How are the twins distinguished?

[Passionflower]

do you increase it as long as there is acceleration in any direction?

No.

Basically when you accelerate with respect to the home clock the inertial clock rate would have to increase to compensate for the natural time dilation and when you decelerate with respect to the home clock the rate would decrease. As soon as you introduce angles the situation becomes increasingly more complicated (first 2D and finally 3D), just as complicated as general 3D velocity additions but then 'backwards'.

 

[JesseM]

It is the same thing but perhaps I was not very clear in my language. Clock rates can only be compared at the same event, the inertial clock will read exactly the time of the home clock as soon as they meet. In flight however the clock will give the time of the home clock as if it moved on a geodesic to the time reading event.

But that's different from the case I was describing where the moving clock adjusts so that it is always synchronized with the home (inertial) clock in the home clock's inertial rest frame, agreed?

See for an overview Minguzzi: http://arxiv.org/abs/physics/0411233

OK, I'll check it out.

 

[starthaus]

This one http://cdsweb.cern.ch/record/999103/files/0611076.pdf?version=1 is interesting too.

Yes, thank you.

 

[Passionflower]

But that's different from the case I was describing where the moving clock adjusts so that it is always synchronized with the home (inertial) clock in the home clock's inertial rest frame, agreed?

Yes, because then you need planes, or heaven forbid, curved surfaces of simultaneity with all the oddities that are discussed in his topic. You call it 'always synchronized' but as you likely know the results are next to bizarre and I much prefer the above described method.

 

[Austin0]

No.

Basically when you accelerate with respect to the home clock the inertial clock rate would have to increase to compensate for the natural time dilation and when you decelerate with respect to the home clock the rate would decrease. As soon as you introduce angles the situation becomes increasingly more complicated (first 2D and finally 3D), just as complicated as general 3D velocity additions but then 'backwards'.

Well as per what you just described then the "home clock" rate would be decreasing from the time of the initiiation of reverse acceleration and would be decreased though the whole homeward inertial phase as the accelerometer would not distinguish between deccelration and what would be increasing relative velocity after the at rest point. Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

 

[Passionflower]

Well as per what you just described then the "home clock" rate would be decreasing from the time of the initiiation of reverse acceleration and would be decreased though the whole homeward inertial phase as the accelerometer would not distinguish between deccelration and what would be increasing relative velocity after the at rest point. Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

Sorry but I really cannot follow what you are saying.

 

[JesseM]

Yes, because then you need planes, or heaven forbid, curved surfaces of simultaneity with all the oddities that are discussed in his topic. You call it 'always synchronized' but as you likely know the results are next to bizarre and I much prefer the above described method.

What's bizarre about the results? This is how GPS clocks work after all (they adjust to maintain a constant rate of ticking in an Earth-centered coordinate system, even as their velocities/heights are changing in this coordinate system), and it would be even simpler if you just wanted some clocks to adjust to maintain a constant rate of ticking in an inertial frame in flat spacetime.

One oddity about your method which occurred to me is that if one twin travels away from the home twin, accelerates to turn around, and and then travels back to the home twin inertially, then during the inertial return trip the adjusted clock will actually keep ticking faster and faster relative to a "normal" clock carried by the traveling twin (one which measures proper time), the ratio between the tick rates of the two clocks won't remain constant despite the fact that no acceleration is happening in this leg. For example, if the traveling twin goes out at 0.8c for 10 years in the home twin's frame, accelerates briefly and then returns at 0.8c for another 10 years, then at the time of acceleration at t=10 in the home frame the adjusted clock should read about 6 years, then at the time the twins reunite at t=20 the adjusted clock should read 20 years, but halfway through the return leg at t=15, the traveling twin will be at a distance of 4 light-years from the home twin so if the home twin took a geodesic path to that point the home twin's velocity would have been (4/15)c, so the time on that geodesic path would be 15*sqrt(1 - (4/15)^2) = 14.46 years, more than halfway between 6 and 20.

 

[Passionflower]

What's bizarre about the results? This is how GPS clocks work after all (they adjust to maintain a constant rate of ticking in an Earth-centered coordinate system, even as their velocities/heights are changing in this coordinate system), and it would be even simpler if you just wanted some clocks to adjust to maintain a constant rate of ticking in an inertial frame in flat spacetime.

Are you at all following what is discussed in this topic?

 

[stevmg]

All right - back to basics.

Twin A is the guy who stays at home

Twin B is the guy who takes a trip and comes back

If the twins are girls or one is a boy and the other is a girl or vice versa this same still holds. Forget about the Turner syndrome female and normal male "identical" twins I once referred to.

If twin B moves at 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

Am I right?

 

[Mentz114]

All right - back to basics.

Twin A is the guy who stays at home

Twin B is the guy who takes a trip and comes back

If the twins are girls or one is a boy and the other is a girl or vice versa this same still holds. Forget about the Turner syndrome female and normal male "identical" twins I once referred to.

If twin B moves at 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

Am I right?

Yes. I get 14 s and 11.2 s. Or, 1 - (0.6)² = 0.64, sqrt(0.64) = 0.80.

 

[stevmg]

That's what I get, too.
That's with just inertial FRs and no acceleration calculations
Anyone else out there have any questions?

 

[Passionflower]

For example, if the traveling twin goes out at 0.8c for 10 years in the home twin's frame, accelerates briefly and then returns at 0.8c for another 10 years, then at the time of acceleration at t=10 in the home frame the adjusted clock should read about 6 years, then at the time the twins reunite at t=20 the adjusted clock should read 20 years, but halfway through the return leg at t=15, the traveling twin will be at a distance of 4 light-years from the home twin so if the home twin took a geodesic path to that point the home twin's velocity would have been (4/15)c, so the time on that geodesic path would be 15*sqrt(1 - (4/15)^2) = 14.46 years, more than halfway between 6 and 20.

Indeed, the difference in age gets larger and larger in time.

Well as per what you just described then the "home clock" rate would be decreasing from the time of the initiiation of reverse acceleration and would be decreased though the whole homeward inertial phase as the accelerometer would not distinguish between deccelration and what would be increasing relative velocity after the at rest point. Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

[strike]Sorry but I really cannot follow what you are saying.[/strike]
Sorry it was rather late when I wrote it (China time)

Remember that the clock rate is adjusted only when the spaceship accelerates. If it accelerates away from the home clock the rate goes up, if it accelerates towards the home clock the rate goes down. Why do you think we cannot determine if we accelerate or decelerate with respect to the home clock?

That's with just inertial FRs and no acceleration calculations

You are incorrect, there is only differential aging if there is acceleration and thus at least one non-inertial frame of reference.

You wrote:

If twin B moves at 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

Instead you should write:

If twin B instantly accelerates to 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

The way you formulated the problem and solution is actually the source of confusion of the opening posting.

 

[Mentz114]

That's what I get, too.
That's with just inertial FRs and no acceleration calculations
Anyone else out there have any questions?

You have to remember that this scenario is unphysical. But if you imagine the drawing represents an infinitesimal part of a big curve, then adding up all the little triangles of the curve gives the right answer.

 

[stevmg]

Indeed, the difference in age gets larger and larger in time.


[strike]Sorry but I really cannot follow what you are saying.[/strike]
Sorry it was rather late when I wrote it (China time)

Remember that the clock rate is adjusted only when the spaceship accelerates. If it accelerates away from the home clock the rate goes up, if it accelerates towards the home clock the rate goes down. Why do you think we cannot determine if we accelerate or decelerate with respect to the home clock?


You are incorrect, there is only differential aging if there is acceleration and thus at least one non-inertial frame of reference.

You wrote:


Instead you should write:

If twin B instantly accelerates to 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

The way you formulated the problem and solution is actually the source of confusion of the opening posting.

You have to remember that this scenario is unphysical. But if you imagine the drawing represents an infinitesimal part of a big curve, then adding up all the little triangles of the curve gives the right answer.

Passionlower - yea that's right - instant acceleration which is NOT survivable. We know that. But JesseM was just illustrating the calculations. It would be true that the proper times for the outward journey of twin B + the inward journey of twin B - even if it were not the same guy but just two different guys who crossed paths in opposite directions at the precise moments would be 20% less than the proper time of twin A sitting fat and pretty at home. This is true in China, too.

The GPS Satellites which are running slower than Earth clocks are not accelerating or decelerating (well, maybe they are at v²/r)

Takes an adjustment chip in them to keep Earth time.

Why did you strike it out? ([strike]Sorry but I really cannot follow what you are saying.[/strike]) Why didn't you just erase it by editing it. I do that all the time.

 

[Passionflower]

It seems you are completely missing the point I was making, it has nothing to do with "instant acceleration is NOT survivable".

 

[stevmg]

Yes, I am missing the point and I ain't in China.

Try again...

 

[Passionflower]

This is true in China, too.

Yes, I am missing the point and I ain't in China.

Excuse me?

Are you a serious person or are you perhaps trolling?

The GPS Satellites which are running slower than Earth clocks are not accelerating or decelerating (well, maybe they are at v²/r)

First of all a GPS satellite clock is going faster than a clock on Earth, second a clock on Earth undergoes proper acceleration while a clock in a GPS satellite undergoes inertial acceleration with respect to a clock on Earth.

 

[Austin0]

Well as per what you just described then the "home clock" rate would be decreasing from the time of the initiiation of reverse acceleration and would be decreased though the whole homeward inertial phase as the accelerometer would not distinguish between deccelration and what would be increasing relative velocity after the at rest point. Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

Remember that the clock rate is adjusted only when the spaceship accelerates. If it accelerates away from the home clock the rate goes up, if it accelerates towards the home clock the rate goes down. Why do you think we cannot determine if we accelerate or decelerate with respect to the home clock?

.

The rocket accelerates away from Earth and the accelerometer increases the rate of the onboard home clock.

At some point the thrust is reversed and the rocket begins to deccelerate wrt Earth and the accelerometer begins to decrease the rate of the home clock. At this point the accelerometer has a constant reading given constant proper acceleration , yes? The accelerometer maintains this constant reading from the point of inertial v [begin decceleration] until reaching inertial (-v) [end acceleration.] The acelerometer does not change its reading at the point of rest wrt Earth where decceleration becomes acceleration wrt Earth and the home clock should begin increasing its rate again. DO you understand?
Without some outside refenence there is no way to determine simply from accelerometer readings when the system is at rest wrt Earth and decceleration becomes acceleration .
Possibly Doppler measurements if there is a constant transmission from earth?
It is an interesting idea but it seems like there could be intrinsic complications.
Eg. It sounds easy to adjust the rate by a calculated amount but the mechanism for implementing this calculation is itself subject to acceleration induced dilation so this implementation itself would require complex differential calculations based on what would be pure assumptions.
Even if you could effect this system so that the final readings agreed with the overlall elapsed time observed upon recolocation this does not neccessarily mean that the intermediate readings during the trip would have any real meaning.
I.e. The initial acceleration could in fact be a decceleration so if the home clock increased in rate wrt the ship proper time htis would not reflect reality as the ship clock could in fact be ticking faster than the Earth clock.
In this case, after turn around the relative acceleration and motion wrt Earth would result in the ship clocks dilating wrt Earth clocks so for this phase the home clock on board should be increased in rate.
Make sense?

 

[Passionflower]

The rocket accelerates away from Earth and the accelerometer increases the rate of the onboard home clock.

At some point the thrust is reversed and the rocket begins to deccelerate wrt Earth and the accelerometer begins to decrease the rate of the home clock. At this point the accelerometer has a constant reading given constant proper acceleration , yes?

Yes but either the rocket turns around, and the gyroscopes would measure it or the acceleration will point in a different direction.

The accelerometer maintains this constant reading from the point of inertial v [begin decceleration] until reaching inertial (-v) [end acceleration.] The acelerometer does not change its reading at the point of rest wrt Earth where decceleration becomes acceleration wrt Earth and the home clock should begin increasing its rate again. DO you understand?

No, I am sorry I do not understand what you are trying to say.

Without some outside reference there is no way to determine simply from accelerometer readings when the system is at rest wrt Earth and dcceleration becomes acceleration .

I do not see why not, all we have to do is record the intensity, direction and duration of the acceleration, that together with 3 gyroscopes for rotations. You realize that deceleration is just acceleration in the opposite direction right?

Even if you could effect this system so that the final readings agreed with the overlall elapsed time observed upon recolocation this does not neccessarily mean that the intermediate readings during the trip would have any real meaning.

They do, as the readings are based on a closed loop!

E.g. the home clock and the inertial spaceship clock would show the same time if we connect a timelike geodesic between the starting event and the current inertial clock reading event in the spaceship and have the home clock travel on it. Think about a cord spanned from the origin to the current location of the traveling twin, this cord is the timelike geodesic between the two events.

What this also makes clear is that one must have acceleration to show differential time aging. Because without acceleration the geodesic would be identical for both travelers. This gives an interesting perspective on the issue of infinite acceleration as well.

I.e. The initial acceleration could in fact be a decceleration so if the home clock increased in rate wrt the ship proper time htis would not reflect reality as the ship clock could in fact be ticking faster than the Earth clock.

Huh? I think you would want to read that again and perhaps conclude you made an error.

 

[Austin0]

The rocket accelerates away from Earth and the accelerometer increases the rate of the onboard home clock..

At some point the thrust is reversed and the rocket begins to deccelerate wrt Earth and the accelerometer begins to decrease the rate of the home clock. At this point the accelerometer has a constant reading given constant proper acceleration , yes?.[/QUOTE]

Yes but either the rocket turns around, and the gyroscopes would measure it or the acceleration will point in a different direction..

Yes this was assumed.

The accelerometer maintains this constant reading from the point of inertial v [begin decceleration] until reaching inertial (-v) [end acceleration.] The accelerometer does not change its reading at the point of rest wrt Earth where decceleration becomes acceleration wrt Earth and the home clock should begin increasing its rate again. DO you understand? .

No, I am sorry I do not understand what you are trying to say..

Without some outside reference there is no way to determine simply from accelerometer readings when the system is at rest wrt Earth and dcceleration becomes acceleration ..

I do not see why not, all we have to do is record the intensity, direction and duration of the acceleration, that together with 3 gyroscopes for rotations. You realize that deceleration is just acceleration in the opposite direction right?.

This is not correct in this context. That is the point that you are not understanding.
In this case the decceleration [i.e. reduction in velocity relative to earth] and the acceleration [increase in velocity relative to earth] are in the same direction . No rotation or adjustment, no change in the accelerometer reading. The thrust is constant in both magnitude and direction and the only difference is passing through the momentary point of 0 velocity wrt earth. At this point decceleration becomes acceleration relative to earth.
But this point has no observable effect within the ship. DO you now understand ?? WOuld you disagree??

Even if you could effect this system so that the final readings agreed with the overlall elapsed time observed upon recolocation this does not neccessarily mean that the intermediate readings during the trip would have any real meaning. .

They do, as the readings are based on a closed loop! .

E.g. the home clock and the inertial spaceship clock would show the same time if we connect a timelike geodesic between the starting event and the current inertial clock reading event in the spaceship and have the home clock travel on it.

What this also makes clear is that one must have acceleration to show differential time aging. Because without acceleration the geodesic would be identical for both travelers. This gives an interesting perspective on the issue of infinite acceleration as well..[/QUOTE]
I certainly have never questioned that acceleration was a necessary condition for observed non-reciprocal elapsed time differential. I don't think anyone else has either.
Your abstract invocation of the clock traveling on a geodesic does carry any information as to how this could be calculated or mechanically implemented with an accelerometer based clock system.

I.e. The initial acceleration could in fact be a decceleration so if the home clock increased in rate wrt the ship proper time htis would not reflect reality as the ship clock could in fact be ticking faster than the Earth clock. .

Huh? I think you would want to read that again and perhaps conclude you made an error..

Not really.
1) The basic concept of differential aging [elapsed time] must also assume an actual change in clock periodicity as a consequence of acceleration.
This also requires an assumption of absolute acceleration.
Given that the initial starting velocity is purely relative with no actual value this implies that that the resulting change in clock rate could be an increase or a decrease.
For example The initial point A, could be traveling at 0.5 c --> +x relative to some other frame Z. If the traveler T accelerates away from A --> (-x) wrt Z then this would be decceleration wrt A so Z would observe T's clocks increasing in rate relative to A up to the point of turn around. Then Z would see T's acceleration in the +x as acceleration wrt to A also, so the rate of T's clock would slow down relative to A's. In this instance the return trip to A would be much longer than the trip out as observed in Z,, so the return with T's clocks running slower than A's would inevitably have a greater final cumulative effect i.e. less elapsed total time.
That this would always be the end result no matter what the state of motion of the initial starting point as the return must in all cases mean more overall spacetime travelled.
DO you disagree with any of this?

 

[Passionflower]

Given that the initial starting velocity is purely relative with no actual value this implies that that the resulting change in clock rate could be an increase or a decrease.

That is simply incorrect, an accelerating loop with respect to an inertial observer will compared to the inertial observer result in a slower clock reading for the accelerating observer. The relative speed of the inertial frame with respect to another frame is completely irrelevant.

Looking at some of the writing I think you have some conceptual problems with acceleration.

 

[Austin0]

That is simply incorrect, an accelerating loop with respect to an inertial observer will compared to the inertial observer result in a slower clock reading for the accelerating observer. The relative speed of the inertial frame with respect to another frame is completely irrelevant.

Looking at some of the writing I think you have some conceptual problems with acceleration.

Notice the loop. I never suggested that, considering the loop.[ i.e. the round trip], that this would not be the case.
More specifically I stated that the final result in ALL cases would be less elapsed proper time for the accelerated observer.
So here is anothe case where someone is responding to a statement I never made and telling me I am incorrect.
You totally ignored all other things I said and simply stated your thought that I have conceptual problems. No actual responce to any specific points, just an amorphous negative insinuation of my lack of understanding.
If I might be wrong, it would not be the first time and I would be glad to improve my understanding, but this kind of non=specific negative responce is not constructive

 

[JesseM]

Yes, because then you need planes, or heaven forbid, curved surfaces of simultaneity with all the oddities that are discussed in his topic. You call it 'always synchronized' but as you likely know the results are next to bizarre and I much prefer the above described method.

What's bizarre about the results? This is how GPS clocks work after all (they adjust to maintain a constant rate of ticking in an Earth-centered coordinate system, even as their velocities/heights are changing in this coordinate system), and it would be even simpler if you just wanted some clocks to adjust to maintain a constant rate of ticking in an inertial frame in flat spacetime.

Are you at all following what is discussed in this topic?

Yes I am. I honestly don't see anything "bizarre" about a clock programmed to tick at a constant rate in some inertial frame regardless of its motion relative to that frame, nor do I see how the previous discussion in this thread pointed to anything "bizarre" about it, this comment seems pretty much unrelated to what has been discussed so far.

 

[Passionflower]

Yes I am. I honestly don't see anything "bizarre" about a clock programmed to tick at a constant rate in some inertial frame regardless of its motion relative to that frame, nor do I see how the previous discussion in this thread pointed to anything "bizarre" about it, this comment seems pretty much unrelated to what has been discussed so far.

You really do not see any issues with planes of simultaneity in examples with acceleration, the main thread of his topic?

Using the method I referenced is an alternative way of looking at differential aging. If you don't like it then just don't use it and stick with planes of simultaneity. At the end it is all calculations and they should come out all the same.

 

[Mentz114]

Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

As the captain of the accelerating ship ( one that travels in the x-direction, stops and returns ) I can monitor my acceleration on the outgoing leg and calculate my relative velocity v wrt to home. At some point I give the order to reverse engines, and apply acceleration until I calculate that v = 0.

 

[JesseM]

You really do not see any issues with planes of simultaneity in examples with acceleration, the main thread of his topic?

Are you talking about a non-inertial frame whose planes of simultaneity are at different angles at different times? ('different angles' relative to planes of simultaneity in some inertial frame, that is) If so, then first of all that has nothing to do with the issue of a clock programmed to adjust so it ticks at a constant rate in an inertial frame, it would only be relevant if you wanted the inertial twin's clock to adjust so it ticks at a constant rate in some non-inertial frame for the traveling twin (in that case the inertial twin's clock would might have to jump forward very rapidly during the section of his worldline that happened simultaneously with the acceleration phase of the other twin in the non-inertial frame). Second of all, I don't think that was actually the main topic of this thread, most of the posts have not been talking about non-inertial frames.

Using the method I referenced is an alternative way of looking at differential aging. If you don't like it then just don't use it and stick with planes of simultaneity.

But your method also requires the onboard computer to be constantly figuring out the clock's position in some frame (which necessarily has requires judgments about simultaneity in that frame) and use that to adjust the rate of ticking, it can't just directly adjust the rate of ticking based on the measurements of the accelerometer, since as I pointed out your method requires that the clock's rate is continually increasing relative to proper time during the entirely inertial inbound leg.

 

[stevmg]

Passionflower -

Apologies for being flippant earlier. Won't happen again.

A)
To any out there, particularly JesseM. I thought that it was shown that disregarding the acceleration/deceleration phases of this round trip twin B makes, that twin B undergoes less aging than twin A while he/she is traveling at the "constant" speeds. Others have answered they agree with this. Is this so?

B)
To Passionflower and anyone else who are of the same thinking. I see six simplistic phases to B's trip (everything in reference to A's inertial frame S:)
1) acceleration from the Earth (let's say it is constant acceleration d²s/dt² = a)
2) trip speed at a constant rate: ds/dt is constant (call it w)
3) deceleration at constant rate from trip velocty to 0 (-d²s/dt² = -a)
4) continuing the negative acceleration in 3) above until a negative constant trip velocity is achieved -ds/dt = -w
5) deceleration or positive acceleration) from the trip velocity d²s/dt² = a, until the actual final speed of B wrt A is zero.

This above is a very simplistic representation of what we are talking about. I want to make sure we are clear here and on the same page (or same sheet of music) so there is no confusion. Is my representation valid for purposes of discussion?

Below I will jump ahead (post #90) with my questions assuming the answer is "yes"

 

[granpa]

The twin paradox is much easier to understand if one imagines a long line of stationary, evenly spaced (say one light sec apart), and synchronized clocks extending from the stationary twin to the point where the other twin turns around. (It may help to imagine that they are simultaneously emitting radio pulses at one sec intervals)

If a rocket starting at rest near the stationary twin accelerates along the line of clocks to velocity v then stops accelerating then the clocks from his point of view will no longer be in synch. this is known as 'relativity of simultaneity' and this is what confuses most beginners.

If the rocket accelerates continuously at 1G the clocks become more and more out of sync from that rockets point of view (due to relativity of simultaneity). That means that the clocks are ticking at different rates from the rockets point of view. Some may even be running backwards. This is non-intuitive and is the source of the confusion surrounding the twins paradox. It also leads to general relativity and gravitational time dilation.

Further, imagine that as these clocks pass the moving twins window a strobe flashes so he can read off the time on that passing clock. Even though the non-moving twins clock (and each of the stationary clocks) seems, to the moving twin, to be ticking at half the rate of his own clock, the elapsed time, as told by the strobe, is passing at twice the rate of his own. More importantly, just before he stops, in order to turn around, the line of clocks are, from his perspective, out of synch but the moment he stops the line of clocks will be perfectly synchronized again which means that the nonmoving twins clock now reads the same as the clock he is next to. That means that his calculation of what the nonmoving twins clock said jumps suddenly while he decelerates.

 

[JesseM]

A)
To any out there, particularly JesseM. I thought that it was shown that disregarding the acceleration/deceleration phases of this round trip twin B makes, that twin B undergoes less aging than twin A while he/she is traveling at the "constant" speeds. Others have answered they agree with this. Is this so?

The amount that twin B ages during the two inertial legs of the trip will be less than the amount twin A ages from the moment of departure to the moment they reunite, if that's what you mean.

 

[stevmg]

1) Do you consider that the differential aging will occur if and only if there are the phases 1,3,4 and 6 (the acceleration/deceleration phases?)
2) If one were to imagine a position in a reference frame S of (0,0) for a person A and a rocket already in motion moves past it at v to the right and travels for time t to a point x so he/she is now at (x,t)
3) Now imagine a second rocket which moves to the left at -v already in motion and crosses (x,t) exactly at t (FR is S) and heads on back to the beginning point. Now, by A's clock has advanced to 2t and the Minkowski position in spacetime would be (0,2t)

I propose that the interval for person A is 2t. The interval for the two rockets added up is
2t\sqrt{(1 - v^2/c^2)}
Am I correct?

Help me out, JesseM and others because that is exactly what was shown to me in the past!