How are the twins distinguished?

[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?

 

[Passionflower]

Notice the loop. I never suggested that, considering the loop.[ i.e. the round trip], that this would not be the case.

Ok so then what do you mean by differential time dilation if there is no loop?

Alright let's talk about acceleration and your potential misconceptions.

This is what you initially wrote:

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.

And then you wrote:

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.

So consider two inertial twins T_home and T_travel at the same location. Explain to me how if T_travel undergoes proper acceleration and T_home stays inertial T_travel's clock rate can increase with respect to T_home's clock rate?

 

[Austin0]

Originally Posted by Austin0

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.

If you take my quote in context it was in responce to the idea that simply through accelerometer readings the clock could be computerized to calibrate the onboard "home" clock.
I didn't suggest that there was no way to calculate when the ship reached a state of rest wrt earth. I in fact explicitly described how this could be done through interferometer Doppler readings of a continuous signal from earth.
Can I assume you do understand that the decceleration/acceleration phase at turn around is actually one continuous acceleration in the direction of earth?
Not two discrete periods as far as onboard thrust and continuous accelerometer readings is concerned. That to determine the abstract relative point of rest wrt Earth requires additional reference whether it is calculation , DOppler or some other outside reference.
That the accelerometer will by itself give no clue as to this point.

This was the point of contention and misunderstanding here.
Would you disagree with any of this?

 

[stevmg]

Again to Passionflower: I have apologized in a prior post (#87) and via private message my apology for flippant statements earlier. As I stated to you (as well as the monitors of PF, if they are watching) that this form of communication will not be repeated to either you or anyone else in the future.

All I understand from all this is that once there is motion, there is a loss of time (time dilation) to the "moving" object in relation to the stationary object. When it is a simple trip at constant velocity, one can postulate that looking from the FR of the moving object, the "stationary" object is moving so the time dilattion occurs to that object (we have been referring to "that object" as "A" while the original moving object was designated "B.")

JesseM has shown that when you have a constant motion (velocity) away from an onject A followed by a constant velocity motion back to A, whether it be from two different objects (call the B and B') traveling at constant velocity but in opposite directions, or the same object B which turns around, with relation to A, the combined proper times of B and B' journey is less than the proper time of the interval that A experiences. This is a consequence of the hyperbolic relation of distance and time given by the Minkowski equations. Even if the same object experiences
1) acceleration
2) constant velocity
3) deceleration to zero
4) acceleration (or continuation, if you will, of the deceleration in 3)
5) constant velocity back to A
6) deceleration to zero so B rejoins A
B experiences less proper time than A. Even with the application of those complicated hyperbolic arcsine equations for proper time in aceleration/deceleration (which I, as yet, cannot do) you still wind up with less time evolved.

JesseM went on to show that choosing any other FR and sticking with it as the "stationary" FR that A still experienced more proper time than B.

Postulating Passionflower's statement (I hope I am correct here) that if you look strictly from B, A would "slow down." I don't know how one uses an FR which itself has acceleration in it, which would be the case if you chose B's FR. Einstein gave the example of the falling elevator and that to the man in it, he doesn't experience any difference than a man sitting still and not moving (no gravity, of course.) One thing is for sure - if you could choose B's FR as your central FR, A would accelerate... and A would be the one who slowed down but I don't know if you can do that. Can one choose an accelerating/decelerating FR and make that one central and make the universe move about it? As it has been stated, in a theoretically gravity free universe, one can do anything as Einstein talked about with his elevator man.

One interesting thing, though, if the man and elevator were accelerating together they would be under the influence of a gravity. A second person in an FR which was stationary would experience the pull of this gravity and the equal and opposite resitance or force as he/she stands on his stationary platfrom. Meanwhile, the elevator man is experiencing no gravity as he is falling and feels weightless. If he were chosen as the central FR, the guy on the platform would appear to accelerate upward and experience the force produced by that upward accelerating motion.

This is weird, though.

So, can one use an FR which is accelerating/decelerating as a central FR?

 

[Passionflower]

To stevemg: I received your message and I replied. As far as I am concerned this mater is resolved and is behind us.

Postulating Passionflower's statement (I hope I am correct here) that if you look strictly from B, A would "slow down."

No.

Assume A and B are inertial at event X.

Then if B undergoes proper acceleration and A stays inertial then it is always true that the spacetime distance between B (at any event on his worldline) and X is larger than the path length from B to X. And a smaller path length means less elapsed time.

 

[granpa]

this has gone on for 8 pages now.
Guys, the twins paradox is not THAT hard.

you've been given good answers and you keep rejecting them and then you just keep asking the same questions over and over. Maybe its time that you started questioning your assumptions.

As I said before, its all about relativity of simultaneity. That is how how each can see the other as aging more slowly when they are moving with respect to each other.
And its also about how relativity of simultaneity changes when one accelerates.
you don't even need to know how it changes WHILE it is accelerating. you can just calculate its state BEFORE and AFTER the acceleration and from that you can see what must have happened over the course of the acceleration.

 

[stevmg]

To stevemg: I received your message and I replied. As far as I am concerned this mater is resolved and is behind us.


No.

Assume A and B are inertial at event X.

Then if B undergoes proper acceleration and A stays inertial then it is always true that the spacetime distance between B (at any event on his worldline) and X is larger than the path length from B to X. And a smaller path length means less elapsed time.

I stand corrected.

Did you mean "it is always true that the spacetime distance between A (at any event on his worldline) and X is larger than the path length from B to X." or

"it is always true that the spacetime distance between B (at any event on his worldline) and X is larger than the path length from A to X?"

Or, is there a difference between path length and spacetime distance for the same object (in this case B) which leaves your original statement as posted, "it is always true that the spacetime distance between B (at any event on his worldline) and X is larger than the path length from B to X." - in which event it is back to the books for me.

 

[yuiop]

If you take my quote in context it was in responce to the idea that simply through accelerometer readings the clock could be computerized to calibrate the onboard "home" clock.
I didn't suggest that there was no way to calculate when the ship reached a state of rest wrt earth. I in fact explicitly described how this could be done through interferometer Doppler readings of a continuous signal from earth.
Can I assume you do understand that the decceleration/acceleration phase at turn around is actually one continuous acceleration in the direction of earth?
Not two discrete periods as far as onboard thrust and continuous accelerometer readings is concerned. That to determine the abstract relative point of rest wrt Earth requires additional reference whether it is calculation , DOppler or some other outside reference.
That the accelerometer will by itself give no clue as to this point.

This was the point of contention and misunderstanding here.
Would you disagree with any of this?

Let us say the traveling twins accelerates for time \tau_{a1} as measured by his own clock and with constant acceleration (a₁).

He can then calculate that the time that elapsed (T_a1) in the Earth frame during the acceleration phase is:

T_{a1} =\frac{c}{a_1}\, sinh(a_{a1} t_{a1}/c)

and that his terminal velocity (v) after the acceleration phase is:

v = \frac{a_1 T_{a1}}{\sqrt{1+(a_1 T_{a1}/c)^2}}

He now cruises for time \tau_{c} as measured by his own clock and can calculate the time elapsed T_c on the Earth clock as

T_c = \frac{\tau_c}{\sqrt{1-v^2/c^2}}

He now decelerates with constant acceleration -a₂. He can calculate the time \tau_{a2} (as measured by his own clock) it will take to come to rest with respect to the Earth as:

t_{a2}=\frac{c}{a_2}\, artanh(v/c)

The above equation is valid even if the rocket does not actually stop but continues reversing direction.

The time that elapses on the Earth clock during the deceleration phase is:

T_{a2} = \frac{c}{a_2}\, sinh(a_2 \tau_{a2}/c)

None of the above calculations require contact with the Earth after take off. All that is required is an onboard accelerometer, a clock and a calculator for the rocket captain to know when he has come to rest with respect to the Earth and how much time has elapsed on the Earth relative to his own clock in the Earth rest frame.

 

[stevmg]

Let us say the traveling twins accelerates for time \tau_{a1} as measured by his own clock and with constant acceleration (a₁).

He can then calculate that the time that elapsed (T_a1) in the Earth frame during the acceleration phase is:

T_{a1} =\frac{c}{a_1}\, sinh(a_{a1} t_{a1}/c)

and that his terminal velocity (v) after the acceleration phase is:

v = \frac{a_1 T_{a1}}{\sqrt{1+(a_1 T_{a1}/c)^2}}

He now cruises for time \tau_{c} as measured by his own clock and can calculate the time elapsed T_c on the Earth clock as

T_c = \frac{\tau_c}{\sqrt{1-v^2/c^2}}

He now decelerates with constant acceleration -a₂. He can calculate the time \tau_{a2} (as measured by his own clock) it will take to come to rest with respect to the Earth as:

t_{a2}=\frac{c}{a_2}\, artanh(v/c)

The above equation is valid even if the rocket does not actually stop but continues reversing direction.

The time that elapses on the Earth clock during the deceleration phase is:

T_{a2} = \frac{c}{a_2}\, sinh(a_2 \tau_{a2}/c)

None of the above calculations require contact with the Earth after take off. All that is required is an onboard accelerometer and clock and a calculator for the rocket captain to know when he has come to rest with respect to the Earth and how much time has elapsed on the Earth relative to his own clock in the Earth rest frame.

Outstanding!

stevmg

 

[Passionflower]

Did you mean "it is always true that the spacetime distance between A (at any event on his worldline) and X is larger than the path length from B to X." or

"it is always true that the spacetime distance between B (at any event on his worldline) and X is larger than the path length from A to X?"

No, neither.

Or, is there a difference between path length and spacetime distance for the same object (in this case B) which leaves your original statement as posted, "it is always true that the spacetime distance between B (at any event on his worldline) and X is larger than the path length from B to X." - in which event it is back to the books for me.

Indeed, the same object.

Consider the above mentioned scenario on a spacetime diagram:
Attach a cord from event X to B. When B starts to accelerate and perhaps later goes inertial or what ever it likes, at each time the cord between X and his current position represents the spacetime distance, which is always the longest possible time, between the two events. However the curve on which B travels is the path that represents B's proper time. Now if you draw it you will see that the curve is at all times longer than the straight line but in relativity we do not use an Euclidean metric but a Minkowski metric and in this metric the line that shows longest is in fact the shortest.

Now since we always can have a hypothetical twin travel inertially on X,B we have a closed loop (e.g. a twin experiment) and thus this proves that at all times we can assert that B's clock must be going slower than than A's clock. B could even have a special clock on-board that at each instant of his trip calculates the proper time of this hypothetical twin, this clock would simply measure the distance between X,B.

For those who are interested: imagine B traveling with various rates of acceleration and various cruising periods, what could you say about the total [strike]volume[/strike] area between X,B and the B's path, what does it represent?

 

[yuiop]

Postulating Passionflower's statement (I hope I am correct here) that if you look strictly from B, A would "slow down." I don't know how one uses an FR which itself has acceleration in it, which would be the case if you chose B's FR. Einstein gave the example of the falling elevator and that to the man in it, he doesn't experience any difference than a man sitting still and not moving (no gravity, of course.) One thing is for sure - if you could choose B's FR as your central FR, A would accelerate... and A would be the one who slowed down but I don't know if you can do that. Can one choose an accelerating/decelerating FR and make that one central and make the universe move about it? As it has been stated, in a theoretically gravity free universe, one can do anything as Einstein talked about with his elevator man.

One interesting thing, though, if the man and elevator were accelerating together they would be under the influence of a gravity. A second person in an FR which was stationary would experience the pull of this gravity and the equal and opposite resistance or force as he/she stands on his stationary platform. Meanwhile, the elevator man is experiencing no gravity as he is falling and feels weightless. If he were chosen as the central FR, the guy on the platform would appear to accelerate upward and experience the force produced by that upward accelerating motion.

This is weird, though.

I think you have it about right. Just to clarify. In the accelerating rocket scenario, the accelerating rocket measures less time between two spatially separated events than the inertial observers. In the elevator scenario, the observers in the falling elevator do not experience proper acceleration. Locally they can treat the falling elevator as an inertial RF. They can place clocks at the top and bottom of the elevator and synchronise them. The "stationary" observer on the platform (who is experiencing proper acceleration) should measure less elapsed time between the events (bottom of the elevator passing the platform) and (top of the platform passing the platform) than the observers in the elevator. Hopefully I have applied the equivalence principle correctly there. The basic rule is that the observers in the RF that have to use two spatially separated clocks to measure the elapsed time will always measure more elapsed time than the observer in the RF that uses a single clock and is present at both events.This is true even if there is no acceleration involved.

 

[yuiop]

For those who are interested: imagine B traveling with various rates of acceleration and various cruising periods, what could you say about the total volume between X,B and the B's path, what does it represent?

I'm interested, but I am too lazy to work it out for myself Presumably you have already worked it out?

 

[Austin0]

Well, you're both right, depending on how you look at it.


But it is also the case that there is a single non-inertial frame in which B is at rest throughout. In that sense B has one frame.

So really it depends what you mean by "frame".

I would also comment that non-inertial frames aren't as easy to define as inertial frames: you have some choice in precisely how to define your non-inertial frame, so there is really more than one non-inertial frame in which B is at rest. For simple linear motion in the absence of gravity, as in this example, however, there is a "natural" choice of frame (where frame simultaneity coincides with co-moving inertial simultaneity). But in other cases, such as rotating frames, problems arise if you want to extend a local frame to be shared by multiple rotating observers. But that's another story...

If the inertial observer on Earth is E and the accelerating traveler is ET
I think we can define a singular reference frame for ET as an elemental system of clocks and ruler.
Given the requisite breakthoughs in energy and propusion needed for an actual enactment of this scenario, it is not a stretch to imagine a fleet of computerized drones arranged along the path of ET that co-accelerate according to a preprogrammed schedule for the entire trip.
This could either be Born rigid or uniform through the fleet.
This kind of singular frame is perhaps the only kind that would conform to reality in the same way that is applicable wrt inertial frames. Providing frame agreed events and rational temporality.
So in this context ET's frame is initially, simply a clone of E's ruler and clocks.
At initiation of acceleration, ET's clock would be passing by E's position at an increasing rate to reach a point of contant rate, and then diminishing rate approaching turn around, when they would be momentarily at rest only to repeat the same sequence in the return phase of the trip.
Now regarding the interim relationship between E and ET's clocks as observed by E , this is subject to various assumptions:

1)We can assume ET's clocks local to E might at points run backwards. Succeeding passing clocks displaying decreasing proper time readings.
But we cannot assume ET's local clock running backwards, so in this case there is no meaningful information regarding the instantaneous relationship of E and ET's proper times possible through the local relationship at E's position.

2) We can assume that at some point or points eg. at the end of the acceleration phases , that the ET clocks are resynchronized by convention.
Assuming magical instantaneous resynch, this could result in significant local changes in ET's clocks. If the procedure is conducted from ET's end this would mean sudden adjustments to the clocks proximate to E at that time. As this is self evidently a mechanical adjustment of the clocks it has no actual temporal meaning wrt the relationship of E and ET at that moment.
Any naive calculations based on this conventional synchronization can also have zero temporal meaning regarding the two spatially separated points.

This highlights a problem with the mathematical structure wrt accelerated systems.
The math provides a temporal interval of the relative local clock readings, but makes no distiction between the inertial and the accelerated system.
Specifically it allows the interpretation that this interval could imply a local change in the proper time of E simply by the assumption of simultaneity in ET.
Clearly this is not consistent with physical reality in this circumstance, as the only change possible is in the dynamic system.
It leads to possible conditions and conclusions that ET's local clocks could be be suddenly colocated, not only with previous points of E's worldline but with previous points of ET's frame's own worldlines. I.e. Two spatial points and clocks from ET's frame simultaneously colocated at some previous point in E's history.

It seems clear that the application of the math based on these assumptions, not only does not contain any meaningful temporal information re: the relationship between E and ET, but is inconsistent, not only with SR, but also with our fundamental understanding of spacetime.
I.e All points of both frames would move forward though time even if at different rates.
To attribute any meaning to these calculations is to divorce them from any possible real world consistent singular coordinate system or any actually implemented system of clocks and rulers.

Now to seriously consider the requirements of a non-inertial frame consistent with reality:
1) The assumption that the clocks in all cases would move forward in time
2) A system of synchronization.
Perhaps a continuous synch signal to update local clocks.
This would obviously require differential calculations to compensate for the difference in velocity at the point of transmission and the point of reception.
The difference in distance between transmission and reception resulting from this velocity difference.
The gamma difference due to the above.
Accelerometers. Etc

Highly complex but not impossible in princple I wouldn't imagine.

If this was either theoretically or practically [in the future] implemented it would seem to be a system where [to an approximation] c would be constant.
It would eliminate all unrealistic jumps of temporality, as even with reversal of direction all changes would be incremental and local and the relationship of simultaneity between frames, that is consistent with reality wrt inertial frames would apply here as well.
Not time traveling clocks etc.

You will notice, I am sure, that I neglected Born rigid acceleration in this picture as well as Rindler coordinates.
Re: Rindler. WHether or not this is accurate within a limited domain it does not seem to lend itself to extended frames. So I am sure it could be included with proper care and complication as could Born acceleration but I left them both out for simplicity.

 

[Austin0]

I think you have it about right. Just to clarify. In the accelerating rocket scenario, the accelerating rocket measures less time between two spatially separated events than the inertial observers. In the elevator scenario, the observers in the falling elevator do not experience proper acceleration. Locally they can treat the falling elevator as an inertial RF. They can place clocks at the top and bottom of the elevator and synchronise them. The "stationary" observer on the platform (who is experiencing proper acceleration) should measure less elapsed time between the events (bottom of the elevator passing the platform) and (top of the platform passing the platform) than the observers in the elevator. Hopefully I have applied the equivalence principle correctly there. The basic rule is that the observers in the RF that have to use two spatially separated clocks to measure the elapsed time will always measure more elapsed time than the observer in the RF that uses a single clock and is present at both events.This is true even if there is no acceleration involved.

I think you are right, in principle the clock at the top would be running faster than the clock at the bottom of the elevator although it would have to be a pretty tall elevator to possibly measure any difference I'm sure

 

[yuiop]

But in other cases, such as rotating frames, problems arise if you want to extend a local frame to be shared by multiple rotating observers. But that's another story...

The other story and I will keep it brief because it slightly off topic, is this.

Consider a rotating ring. Observers on the ring want to synchronise clocks. They assume the speed of light is constant and isotropic and use Einstein's clock synchronisation method to synchronize clocks all the way around the ring. If they start with a master clock at point A and start synchronising clocks in one direction around the ring, they find when they get back to A, the last synchronised clock is not synchronised with A! No matter what they do they can not synchronise all clocks with each around the perimeter. One way they can resolve this issue is to use an alternative synchronise method of using a clock at at the centre of the ring and synchronise all clocks relative to the central master clock. When they do this they find the speed of light is not isotropic. This indicates that in certain accelerating scenarios, the speed of light can not be considered constant and isotropic over extended distances.

 

[Austin0]

The other story and I will keep it brief because it slightly off topic, is this.

Consider a rotating ring. Observers on the ring want to synchronise clocks. They assume the speed of light is constant and isotropic and use Einstein's clock synchronisation method to synchronize clocks all the way around the ring. If they start with a master clock at point A and start synchronising clocks in one direction around the ring, they find when they get back to A, the last synchronised clock is not synchronised with A! No matter what they do they can not synchronise all clocks with each around the perimeter. One way they can resolve this issue is to use an alternative synchronise method of using a clock at at the centre of the ring and synchronise all clocks relative to the central master clock. When they do this they find the speed of light is not isotropic. This indicates that in certain accelerating scenarios, the speed of light can not be considered constant and isotropic over extended distances.

I assume you mean not isotropic wrt measurement with or against direction of rotation?