Room-temperature superconductivity?

Point out news stories, on the net or in mainstream media, related to polywell fusion.

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tomclarke
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Post by tomclarke »

charliem wrote:
tomclarke wrote:
johanfprins wrote: You are avoiding logic and distorting the truth Tom. You yourself have claimed that the ages will be determined by which twin returns to which twin.
This is a loose statement, but yes, if the twins meet up again relative ages depend on there previous movements.
Tom, if I recall correctly you say that Johan's defend the existence of an universal time, and that that doesn't exist.

Doesn't your interpretation implies the existence of privileged FOR?

If so, how is that different to the existence of an universal time?


P.D. Sorry if I repeat questions already answered. This thread is too long and I have not read all of it.
The key is that they meet up again (in the same FOR). In that case the relative age is computed in the original (and final) FOR, which is priviledged because it is their FOR.

You could of course compute ages in some different FOR - but - the result will be the same except for the time dilation (of the age difference) caused by the calculation from a different FOR.

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Post by johanfprins »

tomclarke wrote: This is a loose statement, but yes, if the twins meet up again relative ages depend on their previous movements.
I do not think it is a loose statement. But OK, let us make it less loose.

Two twins have identical spaceships, of identical lengths, engines etc. floating side by side within free space and they have identical clocks which they synchronise at the instant that both spacehips start to move way from one another. In order to start the journey, each twin simultaneously with the other twin, switches on his/her identical engines for exactly the same pre-arranged length of time as measured on their respective clocks; and then switches off their engines. When switching off their engines they coast at a constant speed v relative to one another.

They then both allow their space ships to coast until at a pre-arranged time on both their clocks they both switch on their engines to de-acceleate until they both come to a halt relative to one another. Thus when they start to de-accelerate the time on their clocks must be identical and also when they come to a halt relative to one another

They then simultaneously, as shown by their clocks, gun their engines to identically accelerate towards one another, until at a pre-arranged time on both their clocks they switch off their engines in order to coast towards one another with the same speed v with which they coasted away from one another. Then again at a pre-arranged time on their respective clocks, they switch on the engines to de-accelerate towards one another so that they end up again being stationary relative to one another.

Now answer the following questions:

1. Have their engines been switched on and off simultaneously during the whole journey?

2.Which twin will be younger than the other?

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Post by johanfprins »

charliem wrote:
tomclarke wrote:
johanfprins wrote: You are avoiding logic and distorting the truth Tom. You yourself have claimed that the ages will be determined by which twin returns to which twin.
This is a loose statement, but yes, if the twins meet up again relative ages depend on there previous movements.
Tom, if I recall correctly you say that Johan's defend the existence of an universal time, and that that doesn't exist.

Doesn't your interpretation implies the existence of privileged FOR?
Brilliant questions. Yes Tom's interpretation is doing exactly this. It is really a no-brainer to see that if any inertial reference frame can be used to get the same experimental results, then the time-rate of clocks within all inertial reference frames must be exactly the same.

Time can only change with position within a gravity field. Minkowski space is thus the limiting space-time field when there is no gravity; and this demands that time becomes universal. It is really a trivial case of four-dimensional space-time which dertermines gravity when time does actually change with position. Thus, in the case of Special Relativity all clocks within the universe must keep the same time-rate, since there is no gravity that can alter their time-rates to be different at different positions and speeds.

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Post by tomclarke »

johanfprins wrote:
tomclarke wrote: This is a loose statement, but yes, if the twins meet up again relative ages depend on their previous movements.
I do not think it is a loose statement. But OK, let us make it less loose.

Two twins have identical spaceships, of identical lengths, engines etc. floating side by side within free space and they have identical clocks which they synchronise at the instant that both spacehips start to move way from one another. In order to start the journey, each twin simultaneously with the other twin, switches on his/her identical engines for exactly the same pre-arranged length of time as measured on their respective clocks; and then switches off their engines. When switching off their engines they coast at a constant speed v relative to one another.

They then both allow their space ships to coast until at a pre-arranged time on both their clocks they both switch on their engines to de-acceleate until they both come to a halt relative to one another. Thus when they start to de-accelerate the time on their clocks must be identical and also when they come to a halt relative to one another

They then simultaneously, as shown by their clocks, gun their engines to identically accelerate towards one another, until at a pre-arranged time on both their clocks they switch off their engines in order to coast towards one another with the same speed v with which they coasted away from one another. Then again at a pre-arranged time on their respective clocks, they switch on the engines to de-accelerate towards one another so that they end up again being stationary relative to one another.
Fine. you have now made clear that the movement of the two twins, relative to their initial rest frame, is symmetrical.
Now answer the following questions:

1. Have their engines been switched on and off simultaneously during the whole journey?

Relative to their initial (& final) rest frame, yes. However relative to any other frame no. In this problem it is natural to take the initial rest frame as priviledged, in which case it makes sense to compare times in it.
2.Which twin will be younger than the other?
As I have said many times, in any symmetrical situation as this, when they meet up again the two twins will have equal age.

charliem
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Post by charliem »

tomclarke wrote:
charliem wrote:Tom, if I recall correctly you say that Johan's defend the existence of an universal time, and that that doesn't exist.

Doesn't your interpretation implies the existence of privileged FOR?
The key is that they meet up again (in the same FOR). In that case the relative age is computed in the original (and final) FOR, which is priviledged because it is their FOR.
Two things.

What makes that FOR privileged? Is it acceleration? Its position relative to the rest of the universe?

If acceleration I don't see it. Wouldn't we see the same effect if the traveling twin accelerates very hard or very gently, as long as its final speed is the same for most of his journey?

And if the rest of the universe, would not that imply the existence of an universal FOR?


And the thought experiment Johan just proposed suggested me another one:

Both twins depart from the same FOR (F0) at the same instant (t0, local to F0), travel in opposite directions at the same speed (V, from F0) for the same lapse (local to them). Then come back together again by the same fashion. For clarity I depict in my head F0 as a static point in the center of my screen, Twin1 travels to the left and Twin2 to the right, separate at time t0 and they both do speed V for most of the time (although in opposite directions), and reunit.

When both twins meet at F0 they see that are still of the same age.

Ok, now my variant. Lets add a third agent, one I call Arbiter.

Arbiter also has a ship but he has been traveling at speed V (as seen from F0) well before both twins separate, and in the same direction that Twin2 will take for the first leg of its journey (to the right).

From the very moment Twin2 reaches cruise speed after departing and until Twin2 re-encounter its brother Arbiter mimics Twin2 maneuvers. So for almost the whole journey of Twin2, Arbiter and him share the same FOR (Arbiter, like Twin2, ends static in F0).

But to Arbiter Twin1, the other brother, has always been moving, before t0 at speed V, and after t0 at speed 2.V until the brothers meet at F0, so for Arbiter Twin1 should be younger than Twin2.

But the thing is that when they meet Twin1 and Twin2 say they are the same age according their respective clocks.

Where am I mistaken?
"The problem is not what we don't know, but what we do know [that] isn't so" (Mark Twain)

tomclarke
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Post by tomclarke »

johanfprins wrote:
charliem wrote:
tomclarke wrote: This is a loose statement, but yes, if the twins meet up again relative ages depend on there previous movements.
Tom, if I recall correctly you say that Johan's defend the existence of an universal time, and that that doesn't exist.

Doesn't your interpretation implies the existence of privileged FOR?
Brilliant questions. Yes Tom's interpretation is doing exactly this.
See my reply above. There is a natural reference frame since the twins end up and start in the same frame. If you don't use this, then the precise age difference cannot be determined (unless it is 0).
It is really a no-brainer to see that if any inertial reference frame can be used to get the same experimental results, then the time-rate of clocks within all inertial reference frames must be exactly the same.
No-brainer in the sense of brainless. The statement is extremely ambiguous. What do you mean by time-rate of clocks? If you mean local time-rate relative to physical processes in their rest frame the answer is tautologically yes.

If you mean some sort of comparison between clock rates in different frames the answer is in general no - with the caveat that in clocks in different frames cannot have rates compared except relative to some priviledged frame. Note that without specifying the frame (or implying it, since sometimes a canonical frame is obvious) the comparison does not have a definite answer. (There are specific cases when it does, as when the two clocks are measured at the same spatial position and the time difference is zero).
Time can only change with position within a gravity field.
That is a particularly meaningless comment.
(1) Time is a dimension of (scalar function on) a 4D space.
(2) A unique frame-dependent global time is induced by any frame of reference (the time of a clock at rest in teh frame, extended spatially to clocks at any other point in the frame via Einsteinian synchronisation).
(3) Given such a global time, we can give any event a time. We can also consider the set of all events in space at a given time.
(4) Since, given a priviledged frame and such a global time, time and spatial position are orthogonal, the statement makes no sense. It is obvious that the worldline of any moving object shows time changing with position.
Minkowski space is thus the limiting space-time field when there is no gravity; and this demands that time becomes universal. It is really a trivial case of four-dimensional space-time which dertermines gravity when time does actually change with position.
It is true that MS is limiting case with no gravity, which I stated a few posts ago in response to you saying that MS could only exist in presence of gravity! Glad you have changed your mind. You repeat "time changing with position" which does not make sense.

I think you mean "time rate changing with position" which also makes no sense, see below.
Thus, in the case of Special Relativity all clocks within the universe must keep the same time-rate, since there is no gravity that can alter their time-rates to be different at different positions and speeds.
Your idea here of "time-rate" and of "time-rate" depending on position when their is gravity" is understandable - it seems reasonable in a Newtonian world in which time is separate from space, but it is conceptually wrong.

You have no physical way to define "time-rate". Rate with respect to what? Of course you can define "local time" which by definition is the speed clocks tick in a frame. That ties the local time coordinate to physical processes in the frame. It does not imply anything about time in other frames.

Nor can you compare one "time-rate" with another, except by choosing a priviledged frame.

Some time ago you provided what you claimed was a way to do this, synchronising clocks globally in two frames f0 and F1:

(1) You synchronised clocks over all space in F0 (fine).

(2) You synchronised such clocks in F0 with similar in F1 at a given initial time t0.

(2) contains an implicit priviledged frame. You have to decide whether t0 is measured using F0 or F1 clocks. You get different results since instantaneity in one frame does not imply instantaneity in another.

So your whole conceptual framework, which rests on frame-independent global time, has no physical basis. I await your attempt to mend the above synchronisation experiment with great interest.

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Post by johanfprins »

tomclarke wrote: Fine. you have now made clear that the movement of the two twins, relative to their initial rest frame, is symmetrical.
Unless you have a third party who stays behind, there is no initial reference frame which remains intact to also be final rest frame. If you want to describe the motion of the two space-ships within such a third reference frame, the problem becomes more complex since you have to use the relativistic sum rule to add velocities. The fact is that according to SR, once the space-ships move relative to one another, there is only relative motion between two equivalent inertial reference frames..

There cannot be absolute motion relative to a third stationary rest frame. One must choose one or the other space-ship relative to which you measure the motion of the other space-ship. The space-ship which you choose as being stationary (say twin 1) will see the other spaceship (twin 2) accelerating away, then moving with speed v, then decelerating to stop, then accelerating back, then coasting, then decelerating to stop within the FOR of twin 1. This must surely mean that according to your reasoning twin 2 must be younger. Can you see how absurd your reasoning is?
Now answer the following questions:

1. Have their engines been switched on and off simultaneously during the whole journey?

Relative to their initial (& final) rest frame, yes. However relative to any other frame no. In this problem it is natural to take the initial rest frame as priviledged, in which case it makes sense to compare times in it.
In other words you are claiming that the two spaceships are moving relative to a uniquely stationary reference frame: This is in violation of the postulates on which SR are based.
2.Which twin will be younger than the other?
As I have said many times, in any symmetrical situation as this, when they meet up again the two twins will have equal age.
Twin 1 will see twin 2 move away, turn around and return: This is specifically not a symmetric motion according to twin 1. Thus you argue that twin 2 will now be younger?

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Post by tomclarke »

charliem wrote:
tomclarke wrote:
charliem wrote:Tom, if I recall correctly you say that Johan's defend the existence of an universal time, and that that doesn't exist.

Doesn't your interpretation implies the existence of privileged FOR?
The key is that they meet up again (in the same FOR). In that case the relative age is computed in the original (and final) FOR, which is priviledged because it is their FOR.
Two things.

What makes that FOR privileged? Is it acceleration? Its position relative to the rest of the universe?
Perhaps priviledged is the wrong word. What i mean is that it is the natural frame to use when giving the answer.

The answer is frame-dependent. Use a different frame you get a different answer, however having calculated the answer, a time difference, in the (natural) frame, any other frame will give you this answer multiplied by gamma - the LT time dilation factor between the frames. So for the purpose of knowing whether the ages are the same or different, and which is older, we really do have a frame-independent answer in this case.
If acceleration I don't see it. Wouldn't we see the same effect if the traveling twin accelerates very hard or very gently, as long as its final speed is the same for most of his journey?
You are now asking what determines whether a twin ages more or less. a good question. It is not the acceleration, but the delta-v between inward and outward journeys. The nearer this gets to a max value of c (at which point both inward and outward journeys have speed of c in opposite directions relative to initial rest frame) the larger the time dilation.

You can see this because a light beam emitted at start, reflected at its furthest distance would have delta-v = c, and infinite time dilation.
And if the rest of the universe, would not that imply the existence of an universal FOR?
see above.
And the thought experiment Johan just proposed suggested me another one:

Both twins depart from the same FOR (F0) at the same instant (t0, local to F0), travel in opposite directions at the same speed (V, from F0) for the same lapse (local to them). Then come back together again by the same fashion. For clarity I depict in my head F0 as a static point in the center of my screen, Twin1 travels to the left and Twin2 to the right, separate at time t0 and they both do speed V for most of the time (although in opposite directions), and reunit.

When both twins meet at F0 they see that are still of the same age.

Ok, now my variant. Lets add a third agent, one I call Arbiter.

Arbiter also has a ship but he has been traveling at speed V (as seen from F0) well before both twins separate, and in the same direction that Twin2 will take for the first leg of its journey (to the right).

From the very moment Twin2 reaches cruise speed after departing and until Twin2 re-encounter its brother Arbiter mimics Twin2 maneuvers. So for almost the whole journey of Twin2, Arbiter and him share the same FOR (Arbiter, like Twin2, ends static in F0).

But to Arbiter Twin1, the other brother, has always been moving, before t0 at speed V, and after t0 at speed 2.V until the brothers meet at F0, so for Arbiter Twin1 should be younger than Twin2.

But the thing is that when they meet Twin1 and Twin2 say they are the same age according their respective clocks.

Where am I mistaken?
The two arbiters do see the first leg of their twin's journey as you say. However they can only observe the second leg of their twin, and the whole journey of the other twin, at a distance.

The twin2 arbiter is not in F0, but in the twin2 outward leg rest frame. This is assymetric so he will indeed see different times from the twin1 arbiter.

In fact, the twin1 arbiter will see twin1's outward journey as twin1 does (time T) and backward journey with time dilation at time t << T.

The twin1 arbiter will similarly see twin2's outward journey as taking time t, and inward journey (when relative speed with arbiter is 0) taking time T, the same as twin2.

The arbiter thus agrees with the natural F0 frame that the twins have equal age when the reunite, even though his view of their ages throughout the journey is different from theirs, and of course each twin has a different view of the age difference between the twins throughout the journet until the end, when they all agree.

I'll let you pose a more carefully defined version of this which you think is contradictory, and go through it in more detail, if you continue to worry, since I may have mistaken your example.

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Post by tomclarke »

johanfprins wrote:
tomclarke wrote: Fine. you have now made clear that the movement of the two twins, relative to their initial rest frame, is symmetrical.
Unless you have a third party who stays behind, there is no initial reference frame which remains intact to also be final rest frame.
Reference frames depend on motion of their axes, they do not need a physical embodiment of this.
If you want to describe the motion of the two space-ships within such a third reference frame, the problem becomes more complex since you have to use the relativistic sum rule to add velocities. The fact is that according to SR, once the space-ships move relative to one another, there is only relative motion between two equivalent inertial reference frames..
If you want to do the calculation of this symmetrical problem in an assymmetric frame (say that of the outward leg of one of the twins) that is your priviledge. But note that you then need to use relativistic sum to compute the inner leg speed of this twin.

In the initial rest frame each twin has velocity +V or -V at all times, no sum needed.
There cannot be absolute motion relative to a third stationary rest frame. One must choose one or the other space-ship relative to which you measure the motion of the other space-ship. The space-ship which you choose as being stationary (say twin 1) will see the other spaceship (twin 2) accelerating away, then moving with speed v, then decelerating to stop, then accelerating back, then coasting, then decelerating to stop within the FOR of twin 1. This must surely mean that according to your reasoning twin 2 must be younger. Can you see how absurd your reasoning is?
There certainly can be relative motion relative to any other frame you like. obviously, we can do the claulation consistently in any frame. I just chose the easiest one (the rest frame initially an dfinally of the twins, in which twin velocities are always +V or -V).

I'll leave others to jusge who has corect reasoning in this case.
As I have said many times, in any symmetrical situation as this, when they meet up again the two twins will have equal age.

Twin 1 will see twin 2 move away, turn around and return: This is specifically not a symmetric motion according to twin 1. Thus you argue that twin 2 will now be younger?
if you choose to measure times in the frames of one of the twins I gave the answer in my previous post. Let:
V2 = velocity of each twin relative to the other on outward leg (this is 2*V/(1+V^2/c^2) where V is the outward velocity of each twin relative to the initial rest frame.
T = twin elapsed time on outward leg (in outward leg rest frame).
t = T* gamma(V2)

Measure times relative to Twin1 outward leg frame.

Twin1 outward = T
Twin1 inward = t
Twin2 outward = t
twin2 inward = T

The two twins meet with the same age, as measured in this assymetric frame. that must be the case, since the LT of the time difference to a time in any other frame will be 0 since it is zero (by symmetry) in the symmetrical frame.

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Post by johanfprins »

Tom, as usual you have posted so much nonsense, that, in order not to waste my time, I have to be selective.
tomclarke wrote:
johanfprins wrote:
tomclarke wrote: Fine. you have now made clear that the movement of the two twins, relative to their initial rest frame, is symmetrical.
Unless you have a third party who stays behind, there is no initial reference frame which remains intact to also be final rest frame.
Reference frames depend on motion of their axes, they do not need a physical embodiment of this.
OK, I should have stated this better. So you must thus agree that if you synchronise all possible reference frames simultaneously at the same simultaneous point in space, then clocks within all these reference frames must show the same time at any instant in future that any one of these clocks show at that instant in time. Thus, the time rate is the same within all inertial reference frames; as I have claimed all along.
In the initial rest frame each twin has velocity +V or -V at all times, no sum needed.
You probably mean +V/2 and -V/2. This will only be the case when the Galilean transformation applies, but NOT when the Lorentz transformation applies. Although the speeds will still cancel, the relative speed will not be V.
I'll leave others to jusge who has corect reasoning in this case.
I hope that they are not as clueless as you are!

The fact is that no matter how "assymetrical" the motion is as you claim that it must be to get different ages, there will always be a "preferred" reference frame within which this motion is "symmetrical", and this reference frame, in terms of you own explanation, must be the same as the initial and final reference frame. Thus the ages of the twins cannot EVER differ, as per your own argument.

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Post by tomclarke »

johanfprins wrote:
The fact is that no matter how "assymetrical" the motion is as you claim that it must be to get different ages, there will always be a "preferred" reference frame within which this motion is "symmetrical", and this reference frame, in terms of you own explanation, must be the same as the initial and final reference frame. Thus the ages of the twins cannot EVER differ, as per your own argument.
Johan,

Suppose one twin is stationary in initial rest frame, other twin has speed +V, -V (sorry, I take V to be velocity of both twins relative to rest, not V to be differential velocity).

There is no frame which makes this symmetrical, because the midpoint between the two twins positions has two different velocities, one outward and one inward. Frames must be inertial.

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Post by tomclarke »

johanfprins wrote:Tom, as usual you have posted so much nonsense, that, in order not to waste my time, I have to be selective.
tomclarke wrote:
johanfprins wrote: Unless you have a third party who stays behind, there is no initial reference frame which remains intact to also be final rest frame.
Reference frames depend on motion of their axes, they do not need a physical embodiment of this.
OK, I should have stated this better. So you must thus agree that if you synchronise all possible reference frames simultaneously at the same simultaneous point in space, then clocks within all these reference frames must show the same time at any instant in future that any one of these clocks show at that instant in time. Thus, the time rate is the same within all inertial reference frames; as I have claimed all along.
Johan, you have not read my post above where I dissect your synchronisation argument in detail. "At the same time" is not defined in a frame-independent fashion across different frames. Sure, you can define it for one point, but at any other point it depends on the frame you use.

The synchronisation is obviously different (in different frames) if done at different times. Hence I must not agree with your statement.

You are quite correct that if frame-independent global time exists, your arguments would be true. But all your attempts to show frame-independent global time, like this one, implictly or explictly use "instantaneity" across different frames and space. That is of course the thing you are trying to show, and so circular!

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Post by charliem »

tomclarke wrote:The two arbiters do see the first leg of their twin's journey as you say. However they can only observe the second leg of their twin, and the whole journey of the other twin, at a distance.

The twin2 arbiter is not in F0, but in the twin2 outward leg rest frame. This is assymetric so he will indeed see different times from the twin1 arbiter.

In fact, the twin1 arbiter will see twin1's outward journey as twin1 does (time T) and backward journey with time dilation at time t << T.

The twin1 arbiter will similarly see twin2's outward journey as taking time t, and inward journey (when relative speed with arbiter is 0) taking time T, the same as twin2.

The arbiter thus agrees with the natural F0 frame that the twins have equal age when the reunite, even though his view of their ages throughout the journey is different from theirs, and of course each twin has a different view of the age difference between the twins throughout the journet until the end, when they all agree.

I'll let you pose a more carefully defined version of this which you think is contradictory, and go through it in more detail, if you continue to worry, since I may have mistaken your example.
Sorry Tom, looks like I did not explain myself clearly enough. My arbiter (or two arbiters if you want) don't keep at rest in any inertial FOR.

I propose only one arbiter to try make things a bit less complicated.

Before t0 Arbiter is not at rest in the twins initial FOR, but travels at speed V in the same direction that Twin2 will take.

At t0 Twin2 accelerates until it is at rest in Arbiter's FOR. When Twin2 starts to reverse course Arbiter does the same, and when Twin2 brakes to encounter with his brother so does Arbiter.

So Arbiter and Twin2 share FOR from a bit after departure to the end, and this FOR is not always inertial.

Well, from Arbiter perspective both twins start having the same age but during their travel Twin1 gets younger and younger than Twin2, and that difference never decrease, only grows.

The problem is that when they finally meet the brothers claim to be of the same age according to their respective clocks (having been Twin1 traveling in the opposite direction from the starting point, and back, also at speed V, and for the same local time lapse than its brother).

If my reasoning is flawed, where?
"The problem is not what we don't know, but what we do know [that] isn't so" (Mark Twain)

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Post by tomclarke »

charliem wrote:
tomclarke wrote:The two arbiters do see the first leg of their twin's journey as you say. However they can only observe the second leg of their twin, and the whole journey of the other twin, at a distance.

The twin2 arbiter is not in F0, but in the twin2 outward leg rest frame. This is assymetric so he will indeed see different times from the twin1 arbiter.

In fact, the twin1 arbiter will see twin1's outward journey as twin1 does (time T) and backward journey with time dilation at time t << T.

The twin1 arbiter will similarly see twin2's outward journey as taking time t, and inward journey (when relative speed with arbiter is 0) taking time T, the same as twin2.

The arbiter thus agrees with the natural F0 frame that the twins have equal age when the reunite, even though his view of their ages throughout the journey is different from theirs, and of course each twin has a different view of the age difference between the twins throughout the journet until the end, when they all agree.

I'll let you pose a more carefully defined version of this which you think is contradictory, and go through it in more detail, if you continue to worry, since I may have mistaken your example.
Sorry Tom, looks like I did not explain myself clearly enough. My arbiter (or two arbiters if you want) don't keep at rest in any inertial FOR.

I propose only one arbiter to try make things a bit less complicated.

Before t0 Arbiter is not at rest in the twins initial FOR, but travels at speed V in the same direction that Twin2 will take.

At t0 Twin2 accelerates until it is at rest in Arbiter's FOR. When Twin2 starts to reverse course Arbiter does the same, and when Twin2 brakes to encounter with his brother so does Arbiter.

So Arbiter and Twin2 share FOR from a bit after departure to the end, and this FOR is not always inertial.

Well, from Arbiter perspective both twins start having the same age but during their travel Twin1 gets younger and younger than Twin2, and that difference never decrease, only grows.

The problem is that when they finally meet the brothers claim to be of the same age according to their respective clocks (having been Twin1 traveling in the opposite direction from the starting point, and back, also at speed V, and for the same local time lapse than its brother).

If my reasoning is flawed, where?
The problem is when the arbiter changes its FOR. That will dramatically change its view of the time of the far away twin: as arbiter changes frame, so calculated age of far twin will dramatically increase. Remember, this time is only ever calculated, and frame-dependent, because there is no global time independent of frame.

Johan does not understand this, but the time of far away objects is throroughly frame-dependent.

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Post by charliem »

tomclarke wrote:
charliem wrote:Well, from Arbiter perspective both twins start having the same age but during their travel Twin1 gets younger and younger than Twin2, and that difference never decrease, only grows.
The problem is when the arbiter changes its FOR. That will dramatically change its view of the time of the far away twin: as arbiter changes frame, so calculated age of far twin will dramatically increase. Remember, this time is only ever calculated, and frame-dependent, because there is no global time independent of frame.
With my thought experiment I was trying to elude the necessity of calculating times and ages in different FOR, and only make the comparison at the end, at rest in the initial FOR.

My reasoning is that during most of the experiment Twin2 doesn't move in respect of Arbiter, but Twin1 does. So, as perceived by Arbiter, Twin1 should get younger than Twin2...

(supposed time dilation with speed is objective, measured from the starting FOR, and not a trick of perception that disappears as soon as you brake, and supposed acceleration doesn't play a relevant part in this).
"The problem is not what we don't know, but what we do know [that] isn't so" (Mark Twain)

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