Yet another Twin Paradox thread

[Buckethead]

I've read most of the Paradox within a Paradox thread but rather than hijack it to get to my point I thought I'd start a new thread. First, if this has already been addressed from this angle my apologies, but it is next to impossible to read through all the TP threads to try and find one from this angle.

Examining the wiki entry on the Twin Paradox (which seems very complete) we see it addresses first a jump in simultaneity when the ship turns around and then later addresses the doppler shift and shows two diagrams of pulses from Earth and pulses from the ship. I understand both concepts but still find problems with both. I will only address the first part of the wiki entry and save the second for another time.

My favorite scenario is one that eliminates the acceleration problem thusly: A ship at .866c travels past the earth. At the moment of passage, the ship's stopwatch "high fives" a stopwatch on Earth and this action resets both stopwatches to zero. A year later by ship time the ship "high fives" a second ship also going at .866c but at a heading back to Earth. This contact causes the returning ship to set it's stopwatch to the timeout time on the first ship setting both stopwatches to the same time which would in fact be 1 year ship time.

The second ship arrives at the Earth one year after the ship to ship contact and a photographer takes a picture of the stopwatch on the ship and finds that it has timed out 2 years. The Earth stopwatch on the other hand has timed out exactly 4 years according to the accepted understanding of relativity.

Now according to wiki, if instead of my description one uses a reversal of a single ship the paradox is resolved because the ship is in a gravitational state upon reversal and this causes the Earth time to jump ahead about 3 years in the time it takes for the ship to turn around and get back to speed explaining the 4 year age of the Earth in a 2 year ship time. I can accept this.

But using my scenario where there is no acceleration, this argument cannot be used. Still, I will take a leap of faith here and figure that at the time of ship to ship clock reading, and the shift over to the reference frame of the second ship, the Earth instantaneously jumps into the future by about 3 years. (it's not 2 years like you might think because of the time dilation of Earth time as seen by the ship according to the simultaneity graph in wiki just before and just after the ship to ship contact) Is everyone in agreement that this is what would happen? I can accept this as well as there seems to be no alternative way out of this if I'm to hold true to relativity.

Now here is my problem. A person on Earth at some early time in this 4 year (earth time) time span and after 1 additional year for the light to reach this person, he sees the two ships a light year away making ship to ship contact. And in order to match reality, he will see something much much stranger. For about the next 3 years, he will see that both ships have not moved one single inch, they are frozen in time litereally waiting for the 3 additional year passage of time to pass on Earth and resolve the paradox. After the 3 years of looking, suddently the observer on Earth will once again see the ships traveling on their merry way, one of which will reach the Earth in about a half year time and of course when this happens, the person will have aged 4 years.

Thanks for reading yet another Twin Paradox thread.

 

[ghwellsjr]

I haven't bothered to look at the wiki article but no one ever sees any jumps in time.

An observer on Earth will of course see a steady increase in time spanning four years (going from zero to four years).

An observer on the outgoing spaceship will see a steady increase in time spanning one year (going from zero to one year).

An observer on the returning spaceship will see a steady increase in time spanning one year (going from one to two years).

An observer on Earth will see the outgoing spaceship traveling away at .866c and will observe a relativistic doppler shift of the spaceship's clock "ticks" of a factor of about 0.268 of normal. He will continue to see this until after the ships actually make contact and for some time thereafter. He has to wait for the light signal of the event to reach him.

An observer on the outgoing spaceship will observe the same relativistic doppler shift of the clock "ticks" coming from Earth of the same factor of 0.268 of normal.

An observer on the returning spaceship (let's assume that he also knows the accumulated count of ticks from the Earth as observed by the outgoing spaceship) will observe a relativistic doppler shift of the Earth's clock ticks at a factor of about 3.73 of normal continuing all the way until Earth contact. The number of the ticks from the other spaceship plus his own add up to exactly four years.

Some time after the ships make contact with each other, the observer on the Earth will see the event and will then see the relativistic doppler shift of the returning space ship's clock ticks at a factor of 3.73 of normal and the sum of the outgoing ticks plus the returning ticks add up to two years.

Note that the reciprocal relativistic doppler is the means by which each observer measures the time dilation of the other one but the reason they come up with different total times is because the pair on the ships observes the lower and higher dopplers from Earth for an equal amount of time whereas the Earth observer sees the lower doppler for much longer than the higher doppler.

Simple, huh?

 

[Buckethead]

The wiki article is here:

http://en.wikipedia.org/wiki/Twin_paradox

From the article:

" Just before turnaround, the traveling twin calculates the age of the resting twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnaround, if he recalculates, he'll measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the resting twin. The traveling twin reckons that there has been a jump discontinuity in the age of the resting twin."

With regard to the doppler shift, this does not address the jump in simultaneaty that must occur at the turnaround point caused by the momentary increase in gravity as described by the article here:

"During the turnaround, the traveling twin is in an accelerated reference frame. According to the equivalence principle, the traveling twin may analyze the turnaround phase as if the stay-at-home twin were freely falling in a gravitational field and as if the traveling twin were stationary. A 1918 paper by Einstein presents a conceptual sketch of the idea.[10] From the viewpoint of the traveler, a calculation for each separate leg, ignoring the turnaround, leads to a result in which the Earth clocks age less than the traveler. For example, if the Earth clocks age 1 day less on each leg, the amount that the Earth clocks will lag behind amounts to 2 days. The physical description of what happens at turnaround has to produce a contrary effect of double that amount: 4 days' advancing of the Earth clocks. Then the traveler's clock will end up with a net 2-day delay on the Earth clocks, in agreement with calculations done in the frame of the stay-at-home twin.

The mechanism for the advancing of the stay-at-home twin's clock is gravitational time dilation. When an observer finds that inertially moving objects are being accelerated with respect to themselves, those objects are in a gravitational field insofar as relativity is concerned. For the traveling twin at turnaround, this gravitational field fills the universe."

In other words, in my example, the time on Earth rapidly sweeps across a period of time while the turnaround is taking place, then resumes the time dilation as predicted by SR once the ship resumes. I'm assuming the wiki article has been reviewed and is correct.

 

[Dale]

My favorite scenario is one that eliminates the acceleration problem thusly: A ship at .866c travels past the earth. At the moment of passage, the ship's stopwatch "high fives" a stopwatch on Earth and this action resets both stopwatches to zero. A year later by ship time the ship "high fives" a second ship also going at .866c but at a heading back to Earth. This contact causes the returning ship to set it's stopwatch to the timeout time on the first ship setting both stopwatches to the same time which would in fact be 1 year ship time.

How is this a paradox at all? Surely you are not claiming that a trip involving two separate ships is symmetrical with a trip involving a single ship. Since it is so clearly not symmetrical there is no sense in which it is surprising that the second ship's clock shows a different time than the Earth clock at the end.

 

[JesseM]

The wiki article is here:

http://en.wikipedia.org/wiki/Twin_paradox

From the article:

" Just before turnaround, the traveling twin calculates the age of the resting twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnaround, if he recalculates, he'll measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the resting twin. The traveling twin reckons that there has been a jump discontinuity in the age of the resting twin."

With regard to the doppler shift, this does not address the jump in simultaneaty that must occur at the turnaround point caused by the momentary increase in gravity as described by the article here:

"During the turnaround, the traveling twin is in an accelerated reference frame. According to the equivalence principle, the traveling twin may analyze the turnaround phase as if the stay-at-home twin were freely falling in a gravitational field and as if the traveling twin were stationary. A 1918 paper by Einstein presents a conceptual sketch of the idea.[10] From the viewpoint of the traveler, a calculation for each separate leg, ignoring the turnaround, leads to a result in which the Earth clocks age less than the traveler. For example, if the Earth clocks age 1 day less on each leg, the amount that the Earth clocks will lag behind amounts to 2 days. The physical description of what happens at turnaround has to produce a contrary effect of double that amount: 4 days' advancing of the Earth clocks. Then the traveler's clock will end up with a net 2-day delay on the Earth clocks, in agreement with calculations done in the frame of the stay-at-home twin.

The mechanism for the advancing of the stay-at-home twin's clock is gravitational time dilation. When an observer finds that inertially moving objects are being accelerated with respect to themselves, those objects are in a gravitational field insofar as relativity is concerned. For the traveling twin at turnaround, this gravitational field fills the universe."

In other words, in my example, the time on Earth rapidly sweeps across a period of time while the turnaround is taking place, then resumes the time dilation as predicted by SR once the ship resumes. I'm assuming the wiki article has been reviewed and is correct.

A basic principle in relativity is that you can analyze any situation from the perspective of any frame you like. For example, the simplest approach to the twin paradox is simply to pick an inertial frame and analyze the whole situation from beginning to end from the perspective of this frame--in an inertial frame there will be no "jump in simultaneity" of the kind described above. And even if you want to pick a non-inertial frame where the traveling twin has an unchanging position coordinate, there are an infinite number of ways you could define the "frame of the traveling twin", the analysis above only describes what happens from the perspective of one particular choice of non-inertial frame which is designed to have the property that its definition of simultaneity at each point on the traveling twin's worldline matches up with the definition of simultaneity in the inertial frame where the traveling twin is instantaneously at rest at that point. This may be a particularly simple or elegant way to define a non-inertial rest frame for the traveling twin, but there's nothing physical that compels you to define it this way! You could equally well define a non-inertial frame where the traveling twin is at rest the whole time but the definition of simultaneity always matches that of some specific inertial frame, in which case the stay-at-home twin will age at a constant rate in this frame.

 

[D H]

I'm assuming the wiki article has been reviewed and is correct.

Now why would you assume that?

My Bayesian prior regarding wikipedia articles on an advanced topic (GR certainly qualifies as such) is that while the article might be correct, it will be most likely be poorly written, won't define the nomenclature, and will jump to conclusions. The authors of textbooks and journal articles at least warn you when they are jumping to conclusions via "the derivation is [standard caveat]" where standard caveat is one or more of "obvious", "well-known", "omitted", "derived in [reference]", or "left as an exercise to the reader". But hey, at least they let you know that they did skip over something.

Regarding the twin paradox, there are a number of different ways to look at it. All of them yield the same final answer. Pick the point of view that makes the most sense to you and then for the techniques that don't make as much sense, read up on how to rectify those points of view. The usenet physics FAQ does a nice job of discussing several of these points of view. See http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html.

 

[Buckethead]

How is this a paradox at all? Surely you are not claiming that a trip involving two separate ships is symmetrical with a trip involving a single ship. Since it is so clearly not symmetrical there is no sense in which it is surprising that the second ship's clock shows a different time than the Earth clock at the end.

I'm not sure "symmetrical" is the correct word here, but I will agree that one ship that turns around and two ships that pass each other are not the same phenomenon and will not yield the same results. For one thing a single ship experiences gravity and two ships do not.

If you are referring to symmetry as far as the symmetry between the Earth and a single ship or the symmetry between the Earth and 2 ships again, there is no symmetry in either case.

And this is my point, If you split the experiment into two parts, the ship leaving and the ship returning, then you can say (if SR is correct) that the ship leaving Earth is symetrical to the Earth leaving the ship. You can likewise way that the Earth returning to the ship is symmetrical with the ship returning to the earth. Therefore, in both cases, the ship and the Earth both see the other's clocks moving slower. However when the ships return (in either the 2 ship or 1 ship scenario) the Earth clock has sped up. This means that it was either the turn around time (involving acceleration) or it was at the time the two clocks synchronized that a magic leap in Earth time occurred.

 

[sylas]

As D H has noted, there are various ways to look at the problem, and they all get the same answers unless you make a mistake.

I have considered a concrete case where a traveller moves 6 light years at 60% the speed of light, turns around effectively instantaneously, and then returns at the same speed; all these values being as determined by a twin at rest at the starting point. The description, including some spacetime diagrams, is at msg #46 of Twin Paradox- a quick(ish) question.

Here's a diagram, showing what goes on in three different inertial frames; the twin who is inertial the whole time, the traveler on the outbound leg, and the traveler on the return leg.

The discontinuity shows up at the moment of turn around, it corresponds to the turning twin changing from one frame to another. In the new frame, the light that they are receiving from the start point is coming from four times further away than it was in the old frame.

This would be confirmed by any measurement made by the traveling. In particular, the angular size in the sky of the planet where the stay at home twin is waiting will reduce by a factor of four as the traveling twin turns around; the angular size is an indicator of how far away the object was when it emitted the light that you are "seeing" at the moment.

Cheers -- sylas

 

[Buckethead]

A basic principle in relativity is that you can analyze any situation from the perspective of any frame you like. For example, the simplest approach to the twin paradox is simply to pick an inertial frame and analyze the whole situation from beginning to end from the perspective of this frame--in an inertial frame there will be no "jump in simultaneity" of the kind described above. And even if you want to pick a non-inertial frame where the traveling twin has an unchanging position coordinate, there are an infinite number of ways you could define the "frame of the traveling twin", the analysis above only describes what happens from the perspective of one particular choice of non-inertial frame which is designed to have the property that its definition of simultaneity at each point on the traveling twin's worldline matches up with the definition of simultaneity in the inertial frame where the traveling twin is instantaneously at rest at that point. This may be a particularly simple or elegant way to define a non-inertial rest frame for the traveling twin, but there's nothing physical that compels you to define it this way! You could equally well define a non-inertial frame where the traveling twin is at rest the whole time but the definition of simultaneity always matches that of some specific inertial frame, in which case the stay-at-home twin will age at a constant rate in this frame.

I'm going to stay logical here and assume that there is no "real" sudden jump in simultaneity and that a sudden fast forward in Earth time from the ship is not something the ship would see. But as far as I can see, there is only one thing happening here that can be taken as the gospel truth and that is that in my 2 ship scenario, the traveling stopwatch will show less time elapsing than the stopwatch on earth. In addition, what the ship sees is one of two things in order for logic to prevail. The first is that the ship sees the clock on Earth moving slowly, then at some point (the turnaround or meeting point) must see it speed up rapidly, then slow down again on the way home, or secondly it must see the clock on Earth always moving faster which defies SR. There can be no third alternative. This implies an asymmetry between the ship and the Earth either during the whole trip or only during the turnaround time. Since according to SR there is complete symmetry during the trip out, and again during the trip back, this means that at the instant of signal transfer between the two ships, something profound must happen such as a jump into the future. Now I for one cannot accept this as it defies all logic, so the only other answer is that there is something different about the ship and the Earth that makes it non-symmetrical during the entire trip. And if this is true then SR is not correct in it's prediction of symmetry during the time the ship travels out and again during the time the second ship travels home.

 

[Buckethead]

Regarding the twin paradox, there are a number of different ways to look at it. All of them yield the same final answer. Pick the point of view that makes the most sense to you and then for the techniques that don't make as much sense, read up on how to rectify those points of view. The usenet physics FAQ does a nice job of discussing several of these points of view. See http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html.

Thanks for the link. I think I've found what I think is the heart of the problem. According to the article in the link:

"All well and good, but this discussion at first just seems to sharpen the paradox! Stella sees what Terence sees: a slow clock on the Outbound Leg, a fast clock on the Inbound Leg. Whence comes the asymmetry between Stella and Terence?

Answer: in the duration of the Inbound and Outbound Legs, as seen. For Stella, each Leg takes about a year. Terence maintains that Stella's turnaround takes place at year 7 at a distance of nearly 7 light-years, so he won't see it until nearly year 14. Terence sees an Outbound Leg of long duration, and an Inbound Leg of very short duration."

There is a severe circular logic here. The article assumes that there is asymmetry from the git go, but one cannot make this assumption. It is taking the fact that Stella will age more slowly (which I admit does occur) and applying it to the reason that she ages more slowly, namely that she sees something different due to her slow aging. You can't do this! According to SR, on Stella's trip out, one can say that it is Terrance that is aging more slowly as he is the one that is traveling and that is the reason Terrance is aging more slowly. Circular logic that makes no sense. Experiment bears out that Stella does age more slowly so we can without a doubt say there is some kind of asymmetry here, but we cannot say the asymmetry is due to the fact that Stella ages more slowly!

 

[Buckethead]

As D H has noted, there are various ways to look at the problem, and they all get the same answers unless you make a mistake.

I have considered a concrete case where a traveller moves 6 light years at 60% the speed of light, turns around effectively instantaneously, and then returns at the same speed; all these values being as determined by a twin at rest at the starting point. The description, including some spacetime diagrams, is at msg #46 of Twin Paradox- a quick(ish) question.

Here's a diagram, showing what goes on in three different inertial frames; the twin who is inertial the whole time, the traveler on the outbound leg, and the traveler on the return leg.

The discontinuity shows up at the moment of turn around, it corresponds to the turning twin changing from one frame to another. In the new frame, the light that they are receiving from the start point is coming from four times further away than it was in the old frame.

This would be confirmed by any measurement made by the traveling. In particular, the angular size in the sky of the planet where the stay at home twin is waiting will reduce by a factor of four as the traveling twin turns around; the angular size is an indicator of how far away the object was when it emitted the light that you are "seeing" at the moment.

Cheers -- sylas

Thanks for the graphs, they are very helpful, but I don't think the heart of what I'm getting at is simply a matter of changing perspective. I've clarified this in my posts above.

 

[D H]

There is a severe circular logic here. The article assumes that there is asymmetry from the git go, but one cannot make this assumption.

Read the articles and do the math. The asymmetry is derived, not assumed.

 

[sylas]

Thanks for the graphs, they are very helpful, but I don't think the heart of what I'm getting at is simply a matter of changing perspective. I've clarified this in my posts above.

Thanks. I am glad the diagram helped. This time I will quote a number of posts, and try to hone in further on the problem.

I'm not sure "symmetrical" is the correct word here, but I will agree that one ship that turns around and two ships that pass each other are not the same phenomenon and will not yield the same results. For one thing a single ship experiences gravity and two ships do not.

Symmetrical is the correct word; the situation of the two twins is not symmetrical. Also, two ships that pass each other do give the same phenomenon and the same result. All you need is to have the two ships synchronize clocks at the instant they pass each other, and it all reduces to the same problem, with the same answers.

You can also consider a case of someone using a Star Trek Teleporter as the ships pass, to move from one ship to another. As has been seen when they teleport to a planet, there's no problems with acceleration.

Or you can consider a weird twist in spacetime which means that the ship reverses direction without acceleration.

Gravity, and acceleration, are not the crucial point. It is the change of frame which is crucial. Of course, conventionally the only way we change frames in practice is by accelerating, but if you propose any other instantaneous change in frame at the turn point, as I have done in the examples above, it all gives the same result, because it is the change of frame that really matters.

And this is my point, If you split the experiment into two parts, the ship leaving and the ship returning, then you can say (if SR is correct) that the ship leaving Earth is symetrical to the Earth leaving the ship. You can likewise way that the Earth returning to the ship is symmetrical with the ship returning to the earth. Therefore, in both cases, the ship and the Earth both see the other's clocks moving slower. However when the ships return (in either the 2 ship or 1 ship scenario) the Earth clock has sped up. This means that it was either the turn around time (involving acceleration) or it was at the time the two clocks synchronized that a magic leap in Earth time occurred.

As I was trying to show in the diagram, it is the turn point which involves a "leap" in Earth time, and this is because of a change in frame, whether that change in frame is due to an acceleration or due to some other mechanism we might propose with no acceleration or gravity involved.

I'm going to stay logical here and assume that there is no "real" sudden jump in simultaneity and that a sudden fast forward in Earth time from the ship is not something the ship would see.

"Logic" in these discussions is nearly always an example of someone holding onto an invalid assumption that is actually incorrect.

In this case, it's a bit subtle. What the ship "sees", of course, is light that left Earth a long time ago. How long ago depends only on how far away the Earth was at the time the light was emitted.

That DOES change instantaneously, at the turning point, and the turning ship DOES see it. What they see is that the home planet is suddenly (instantaneously) much much further away. Specifically, in the case where an observer moving at 60% light speed suddenly reverses (whether by acceleration or time warps or teleporting into another passing ship) to be 60% light speed in the other direction sees the remote planet suddenly four times further away. It's a real observation. The remote planet really will be four times smaller in the sky. The brightness will be affected by doppler shifts as well, but taking that into account still gives the same result... the remote planet is suddenly seen to be four times further away.

This is seeing via the light that left the remote planet 3 years ago (in the first frame) or 12 years ago (in the second frame).

... In addition, what the ship sees is one of two things in order for logic to prevail. The first is that the ship sees the clock on Earth moving slowly, then at some point (the turnaround or meeting point) must see it speed up rapidly, then slow down again on the way home, or secondly it must see the clock on Earth always moving faster which defies SR. There can be no third alternative. This implies an asymmetry between the ship and the Earth either during the whole trip or only during the turnaround time.

Forget logic. Use maths and relativity. As the ship and planet are moving apart, so that each one sees the other receding at 60% light speed, each one sees the other clock advancing at half the speed of their own clock. This is not the same as the dilation factor, because you also have light taking longer to get from the remote clock to an observer (either observer).

Similarly, as the ship and planet are moving together, so that each one sees the other approaching at 60% light speed, each one sees the other clock advancing at twice the speed of their own clock. Again, this is not only dilation; because as the clocks approach it takes less time for a signal to arrive.

The difference is in HOW LONG each observer sees the other one approaching/receding.

I describe this also in the thread I linked earlier. In the case of travel at 60% light speed to a star 6 light years away, the stay at home twin sees the ship receding for 16 years, and sees it returning for 4 years. The elapsed time seen passing on the remote shipboard clock is 16/2 + 4*2 = 8+8 = 16 years. All this takes place over 20 years, and indeed the traveler has aged only 16 years when they return 20 years after setting out.

For the ship, they see Earth receding for 8 years, then there is the turn around instant and they see the Earth approaching for 8 years. The time seen elapsing on the Earth clock is thus 8/2 + 8*2 = 4 + 16 = 20.

And indeed, on return after 16 years of traveling, the spacefarer sees everyone who remained at home is 20 years older.

Thanks for the link. I think I've found what I think is the heart of the problem. According to the article in the link:

"All well and good, but this discussion at first just seems to sharpen the paradox! Stella sees what Terence sees: a slow clock on the Outbound Leg, a fast clock on the Inbound Leg. Whence comes the asymmetry between Stella and Terence?

Answer: in the duration of the Inbound and Outbound Legs, as seen. For Stella, each Leg takes about a year. Terence maintains that Stella's turnaround takes place at year 7 at a distance of nearly 7 light-years, so he won't see it until nearly year 14. Terence sees an Outbound Leg of long duration, and an Inbound Leg of very short duration."

There is a severe circular logic here. The article assumes that there is asymmetry from the git go, but one cannot make this assumption.

This is not circular logic. You already know that the case is asymmetrical, and this is explaining aspects of the asymmetry.

I'm happy to explain further if it helps; but do be aware that this is well understood physics, and the issue is really just helping students comes to grips with it. The article is not circular; and it is cited as something that may help you understand it a bit better.

Cheers -- sylas

 

[JesseM]

I'm going to stay logical here and assume that there is no "real" sudden jump in simultaneity and that a sudden fast forward in Earth time from the ship is not something the ship would see. But as far as I can see, there is only one thing happening here that can be taken as the gospel truth and that is that in my 2 ship scenario, the traveling stopwatch will show less time elapsing than the stopwatch on earth. In addition, what the ship sees is one of two things in order for logic to prevail. The first is that the ship sees the clock on Earth moving slowly, then at some point (the turnaround or meeting point) must see it speed up rapidly, then slow down again on the way home, or secondly it must see the clock on Earth always moving faster which defies SR. There can be no third alternative. This implies an asymmetry between the ship and the Earth either during the whole trip or only during the turnaround time. Since according to SR there is complete symmetry during the trip out, and again during the trip back, this means that at the instant of signal transfer between the two ships, something profound must happen such as a jump into the future. Now I for one cannot accept this as it defies all logic, so the only other answer is that there is something different about the ship and the Earth that makes it non-symmetrical during the entire trip. And if this is true then SR is not correct in it's prediction of symmetry during the time the ship travels out and again during the time the second ship travels home.

Buckethead, just to be clear do you understand that there's a difference between optical appearances, i.e. what either twin "sees" visually, and facts about simultaneity in a coordinate system for either twin? For example, in the rest frame of the inertial twin the clock of the non-inertial twin always runs slow, but that's not what the inertial twin sees in a visual sense, instead the inertial twin sees the clock of the non-inertial twin appear to be running faster than his own clock as the non-inertial twin travels back towards him (a consequence of the Doppler effect).

Second, do you understand that although inertial coordinate systems must be defined in a particular characteristic way, there are an infinite number of ways to define a non-inertial coordinate system, since in general a "coordinate system" is just an arbitrary way of labeling events with position and time coordinates? For example, one could define a non-inertial frame where the non-inertial twin was at rest, and where the definition of simultaneity was a strange one where as the non-inertial twin's clock ticked forward, the clock of the inertial twin would alternate between periods of running fast and running slow, such that if you graphed the fast vs. slow periods on a chart they would spell out "DONALD DUCK" in morse code. Really any continuous coordinate system is allowed if you don't restrict yourself to inertial coordinate systems, this is spelled out more clearly by the principle of "diffeomorphism invariance" in general relativity, see this article for example.

 

[Buckethead]

Read the articles and do the math. The asymmetry is derived, not assumed.

In order to derive an asymmetry, you must first start from a frame of reference, either Stella's or Terence. If Terence (as in the article) you will come up with an asymmetry that is biased toward Stella aging more slowly, if from Stella you will derive an asymmetry that is biased toward Terence aging more slowly. Therefore the situation is symmetrical, not asymmetrical. This symmetry is broken of course at the time of the turnaround, but from the start of the experiment until then, the situation is symmetrical.

 

[Buckethead]

Gravity, and acceleration, are not the crucial point. It is the change of frame which is crucial. Of course, conventionally the only way we change frames in practice is by accelerating, but if you propose any other instantaneous change in frame at the turn point, as I have done in the examples above, it all gives the same result, because it is the change of frame that really matters.

Agreed. The change in frame matters. Experiment bears out that the ship ages more slowly and since there is perfect symmetry at all times (according to SR) except during the turnaround point, it is this point where all the fun happens.



As I was trying to show in the diagram, it is the turn point which involves a "leap" in Earth time, and this is because of a change in frame, whether that change in frame is due to an acceleration or due to some other mechanism we might propose with no acceleration or gravity involved.

"Logic" in these discussions is nearly always an example of someone holding onto an invalid assumption that is actually incorrect.

In this case, it's a bit subtle. What the ship "sees", of course, is light that left Earth a long time ago. How long ago depends only on how far away the Earth was at the time the light was emitted.

That DOES change instantaneously, at the turning point, and the turning ship DOES see it. What they see is that the home planet is suddenly (instantaneously) much much further away. Specifically, in the case where an observer moving at 60% light speed suddenly reverses (whether by acceleration or time warps or teleporting into another passing ship) to be 60% light speed in the other direction sees the remote planet suddenly four times further away. It's a real observation. The remote planet really will be four times smaller in the sky. The brightness will be affected by doppler shifts as well, but taking that into account still gives the same result... the remote planet is suddenly seen to be four times further away.

This is seeing via the light that left the remote planet 3 years ago (in the first frame) or 12 years ago (in the second frame).[/QUOTE]

I agree that the apparent distance will change at turnaround and this is an optical phenomenon. The distance appears closer when moving away because the angle that the light hits the ship changes (like walking in the rain) making it appear close when moving away or far when moving toward. However I don't agree with your last sentence. The light will have traveled x number of light years just before or just after the turnaround point. You are basing this on your observation of the stars suddenly "appearing" close then "appearing" far but again this is simply an optical illusion. In reality, the light has traveled a greater number of light years in both cases then the ship captain will think it has due to time dialation.

Forget logic. Use maths and relativity. As the ship and planet are moving apart, so that each one sees the other receding at 60% light speed, each one sees the other clock advancing at half the speed of their own clock. This is not the same as the dilation factor, because you also have light taking longer to get from the remote clock to an observer (either observer).

Similarly, as the ship and planet are moving together, so that each one sees the other approaching at 60% light speed, each one sees the other clock advancing at twice the speed of their own clock. Again, this is not only dilation; because as the clocks approach it takes less time for a signal to arrive.

The difference is in HOW LONG each observer sees the other one approaching/receding.

I describe this also in the thread I linked earlier. In the case of travel at 60% light speed to a star 6 light years away, the stay at home twin sees the ship receding for 16 years, and sees it returning for 4 years. The elapsed time seen passing on the remote shipboard clock is 16/2 + 4*2 = 8+8 = 16 years. All this takes place over 20 years, and indeed the traveler has aged only 16 years when they return 20 years after setting out.

For the ship, they see Earth receding for 8 years, then there is the turn around instant and they see the Earth approaching for 8 years. The time seen elapsing on the Earth clock is thus 8/2 + 8*2 = 4 + 16 = 20.

And indeed, on return after 16 years of traveling, the spacefarer sees everyone who remained at home is 20 years older.

All of the above assumes that the clock in the ship is aging more slowly, which is fine as experiment bears this out, but again this is circular logic. If at the start of the experiment everything is symmetrical and remains symmetrical up until the turnaround point, then you can also say that in reality the Earth is aging more slowly which of course it does not when the experiment is over. The reason is because there is something unusual that either happens when the ship turns around or there is something asymmetrical about the entire experiment.

What is additionally interesting is what is seen from Earth at the time of turnaround. It would seem that as the ship is traveling away it will appear closer than it really is and after it turns around it should appear farther than it really is. But does this really happen? If it doesn't then this is another example of asymmetry. But according to SR it does because of the inherent symmetry both when the ship is moving away and again when it is returning.

 

[Buckethead]

Buckethead, just to be clear do you understand that there's a difference between optical appearances, i.e. what either twin "sees" visually, and facts about simultaneity in a coordinate system for either twin? For example, in the rest frame of the inertial twin the clock of the non-inertial twin always runs slow, but that's not what the inertial twin sees in a visual sense, instead the inertial twin sees the clock of the non-inertial twin appear to be running faster than his own clock as the non-inertial twin travels back towards him (a consequence of the Doppler effect).

Yes, I understand this.

Second, do you understand that although inertial coordinate systems must be defined in a particular characteristic way, there are an infinite number of ways to define a non-inertial coordinate system, since in general a "coordinate system" is just an arbitrary way of labeling events with position and time coordinates? For example, one could define a non-inertial frame where the non-inertial twin was at rest, and where the definition of simultaneity was a strange one where as the non-inertial twin's clock ticked forward, the clock of the inertial twin would alternate between periods of running fast and running slow, such that if you graphed the fast vs. slow periods on a chart they would spell out "DONALD DUCK" in morse code. Really any continuous coordinate system is allowed if you don't restrict yourself to inertial coordinate systems, this is spelled out more clearly by the principle of "diffeomorphism invariance" in general relativity, see this article for example.

Yes, I understand this as well, and this is just fine as far as it goes. But the paradox still remains. Let me whittle it down as clearly as I can by stating what I see as the facts.

FACT 1: Stella passes by the Earth at .866c and returns 4 years later Earth time having actually aged only 2 years. An asymmetrical situation.

FACT 2: From the moment Stella passes by Earth up until the moment she turns around and according to SR, there is complete symmetry between the Earth's frame and the ships frame. In other words, if Stella is aging more slowly while traveling out, then her brother is simultaneously aging more slowly as well. This is also the case for the return trip after turnaround.

FACT 3: Since Stella is indeed 2 years younger when she returns, and since this is NOT an optical illusion then her brother in reality must be aging more slowly as well, but only during the trip out and the trip back as these are the only two times the situation is symmetrical.

CONCLUSION: Since both Stella and Terrence are aging more slowly during the symmetrical phases of the trip, and since at the end of the trip Stella is the only one that is really younger, at the moment of turnaround, Terrence must age considerably, in the area of about 3 years.

 

[JesseM]

Yes, I understand this as well, and this is just fine as far as it goes. But the paradox still remains. Let me whittle it down as clearly as I can by stating what I see as the facts.

FACT 1: Stella passes by the Earth at .866c and returns 4 years later Earth time having actually aged only 2 years. An asymmetrical situation.

FACT 2: From the moment Stella passes by Earth up until the moment she turns around and according to SR, there is complete symmetry between the Earth's frame and the ships frame. In other words, if Stella is aging more slowly while traveling out, then her brother is simultaneously aging more slowly as well. This is also the case for the return trip after turnaround.

There is no single pair of inertial frames where there is "complete symmetry" in this sense. Suppose frame A is Terence's rest frame, frame B is the frame where Stella was at rest during the trip out (but not during the return trip), and frame C is the frame where Stella was at rest during the return trip (but not the trip out). Then it's true that during the trip out, there is symmetry between frame A and frame B, but both frame A and frame B agree that Stella is aging more slowly during the return trip; likewise, it's true that during the return trip, there is symmetry between frame A and frame C, but both frames agree that Stella was aging more slowly during the trip out. All three frames will get the same answer for the total aging of each twin if you use them to analyze the entire experiment from beginning to end, and none of them say that Terence aged rapidly when Stella was accelerating.

FACT 3: Since Stella is indeed 2 years younger when she returns, and since this is NOT an optical illusion then her brother in reality must be aging more slowly as well, but only during the trip out and the trip back as these are the only two times the situation is symmetrical.

No, that doesn't follow, see above. There is no "reality" about who was aging more slowly during any particular phase of the trip--the outbound leg, the inbound leg, or the acceleration phase--all of these things are frame-dependent. Only the total aging over the entire trip is frame-independent, and if you analyze the entire trip from start to finish using any given choice of frame, you'll get the same answer for the total aging as if you analyze it from start to finish using a different frame. You may find it helpful to think in terms of the [post=2972720]geometric analogy I gave in this post[/post] (in terms of the analogy, total aging is analogous to the total length of a path from beginning to end as measured by a car driving along the path with its odometer running, while 'rate of aging in some frame' is analogous to the rate at which the odometer increases for a given increase in x-coordinate, which is of course relative as you can orient the x-axis of a Cartesian coordinate system on the plane in any direction you like)

 

[D H]

This symmetry is broken of course at the time of the turnaround, but from the start of the experiment until then, the situation is symmetrical.

Also note that the situation is also symmetric near the end of the journey.

Personally I like the Doppler explanation best. The asymmetry is easily explained here. For Stella, the switch from seeing Terence's clock as running slow to seeing Terence's clock as running fast happens exactly at the turnaround point. For Terence, the switch from seeing Stella's clock as running slow to seeing Stella's clock as running fast happens much later.

 

[sylas]

In order to derive an asymmetry, you must first start from a frame of reference, either Stella's or Terence. If Terence (as in the article) you will come up with an asymmetry that is biased toward Stella aging more slowly, if from Stella you will derive an asymmetry that is biased toward Terence aging more slowly. Therefore the situation is symmetrical, not asymmetrical. This symmetry is broken of course at the time of the turnaround, but from the start of the experiment until then, the situation is symmetrical.

Wrong again.

The simple fact of the matter is that only one twin has an inertial frame of reference. The other twin has TWO inertial reference frames... one for outgoing, and one for coming back.

That is why the situation is asymmetrical. This is given you right from the very start, for heavens sake. It is the very problem statement.

Everything else you have said about this being "assumed" in your other post continues to be incorrect. All you do is apply relativity to the problem AS GIVEN, and the problem AS GIVEN states that one twin goes to a star and COMES BACK.

You are getting this wrong, and until you stop trying to teach physics to the people who are explaining it for you, you will continue to get it wrong.

I've been a little bit more blunt this time, but it is still with the aim of helping. I'm happy to keep trying, but you really really to grasp this as a part of getting a real understanding of the matter.

Added in edit... going into specifics of your earlier post #16:

Agreed. The change in frame matters. Experiment bears out that the ship ages more slowly and since there is perfect symmetry at all times (according to SR) except during the turnaround point, it is this point where all the fun happens.

In this, and in much else, you appear to be trying to hold onto a some kind of "absolute" framework where you can identify what things "really" are. The truth is that you can approach the problem in all kinds of ways in all kinds of different co-ordinate systems, including non-inertial ones (in which case you need general relativity). In the specific method I chose for explaining the problem, you can single out turn around as where the fun happens. In ANY method, you can also derive a discontinuity in what is being observed by the traveler at the turn around point; and for the stay at home twin when they observe the turn around point.

I agree that the apparent distance will change at turnaround and this is an optical phenomenon.

Huh?

It's not an "optical" phenomenon in the sense of some illusion or something that is specific to light. At the turn around the reference frame changes to an inertial frame in which the planet REALLY IS further away, in any way you could possibly determine. It's as real as anything else you can test or measure.

Distance is not some kind of absolute thing, for which there is one "real" value and everything else is some kind of illusion. Distance is something that turns out to be relative to a particular frame of reference. And when you change your reference frame, the distance changes.

However I don't agree with your last sentence. The light will have traveled x number of light years just before or just after the turnaround point. You are basing this on your observation of the stars suddenly "appearing" close then "appearing" far but again this is simply an optical illusion. In reality, the light has traveled a greater number of light years in both cases then the ship captain will think it has due to time dialation.

Shrug. I am telling you some basic consequences of relativity applied to a simple problem which confuses a lot of people. But it really isn't a matter that is up to debate. If you want to try and understand better, I can help, and so can many others here. It's going to be that much harder for you to learn the basics of relativity while you think of this as a debate where people might disagree over basic results.

All of the above assumes that the clock in the ship is aging more slowly, which is fine as experiment bears this out, but again this is circular logic.

As noted above, all this does is apply basic relativity to the problem as given. If you want to say that we "assume" relativity, ok. I do. A consequence of this is that neither clock is aging more slowly than the other in any absolute sense.

What is additionally interesting is what is seen from Earth at the time of turnaround. It would seem that as the ship is traveling away it will appear closer than it really is and after it turns around it should appear farther than it really is. But does this really happen? If it doesn't then this is another example of asymmetry. But according to SR it does because of the inherent symmetry both when the ship is moving away and again when it is returning.

No, again. There is no "really is". The change in distance is as real as distance ever gets, because distance depends on the reference frame in which it is defined. Grasping this is essential to understanding relativity. Distance is not some absolute quantity which has a single "real" value.

Cheers -- sylas

 

[John232]

I think the reason why we don't have people that suffer from both being younger than the other at the same time is because the twin paradox requires two observers that have always traveled at a constant speeed in the past. In the universe this is impossible, from the big bang everything would have accelerated away from each other and back to each others frame of reference. Then one object would always have more or less of a age than any other object.

 

[John232]

Two twins always traveled at a constant speed relative to each other forever into the past. The two twins both say they are younger than the other, big deal, it never happened.

 

[JesseM]

I think the reason why we don't have people that suffer from both being younger than the other at the same time is because the twin paradox requires two observers that have always traveled at a constant speeed in the past.

No, you don't need any observers that have always been at rest in an inertial frame to analyze things from the perspective of that inertial frame, inertial frames are just coordinate systems and you're free to use a given coordinate system even if there is no physical "observer" at rest in it. And it's not clear what you mean by "both being younger than the other at the same time", if you're talking about analyzing things from the perspective of two different inertial frames when the twins are far apart then it is possible the frames can disagree about who is younger, but if the twins reunite at a single point in space then there can never be such a disagreement, all frames will make the same prediction about which twin is younger when they reunite.

 

[Mike_Fontenot]

[...]
Specifically, in the case where an observer moving at 60% light speed suddenly reverses to be 60% light speed in the other direction sees the remote planet suddenly four times further away.
[...]

You need to re-think that conclusion. Try applying the standard length-contraction result. Or, even easier, try using the time-dilation result, combined with the fact that during the outbound leg, the outbound inertial frame and the "home" inertial frame will both agree about their relative velocity. Likewise for the inbound inertial frame and the "home" inertial frame.

Mike Fontenot

 

[JesseM]

You need to re-think that conclusion. Try applying the standard length-contraction result. Or, even easier, try using the time-dilation result, combined with the fact that during the outbound leg, the outbound inertial frame and the "home" inertial frame will both agree about their relative velocity. Likewise for the inbound inertial frame and the "home" inertial frame.

Mike Fontenot

I think sylas was talking about optical appearance which take into account light delays, not what is true "instantaneously" in the inertial frames where the accelerating object is at rest immediately before and immediately after acceleration. For example, suppose I am traveling away from the Earth at 0.6c, and then when I am 12 light years from Earth in the Earth's frame, when the Earth's clock reads a time of t=100 years, I instantaneously accelerate so I am traveling towards the Earth at 0.6c. Since in the Earth's frame I was 12 light-years away from the Earth both immediately before and immediately after acceleration, visually I will be seeing the Earth as it was when the Earth's clock read t=88 years. And if I have an inertial ruler at rest relative to myself before the turnaround, such that before the turnaround I am at the x=0 l.y. mark on the ruler, then when the Earth's clock read t=88 years it must have been passing the x=6 l.y. mark on this ruler.* But if there is a second inertial ruler which is moving in such a way that I come at rest relative to it after the turnaround, again at the x=0 l.y. mark on this ruler, then when the Earth's clock read t=88 years it must have been passing the x=24 l.y. mark on this second ruler.** So, immediately before the turnaround I will see the Earth as 6 light-years away on a ruler at rest relative to myself, but immediately after the turnaround I will see the Earth as 24 light-years away on a ruler at rest relative to myself, so the visual distance has seemed to increase by a factor of four.

*This ruler is moving away from the Earth at 0.6c, so if the x=0 l.y. mark on this ruler is 12 l.y. away from the Earth at the moment I turn around (which in the Earth's frame is simultaneous with the Earth's clock reading t=100 years), then the x=0 l.y. mark must have been passing the Earth 12/0.6=20 years earlier in the Earth's frame, when the Earth's clock read t=80 y. In the ruler's frame the Earth's clock is running slow by a factor of 0.8, so a time of (88-80)/0.8 = 10 years must have passed in this frame between the x=0 l.y. mark passing the Earth and the Earth's clock reading t=88 y, so since the Earth is moving at 0.6c in the ruler's frame the Earth must have been at position x=6 l.y. on the ruler when its clock read t=88 years.

**This ruler is moving towards the Earth, so the x=0 l.y. mark on this ruler will finally reach Earth after a time of 12/0.6=20 years in the Earth's frame, when the Earth's clock reads t=120 y; and in the ruler's frame the Earth's clock is running slow by a factor of 0.8, so the Earth's clock must have read t=88 years a time interval of (120 - 88)/0.8 = 40 years before reaching the x=0 l.y. mark, so since the Earth is moving at 0.6c in the ruler's frame it must have been at position x=40*0.6 = 24 light-years at the time it read t=88 years.

 

[sylas]

You need to re-think that conclusion. Try applying the standard length-contraction result. Or, even easier, try using the time-dilation result, combined with the fact that during the outbound leg, the outbound inertial frame and the "home" inertial frame will both agree about their relative velocity. Likewise for the inbound inertial frame and the "home" inertial frame.

Mike Fontenot

I am explaining here the consequence of standard results. See the conventional spacetime diagrams I already provided earlier in this thread, at msg #8.

It's much safer to apply Lorentz transformations, rather than attempt to do length contraction, as length contraction is not all that is involved here. If you try to use length contraction without full consideration of the problem you are very likely to get the wrong answer.

Consider again the case of travel at 60% light speed to a star 6 light years from Earth. For the twin remaining on Earth, they conclude that 10 years after launch, the ship will be arriving at the distant star. Hence, to send a radio message of congratulations, they need to send it 4 years after launch, so that after another 6 years the radio message will arrive at the star simultaneously with the traveler.

Now, using a simple case of Lorentz transformations with time and with one space dimension. You can do this with launch as the origin (0,0), but it works slightly easier using the arrival at the star of the message and ship as the origin. That is because we are speaking of the distances from this event in two different frames.

The "star frame" is the frame of the star six light years from Earth, which takes times 0 when the traveler and radio message arrives. Earth is at location -6 in this frame. The "outward frame" is the frame of a ship moving from Earth towards to the star at 0.6c (γ = 1.25) . Time 0 is when arriving at the star. The "inward frame" is the frame of a ship moving from the star back to Earth at -0.6c. Time 0 is when leaving the star.

The Lorentz transformations, simplified for units where c=1, are as follows:

$$(t',x') = \gamma ( t - xv, x - tv )$$​

Here are the events in three frames.

$$\begin{array}{l|l|rl|rl} \text{event (t,x)} & \text{Star frame} & \multicolumn{2}{l}{\text{Outward frame}} & \multicolumn{2}{|l}{\text{Inward frame}} \\ \hline \text{Launch from Earth} & (-10,-6) & 1.25(-10+3.6, -6+6) & = (-8, 0) & 1.25(-10-3.6, -6-6) & = (-17, -15) \\ \text{Send radio} & (-6,-6) & 1.25(-6+3.6, -6+3.6) & = (-3, -3) & 1.25(-6-3.6, -6-3.6) & = (-12, -12) \\ \text{Ship arrives at star} &(0,0) & (0, 0) && (0,0) \\ \text{Ship back at Earth} & (10,-6) & 1.25(10+3.6, -6-6) & = (17, -15) & 1.25(10-3.6, -6+6) & = (8, 0) \end{array}$$​

Note in particular the event “send radio transmission”, transforms from (-3,-3) in the outward frame to (-12,-12) in the inward frame. That’s four times further away, also as shown in the spacetime diagrams.

This really is the correct answer.

Cheers – sylas

 

[JesseM]

I am explaining here the consequence of standard results. See the conventional spacetime diagrams I already provided earlier in this thread, at msg #8.

It's much safer to apply Lorentz transformations, rather than attempt to do length contraction, as length contraction is not all that is involved here. If you try to use length contraction without full consideration of the problem you are very likely to get the wrong answer.

But your calculations depend on what physical question you're trying to answer. You are trying to answer the question of the distance to the event of Earth emitting the light that the traveler is seeing at the moment of the turnaround, in both the inertial rest frame before the turnaround and the inertial rest frame after. Mike was presumably not thinking about the optical question of what is seen at the turnaround, but rather the question of what events are actually simultaneous with the event of the turnaround in both the inertial frame where the traveler was at rest before the turnaround, and the inertial frame where the traveler was at rest afterwards. In this sense, if the traveler's speed relative to Earth is 0.6c before and after, then the distance of the Earth to themselves in their current rest frame will remain unchanged before and after.

 

[sylas]

But your calculations depend on what physical question you're trying to answer. You are trying to answer the question of the distance to the event of Earth emitting the light that the traveler is seeing at the moment of the turnaround, in both the inertial rest frame before the turnaround and the inertial rest frame after.

Of course. I've been pretty clear about that all along, and even the extract Mike quoted spoken of what was being seen. The context of the post he quoted also says:

In this case, it's a bit subtle. What the ship "sees", of course, is light that left Earth a long time ago. How long ago depends only on how far away the Earth was at the time the light was emitted.

...

This is seeing via the light that left the remote planet 3 years ago (in the first frame) or 12 years ago (in the second frame).

...Mike was presumably not thinking about the optical question of what is seen at the turnaround, but rather the question of what events are actually simultaneous with the event of the turnaround in ...

Yes, quite likely; he can confirm himself if he likes what problem he was considering.

It's easy to get mixed up in these discussions. However, I do think I have been careful to describe precisely what I am quantifying. I don't mind explaining it again as required; all Mike said was that I needed to rethink, so I explained again the calculations, and -- as always -- described carefully what I was quantifying.

Cheers -- sylas

 

[michelcolman]

Let's get back to the original question:
A ship flies out at 0.866 c for one year (I assume that's half the square root of 3), meets a ship going in the other direction, high-fives it so the clock of the inbound ship is set to the time of the outbound ship. The inbound ship arrives when its clock shows 2 years, but on Earth 4 years have passed.

All you have to do now, is look at this from all the different points of view:

1. From Earth's point of view, the outbound ship high-fives the inbound one after two years (as measured by Earth's clocks in Earth's reference frame), at a point 1.732 light years away, the return trip takes two years as well, but due to time dilation both the inbound and outbound clocks have been ticking at half speed so they only show two years total.

2. From the point of view of the inbound ship, at the time of the high five, the Earth is 0.866 light years away and approaching at 0.866c, and the outbound ship is approaching at 0.990c (relativistic addition of speeds). Clocks on Earth are ticking at half speed, clocks on the outbound ship are ticking at 1/7 the normal speed. Since the outbound clock is indicating 1 year, it must have left seven years ago. That means the Earth's clock is, right now, indicating three and a half years since it's ticking only at half speed. Now the return trip takes one year, which is of course only half a year on the slower Earth clocks. This means that Earth's clock will indicate 4 years, the high-fived onboard clock will indicate two years, but the entire thing will "really" have taken 8 years according to the inbound ship.

3. From the point of view of the outbound ship (that continues its journey away), it reaches the high five point after one year, the Earth is moving away at a speed of 0.866c and is 0.866 light years away, and Earth's clocks are indicating half a year after departure (note that everybody disagrees on this!). The inbound ship is flying at 0.990c, so it's trying to catch up with the Earth at a relative speed of 0.1237c. At this speed, it takes seven years to close the 0.866 light year gap. That means three and a half years on Earth's clocks (so they will indicate 4 years total), and only one year on the inbound clock (so it will indicate two years total).

Basically, the whole paradox is solved by the lack of objective simultaneity: At the moment of the high-five, the clocks on Earth were either indicating half a year, two years or three and a half years depending on who you ask.

In the original paradox, with just one ship turning back, the ship suddenly jumps from the outbound into the inbound reference frame, and "current" time on Earth therefore jumps three years ahead from half a year to three and a half years after departure (as measured by Earth's clocks). This can be seen by interpreting the incoming images from earth, too: for the outbound ship, since light is traveling at light speed relative to the ship, the images must have been sent when the Earth was a lot closer than where it is now. But for the inbound ship, the images must have been sent when the Earth was a lot further away, so more time must have passed between then and "now", and it must therefore be a lot later "now".

I love this kind of thing, no matter how you look at it, you get completely different representations that somehow end up in being totally consistent after all! No matter how hard you try :-)

 

[Mike_Fontenot]

I didn't do a good job in picking the quote of Sylas' that I gave ... sorry. The one I was actually reacting to was this one:

Sylas wrote:

"At the turn around the reference frame changes to an inertial frame in which the planet REALLY IS further away, in any way you could possibly determine."

That quote comes from this Sylas post:

https://www.physicsforums.com/showpost.php?p=2985554&postcount=20

If (immediately before or immediately after your turnaround), you look at your home twin through a telescope, you will of course be looking at an image from the past. If you try to deduce the current distance to your home twin from the solid angle she subtends in the image, you must take into account the fact that you're looking at an old image. If you do that correctly, you will get the result given (much more easily) by the Lorentz equations (or, even easier, by the length-contraction result, or by the time-dilation result (combined with the "velocity reciprocity" result), which of course are both derived from the Lorentz equations). And in that result you get, the distance to your home twin will be the same before and after your turnaround (in the case where the magnitude of the outbound and inbound velocities are the same).

Mike Fontenot

 

[sylas]

I didn't do a good job in picking the quote of Sylas' that I gave ... sorry. The one I was actually reacting to was this one:

Sylas wrote:

"At the turn around the reference frame changes to an inertial frame in which the planet REALLY IS further away, in any way you could possibly determine."

That quote comes from this Sylas post:

https://www.physicsforums.com/showpost.php?p=2985554&postcount=20

If (immediately before or immediately after your turnaround), you look at your home twin through a telescope, you will of course be looking at an image from the past. If you try to deduce the current distance to your home twin from the solid angle she subtends in the image, you must take into account the fact that you're looking at an old image. If you do that correctly, you will get the result given (much more easily) by the Lorentz equations (or, even easier, by the length-contraction result, or by the time-dilation result (combined with the "velocity reciprocity" result), which of course are both derived from the Lorentz equations). And in that result you get, the distance to your home twin will be the same before and after your turnaround (in the case where the magnitude of the outbound and inbound velocities are the same).

Mike Fontenot

Quite correct, Mike. I had intended this phrase to refer to the planet at the time of emission of radio signal, meaning that the radio signal really is coming from further way. However, the phrasing in that post, taken by itself, does not identify this clearly enough.

An interesting twist is that speaking of the Earth "simultaneous" with the turn around is no longer identifying a unique event. For the outbound frame, the arrival is 8 years after the launch, and so Earth clocks (due to time dilation) "now" indicate 6.4 years after the launch. However, in the inbound frame, the Earth is "now" 17 years after the launch event, and the clocks there "now" show 13.6 years. In either case, the signals from Earth now arriving at the star were emitted back when Earthbound clocks indicated 4 years after launch.

BTW... nice post also by michelcolman, with a return to the original question of the thread.

Cheers -- sylas

 

[Dale]

It's much safer to apply Lorentz transformations, rather than attempt to do length contraction, as length contraction is not all that is involved here.

I agree. I have made it a personal policy to never use length contraction or time dilation formulas. The Lorentz transform automatically simplifies when appropriate, so it doesn't really save any effort, and using it inappropriately causes lots of problems.

 

[John232]

No, you don't need any observers that have always been at rest in an inertial frame to analyze things from the perspective of that inertial frame, inertial frames are just coordinate systems and you're free to use a given coordinate system even if there is no physical "observer" at rest in it. And it's not clear what you mean by "both being younger than the other at the same time", if you're talking about analyzing things from the perspective of two different inertial frames when the twins are far apart then it is possible the frames can disagree about who is younger, but if the twins reunite at a single point in space then there can never be such a disagreement, all frames will make the same prediction about which twin is younger when they reunite.

The big problem everyone sees with relativity is the twin paradox and how each twin sees himself younger than the other. Why does it matter if they are close together at rest or traveling past each other at a high speed, how are they both younger than each other?

If they are both traveling with each other at the same constant speed then accelerate away from each other at the same rate and then travel the same speed then they will reunite at the same age. Then if one accelerates more than the other one will age more than the other.

But, if they have always traveled at a constant speed relative to each other then they would both be younger than the other and that seems impossible weither they are stopped next to each other or not. I think it is impossible because the situation itself can never happen.

I don't think satellites that use the theory to measure time with higher accuracy suffer from the problem of the satellites also measureing Earths time to be slower at the same time we measure their clocks to be slower.

They did the acceleration away from Earth to travel to orbit at a high speed and we measure their time to be slower. How could we set them correctly if they measured our clocks to be slower too? They would still be set with a slow clock saying their clock was too slow at the same time.

Acceleration is a preffered frame of reference, you can prove that you are accelerating because of the forces acting on you as long as you don't confuse it with gravitation. Only constant frame of reference are indistinguishable, but what in the universe actually travels at a constant speed relative to each other? Everything is in orbit around somthing else or exponentially expanding away from everything else in the universe...

 

[sylas]

If they are both traveling with each other at the same constant speed then accelerate away from each other at the same rate and then travel the same speed then they will reunite at the same age. Then if one accelerates more than the other one will age more than the other.

The only relevance of acceleration is that it changes velocity. Time dilation is determined by velocity; not acceleration.

To prove this, consider a case where two twins travel from Earth to a distant star. Each one accelerates from 0 to 0.2c, and then later on from 0.2c to 0.4c, and then later on from 0.4c back to 0, all velocities in the frame where the origin and destination of the journey are at rest. The remainder of the journey is coasting at constant velocity. Further assume that in each case the accelerations are identical, with a proper acceleration of 10 m/s/s. (Proper acceleration means the acceleration experienced by the ship itself in its own frame. That is, the acceleration is such that the passengers experience a constant virtual force of just over 1g during these accelerations.)

Despite the fact that the accelerations are of equal magnitudes, the time dilations will be different, depending simply on how long the ship coasts at 0.2c, and how long it costs at 0.4c. To calculate the time dilation, you integrate proper time along the ships path in an inertial frame, using the dilation factor determined by velocity at each point.

But, if they have always traveled at a constant speed relative to each other then they would both be younger than the other and that seems impossible weither they are stopped next to each other or not. I think it is impossible because the situation itself can never happen.

Each twin will be older than the other twin considered in their own frame of reference. What is impossible is to have them both at constant velocity (not speed) and end up at the same point to synchronize clocks at a common location. Without synchronization, there's no paradox.

I don't think satellites that use the theory to measure time with higher accuracy suffer from the problem of the satellites also measureing Earths time to be slower at the same time we measure their clocks to be slower.

This is because the situation is different, involving circular motions and gravitational effects as well. It is a non sequitur, of no relevance to the actual reasons that the traveling twins are not a paradox.

Cheers -- sylas

 

[John232]

Then what of the equivalence between gravity and acceleration? I thought since the gravitational effect caused spacetime dilation then acceleration would also cause spacetime dialation. An object at rest in a gravitational field will have its clock slow down.

You will have to excuse me, because I consider myself a professional laymen, since I have read about 60 books on the subject but haven't really gotten my hands dirty with the mathmatics, even so I find it hard to see how I could have got this confused.

But, say you had a light clock set up around a planet. The path of the light would curve and then the clock would read a slower time. Then say you set up a light clock in an accelerating ship, the light in the clock would curve giving an increasingly longer measurment of time. In SR the light clock only has straight lines, I would think adding the curvature would cause the clock to run increasingly slower due to the curvature itself.