From Luigi Fortunati@21:1/5 to All on Thu May 21 07:36:20 2020
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH
twins) lasts 16 years, 8 for the outward journey and 8 for the return.

Both will be 16 years old!

Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

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• From Douglas Eagleson@21:1/5 to Luigi Fortunati on Fri May 22 06:33:38 2020
On Thursday, May 21, 2020 at 3:36:24 AM UTC-4, Luigi Fortunati wrote:
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.

Both will be 16 years old!

Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

You can tell which one moved when the mover
returns and is younger than the other.

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• From richalivingston@gmail.com@21:1/5 to Luigi Fortunati on Fri May 22 06:33:38 2020
On Thursday, May 21, 2020 at 2:36:24 AM UTC-5, Luigi Fortunati wrote:
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.

Both will be 16 years old!

Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

You are ignoring the effects of acceleration when one twin reverses
direction. You can't ignore that, it is the key to understanding the "paradox".

Rich L.

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• From Sylvia Else@21:1/5 to Luigi Fortunati on Fri May 22 06:33:38 2020
On 21-May-20 5:36 pm, Luigi Fortunati wrote:
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.

Both will be 16 years old!

Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

One twin has to accelerate in order to return. So the situation is not symmetrical. There is no reason to expect a symmetrical outcome, and SR
does not predict one.

Sylvia.

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• From Torn Rumero DeBrak@21:1/5 to All on Fri May 22 06:34:50 2020
Am 21.05.2020 um 09:36 schrieb Luigi Fortunati:
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.

Wrong. In the reference of A the twin B returns to A and not to 0.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

Wrong. B does not stay in 0 but in B and the twin returns to B.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.
Both will be 16 years old!

Right for the same age of A and B, but the ages will not be 16 years older.
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

Why should there be a conceptual difference?

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• From Tom Roberts@21:1/5 to Luigi Fortunati on Fri May 22 19:15:36 2020
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...]
Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

The difference between the twins comes from the simplicity that applies
only to inertial frames.

If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate

T = \integral sqrt(1-v^2/c^2) dt

where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame. Note that neither position nor acceleration
appear in this equation; all that matters is the speed of the twin
relative to the inertial frame being used to calculate.

For twin A, v = 0, giving an age that is simply the frame's total
coordinate time of the scenario.

For twin B, 0<v<c so the integrand is strictly less than 1, and it is
clear that for any value of t, B's age will be less than the value of t.
So if twin B follows a path that leaves and then returns to A, when they
rejoin B will have aged less than A.

If you want to calculate what happens using twin B as a reference, there
is a problem: B does not remain at rest in any inertial frame, and the
above equation does not apply. B must necessarily accelerate in order to
return to A. It is possible to use SR to calculate using the accelerated coordinates of B; this is done in some textbooks, but is beyond the
scope of a newsgroup post. When done correctly (which is non-trivial),
this calculation yields the same answer as the much easier calculation
using A's inertial frame.

Tom Roberts

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• From Hendrik van Hees@21:1/5 to Tom Roberts on Fri May 22 19:22:18 2020
One should note that the proper time is an invariant and as such
completely independent of the coordinates and the parametrization you
choose to calculate it. You can introduce arbitrary generalized
coordinates (most of which then correspond to the description of a
non-inertial observer) and an arbitrary parameter to parametrize the
worldline. As in Euclidean space the length of a curve also proper
time, which is in a somewhat generalized sense nothing else than the
length of a time-like worldline, is independent of these choices.

On 22/05/2020 21:15, Tom Roberts wrote:
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...]
Where is the conceptual difference that should make the time of a twin
different from that of the other, if we are talking about Special
Relativity ONLY?
The difference between the twins comes from the simplicity that applies
only to inertial frames.

If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate

T = \integral sqrt(1-v^2/c^2) dt

where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame. Note that neither position nor acceleration
appear in this equation; all that matters is the speed of the twin
relative to the inertial frame being used to calculate.

For twin A, v = 0, giving an age that is simply the frame's total
coordinate time of the scenario.

For twin B, 0<v<c so the integrand is strictly less than 1, and it is
clear that for any value of t, B's age will be less than the value of t.
So if twin B follows a path that leaves and then returns to A, when they rejoin B will have aged less than A.

If you want to calculate what happens using twin B as a reference, there
is a problem: B does not remain at rest in any inertial frame, and the
above equation does not apply. B must necessarily accelerate in order to return to A. It is possible to use SR to calculate using the accelerated coordinates of B; this is done in some textbooks, but is beyond the
scope of a newsgroup post. When done correctly (which is non-trivial),
this calculation yields the same answer as the much easier calculation
using A's inertial frame.

Tom Roberts

--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/

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• From Phillip Helbig (undress to reply)@21:1/5 to fortunati.luigi@gmail.com on Fri May 22 13:57:20 2020
In article <ra3mu2$1gre$1@gioia.aioe.org>, Luigi Fortunati <fortunati.luigi@gmail.com> writes:

Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then returns to O.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

This is, of course, the "twin paradox". As you hint, special relativity
is sufficient to explain it. (One might think that GR has something to
do with it, since acceleration is involved, but that is not the case.
Note that the time dilation depends on the length of the journey but not
on the amount of acceleration.) There is a huge literature on this
topic, and it has certainly been answered better than is possible in a newsgroup post.

The basic explanation is that the twin who accelerates moves through
space while the other doesn't. While the acceleration is not the CAUSE
of the difference, it does allow one to say which twin really moves and
which doesn't. In other words, the motion is NOT relative in this case.
(Why there is such a thing as absolute acceleration is, as far as I
know, an unsolved problem. Some believe that Mach's Principle explains
it, i.e. acceleration is relative to the bulk of the matter in the
universe. That implies that if there were no matter in the universe
other than the two twins, then the resolution of the "paradox" might be different.)

[[Mod. note -- As the author notes, there is a huge literature on this problem/paradox. For anyone wanting to explore, Wikipedia provides a
nice introduction & overview:
-- jt]]

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• From richalivingston@gmail.com@21:1/5 to All on Sat May 23 11:10:51 2020
On Friday, May 22, 2020 at 3:57:23 PM UTC-5, Phillip Helbig (undress to reply) wrote:
...
(Why there is such a thing as absolute acceleration is, as far as I
know, an unsolved problem. Some believe that Mach's Principle explains
it, i.e. acceleration is relative to the bulk of the matter in the
universe. That implies that if there were no matter in the universe
other than the two twins, then the resolution of the "paradox" might be different.)
...

If you combine some simple ideas from QM and SR you can show that
acceleration necessarily involves an addition of energy in some
frame, and energy-momentum in all other frames:

-consider the QM wave function of a particle of mass M at rest.
Per SR it has an energy Mc^2. The wave function has zero momentum,
therefore k=0, and therefore the particle is not localized but has
a frequency w=2pi Mc^2 /hbar.

-Now consider an observer moving at speed v past this location (i.e.
in a differeent inertial frame). The wave function will be transformed
by the Lorentz transform. Some simple math shows that this observer
will see a frequency w' = 2 pi gamma Mc^2/hbar and wave number (i.e.
momentum) k'= 2 pi gamma beta Mc^2/hbar.

-i.e., the particle in the "moving frame" has a higher energy as
well as higher momentum. You much add energy to a particle to get
it boosted to a moving frame (which is not a surprise...).

-The moving particle has non-zero k. The way to get that is an
acceleration. Per GR, an accelerating frame will show "gravitational"
red shift. A particle initially stationary in such a frame will
have different frequencies at different locations. As a result the
wave function at different locations will get out of phase, which
results in the wavenumber changing (i.e. change in momentum). This
is an easy conceptual way to understand how GR metrics, such as
Schwartzschild metric, result in gravitational acceleration.

-This argument can be carried further to show how the classical
action is related to the change in phase of the wave function over
some path. The path of least action is the path with the maximum
change in wavefunction phase.

Rich L.

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• From Nicolaas Vroom@21:1/5 to Hendrik van Hees on Mon May 25 13:38:44 2020
On Friday, 22 May 2020 21:22:21 UTC+2, Hendrik van Hees wrote:
On 22/05/2020 21:15, Tom Roberts wrote:

If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate

T = \integral sqrt(1-v^2/c^2) dt

where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame.

One should note that the proper time is an invariant and as such
completely independent of the coordinates and the parametrization you
choose to calculate it. You can introduce arbitrary generalized
coordinates (most of which then correspond to the description of a non-inertial observer) and an arbitrary parameter to parametrize the worldline. As in Euclidean space the length of a curve also proper
time, which is in a somewhat generalized sense nothing else than the
length of a time-like worldline, is independent of these choices.

Is the above text not much too complicated?
t is the time of the clock at rest.
T is the time of the moving clock.
I personally would prefer to write:
t is the time calculated based on the observed clock counts of a clock.
Why calling T an invariant?

The whole issue is if the function or equation is correct.
That means is the prediction using this equation in accordance
with an actually performed experiment.
What makes this equation tricky is that it uses two parameters v and c.
Both of these parameters have to be calculated based on observations
as an integral part of this experiment.
Declaring c a constant makes this problem slightly simpler but does
not solve the physical issue.

Nicolaas Vroom

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• From Nicolaas Vroom@21:1/5 to Luigi Fortunati on Mon May 25 13:38:45 2020
On Thursday, 21 May 2020 09:36:24 UTC+2, Luigi Fortunati wrote:
Twins A and B are stationary at point O of space: (-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.

If one twin physical moves away from point O (and returns back to point O)
it is only his clock that will show a different clock count compared to
a reference clock which stayed at point O.
(if that is what actually is measured when he returns back to point O.).

So, in the reference of the twin A, it is B goes, arrives at point +4 (4 light-years away) and then return to O.

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.

Reference frames are not important. What physical happens is important.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH twins) lasts 16 years, 8 for the outward journey and 8 for the return.

Both will be 16 years old!

Only in the case that both twins physical move away, it is possible that
the clock count of both clocks don't coincide with the reference clock
(count) at point O.

In that case what the clock at O shows is also important.

Where is the conceptual difference that should make the time of a twin different from that of the other, if we are talking about Special
Relativity ONLY?

In order to understand better, the exact details of the physical experiment
are most important to known and to predict the outcome and to compare
if the prediction is correct.
If the comparison between prediction and actual are wrong the error
can be both in the prediction (calculation) or in the experiment.

Nicolaas Vroom.

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• From Nicolaas Vroom@21:1/5 to Tom Roberts on Mon May 25 13:38:44 2020
On Friday, 22 May 2020 21:15:39 UTC+2, Tom Roberts wrote:
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...]
Where is the conceptual difference that should make the time of a twin different from that of the other if we are talking about Special
Relativity ONLY?

The difference between the twins comes from the simplicity that applies
only to inertial frames.

What is the reason that you cannot use one reference frame for the whole experiment from start to finish i.e. when both clocks again can be
compared at point O?

If twin A remains at rest in an inertial frame, it is easy to calculate
the age (elapsed proper time) of each twin: simply integrate

T = \integral sqrt(1-v^2/c^2) dt

where T is the elapsed proper time, the integral is taken over the path
of the twin relative to A's rest frame, v is the speed of the twin
relative to that frame (as a function of t), and t is the time
coordinate of the frame.

My understanding is that t represents the time (clock count) of a clock
at rest and T the (time) clock count of a moving clock.

The question is how is v (the speed of the clock) measured?
IMO in order to do that, you need a reference rod in the frame at rest with
two clocks at backend x1 and the frontend x2 of this rod, also at rest.
When you do that you can calculate the speed v by applying
the following formula: v = (x2-x1) / (t2-t1)
where t1 is the time that the moving clock coincides with the backend x1
and t2 the time that the clock coincides with the front end x2 of the
clock.

In fact, if you want to do this whole calculation correct you
need clocks all along the path from point O to point +4.

Note that neither position nor acceleration
appear in this equation; all that matters is the speed of the twin
relative to the inertial frame being used to calculate.

That is correct.
But if the speed varies along this path you need more clocks
to calculate the time T correctly.
In fact, you need extra clocks near point O and point +4

But there is another issue:
How do you know that the equation to calculate T is correct?
The only way is when the calculated T at the end can be verified by
means of an actual experiment.
IMO experiments are the only way to derive this formula.
Using different experiments you can also test if all clocks
with a different internal physical construction behave the same
i.e. show the same number of clock counts.

If you want to calculate what happens using twin B as a reference, there
is a problem: B does not remain at rest in any inertial frame, and the
above equation does not apply. B must necessarily accelerate in order to return to A. It is possible to use SR to calculate using the accelerated coordinates of B; this is done in some textbooks, but is beyond the
scope of a newsgroup post. When done correctly (which is non-trivial),
this calculation yields the same answer as the much easier calculation
using A's inertial frame.

This conclusion follows the same rule I always use when I want to
study any system which includes different objects.
My golden rule is to use as a reference point a point that moves the least,
in order to keep things simple.
In practice, this means I study only one reference frame.

Nicolaas Vroom

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• From Nicolaas Vroom@21:1/5 to Phillip Helbig on Tue May 26 13:43:46 2020
On Friday, 22 May 2020 22:57:23 UTC+2, Phillip Helbig wrote:
In article <ra3mu2$1gre$1@gioia.aioe.org>, Luigi Fortunati <fortunati.luigi@gmail.com> writes:

Where is the conceptual difference that should make the time of a twin different from that of the other if we are talking about Special
Relativity ONLY?

This is, of course, the "twin paradox". As you hint, special relativity
is sufficient to explain it.

Maybe this is a tricky discussion.
A clock is a physical device. The most interesting part is the construction. Why not explain a clock strictly based on how they operate i.e.
their internal functioning?
For example, a clock based on light signals consists of two mirrors. Inbetween these mirrors a light pulse operates as an oscillator i.e. counter.
For a clock at rest, this means that a light pulse moves in a vertical direction.
For a moving clock, in the same frame as the clock at rest,
this is a zigzag pattern. The result is that the lightpath (between two
clock counts) is much longer as in the case of the clock at rest.
This explains why the moving clock ticks much slower.
For the mathematics see: https://en.wikipedia.org/wiki/Time_dilation par 2.1

There is a huge literature on this topic, and it has certainly been
answered better than is possible in a newsgroup post.

Please study page 30 of the book
Introducing Einstein's Relativity by Ray d'Inverno.
At the bottom of that page he writes:
cost* = 1/sect* = 1/sqr (1+ tan^2t*) = 1/sqr(1-v^2/c^2)
At page 31 the Lorentz group is discussed.
My impression is that all of this is (mathematically) not simple.

The basic explanation is that the twin who accelerates moves through
space while the other doesn't.

Yes and no. This becomes apparent when you consider a second completely identical experiment performed at a different (moving) location O.

While the acceleration is not the CAUSE of the difference,
it does allow one to say which twin really moves and which doesn't.

Which of the objects in our solar system does not move?
Don't they all move? Newton's law uses only distances.

In other words, the motion is NOT relative in this case.

This is a tricky sentence to understand.

(Why there is such a thing as absolute acceleration is, as far as I
know, an unsolved problem. Some believe that Mach's Principle explains
it, i.e. acceleration is relative to the bulk of the matter in the
universe. That implies that if there were no matter in the universe
other than the two twins, then the resolution of the "paradox" might be different.)

Why should there be something like absolute acceleration?
Acceleration is a calculated parameter the same as velocity.
Both are calculated based on observations ie measurements.
As such you should never add or subtract velocities.

Nicolaas Vroom

[[Mod. note -- In Newtonian mechanics and in special relativity,
there is indeed a useful concept of an absolute acceleration, namely acceleration relative to an inertial reference frame. (It doesn't
matter *which* inertial reference frame, since all inertial reference
frames are non-acclerating with respect to each other.)
-- jt]]

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• From Hendrik van Hees@21:1/5 to Nicolaas Vroom on Tue May 26 13:39:25 2020
Once more: The twin paradox is very easily resolved by the math
underlying special relativity. A moving (ideal) clock shows its
proper time, which can be calculated in any frame of reference. The
most simple choice is to use an inertial frame of reference.

Usually in the twin paradox you assume that one twin (say twin A)
is at rest in some inertial frame of referenc, while twin B moves
on an arbitrary timelike world line. This world line can be
parametrized with A's time t as a parameter, and then the time shown
by B's clock at any time is

tau=\int_0^t dt' \sqrt{1-v^2(t')/c^2}

Now in order be able to compare B's clock reading with A's clock
reading, B has to make a round trip, comming back to A to be able
to compare the clocks.

Concerning the experimental evidence there is a plethora of
confirmations of this assumptions, among them very accurate ones.
One recent example is the comparison of the lifetime of an atomic
state of Li ions running around in a storage ring at GSI with their
lifetime when they are at rest:

https://arxiv.org/abs/1409.7951

On 25/05/2020 15:38, Nicolaas Vroom wrote:
On Thursday, 21 May 2020 09:36:24 UTC+2, Luigi Fortunati wrote:
Twins A and B are stationary at point O of space:
(-4)----(-3)----(-2)----(-1)----O----(+1)----(2+)----(+3)----(+ 4)

If a twin moves away at constant relativistic speed, there is no
difference between them, in the sense that twin A can safely say that it
is B that is moving away from him, just as twin B can say (just as
safely) which is A who is moving away.
If one twin physical moves away from point O (and returns back to point O)
it is only his clock that will show a different clock count compared to
a reference clock which stayed at point O.
(if that is what actually is measured when he returns back to point O.).

So, in the reference of the twin A, it is B goes, arrives at point +4 (4

Instead, in the reference of the twin B, it is he who remains in O while
the twin A goes, arrives at the point -4 and then returns.
Reference frames are not important. What physical happens is important.

If the speed corresponds to gamma=2, the travel the other twin (for BOTH
twins) lasts 16 years, 8 for the outward journey and 8 for the return.

Both will be 16 years old!
Only in the case that both twins physical move away, it is possible that
the clock count of both clocks don't coincide with the reference clock (count) at point O.

In that case what the clock at O shows is also important.

Where is the conceptual difference that should make the time of a twin
different from that of the other, if we are talking about Special
Relativity ONLY?
In order to understand better, the exact details of the physical experiment are most important to known and to predict the outcome and to compare
if the prediction is correct.
If the comparison between prediction and actual are wrong the error
can be both in the prediction (calculation) or in the experiment.

Nicolaas Vroom.

--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/

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• From Tom Roberts@21:1/5 to Nicolaas Vroom on Thu May 28 11:13:59 2020
On 5/25/20 8:38 AM, Nicolaas Vroom wrote:
On Friday, 22 May 2020 21:15:39 UTC+2, Tom Roberts wrote:
On 5/21/20 2:36 AM, Luigi Fortunati wrote:
[...] Where is the conceptual difference that should make the
time of a twin different from that of the other if we are

The difference between the twins comes from the simplicity that
applies only to inertial frames.

What is the reason that you cannot use one reference frame for the
whole experiment from start to finish i.e. when both clocks again
can be compared at point O?

There is no such reason, and my discussion explicitly showed using the
inertial frame of A for the entire analysis.

If twin A remains at rest in an inertial frame, it is easy to
calculate the age (elapsed proper time) of each twin: simply
integrate

T = \integral sqrt(1-v^2/c^2) dt

where T is the elapsed proper time, the integral is taken over the
path of the twin relative to A's rest frame, v is the speed of the
twin relative to that frame (as a function of t), and t is the time
coordinate of the frame.

My understanding is that t represents the time (clock count) of a
clock at rest and T the (time) clock count of a moving clock.

Not quite. t is the time coordinate of the inertial frame, and T is the
elapsed proper time of the clock whose velocity profile is v(t). The
equation applies to EVERY clock, and can be used for multiple clocks
with different velocity profiles.

[[Mod. note -- Note that each clock has its own individual velocity
profile v(t) and hence its own individual elapsed proper time T .
-- jt]]

One can apply the equation to a clock at rest in the inertial frame: v=0
so the integral is trivial: T=t. So it is clear that the time coordinate
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptually
different but numerically equal.

A clock can only indicate its proper time, and only
along its worldline; the time coordinate of an inertial
frame can indicate the frame's coordinate time anywhere
the frame is valid. Of course such coordinates are
models, and to implement them requires physical clocks
located where events of interest occur; all such clocks
must of course be synchronized to the coordinate time.

In practice, large experiments place detectors where
events of interest might occur; they connect the
detectors with cables to a common place where timing can
be applied (e.g. a data-acquisition computer); of course
they must correct for the time delays in the cables,
each of which must be measured.

The question is how is v (the speed of the clock) measured?

In the usual way: v=dx/dt, where x(t) is the position of the object at
time t, and t is the time coordinate of the inertial frame.

IMO in order to do that, you need a reference rod in the frame at
rest with two clocks at backend x1 and the frontend x2 of this rod,
also at rest. When you do that you can calculate the speed v by
applying the following formula: v = (x2-x1) / (t2-t1) where t1 is
the time that the moving clock coincides with the backend x1 and t2
the time that the clock coincides with the front end x2 of the
clock.

That works, too.

Note this is a gedanken, and we implicitly presume that when we use a
given inertial frame, we can identify its coordinate values for all
events of interest. This, of course, includes the events occupied by the
clock as it moves around.

Note that neither position nor acceleration appear in this
equation; all that matters is the speed of the twin relative to
the inertial frame being used to calculate.

That is correct. But if the speed varies along this path you need
more clocks to calculate the time T correctly.

Hmmm. I repeat: this is a gedanken, and we implicitly presume that when
we use a given inertial frame, we can identify its coordinate values for
all events of interest. This, of course, includes the events occupied by
the clock as it moves around.

If you want to place clocks at rest in the inertial frame, located along
the path of the moving clock, that is fine, but it provides no
additional information than is contained in the coordinates themselves.

But there is another issue: How do you know that the equation to
calculate T is correct?

This is NOT an issue. We are using SR to model this gedanken, and that
is the correct equation of SR. We know this because it has been derived
many times in many textbooks.

The only way is when the calculated T at the end can be verified by
means of an actual experiment.

Hmmmm. The model itself has no need of experiments, because it is merely
a set of postulates and theorems, plus definitions of the symbols that
appear in them. Testing the model and validating it certainly does
require experiments. For this equation of SR, that has been done many
times, for many different types of clocks and many different physical situations.

IMO experiments are the only way to derive this formula.

No. The equation is mathematically derived from the postulates of the
theory. Experiments are used to test and verify that the resulting
theory is in agreement with the world we inhabit. These are quite
different -- the first just uses math, while the second compares that
math to experimental results.

Using different experiments you can also test if all clocks with a
different internal physical construction behave the same i.e. show
the same number of clock counts.

Hmmm. If you can apply different types of clocks in the same physical situation, yes. But in general that is not practical, and what is
required is to show agreement with the theory; after all, it is the
theory we are testing.

Tom Roberts

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• From Nicolaas Vroom@21:1/5 to Tom Roberts on Sat May 30 14:17:44 2020
On Thursday, 28 May 2020 20:14:03 UTC+2, Tom Roberts wrote:
On 5/25/20 8:38 AM, Nicolaas Vroom wrote:

What is the reason that you cannot use one reference frame for the
whole experiment from start to finish i.e. when both clocks again
can be compared at point O?

There is no such reason, and my discussion explicitly showed using the inertial frame of A for the entire analysis.

In fact, there are two issues:
1) Is it wise to select only one reference frame?
2) Which is the best reference frame to select?
Apparently you agree with issue 1.
I would prefer point O, as indicated in the text.

T = \integral sqrt(1-v^2/c^2) dt

My understanding is that t represents the time (clock count) of a
clock at rest and T the (time) clock count of a moving clock.

Not quite. t is the time coordinate of the inertial frame,

This raises a new question: What is the difference my 't represents the
time of a clock in a reference frame' versus 'the time coordinate of
the inertial frame'

So it is clear that the time coordinate
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptual
different but numerically equal.

My philosophy is to use as many concepts as possible

A clock can only indicate its proper time, and only
along its worldline; the time coordinate of an inertial
frame can indicate the frame's coordinate time anywhere
the frame is valid. Of course, such coordinates are
models, and to implement them requires physical clocks
located where events of interest occur; all such clocks
must, of course, be synchronized to the coordinate time.

That means IMO that all the clocks should be synchronized with
one standard clock. IMO at the origin O

The question is how is v (the speed of the clock) measured?

In the usual way: v=dx/dt, where x(t) is the position of the object at
time t, and t is the time coordinate of the inertial frame.

In practice, this means that t is the time of the nearest physical clock.

IMO in order to do that, you need a reference rod in the frame at
rest with two clocks at backend x1 and the frontend x2 of this rod,
also at rest. When you do that you can calculate the speed v by
applying the following formula: v = (x2-x1) / (t2-t1) where t1 is
the time that the moving clock coincides with the backend x1 and t2
the time that the clock coincides with the front end x2 of the
clock.

That works, too.

Note this is a gedanken, and we implicitly presume that when we use a
given inertial frame, we can identify its coordinate values for all
events of interest. This, of course, includes the events occupied by the clock as it moves around.

IMO this is not a gedanken.
It should be a description (a plan) of a possible experiment,
as detailed as possible.
When you start such a description with one reference frame
which consists of a 3D grid and at each crossing point a clock,
'everything' becomes much simpler to understand.

Note that neither position nor acceleration appears in this
equation; all that matters is the speed of the twin relative to
the inertial frame being used to calculate.

That is correct. But if the speed varies along this path you need
more clocks to calculate the time T correctly.

Hmmm. I repeat: this is a gedanken, and we implicitly presume that when
we use a given inertial frame, we can identify its coordinate values for
all events of interest. This, of course, includes the events occupied by
the clock as it moves around.

IMO it is not clear what you really mean with 'a Gedanken'.
What is in my mind and within the mind of anybody else can be completely different.
Each person, in general, can have its own experience and its own
understanding. That implies to compare, what different people mean,
is difficult. That is why I prefer to start with experiments is IMO
much more practical and to wait with mathematics and any theory.

If you want to place clocks at rest in the inertial frame, located along
the path of the moving clock, that is fine, but it provides no
additional information than is contained in the coordinates themselves.

IMO to discuss a real experiment first and then to explain, that the same information also is 'contained in the coordinates themselves', seems
to me, much more logical.
IMO often by only studying an experiment is enough to derive the mathematics that explains what is observed.
When you do that is a good start to unravel the laws that describe
similar experiments.

But there is another issue: How do you know that the equation to
calculate T is correct?

This is NOT an issue. We are using SR to model this Gedanken, and that
is the correct equation of SR. We know this because it has been derived
many times in many textbooks.

The issue is what these thoughts are about and if all textbooks have the
same thoughts in mind.

The only way is when the calculated T at the end can be verified by
means of an actual experiment.

Hmmmm. The model itself has no need of experiments because it is merely
a set of postulates and theorems, plus definitions of the symbols that
appear in them.

hmmm. 'The' model is always a model of something. Its that something that
is the most important.

Testing the model and validating it certainly does require experiments.

For this equation of SR, that has been done many
times, for many different types of clocks and many different physical situations.

It is important at least to describe one clock in detail because
each different type could require its own mathematics.

Using different experiments etc

Hmmm. If you can apply different types of clocks in the same physical situation, yes. But in general, that is not practical, and what is
required is to show agreement with the theory; after all, it is the
theory we are testing.

You cannot test a theory on its own. You must have an example
which require the theory to make predictions.
You can use Newton's Law to verify the movement of the planet Mercury.
You will fail. The next possibility is to try GR.

There are different types of clocks i.e. clocks based on time signals,
atomic clocks or nuclear optical clocks.
The above-mentioned equation can be verified with a clock based
on light signals. Of such a clock there are two types:
One in which the light signal at rest moves vertical and a different
one in which the light signal moves horizontally. The mathematics
that describes each is different.

Nicolaas Vroom

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• From Hendrik van Hees@21:1/5 to Nicolaas Vroom on Sat May 30 14:38:15 2020
On 30/05/2020 16:17, Nicolaas Vroom wrote:
On Thursday, 28 May 2020 20:14:03 UTC+2, Tom Roberts wrote:
On 5/25/20 8:38 AM, Nicolaas Vroom wrote:

What is the reason that you cannot use one reference frame for the
whole experiment from start to finish i.e. when both clocks again
can be compared at point O?
There is no such reason, and my discussion explicitly showed using the
inertial frame of A for the entire analysis.
In fact, there are two issues:
1) Is it wise to select only one reference frame?
2) Which is the best reference frame to select?
Apparently you agree with issue 1.
I would prefer point O, as indicated in the text.
Ad 1) It is wise to formulate a problem in a manifestly covariant
way and then use the most convenient reference frame to do the
calculations.

The obviously Lorentz-invariant quantity to be calculated here is
the proper time tau of the moving twin as a function of the
proper time of the twin staying in his inertial frame at rest, which
is the coordinate time, t, of this inertial frame.

Ad 2) In this case of course the inertial frame (the restframe of
the twin staying at home all the time).

The unique Lorentz-invariant solution is
\tau = \int d t \sqrt{1-\beta^2(t)},
where \beta=|\vec{v}|/c, where \vec{v} is the three-velocity of the
moving twin as measured in the inertial frame of the resting twin.

That completely solves the task and shows that \tau<t, i.e., the traveling
twin aged less than the resting twin, which they can compare by looking at their clocks when meeting again at point 0.

T = \integral sqrt(1-v^2/c^2) dt
My understanding is that t represents the time (clock count) of a
clock at rest and T the (time) clock count of a moving clock.
Not quite. t is the time coordinate of the inertial frame,
This raises a new question: What is the difference my 't represents the
time of a clock in a reference frame' versus 'the time coordinate of
the inertial frame'
t is the proper time of the resting twin, which is of course identical
with the inertial
restframe's coordinate time.

So it is clear that the time coordinate
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptual
different but numerically equal.
My philosophy is to use as many concepts as possible
My philosophy is to use as simple as possible concepts, and in
relativity the most simple
concept is to describe a physical situation in terms of
Lorentz-invariant quantities. Here
we compare two Lorentz invariant proper times: the proper time of the
twin at rest and the
proper time of the traveling twin the round-trip of the traveling twin
takes.

This resolves all paradoxes with different frames of reference, because
the results do not depend
on any frame of reference, and it's allowed to use any frame of
reference to evaluate the quantities.
For that I choose the most convenient reference frame.

A clock can only indicate its proper time, and only
along its worldline; the time coordinate of an inertial
frame can indicate the frame's coordinate time anywhere
the frame is valid. Of course, such coordinates are
models, and to implement them requires physical clocks
located where events of interest occur; all such clocks
must, of course, be synchronized to the coordinate time.
That means IMO that all the clocks should be synchronized with
one standard clock. IMO at the origin O
Of course the clocks have to be synchronized at one instant, i.e., in
this case on the
beginning of the trip, where both twins are at the location 0. That's
another important
concept: use as long as you can simple local events to describe a
situation in relativity.

The question is how is v (the speed of the clock) measured?
In the usual way: v=dx/dt, where x(t) is the position of the object at
time t, and t is the time coordinate of the inertial frame.
In practice, this means that t is the time of the nearest physical clock.
There is not one "nearest physical clock". As this example shows, what a
clock once
synchronized with another clock, when both clocks have been at the same
place, shows after
some time when you compare these clocks (again at a common place!)
depends on
the motion of these clocks.

In SR it's pretty easy to synchronize a set of clocks in one inertial
frame, because you can
use light signals and the univerality of the speed of light. That's how Einstein's clock synchronization
works for an inertial frame.

In accerated reference frames or even relativity, clock synchronization
and the definition of "distance"
becomes a complicated issue, but that's not related to the

--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/

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• From Nicolaas Vroom@21:1/5 to Hendrik van Hees on Tue Jun 2 23:18:01 2020
On Saturday, 30 May 2020 16:38:18 UTC+2, Hendrik van Hees wrote:
On 30/05/2020 16:17, Nicolaas Vroom wrote:
Apparently you agree with issue 1.
I would prefer point O, as indicated in the text.
Ad 1) It is wise to formulate a problem in a manifestly covariant
way and then use the most convenient reference frame to do the
calculations.

Let us first study a certain experiment.
Consider I'm located at a spherical-shaped object A (with a clock).
The world around me seems empty with one exception: there is a second
observer B located at a spherical-shaped iron object B (with a clock).
All of a sudden there is a series of incidents:
I observe B's object. I observe that object B approaches me. There
is a collision. Object B moves away in a different direction.
However, there is something more: As part of this collision there
is also a light flash which propagates in a sphere.
I expect that this example can be presented like a Feynmann diagram.
There are two straight lines (starting from the bottom) like a ^
They meet each other at the centre.
Thereafter both lines continue (like a V) towards the top.

IMO the most convenient reference frame is the meeting point.
This point represents the origin of the light flash and
is the most obvious point to consider at rest.
It is also the most obvious point to place your reference clock.
The problem is when you do that neither A nor B can now
be considered at rest.

The obviously Lorentz-invariant quantity to be calculated here is
the proper time tau of the moving twin as a function of the
proper time of the twin staying in his inertial frame at rest, which
is the coordinate time, t, of this inertial frame.

In the example, the clock of A before and after the collision
will have a different speed. The same with the clock of B.
(All (?) clocks are considered moving clocks)

Ad 2) In this case, of course, the inertial frame (the restframe of
the twin staying at home all the time).

t is the proper time of the resting twin, which is, of course, identical
with the inertial restframe's coordinate time.

So it is clear that the time coordinate
of the inertial frame is the same as the elapsed proper time of a clock
at rest in it (reset the clock to zero at t=0). They are conceptual
different but numerically equal.
My philosophy is to use as many concepts as possible

This sentence is wrong.

My philosophy is to use concepts (the least the better) and assumptions.
I consider two types of concepts: basic concepts and derived concepts.
Basic concepts serve like stand-alone concepts.
Derived concepts use basic concepts.
Assumptions are used to describe the universe (experiments) simpler
as the reality. But not too simple.

My philosophy is to use as simple as possible concepts, and in
relativity the most simple concept is to describe a physical situation
in terms of Lorentz-invariant quantities.
Here we compare two Lorentz invariant proper times:
the proper time of the twin at rest and the proper time of the travelling twin the round-trip of the travelling twin takes.

My philosophy is always to use (to start from) one reference frame.

This resolves all paradoxes with different frames of reference, because
the results do not depend on any frame of reference, and it's allowed
to use any frame of reference to evaluate the quantities.
For that I choose the most convenient reference frame.

My philosophy is to start from experiments. Experiments can never
give rise to any paradox. Experiments can lead to unexpected new results
or be used to improve old experiments.

That means IMO that all the clocks should be synchronized with
one standard clock. IMO at the origin O
Of course, the clocks have to be synchronized at one instant, i.e., in
this case on the beginning of the trip, where both twins are at the
location 0.

That is correct.

That's another important concept: use as long as you can simple local
events to describe a situation in relativity.

Hmmm. See below.
Strange as it sounds I also would not use the word relativity in this sentence. You can still do that but when you do that you must explain the difference
in a (physical) situation with (relativity) and without relativity.

For gravity this is simple: In all physical experiments always
gravity is involved. At the same time, you can also make the assumption
that in a certain experiment or theory all gravitational influences
are excluded.

In the usual way: v=dx/dt, where x(t) is the position of the object at
time t, and t is the time coordinate of the inertial frame.
In practice, this means that t is the time of the nearest physical clock.

There is not one "nearest physical clock". As this example shows, what
a clock once synchronized with another clock, when both clocks have been
at the same place, shows after some time when you compare these clocks
(again at a common-place) depends on the motion of these clocks.

That is correct. But in order to know what you call "the time coordinate of
the inertial frame" you need more clocks.
See next comment.

In SR it's pretty easy to synchronize a set of clocks in one inertial
frame because you can use light signals and the universality of the
speed of light. That's how Einstein's clock synchronization
works for an inertial frame.

I prefer a simpler system.
1) Use one reference point (RP) which emits two light flashes in both directions towards two clocks. The RP and the two clocks are all in one straight line. The RP is at an equal distance between the two clocks
2) Replace each clock by reflecting mirrors to synchronize 4 clocks.
3) Replace each clock by reflecting mirrors to synchronize 8 clocks.
That means all the 8 clocks are synchronised from the same source
in one common frame and show the time t. (See above)
In fact, these 8 clocks indicate coordinate time.

The eight clocks are marked as #1 towards the left and #8 towards the right. Next place an extra clock #9 near #1 and synchronise #9 with #1.
Next move #9 towards the right.
When you do that and when #9 meets #2 you can calculate the speed v
and the proper time T of clock #9 and evaluate if this agrees with
the time observed with #9.

In accelerated reference frames or even relativity, clock synchronization
and the definition of "distance"
becomes a complicated issue, but that's not related to the special-relativistic twin paradox at all.

You can perform this whole experiment going from #1 to #8 both going
towards the right and back towards the left, towards #1.
The speed v does not have to be constant.
The more the speed varies the more clocks in the reference frame are
required.

It is assumed that all clocks used, use light signals to function.
However, clock #9 can also be an atomic clock.
When that is the case after #9 is synchronised with #1, an extra
step is involved. This means to observe that both clocks run synchronize
before #9 is moved towards the right.
When that is the case the same can be done as explained as above.

What my whole point is that it is not necessary to use concepts
like covariant, Lorentz invariant and inertial frame.
IMO the most important assumption is that the speed of light
in the selected reference frame is the same in both directions.
If that is the case it is possible to synchronise all clocks
involved and calculate the clock readings of moving clocks
and compare them with the physical reality ie of what is
observed.

An even more important issue is the relation between moving
clocks versus moving objects in the solar system.
IMO very little.
When studying the solar system IMO it makes sense to use
only one reference frame and assume that that reference frame
is at rest and the speed of light is the same in all directions.

The advantage is also that (in principle) you can place
as many clocks as you like inside this reference frame
and synchronise all these clocks from one point.
When you do that the positions of all the objects can be
monitored simultaneously. etc etc.

Nicolaas Vroom

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• From Douglas Eagleson@21:1/5 to All on Tue Jun 2 23:17:53 2020
I found this reference by Rindler. Page 34