Can the Symmetric Twin Paradox be Tested with Atomic Clocks?

[quadraphonics]

So, I was thinking about a variation on the Twin Paradox, and was hoping someone could help me work through it. The motivation is the usual explanation for the Twin Paradox, namely that one twin accelerates and so breaks the symmetry. This begs the question of what happens when both twins ride in rockets, traveling in opposite directions, then stop, reverse course, and come back to meet up again. It seems to me that while they're moving at constant relative speed, each one "sees" the other one aging at a different rate than himself, so I'm assuming that the effects of reversing course end up canceling this out? Anyhow, my apologies if this has been discussed before, but I'd appreciate any insight.

 

[pam]

The TP is best dispensed with by plotting the trajectories on a space-time diagram.
In your case it is clear that they each have the same overall distance, defined as
[tex]\int\sqrt{dt^2-dx^2}[/tex].

 

[Ich]

When he says "distance", pam means "elapsed proper time", so you're right, there is no net effect.

 

[quadraphonics]

I'm intuitively comfortable with the answer that each twin ends up the same age at the end, due to the symmetry. For example, you could modify the "paradox" so that one twin travels, comes back younger, and then the second twin travels and evens out the situation. Likewise, the space-time trajectories seem to make this rigorous. However, I suppose what I'm really after is a more intuitive explanation. Given that each twin "sees" the other aging at a different rate during the constant-velocity portion of their trip (right?), at some point (acceleration phase, presumably), they "see" some different phenomenon that cancels this effect out. Hopefully someone can explain to me what it is, exactly, that they "see?"

 

[Ich]

Hopefully someone can explain to me what it is, exactly, that they "see?"

You know that you open a can of worms with this question?
Maybe you try http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html" , it gives quite comprehensive answers.

 

[Fredrik]

Given that each twin "sees" the other aging at a different rate during the constant-velocity portion of their trip (right?),

Yes.

at some point (acceleration phase, presumably), they "see" some different phenomenon that cancels this effect out. Hopefully someone can explain to me what it is, exactly, that they "see?"

The other effect is the same effect that solves the standard twin paradox problem. Check out the http://web.comhem.se/~u87325397/Twins.PNG for the twin paradox. When the rocket is moving away, it considers events on a blue line simultaneous. When it's on its way back, it considers events on a red line simultaneous.

So if you draw the world line of the other twin doing moving in the opposite direction, you'll see that the blue lines intersect his world line at times soon after he started, and the red lines intersect his world line at times just before he came home. From twin B's point of view, twin A is aging slower the whole trip, except when twin B turns around. During the turnaround phase, twin B's "point of view" is undefined. (The end result will be the same even if the acceleration isn't constant during the turnaround phase, and it's only in the case of constant acceleration that it makes some sense to point at events on A's world line and say that they are simultaneous with event's on B's world line). After the turnaround, B will be co-moving with a different frame, and in this frame, his brother is much older than him.

 

[Hurkyl]

Hopefully someone can explain to me what it is, exactly, that they "see?"

A Lorentz transformation.

 

[quadraphonics]

Thank you, Frederick and Ich, for your helpful responses. And, yes, I probably should have specified whether I was interested in what they actually *see*, or what they calculate after compensating for Doppler shift, in order to avoid any cans of worms. Although both aspects are informative, if you ask me. It seems to me that, once Doppler shifts have been accounted for, each twin considers the other to be aging slowly during the constant velocity segments, and so they must then consider them to have aged very quickly during the turnaround phase, right? I'm a bit put off by the assertion that their POV is not defined there; if Fredrick (or whomever) would care to expand on this point, I'd appreciate it.

No thanks to Hurkyl for the flip response.

 

[Fredrik]

I'm a bit put off by the assertion that their POV is not defined there; if Fredrick (or whomever) would care to expand on this point, I'd appreciate it.

This is kind of difficult. One way to see it is to realize that the usual procedure to define simultaneity won't work:

(Copied from a post I made in another thread...)

Suppose that there's a mirror at some point along the x axis. Suppose also that you emit light from x=0, in the positive x direction, at t=-T, and that it's reflected by the mirror and returns to x=0 at t=T. Now the reflection event must have been simultaneous with the event t=0,x=0. That implies that we must assign time coordinate 0 to the reflection event, and the fact that the speed of light is c implies that we must assign the x coordinate cT to the reflection event.

Can you see why this won't work when you're accelerating? Which events are simultaneous in your accelerating frame with "right here, right now"? If we use the usual procedure to answer that, the answer depends on what you're going to do in your future.

 

[Hurkyl]

No thanks to Hurkyl for the flip response.

How is that flip?

I've been assuming this whole time that by "what the twin 'sees'" you mean the appearance of an inertial coordinate chart centered on the twin. If so, then a Lorentz transform is exactly the net effect that each twin 'sees' during the acceleration phase. And since I presume you know what a Lorentz transform is, that should be enough information for you to answer any question you have. (And it should also be clear why their POV is ill-defined -- there cannot exist an inertial coordinate chart centered on the twin while he's accelerating)

If that is not what you meant, then you'll have to go into more detail about what you mean by 'sees'. For example, what the heck do you mean by 'correcting for Doppler shift', if you aren't using some coordinate chart which you use to determine relative velocities? (If your response involves any sort of 'distance' between the twins, then again you need a coordinate chart for that)


The only thing each twin really and truly sees (i.e. by observing through his telescope) is:
(1) During the outbound trip, he sees his twin age slowly
(2) During the first part of the inbound trip, he sees his twin age normally
(3) During the second part of the inbound trip, he sees his twin age rapidly

Anything other than that requires some conventional choice for reinterpreting this observational data. i.e. your talk about 'seeing' requires some method for turning this data into some other sort of description.



Wait a minute -- it just strikes me that "correcting for Doppler shift" would mean calculating that his twin really is aging normally. But I've been assuming this whole time that you really meant "correcting for the propagation delay of the light".

 

[DrGreg]

It seems to me that, once Doppler shifts have been accounted for, each twin considers the other to be aging slowly during the constant velocity segments, and so they must then consider them to have aged very quickly during the turnaround phase, right?

Yes.

I'm a bit put off by the assertion that their POV is not defined there; if Fredrick (or whomever) would care to expand on this point, I'd appreciate it.

Fredrik isn't exactly right. It is possible for an accelerating observer to define simultaneity (except in the physically impossible situation of infinite acceleration with an instant turnaround).

Fredrik is right to say, in post #9, that the "radar" method of simultaneity can't work for an accelerated observer. But there's another method, using a "comoving inertial observer". At the moment you want to define simultaneity, you consider another inertial observer who happens to be stationary relative to you at that moment (but not before or after). Use that observer's definition of simultaneity.

This definition isn't much use in practice because you'd need many different inertial observers (in theory, an infinite number) to calculate simultaneity throughout the acceleration, but mathematically the definition works just fine and gives you a (locally) well-defined coordinate system to assign time coordinates.

(Although the coordinate system is well-defined locally, it doesn't behave too well further away. It's possible to "see" someone else's clock (a long, long way away) stop or even go backwards! I stress that I'm using the word "see" as it has been used throughout this thread, what you calculate according to your definition of simultaneity, not what you see with your eyes. Your eyes will always see clocks moving forwards. This illustrates that "simultaneity" is an artificial construct, absolutely vital to perform calculations but of no physical significance.)
________

For those who are interested, an example is (I think, if I've done it right)

[tex]x = X \cosh \frac{aT}{c} [/tex] (1)
[tex]t = \frac {X}{c} \sinh \frac{aT}{c} [/tex] (2)​

where [itex](t, x)[/itex] are inertial coordinates, [itex](T, X)[/itex] are accelerated coordinates.

In these coordinates, the accelerated observer lies at the location [itex]X = c^2 / a[/itex] and moves with constant proper acceleration of [itex]a[/itex]. For that value of [itex]X[/itex] only, [itex]T[/itex] is the proper time of the accelerated observer -- i.e. the observer's clock; for all other values it represents simultaneity with the observer's clock according to the comoving inertial observer.

According to the accelerated observer, inertial clocks stop at [itex]X = 0[/itex] and go backward for [itex]X < 0[/itex].​

 

[quadraphonics]

Thanks for the input Dr. Greg, although I'm not sure whether it leaves me more or less confused. Specifically, if we imagine defining a continuum of comoving inertial observers for the acceeleration phase, it seems to me that any of those inertial observers is still going to perceive the other twin's clock as running slowly, due to their relative velocity. So, I don't see how it is that when we get to the end of the turnaround phase, that all adds up to the other twin having aged rapidly? Clearly I must be missing some aspect of how this fits together.

 

[Hurkyl]

Thanks for the input Dr. Greg, although I'm not sure whether it leaves me more or less confused. Specifically, if we imagine defining a continuum of comoving inertial observers for the acceeleration phase, it seems to me that any of those inertial observers is still going to perceive the other twin's clock as running slowly, due to their relative velocity. So, I don't see how it is that when we get to the end of the turnaround phase, that all adds up to the other twin having aged rapidly? Clearly I must be missing some aspect of how this fits together.

There are two sources of time dilation here: one is due to the relative velocity between the remote twin and the inertial observers, as you've identified. The other is that your inertial observers each ascribe different coordinates to the same event. As you change observers, you change how they assign time coordinates¹, which results in a second contribution to time dilation.

1. Given, of course, by a Lorentz transformation

 

[Fredrik]

Specifically, if we imagine defining a continuum of comoving inertial observers for the acceeleration phase, it seems to me that any of those inertial observers is still going to perceive the other twin's clock as running slowly, due to their relative velocity. So, I don't see how it is that when we get to the end of the turnaround phase, that all adds up to the other twin having aged rapidly? Clearly I must be missing some aspect of how this fits together.

What you're missing is that a co-moving inertial observer just before the rocket turns around thinks of the upper blue line as "space" at that time (I'm referring to the same http://web.comhem.se/~u87325397/Twins.PNG as before), and that a co-moving inertial observer just after the rocket turns around thinks of the lower red line as space at that time. If you consider a sequence of co-moving frames, the Earth twin will be aging slower in all of them, but each step from one frame in the sequence to the next tilts the simultaneity line so that the event on the rocket is simultaneous with a later time on Earth.

Greg was right to correct me by the way.

 

[quadraphonics]

What you're missing is that a co-moving inertial observer just before the rocket turns around thinks of the upper blue line as "space" at that time (I'm referring to the same http://web.comhem.se/~u87325397/Twins.PNG as before), and that a co-moving inertial observer just after the rocket turns around thinks of the lower red line as space at that time. If you consider a sequence of co-moving frames, the Earth twin will be aging slower in all of them, but each step from one frame in the sequence to the next tilts the simultaneity line so that the event on the rocket is simultaneous with a later time on Earth.

Okay, this is starting to make sense to me. If I can hazard a rough paraphrase, might we say that, while each co-moving inertial observer sees the other twin aging slowly, said twin "starts out" much older in each successive co-moving frame than he was in the last? The net effect of that being that the other twin's overall aging is faster than normal during turnaround. And so if we consider a continuous succession of co-moving frames, the former effect is swamped out, and we just see the other twin aging rapidly, right?

 

[Fredrik]

Yes, that's right.

 

[quadraphonics]

Great; thanks for your help.

 

[HarryWertM]

Looking up [googling] various SR experiments and found this ancient thread. Has anyone attempted any direct test of this interesting "symmetric twin" problem? I mean, anyone sent precise atomic clocks off in different directions; brought them to relative rest; and compared their readouts?