

[vanhees71]

Then you cannot compare the age of the twins to begin with and the question doesn't make sense at all, and it's not a twin paradox at the usual sense.

 

[PeroK]

Then you cannot compare the age of the twins to begin with and the question doesn't make sense at all, and it's not a twin paradox at the usual sense.

In other words:

Hey, maybe them twins ain't twins!

 

[Ibix]

Then you cannot compare the age of the twins to begin with and the question doesn't make sense at all, and it's not a twin paradox at the usual sense.

Indeed. It's actually nearer the cosmic ray muons experiment.

 

[vanhees71]

The cosmic-ray muons experiment is an interesting example for a "one-way twin paradox"-like setting, which is well defined, because what's fixed here is the travel distance of the muons. The experiment is simply measuring the rate of cosmic muons as function of their momentum, or their $\gamma=E/(mc^2)$ factor, relative to Earth at two different heights, i.e., at some fixed distance $L$ (as measured relative to Earth). The travel time (measured relative to Earth) is of course $t=L/v=L c p/E$. The lifetime of the muon is $\tau$ and by defnition measured in their rest frame, i.e., it's measured in terms of the muons' proper time. The SRT prediction then is that the lifetime as measured with respect to Earth is $\tau'=\gamma \tau$, i.e., one should get
$$N=N_0 \exp[-t/(\gamma \tau)].$$
muons, and that's confirmed by experiment.

https://en.wikipedia.org/wiki/Experimental_testing_of_time_dilation#Frisch–Smith_experiment

 

[Dale]

The proper time of each twin is a scalar and thus frame independent, right?

Yes, but the clocks are spatially separated so the initial ages depend on the frame’s definition of simultaneity, and the rate of aging is a frame-variant ratio of coordinate time and proper time

If you want to compare clocks of two observers, this must be done at one event, i.e., when they meet.

At the beginning of this scenario the twins are spatially separated. You certainly can compare such clocks (that is indeed the point of a simultaneity convention), but the result is frame-variant as I said.

 

[vanhees71]

Ok, but then we are not discussing a twin paradox anymore.

 

[Dale]

Ok, but then we are not discussing a twin paradox anymore.

Agreed. Hence @PeroK’s funny comment

Hey, maybe them twins ain't twins!

 

[mitochan]

Say twin pilot A and B are on board the same product of VERY LONG spaceship with synchronized clocks in each way outfitted in everywhere in ship. They are at cockpits situated at the top end of the ships. Along the same line course they approach with inertial motion, pass nearby. A and B observe they share cokpit clock time = 0 when the cockpits are nearby.

The cockpit of B ship pass the tail end of A ship, then and there the B top cockpit clock time < the A tail end clock time
A judges "If now B recoils back with same speed to see me again, then I will be older than B by 2* ( RHS - LHS of the above inequality ) when we meet."

The cockpit of A ship pass the tail end of B ship, then and there the A top cockpit clock time < the B tail end clock time
B judges "If now A recoils back with same speed to see me again, then I will be older than A by 2* ( RHS - LHS of the above inequality ) when we meet."

Such a scenario comes to my mind inspired by the title of OP. If which pilot will turn back is a kind of chicken race, the looser who recoils back is younger when they meet. If they both do not want to lose, they both keep going and will not meet again forever so the race is a draw or no match.