How are the twins distinguished?

[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!

 

[JesseM]

Imagine a rocket starting at rest beside a long line of synchronized clocks one light sec apart (they tick simultaneously). 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.

This is true if you use a particular type of non-inertial frame for the rocket, namely one where the plane of simultaneity at any point on the rocket's worldline happens to agree with the plane of simultaneity in the inertial frame where the rocket has an instantaneous velocity of zero at that moment. However there are an infinite number of different ways to define a non-inertial frame where the rocket is at rest, there's no compelling reason why we must define the rocket's "point of view" in this way.

 

[stevmg]

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.

You got ahead of me, JesseM but that was EXACTLY what I meant. Thank you.

It looks like we added the proper times of the outbound and inbound trips for B and compared this total to the proper time of the interval of the non-moving twin (wrt itself.)

Now for a more precise answer, I would need to know how to get proper times for the acceleration/deceleration phases (there are four of them) to get an "honest" answer.

Now I think this keeps it simple and intuitive without worrying about "in-phase" and strobes and the like. I can grasp the concept of distance and time do work on each other and you get one by sacrificing (by Minkowski or Lorentz rules) some of the other.

 

[granpa]

This is true if you use a particular type of non-inertial frame for the rocket, namely one where the plane of simultaneity at any point on the rocket's worldline happens to agree with the plane of simultaneity in the inertial frame where the rocket has an instantaneous velocity of zero at that moment. However there are an infinite number of different ways to define a non-inertial frame where the rocket is at rest, there's no compelling reason why we must define the rocket's "point of view" in this way.

not at all. just imagine that the rocket accelerates is short millisec bursts and glides for a millisec between these bursts. the 'rockets point of view' is simply its view when it is gliding between the bursts of acceleration. Then take the limit as the bursts and intervals become smaller and smaller.

there is no need to introduce 'noninertial' reference frames. special relativity is all you need here.

its very very simple and you are just confusing the whole issue needlessly

 

[JesseM]

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?)

There were actually only 5 phases you listed, probably you meant 5) to be a constant-velocity return phase and 6) to be the deceleration phase. You don't actually need phases 1 and 6, since you could imagine the ship just travels inertially past the inertial twin and they compare clocks as they pass, and likewise passes the inertial twin again moving inertially on the return trip, instead of starting and ending at rest relative to the inertial twin. 3 and 4 are needed, but in a theoretical analysis you are free to consider the limit as the time spent accelerating in phases 3 and 4 approaches zero. Either way, it is always true that differential aging happens if one twin moves inertially while the other's path involve some change in velocity, and it's always true that the twin that moved inertially between their two meetings has aged more than the one that didn't. This is mathematically very closely analogous to the fact that in ordinary 2D plane geometry, if you pick two points and draw a straight line between them, then draw some other non-straight-line path between the same two points, the straight line path always has a shorter distance.

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?

Yes, that's right.

 

[starthaus]

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!

The answer that you want, incorporating realistic motion and not cutting any corners can vbe found here.

 

[stevmg]

There were actually only 5 phases you listed, probably you meant 5) to be a constant-velocity return phase and 6) to be the deceleration phase. You don't actually need phases 1 and 6, since you could imagine the ship just travels inertially past the inertial twin and they compare clocks as they pass, and likewise passes the inertial twin again moving inertially on the return trip, instead of starting and ending at rest relative to the inertial twin. 3 and 4 are needed, but in a theoretical analysis you are free to consider the limit as the time spent accelerating in phases 3 and 4 approaches zero. Either way, it is always true that differential aging happens if one twin moves inertially while the other's path involve some change in velocity, and it's always true that the twin that moved inertially between their two meetings has aged more than the one that didn't. This is mathematically very closely analogous to the fact that in ordinary 2D plane geometry, if you pick two points and draw a straight line between them, then draw some other non-straight-line path between the same two points, the straight line path always has a shorter distance.

Yea... I've been drinking again, or maybe not enough. Actually, phase 5 was the constant speed on the return trip phase of -v and phase 6 was the deceleration (or positive acceleration phase.) But it seems that even without considering acceleration/deceleration you STILL get an age difference (mover < stationary.) But I get where you are going. With a change in velocity (which always means acceleration/deceleration) comes a decrease in aging. Of course the Minkowski equations of c²t² - x² - y² - z² = c'²t'² - x'² - y'² - z'² are counter- intruitive to Euclidean Geometry but who cares?

As I said, I guess I haven't been drinking enough.

Yes, that's right.

Again, thanks for the support as at least I made myself clear to the others that I am not wrong, although maybe not complete in my understanding.

 

[yuiop]

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 v)
3) deceleration at constant rate from trip velocity to 0 (-d²s/dt² = -a)
4) continuing the negative acceleration in 3) above until a negative constant trip velocity is achieved -ds/dt = -v
5) deceleration or positive acceleration) from the trip velocity d²s/dt² = a, until the actual final speed of B wrt A is zero.

You have missed out one of the 6 phases in the above list. It should be more like this:

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 v)
3) deceleration at constant rate from trip velocity to 0 (-d²s/dt² = -a)
4) continuing the negative acceleration in 3) above until a negative constant trip velocity is achieved -ds/dt = -v
5) trip speed at a constant rate: ds/dt is constant (-v)
6) deceleration or positive acceleration) from the trip velocity d²s/dt² = a, until the actual final speed of B wrt A is zero.

This makes it consistent with the scenario described in this Wikipedia page:

http://en.wikipedia.org/wiki/Twin_p...lt_of_differences_in_twins.27_spacetime_paths

The Wiki page gives a formula for the 2 constant velocity phases and the 4 accelerating phases.

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a}\, asinh(aT_a/c)$$

where $T_a$ is the coordinate time of a single accelerating phase and $T_a$ is the coordinate time of a single cruising phase.

The whole thing can be expressed in terms of velocity as:

$$d \tau=2T_c/\gamma+4\frac{cT_a}{v \gamma}\, asinh(v\gamma/c)$$

where v is the cruising velocity.

Both the equations above assume all the acceleration phases last the same time and the cruising phases last the same time.

 

[JesseM]

not at all. just imagine that the rocket accelerates is short millisec bursts and glides for a millisec between these bursts. the 'rockets point of view' is simply its view when it is gliding between the bursts of acceleration.

But "point of view" doesn't reflect what the rocket actually sees, it reflects what is measured in some inertial frame. It is merely a matter of linguistic convention that physicists sometimes use the shorthand "point of view" to refer to the inertial rest frame of an inertial observer, but this shorthand becomes too ambiguous when talking about non-inertial observers so I don't think physicists would normally talk that way. What's more, you just have a series of different frames there, if you want to talk about how the rates of different clocks are changing over a period of time covering multiple "bursts", then you are combining the different inertial frames into a single non-inertial frame.

Think of it this way, inertial coordinate systems in special relativity are understood physically in terms of local readings on inertial ruler/clock systems of the type I illustrated on this thread. So if the rocket's velocity after each "burst" are represented by v₁, v₂, v₃, etc., suppose we have a bunch of lines of rulers and clocks which are eternally moving inertially at different speeds, the first always moving at v₁, the second always moving at v₂, the third always moving at v₃, etc. Assume also that the synchronized clocks on each ruler/clock system have been set so that whenever the rocket is at rest relative to one of these ruler/clock systems, the clock next to him on that ruler/clock system reads the same time as his own clock. Then it's true that if the rocket moves at speed v₁ between times T_1A and T_1B, he can say "for any event which happens next to a clock on the ruler/clock system moving at v₁ when that clock reads a time T which is between T_1A and T_1B, I choose to say that this event also happened simultaneously with my own clock reading T". Likewise, if he is moving at speed v2 between times T_2A and T_2B, he can say "for any event which happens next to a clock on the ruler/clock system moving at v₂ when that clock reads a time T which is between T_2A and T_2A, I choose to say the event happened simultaneously with my own clock reading T". But you can see this is a very abstract procedure, all the ruler/clock systems are there all the time and it's the rocket guy's choice which you use to define simultaneity, and in fact the rocket observer may not see the light from some distant event whose time he assigns using the ruler/clock system moving at v₁ until after he has already accelerated (perhaps more than once) to some velocity other than v₁.

there is no need to introduce 'noninertial' reference frames. special relativity is all you need here.

The modern point of view is that even if one uses a non-inertial frame, as long as spacetime is flat you are still dealing with "special relativity", special relativity doesn't necessarily denote

its very very simple and you are just confusing the whole issue needlessly[/QUOTE]

 

[Mike_Fontenot]

[...]
But you can see this is a very abstract procedure, all the ruler/clock systems are there all the time and it's the rocket guy's choice which you use to define simultaneity,
[...]

He actually DOESN'T have a choice of which particular set, of all those sets of (conceptual) clocks and rulers, to use. There is only ONE such set that agrees with his own potential measurements. If he decides to stop accelerating at that instant, and remain inertial thereafter, he can immediately begin to make his own elementary observations, and elementary, first-principle calculations, to determine the home twin's current age. If he does that, he will find that he agrees with the inertial frame with which he was stationary at the instant he quit accelerating (his "MSIRF" at that instant). And he will agree with that MSIRF from the instant he quits accelerating, and at all times thereafter, as long as he remains unaccelerated.

I describe in detail the required elementary measurements and calculations that the traveler must make, and the required limiting arguments to effect the proof, in my paper:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

 

[JesseM]

He actually DOESN'T have a choice of which particular set, of all those sets of (conceptual) clocks and rulers, to use. There is only ONE such set that agrees with his own potential measurements.

What are "his own potential measurement", if not measurements on some set of rulers and clocks? Why can't he use rulers and clocks in motion relative to himself to make measurements? Also, for an event far away from him, even if the event was simultaneous with an event on his worldline when he was moving at speed v₁, he may not actually get light from the event until his is moving at v₂, so if the event was next to a clock on the ruler/clock system moving at v₁ when that clock read T₁, and also next to a clock on the ruler/clock system moving at v₂ when that clock read T₂, which time would qualify as "his measurement"? It seems to me that this sort of thing is just a matter of arbitrary convention.

 

[Mike_Fontenot]

What are "his own potential measurements", if not measurements on some set of rulers and clocks?

His measurements use only his own wristwatch.

[...] Also, for an event far away from him, even if the event was simultaneous with an event on his worldline when he was moving at speed v₁, he may not actually get light from the event until his is moving at v₂,
[...]

That's why I used the term "potential measurements". The proof involves showing what he WOULD measure IF he chose to permanently remain inertial at the speed v₁ ... there never would BE a different speed v₂ in that case. If the home twin constantly sends messages reporting her age, the traveler will be receiving messages from the instant he stops accelerating (and also before then, of course, but those aren't involved in the calculations). The traveler can take all the time he needs to make his measurements and do his calculations.

Mike Fontenot

 

[JesseM]

His measurements use only his own wristwatch.

Well, how does an accelerating observer define the time of distant events using his own wristwatch? Does look at the distance on some ruler, and then subtract (distance)/c from the time he got a signal from the event? If so, if there are multiple inertial rulers moving at different velocities, which does he choose? The one that's at rest relative to him when he receives the signal, or the one which has the property that when you subtract (distance)/c from the time he received it, it was at rest relative to him at that time on his watch? Again it seems like a matter of arbitrary convention.

That's why I used the term "potential measurements". The proof involves showing what he WOULD measure IF he chose to permanently remain inertial at the speed v₁ ... there never would BE a different speed v₂ in that case.

So if he receives a signal at time T on his watch when he is moving at v₁, does he calculate the time the event would have occurred in his frame if he had been moving at v₁ forever, or something else?

 

[Mike_Fontenot]

Well, how does an accelerating observer define the time of distant events using his own wristwatch? Does he look at the distance on some ruler, [...]
[...]

He doesn't use ANY ruler.

[...]
So if he receives a signal at time T on his watch when he is moving at v₁, does he calculate the time the event would have occurred in his frame if he had been moving at v₁ forever, or something else?

To understand exactly what calculations and observations the (formerly) accelerating observer must carry out, it's first necessary to answer the question "After the accelerating observer stops accelerating, how soon does he become 'an inertial observer', and exactly how soon can he legitimately perform the same calculations that a perpetually inertial observer can perform?". To do that requires a fairly lengthy argument that isn't appropriate or practical in this venue ... but it's all in the paper, for anyone who really wants to know.

Mike Fontenot

 

[stevmg]

You have missed out one of the 6 phases in the above list. It should be more like this:

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 v)
3) deceleration at constant rate from trip velocity to 0 (-d²s/dt² = -a)
4) continuing the negative acceleration in 3) above until a negative constant trip velocity is achieved -ds/dt = -v
5) trip speed at a constant rate: ds/dt is constant (-v)
6) deceleration or positive acceleration) from the trip velocity d²s/dt² = a, until the actual final speed of B wrt A is zero.

This makes it consistent with the scenario described in this Wikipedia page:

http://en.wikipedia.org/wiki/Twin_p...lt_of_differences_in_twins.27_spacetime_paths

The Wiki page gives a formula for the 2 constant velocity phases and the 4 accelerating phases.

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a}\, asinh(aT_a/c)$$

where $T_a$ is the coordinate time of a single accelerating phase and $T_a$ is the coordinate time of a single cruising phase.

The whole thing can be expressed in terms of velocity as:

$$d \tau=2T_c/\gamma+4\frac{cT_a}{v \gamma}\, asinh(v\gamma/c)$$

where v is the cruising velocity.

Both the equations above assume all the acceleration phases last the same time and the cruising phases last the same time.

We fixed that error (the returning constant velocity phase missing) in a prior post. We downloaded the Wikipedia article you referred to and we haven't got a clue on how to interpret it.

But, things being as they are, we will let that slide and just know that the "moving" traveler ages slower. That's good enough for me for now.

Thanks for all your help.

 

[Passionflower]

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.

I you read the article you would understand that the 'inertial clock' is not based on a frame of simultaneity. Why do you think it prompted me to attend members of this possibility if not for the difficulties in using frames of simultaneity in accelerating situations? I am just trying to help and thought it would be interesting for my fellow members to know about this alternative. Perhaps it is wasted on you, as you seem to be rather negative about it. No offense you can always ignore it!

Yes an on-board computer would be necessary, but so what? What's your point? You think it is too advanced for a spaceship that travels relativistically to have an on-board computer?