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quadraphonics

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- Thread starter quadraphonics
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quadraphonics

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In your case it is clear that they each have the same overall distance, defined as

[tex]\int\sqrt{dt^2-dx^2}[/tex].

- #3

Ich

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

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quadraphonics

- #5

Ich

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You know that you open a can of worms with this question?Hopefully someone can explain to me what it is, exactly, that they "see?"

Maybe you try http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html" [Broken], it gives quite comprehensive answers.

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- #6

Fredrik

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

The other effect is the same effect that solves the standard twin paradox problem. Check out the http://web.comhem.se/~u87325397/Twins.PNG [Broken] 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.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?"

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

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- #7

Hurkyl

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

- #8

quadraphonics

No thanks to Hurkyl for the flip response.

- #9

Fredrik

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

(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.

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Hurkyl

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How is that flip?No thanks to Hurkyl for the flip response.

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

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

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".

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- #11

DrGreg

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Yes.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?

Fredrik isn't exactly right. ItI'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 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

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

________

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)

[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].

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quadraphonics

- #13

Hurkyl

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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 coordinates

1. Given, of course, by a Lorentz transformation

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Fredrik

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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 [Broken] 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.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.

Greg was right to correct me by the way.

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- #15

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 [Broken] 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?

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Fredrik

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Yes, that's right.

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quadraphonics

Great; thanks for your help.

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