The Earth takes 24 hours of the Earth's twin time for one complete
rotation on its axis.

How much time does the traveling twin (v=0.866c, gamma=2) take for the
same rotation?

Does it take 12 hours (24/2) or 48 hours (24*2)?

[[Mod. note -- A couple of comments:
1. I presume that in the 2nd sentence, the author actually meant to ask
how much time the travelling time *measures* for the same rotation.

2. The Earth's rotation period with respect to an inertial reference frame
is actually about 23 hours and 56 minutes. Because the Earth is also
orbiting about the Sun in the same direction as its rotation
(counterclockwise when looking down from above the North pole), the
mean time from noon to noon is slightly longer, namely 24 hours.
The first image in
https://en.wikipedia.org/wiki/Sidereal_time
shows this nicely; in this context "the fixed stars" means an inertial
reference frame.
-- jt]]

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Luigi Fortunati <fortunati.luigi@gmail.com> writes:

The Earth takes 24 hours of the Earth's twin time for one complete
rotation on its axis.
How much time does the traveling twin (v=0.866c, gamma=2) take for the
same rotation?
Does it take 12 hours (24/2) or 48 hours (24*2)?

(I take the rotating Earth to be a clock that flashes every
24 hours.)

When a person travels directly away from or directly towards
the Earth with a speed of v=0.866c, then gamma is indeed 2,
and one rotation of the Earth takes 48 hours for the travelling
person. (However, as long as the travelling person travels with
a constant velocity all the time, it's not the twin paradox.)

Think of the clock as a resting muon that lives 2.197 microseconds
in its own system. An observer on Earth is travelling with
0.866 c towards the muon. Seen from him, the muon lives
4.394 microseconds, which extension of lifetime indeed is
observed on muons from cosmic rays.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Il giorno domenica 16 luglio 2023 alle 09:58:06 UTC+2 Stefan Ram ha scritto:

Luigi Fortunati <fortuna...@gmail.com> writes:

The Earth takes 24 hours of the Earth's twin time for one complete
rotation on its axis.
How much time does the traveling twin (v=0.866c, gamma=2) take for the
same rotation?
Does it take 12 hours (24/2) or 48 hours (24*2)?

(I take the rotating Earth to be a clock that flashes every
24 hours.)

When a person travels directly away from or directly towards
the Earth with a speed of v=0.866c, then gamma is indeed 2,
and one rotation of the Earth takes 48 hours for the travelling
person. (However, as long as the travelling person travels with
a constant velocity all the time, it's not the twin paradox.)

It is not (yet) the paradox of the twins, because the return of the traveling twin is missing.

So let's make him come back at the same speed, so that it takes him another 48 hours to come back, while the Earth makes another rotation.

In doing so, the traveler twin travels for 96 hours (48 to go and 48 to return) and ages 96 hours (4 days).

But on Earth (which rotated only twice) the terrestrial twin aged only 2 days and, therefore, remained younger than his traveler brother!

How is it possible?

--- SoupGate-Win32 v1.05
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Luigi Fortunati <fortunati.luigi@gmail.com> writes:

So let's make him come back at the same speed, so that it
takes him another 48 hours to come back, while the Earth
makes another rotation.
In doing so, the traveler twin travels for 96 hours (48 to go
and 48 to return) and ages 96 hours (4 days).
But on Earth (which rotated only twice) the terrestrial twin
aged only 2 days and, therefore, remained younger than his
traveler brother!
How is it possible?

There is an additional time of aging on Earth that you have not
taken into account.

For the sake of simplicity, I assume that the traveler reverses
his speed instantaneously during the turnaround.

The remaining twin ages during three intervals:

1. During the first phase of the journey, the remaining person
experiences points in time which the traveler moving away from
him perceives as simultaneous to himself (to the traveler).

2. Then, the remaining person experiences points in time that
the traveler never perceives as simultaneous with himself
(with the traveler), neither during the outward journey nor
during the return journey.

3. Finally, during the second and last part of the journey, the
remaining person experiences points in time which the traveler
moving towards him perceives as simultaneous with himself
(with the traveler).

During "1." and "3." the traveling twin ages 96 hours and the
remaining person ages only 48 hours, insofar you were right,
but during "2." the remaining person ages so much more that
at the reunion he is older than the traveler!

Here is a diagram (not true to the angle); it is intended to
be viewed with a monospaced font, but you may get the gist
of it with other fonts as well. (Moderation may omit it if
diagrams of this kind are not allowed in this newsgroup.)

The lines A, B, C, and D, E, F each consist of events which
are simultaneous for the traveling twin. One can see the
large interval "2." with events on Earth for which there
are no simultaneous events for the traveler.

Earth
time ^ t (Earth)
-
-
-
..:. reunion
.- =-,
| - .-.
| =. ,-
| =-. .-,
3.| = ,. ,,
| =. F ,-,
| =-. . ,=.
| - ,. ,=-
| =. .. ,-.
'-..:-. ,. ,-
.- = ,. ,. .-,
| - .. E., ,-
| - ,. . ..-,
| - ,. ,. ,-. inward journey
| = ., .,-,
| = D .. ,:.
| - ,. .=-
| - ,, .-,
| - ., ,-
| ..= .. . -,
| ., .-.
| - .,.-,
| - -=.
| - ..--
| ,=-
2.| = ,== point of reversal
| - .-=.
| - -=.
| - .,.-,
| = .. .-.
| ..= . .. -,
| - ., ,-.
| - ,, .-,
| - .. .--
| = C. ,:,
| ., .,-,
| - ,. ,. ,-.
| - . . . . .-, outward journey
| - .. ., ,-.
'- = ,. ,.B .-,
.-..:,. ,. ,-
| =. .. .-.
| - ,. ,=,
| =-. . ,=.
| =. ,-,
1.| = ,. ,-.
| -,.A .-,
| =, ,-
| - .-.
| =-- Start location
'- ,-..,.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,.,.>
x (Earth)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Monday, July 17, 2023 at 4:38:49 AM UTC-5, Luigi Fortunati wrote:

Il giorno domenica 16 luglio 2023 alle 09:58:06 UTC+2 Stefan Ram ha scritto:

Luigi Fortunati <fortuna...@gmail.com> writes:

The Earth takes 24 hours of the Earth's twin time for one complete
rotation on its axis.
How much time does the traveling twin (v=0.866c, gamma=2) take for the
same rotation?
Does it take 12 hours (24/2) or 48 hours (24*2)?

(I take the rotating Earth to be a clock that flashes every
24 hours.)

When a person travels directly away from or directly towards
the Earth with a speed of v=0.866c, then gamma is indeed 2,
and one rotation of the Earth takes 48 hours for the travelling
person. (However, as long as the travelling person travels with
a constant velocity all the time, it's not the twin paradox.)

It is not (yet) the paradox of the twins, because the return of the traveling twin is missing.

So let's make him come back at the same speed, so that it takes him another 48 hours to come back, while the Earth makes another rotation.

In doing so, the traveler twin travels for 96 hours (48 to go and 48 to return) and ages 96 hours (4 days).

But on Earth (which rotated only twice) the terrestrial twin aged only 2 days and, therefore, remained younger than his traveler brother!

How is it possible?

You are failing to take into account the acceleration when the traveling brother turns around. There is a sudden shift in "now" on earth for the traveling brother during that acceleration.

Rich L.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Luigi Fortunati <fortunati.luigi@gmail.com> writes:

So let's make him come back at the same speed, so that it
takes him another 48 hours to come back, while the Earth
makes another rotation.
In doing so, the traveler twin travels for 96 hours (48 to go
and 48 to return) and ages 96 hours (4 days).

In fact, he ages even more than 96 hours during this time.
But since this aspect is less relevant than the one given
below, I will not delve into this.

But on Earth (which rotated only twice) the terrestrial twin
aged only 2 days and, therefore, remained younger than his
traveler brother!
How is it possible?

There is an additional time of aging on Earth that you have not
taken into account.

For the sake of simplicity, I assume that the traveler reverses
his speed instantaneously during the turnaround.

The remaining twin ages during three intervals:

1. During the first phase of the journey, the remaining person
experiences points in time which the traveler moving away from
him perceives as simultaneous to himself (to the traveler).

2. Then, the remaining person experiences points in time that
the traveler never perceives as simultaneous with himself
(with the traveler), neither during the outward journey nor
during the return journey.

3. Finally, during the second and last part of the journey, the
remaining person experiences points in time which the traveler
moving towards him perceives as simultaneous with himself
(with the traveler).

During "1." and "3." the traveling twin ages a little more than
96 hours and the remaining person ages only 48 hours, insofar
you were right, but during "2." the remaining person ages so
much more that at the reunion he is older than the traveler!

Here is a diagram (not true to the angle); it is intended to
be viewed with a monospaced font, but you may get the gist
of it with other fonts as well. (Moderation may omit it if
diagrams of this kind are not allowed in this newsgroup.)

The lines A, B, C, and D, E, F each consist of events which
are simultaneous for the traveling twin. One can see the
large interval "2." with events on Earth for which there
are no simultaneous events for the traveler.

Earth
time ^ t (Earth)
-
-
-
..:. reunion
.- =-,
| - .-.
| =. ,-
| =-. .-,
3.| = ,. ,,
| =. F ,-,
| =-. . ,=.
| - ,. ,=-
| =. .. ,-.
'-..:-. ,. ,-
.- = ,. ,. .-,
| - .. E., ,-
| - ,. . ..-,
| - ,. ,. ,-. inward journey
| = ., .,-,
| = D .. ,:.
| - ,. .=-
| - ,, .-,
| - ., ,-
| ..= .. . -,
| ., .-.
| - .,.-,
| - -=.
| - ..--
| ,=-
2.| = ,== point of reversal
| - .-=.
| - -=.
| - .,.-,
| = .. .-.
| ..= . .. -,
| - ., ,-.
| - ,, .-,
| - .. .--
| = C. ,:,
| ., .,-,
| - ,. ,. ,-.
| - . . . . .-, outward journey
| - .. ., ,-.
'- = ,. ,.B .-,
.-..:,. ,. ,-
| =. .. .-.
| - ,. ,=,
| =-. . ,=.
| =. ,-,
1.| = ,. ,-.
| -,.A .-,
| =, ,-
| - .-.
| =-- Start location
'- ,-..,.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,.,.>
x (Earth)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

ram@zedat.fu-berlin.de (Stefan Ram) writes:

In fact, he ages even more than 96 hours during this time.
But since this aspect is less relevant than the one given
below, I will not delve into this.

This statement was probably based on a misunderstanding.
I had then written a new post without it, which appeared
as the first in the newsgroup.

--- SoupGate-Win32 v1.05
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Il giorno marted=C3=AC 18 luglio 2023 alle 09:28:25 UTC+2 Stefan Ram ha scritto:

There is an additional time of aging on Earth that you have not
taken into account.

Il giorno marted=C3=AC 18 luglio 2023 alle 09:28:25 UTC+2 Richard Livingston ha scritto:

You are failing to take into account the acceleration when the traveling

brother turns around. There is a sudden shift in "now" on earth for the traveling brother during that acceleration.

The sudden change of time on Earth is not painless.

If in one minute of the spaceship the time on Earth changes by 24 hours,
it means that in that minute of the spaceship the Earth becomes a crazed
top that makes a complete revolution on itself in one minute.

And do you want to know how much Earth time jumps forward when the
spaceship reverses?

I'll do the calculations.

The traveling twin ages 48 hours outward and 48 hours back, so he returns
to Earth aged 96 hours (4 days) and should find the Earth twin aged twice
as much (8 days, 192 hours).

But since during the journey the terrestrial twin ages by only 48 hours
(24 hours during the outward journey and 24 during the return journey),
all the remaining aging (6 days and 6 almost instantaneous rotations!)
should take place during the U-turn.

Is this what you two think is happening?

Is it during the U-turn that (in the reference of the spaceship) the Earth rotates madly (6 almost instantaneous rotations)?

[[Mod. note -- No, during the U-turn the spaceship observer *observes*
the Earth to rotate madly. But at the same time an observer stationary
with respect to the Earth *observes* the Earth rotating at its normal
angular velocity.

This is sort of like taking a video of something, then having two
observers play back the video at differing rates -- one observer
playing back the video at a very high frame/second rate observes the
action happening very fast, while the other observer playing back the
video at a normal frame/second rate observes the action happening
normally.
-- jt]]

--- SoupGate-Win32 v1.05
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Luigi Fortunati <fortunati.luigi@gmail.com> writes:

If in one minute of the spaceship the time on Earth changes by 24 hours,
it means that in that minute of the spaceship the Earth becomes a crazed
top that makes a complete revolution on itself in one minute.

...

But since during the journey the terrestrial twin ages by only 48 hours
(24 hours during the outward journey and 24 during the return journey),
all the remaining aging (6 days and 6 almost instantaneous rotations!)
should take place during the U-turn.
Is this what you two think is happening?

From the traveler's perspective, it looks like this.
But an observer on Earth does not observe anything special.

And the traveler would /not/ observe a jump in the Earth's
rotation during his acceleration if he used a telescope to
observe the Earth, because this simultaneous moment on the
Earth only results after the traveler's calculations, /in which
he takes the effect of signal travel times into account/.

For example, if I see a flash coming from a lamp one light-year away,
I know that the flash there was not produced simultaneously with my
observation, but one year before it, while I will not be able to see
what is happening there simultaneously with me for another year.

If the traveler used a telescope to observe the Earth, he would
find that the Earth rotates slower in the first half of his
journey and faster in the second half (without a jump in the
middle). But because of the time it takes light to travel, he would
/not/ observe moments that are simultaneous for him in his frame.

Therefore, what is seen in the telescope should not be equated
with what is happening simultaneously!

One can understand the whole situation with special relativity
alone. But one can also wonder how the traveler interprets the
passage of time if the acceleration at the reversal point does
not happen instantaneously but takes a finite amount of time.

Albert Einstein addressed this in an article in 1918, "Dialog ueber
Einwaende gegen die Relativitaetstheorie" ("Dialogue on Objections to
the Theory of Relativity"). According to him, for the traveler during
his (constant) acceleration during the reversal, the acceleration
would be equivalent to a homogeneous gravitational field that
fills the entire space.

In this case, the traveler considers himself to be at rest,
since the propulsion of his rocket prevents him from falling
within this gravity field. The Earth and the spacecraft are at
completely different altitudes in this field (and the difference
is the greater the further away the spacecraft is from the Earth).
These different gravitational potentials explain to the traveller
why time passes faster on Earth during his acceleration.

In this sense, the Earth would rotate quite fast for the
traveler during his turn, but he would not see this in a
telescope, and nothing special would be noticed on Earth.

--- SoupGate-Win32 v1.05
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Il giorno mercoledì 19 luglio 2023 alle 09:07:26 UTC+2 Stefan Ram ha scritto:

For example, if I see a flash coming from a lamp one light-year away,
I know that the flash there was not produced simultaneously with my observation, but one year before it...

Of course, everything we see concerns phenomena that occurred in the past, even our Sun we see as it was 8 minutes ago and not as it is now.

But special relativity doesn't concern what is observed, it concerns what happens in the references in motion with respect to the observer.

And what happens in the spaceship frame (where the Earth is in motion) is this: 1) The Earth rotates once on the way out and once on the way back (2 rotations in all).
2) At the end of the journey, the terrestrial twin aged 8 days (twice the traveling twin) and, therefore, the Earth rotated 8 times.
3) The remaining 6 terrestrial rotations that are missing and that did not occur during the outward and return journeys must have occurred (necessarily) during the U-turn.

What I want to point out is that in (1) there is the *dilation* of terrestrial time (which runs slower in the spaceship frame) and in (3) there is the *contraction* of terrestrial time (which runs faster in the spaceship frame).

--- SoupGate-Win32 v1.05
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Luigi Fortunati <fortunati.luigi@gmail.com> writes:

What I want to point out is that in (1) there is the
*dilation* of terrestrial time (which runs slower in the
spaceship frame) and in (3) there is the *contraction* of
terrestrial time (which runs faster in the spaceship frame).

This is true from the point of view of the space traveler if
he considers himself to be at rest. (But which traveler would
seriously consider himself to be at rest?)

However, if we do not look at this from a special coordinate system,
we can say that each twin travels a different world line (travel
route) in spacetime between the start of the spacecraft and its
landing. The length of each of these world lines, measured in the
metric of spacetime, gives the the time that has passed for each
of them. And these lengths are independent of any particular
reference system, so it is actually easier to consider this invariant
point of view than "simultaneities" that depend on reference systems.

Quantitatively, let dS and dT denote the length (elapsed proper
time) of the world line of the traveller and the other twin,
respectively, between the launch of the rocket and its return.
According to the metric, ( c dS )^2 = ( c dt )^2 - dx^2, where
dt is the difference of the times of launch and landing in
coordinate time of the resting twin, and dx is the total length
traveled by the rocket in the frame of the resting twin, so
dS = sqrt( 1 -( v/c )^2 )dt, where v = dx/dt, the speed of the
rocket. (For simplification, I assumed that v^2 is constant and
I did not distinguish between v and -v, as it does not matter
for terms where v is squared. Actually, the calculation should
be split into one summand for the outward and one summand for
the return trip, but the result would be the same.) For the
twin resting on Earth dT is simply dt, i.e., his proper time is
the coordinate time as he is not moving. So, when the twin on
Earth at the time of the reunion has aged by 1 unit since the
launch, dT=1, dt=1, and dS for the twin travelling with v is
sqrt( 1 -( v/c )^2 ), i.e., 1/2 given v=0.866c.

--- SoupGate-Win32 v1.05
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On Monday, 24 July 2023 at 09:06:24 UTC+2, Stefan Ram wrote:

The length of each of these world lines, measured in the
metric of spacetime, gives the the time that has passed for each
of them.

No, it gives the time *it takes* to each of them, whence they do not in
fact get to the "rendez-vous" point at the same time, the travelling
twin getting there "earlier" and meeting *a future (along their worldline, whence older) version of* the twin that stayed, and, conversely, the twin
that stayed getting there "later" and meeting *a past (whence younger)
version of* the twin that travelled.

And I understand that that is not "orthodox", but please let me propose
that it is the correct reading of Einsteinian Relativity, under the notion
of "proper time" as "locally universal" time, together with the "clock postulate", i.e. that all working clocks indeed tick at the same rate (the proper time rate) in their own frame of reference.

Julio

--- SoupGate-Win32 v1.05
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[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]

On Sunday, July 23, 2023 at 4:59:57=E2=80=AFAM UTC-5, Luigi Fortunati wrote:

Il giorno mercoled=C3=AC 19 luglio 2023 alle 09:07:26 UTC+2
Stefan Ram ha scritto:

...

What I want to point out is that in (1) there is the *dilation* of terrestrial time
(which runs slower in the spaceship frame) and in (3) there is the
*contraction
* of terrestrial time (which runs faster in the spaceship frame).

I think one issue that causes confusion is the difference between
what each observer SEES (with their eyes via light) and what SR says
is happening "now" for each observer.
-For the traveling twin, what he SEES:
-During outbound trip using a telescope to observe the earth,
time on earth appears to go very slow, even slower than SR says.
This is because in addition to the time dilation of SR there is
also the Doppler shift due to the twin receding rapidly at nearly
the speed of light.
-During the turn-around, which we typically assume is almost
instantaneous, there is little change visible on earth.
-During the return trip, the traveling twin is encountering all
the light that has been leaving earth during the outbound trip.
As a result the traveling twin sees earth spinning rapidly through
the telescope, again due to the Doppler effect.
-When the traveling twin returns to earth, many years have passed,
just as SR calculates.
-For the traveling twin, what SR calculates for "now":
-During the outbound trip, the calculated "now" on earth advances
slowly due to time dilation of the Lorentz transform. This slow
down in apparent time is not as great as what they see in the
telescope, however.
-During turn-around, the "now" calculated for earth advances
almost instantaneously many years. This is not seen in the
telescope, however. We do not see "now" at a distant object,
we only see that object on our past light cone.
-During the return trip, time on earth is calculated to advance
slowly per the Lorentz time dilation. The telescope, however
shows earth time advancing rapidly.
-For the people on earth:
-During the outbound trip, the traveling twin does not appear
to age at all when viewed through the telescope. The calculated
"now" however has the twin aging slower than "normal" but faster
than shown through the telescope.
-During turnaround, the traveling twin does not change at all.
-During the return trip, the traveling twin appears to age
rapidly due to the Doppler effect on the light from the twin.
This aging rate is faster than calculated for the "now" time
of the twin.
-When the traveling twin arrives back on earth, they have aged
as SR predicts and is younger than the earth twin.

Rich L.

--- SoupGate-Win32 v1.05
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Il giorno luned=C3=AC 24 luglio 2023 alle 09:06:24 UTC+2 Stefan Ram ha scri= tto:

What I want to point out is that in (1) there is the=20
*dilation* of terrestrial time (which runs slower in the=20
spaceship frame) and in (3) there is the *contraction* of=20
terrestrial time (which runs faster in the spaceship frame).

This is true from the point of view of the space traveler if=20
he considers himself to be at rest. (But which traveler would=20
seriously consider himself to be at rest?)=20

I have an answer and it seems extraordinarily reasonable to me.

The traveling twin can (rightly) be considered at rest if his
spaceship travels in space with the engine off and cannot be
considered at rest if the engines are running.

If this is the case (and it seems to me that this is the case),
during the outward and return journeys the traveling twin is at
rest and special relativity is usable.

Instead, during the reversal, the engine is running, the motion is
not inertial, the traveling twin is not at rest and special relativity
cannot be used because it does not apply in accelerated references.

So, if special relativity isn't usable, what theory can we use to
evaluate the behavior of Earth's time in the spaceship frame during
U-turn?

However, if we don't look at this from a special coordinate system...

This goes beyond my question which concerns exclusively the traveler
twin's reference system.

--- SoupGate-Win32 v1.05
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Il giorno luned=C3=AC 24 luglio 2023 alle 18:52:50 UTC+2 Richard Livingston ha scritto:

[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]

On Sunday, July 23, 2023 at 4:59:57=E2=80=AFAM UTC-5, Luigi Fortunati wrote:

Il giorno mercoled=C3=AC 19 luglio 2023 alle 09:07:26 UTC+2
Stefan Ram ha scritto:

...

What I want to point out is that in (1) there is the *dilation* of terrestrial time
(which runs slower in the spaceship frame) and in (3) there is the
*contraction
* of terrestrial time (which runs faster in the spaceship frame).

I think one issue that causes confusion is the difference between
what each observer SEES (with their eyes via light) and what SR says
is happening "now" for each observer.
-For the traveling twin, what he SEES:
-During outbound trip using a telescope to observe the earth,
time on earth appears to go very slow, even slower than SR says.
This is because in addition to the time dilation of SR there is
also the Doppler shift due to the twin receding rapidly at nearly
the speed of light.
-During the turn-around, which we typically assume is almost
instantaneous, there is little change visible on earth.

What are you saying? During the turn-around the spacecraft has to brake
hard and then accelerate just as hard backwards and you say that in the
images of his telescope, at this stage, there is little change visible on earth?

No! There is *big* change visible on earth, because the light rays from
the Earth that previously had a hard time reaching the telescope (and
lagged behind), now come all at it at once.

Luigi.

[[Mod. note -- By definition, the travelling twin looking through their telescope at Earth sees the event where Earth's worldline intersects the travelling twin's backwards light cone. The travelling twin's backwards
light cone depends only on the event at its apex, NOT on the travelling
twin's velocity. Therefore, the intersection of that light cone with
Earth's worldline (i.e., the event that the travelling twin sees when
looking at Earth through their telescope) depends on the travelling
twin's position but NOT on the travelling twin's velocity. Therefore,
the travelling twin sees little visible change when looking at Earth
during their (the travelling twin's) turnaround.
-- jt]]

--- SoupGate-Win32 v1.05
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On 7/24/23 10:22 AM, Julio Di Egidio wrote:

On Monday, 24 July 2023 at 09:06:24 UTC+2, Stefan Ram wrote:

The length of each of these world lines, measured in the metric of
spacetime, gives the the time that has passed for each of them.

No, [...]

Yes. That path length is the elapsed proper time of each twin, because
these are timelike paths through spacetime.

it gives the time *it takes* to each of them,

That's just another way of saying the same thing. The elapsed proper
time along a path is indeed the time it takes along that path.

whence they do not in fact get to the "rendez-vous" point at the same
time,

PUN ALERT -- you did not specify what you mean by "time". This is
apparently the source of your confusion.

Note that to twin 1, twin 2 arrives at their rendezvous simultaneously
with their own arrival. Ditto for 1<=>2. Ditto for any other observer or coordinate system. Because that rendezvous is a single event in spacetime.

the travelling twin getting there "earlier" and meeting *a future
(along their worldline, whence older) version of* the twin that
stayed, and, conversely, the twin that stayed getting there "later"
and meeting *a past (whence younger) version of* the twin that
travelled.

This is just plain wrong. You are attempting to compare values in two
different coordinate systems as if they were comparable. They aren't.
This is directly related to the PUN above -- you are using two different meanings for "time" (one for each twin) and then comparing them. That's invalid.

The twins reunite at a single event in spacetime. That rendezvous is necessarily simultaneous to any observer/coordinates. There is no sense
in which one twin arrives "earlier" or "later" than the other, because
they arrive TOGETHER.

Yes, the elapsed proper times along their paths are different. That is
the timelike version of the difference in spatial path lengths between
(path 1) Chicago to New York and (path 2) Chicago to New Orleans to New
York.

And I understand that that is not "orthodox", but please let me
propose that it is the correct reading of Einsteinian Relativity,
under the notion of "proper time" as "locally universal" time,

I have no idea what you are talking about. Making up new concepts
("locally universal") is not useful, especially in a theory as well
known as SR. But no matter -- the twins reunite at a single event.

Moreover, the twin paradox shows that proper time is not "universal".

together with the "clock postulate", i.e. that all working clocks
indeed tick at the same rate (the proper time rate) in their own
frame of reference.

That is as much a definition as an hypothesis. But it seems unrelated to
your claims of "earlier" and "later".

Tom Roberts

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[[Mod. note -- By definition, the travelling twin looking through their telescope at Earth sees the event where Earth's worldline intersects the travelling twin's backwards light cone. The travelling twin's backwards
light cone depends only on the event at its apex, NOT on the travelling twin's velocity. Therefore, the intersection of that light cone with
Earth's worldline (i.e., the event that the travelling twin sees when
looking at Earth through their telescope) depends on the travelling
twin's position but NOT on the travelling twin's velocity. Therefore,
the travelling twin sees little visible change when looking at Earth
during their (the travelling twin's) turnaround.
-- jt]]

I have reflected intensely on the replies received from all of you and I thank you from the bottom of my heart.

I realized that I had made a mistake: that of having (me too) considered the change of direction to be almost instantaneous.

Apart from the fact that one does not pass from the speed v=0.866c to the speed v=-0.866c in an instant, it must be said that, if it were possible to do so, none of what I said could happen during the reversal of direction because nothing happens in zero
time.

So, I'll have to revise my example by assigning a precise time duration to the reverse.

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Op 25/07/2023 om 11:23 schreef wugi:

Actually a *space zoom* occurs, as someone called it here earlier. For
those who prefer to not believe that, you're welcome to do your own calculations, but properly.
I did mine (and indeed at the time it came as a surprise "even to
me"!;) with these results:

(instantaneous turnback event transitions in magenta) https://wugi.be/animgif/RelaSee_ObsvTrav.gif https://wugi.be/animgif/TwinSee_ObsvTrav.gif https://wugi.be/MySRT/TravtwinSee%20pt-of-vw.PNG

Desmos file:

A humbling experience: my bad! (some wrong time display calculations,
due to translating case TP2 into TP4). Hopefully put it right in the new
link. Sorry for that.

Bad:

https://www.desmos.com/calculator/ctr5flpzjg?lang=nl

Correct:

https://www.desmos.com/calculator/aoacey9t1v?lang=nl

(choose TP4 = what the traveller _sees_)


--

guido wugi

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Op 25/07/2023 om 9:38 schreef Luigi Fortunati:

Il giorno luned=C3=AC 24 luglio 2023 alle 18:52:50 UTC+2 Richard Livingston ha scritto:

[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]

On Sunday, July 23, 2023 at 4:59:57=E2=80=AFAM UTC-5, Luigi Fortunati wrote: >>> Il giorno mercoled=C3=AC 19 luglio 2023 alle 09:07:26 UTC+2

Stefan Ram ha scritto:

...

What I want to point out is that in (1) there is the *dilation* of
terrestrial time
(which runs slower in the spaceship frame) and in (3) there is the
*contraction
* of terrestrial time (which runs faster in the spaceship frame).

I think one issue that causes confusion is the difference between
what each observer SEES (with their eyes via light) and what SR says
is happening "now" for each observer.
-For the traveling twin, what he SEES:
-During outbound trip using a telescope to observe the earth,
time on earth appears to go very slow, even slower than SR says.
This is because in addition to the time dilation of SR there is
also the Doppler shift due to the twin receding rapidly at nearly
the speed of light.
-During the turn-around, which we typically assume is almost
instantaneous, there is little change visible on earth.

What are you saying? During the turn-around the spacecraft has to brake
hard and then accelerate just as hard backwards and you say that in the images of his telescope, at this stage, there is little change visible on earth?

No! There is *big* change visible on earth, because the light rays from
the Earth that previously had a hard time reaching the telescope (and
lagged behind), now come all at it at once.

That's about the difference between the outbound and inbound trip. But
indeed, it must also imply a difference *at the very turn-about event*
itself!
Because of
- symmetry: traveller sees Earth moving away and then back during _equal time-intervals_ in outbound and homebound trip. And
- asymmetry: Earth is seen "lagging behind" in outbound trip, and
"hurrying back" in homebound trip.
This cannot happen over _equal distances_ in equal time-intervals!
See below.

Luigi.

[[Mod. note -- By definition, the travelling twin looking through their telescope at Earth sees the event where Earth's worldline intersects the travelling twin's backwards light cone. The travelling twin's backwards light cone depends only on the event at its apex, NOT on the travelling twin's velocity. Therefore, the intersection of that light cone with
Earth's worldline (i.e., the event that the travelling twin sees when
looking at Earth through their telescope) depends on the travelling
twin's position but NOT on the travelling twin's velocity. Therefore,
the travelling twin sees little visible change when looking at Earth
during their (the travelling twin's) turnaround.
-- jt]]

Except that this is wrong.
In the limit case, an instantaneous turn-back, admittedly the traveller
sees the _same_ *Earth* event in the _same_ split-moment of his
*turnback*... yet in _different_ *inertial systems*, the outbound and
the homebound.
This accounts for a _real_ and, possibly, a big _difference in seeing_
the (same!) event, just before and just after turn-back.
Actually a *space zoom* occurs, as someone called it here earlier. For
those who prefer to not believe that, you're welcome to do your own calculations, but properly.
I did mine (and indeed at the time it came as a surprise "even to me"!;)
with these results:

(instantaneous turnback event transitions in magenta) https://wugi.be/animgif/RelaSee_ObsvTrav.gif https://wugi.be/animgif/TwinSee_ObsvTrav.gif https://wugi.be/MySRT/TravtwinSee%20pt-of-vw.PNG

Desmos file:
https://www.desmos.com/calculator/ctr5flpzjg?lang=nl
(choose TP4, what the traveller _sees_)

Swapping inertial frames is not a trivial feature in SRT. Or inertial
frames themselves: different inertial observers crossing the turn-back
event of our traveller will see the same Earth event, but differently each!

--
guido wugi

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On Tuesday, 25 July 2023 at 09:38:07 UTC+2, Luigi Fortunati wrote:

Il giorno luned=C3=AC 24 luglio 2023 alle 18:52:50 UTC+2 Richard Livingston ha scritto:

-During the turn-around, which we typically assume is almost
instantaneous, there is little change visible on earth.

What are you saying? During the turn-around the spacecraft has to brake
hard and then accelerate just as hard backwards and you say that in the images of his telescope, at this stage, there is little change visible on earth?

No, there is in fact no need for any of that: for the simplest form of twins experiment just take SR in flat spacetime and compute the proper length
of the two path segments, and there already you have it: one of the two
paths is *shorter* (in terms of proper time/distance) than the other, and that's it. And that experiment can even be really simulated, with *two* rockets appropriately coordinated, one for each leg of the journey. So, overall, it is true that acceleration concerns and even more so gravitational effects are just *not germane to the twin paradoxes*.

No! There is *big* change visible on earth,

No, there is no change at all on Earth expect for effects occurring
after any light goes from the travelling twin back to the one on Earth.
OTOH, the change of direction for the travelling twin does effect an instantaneous change (for him), which is rather related to *rotation*
re the (his own) inertial path. There is a nice clip, by Brian Greene
IIRC, "the slices of bread", with the traveller going in a direction and
seeing a planet with its advanced technology, then the traveller inverts direction and now sees the same planet in the middle ages: Greene
shows in pictures how the whole "jump in time" thing happens while
rotating (in space time), not earlier, not later...

You are totally grasping at straws, and necessarily so since you
have systematically disregarded all advice and corrections, all the
more so the methodological ones...

Julio

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Il giorno gioved=C3=AC 27 luglio 2023 alle 22:38:59 UTC+2 wugi ha scritto:

Except that this is wrong.
In the limit case, an instantaneous turn-back, admittedly the traveller
sees the _same_ *Earth* event in the _same_ split-moment of his
*turnback*... yet in _different_ *inertial systems*, the outbound and
the homebound.
This accounts for a _real_ and, possibly, a big _difference in seeing_
the (same!) event, just before and just after turn-back.
Actually a *space zoom* occurs, as someone called it here earlier. For
those who prefer to not believe that, you're welcome to do your own calculations, but properly.
I did mine (and indeed at the time it came as a surprise "even to me"!;)
with these results:

(instantaneous turnback event transitions in magenta) https://wugi.be/animgif/RelaSee_ObsvTrav.gif https://wugi.be/animgif/TwinSee_ObsvTrav.gif https://wugi.be/MySRT/TravtwinSee%20pt-of-vw.PNG

Desmos file:
https://www.desmos.com/calculator/ctr5flpzjg?lang=nl
(choose TP4, what the traveller _sees_)

Swapping inertial frames is not a trivial feature in SRT. Or inertial
frames themselves: different inertial observers crossing the turn-back
event of our traveller will see the same Earth event, but differently each!

Your animations are beautiful to look at but in the tangle of points
and lines it is not clear where the Earth is and where the spaceship
is because the viewer does not see clearly identified bodies that
go straight but generic points that move along inclined lines.

Instead, in my animation
https://www.geogebra.org/classic/cnkh3wpu
I show exactly how the spaceship moves in the Earth frame and how
the Earth moves in the spaceship frame.

It's a *thought* experiment in which the Earth has a red tail that
is 0.866 light-days long and the spaceship has a blue tail of the
same length.

In the Earth reference:
- The Earth stands still and the spaceship moves to the right at
the speed v=0.866c
- the tail of the spaceship (in motion) is compressed by half, i.e.
to 4.33 light-days
- the spaceship travels for 24 hours, after which it reaches point
D in 24-hour Earth time and 12-hour spaceship time

In the spaceship reference:
- The spaceship stands still and the Earth (with its tail) moves
to the left at the speed v=0.866c
- the tail of the Earth (in motion) contracts to half, i.e. to
4.33 light-days
- point D travels until it reaches the spaceship.

Can anyone help me find the correct times of the spaceship and Earth
in the spaceship reference during the journey and at the moment it
meets point D?

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Stefan Ram a =C3=A9crit=C2=A0:

Luigi Fortunati <fortuna...@gmail.com> writes:

What I want to point out is that in (1) there is the
*dilation* of terrestrial time (which runs slower in the
spaceship frame) and in (3) there is the *contraction* of
terrestrial time (which runs faster in the spaceship frame).

This is true from the point of view of the space traveler if
he considers himself to be at rest. (But which traveler would
seriously consider himself to be at rest?)

However, if we do not look at this from a special coordinate system,
we can say that each twin travels a different world line (travel
route) in spacetime between the start of the spacecraft and its
landing. The length of each of these world lines, measured in the
metric of spacetime, gives the the time that has passed for each
of them. And these lengths are independent of any particular
reference system, so it is actually easier to consider this invariant
point of view than "simultaneities" that depend on reference systems.

Quantitatively, let dS and dT denote the lengtStefan Ram a =C3=A9crit=C2=A0: Luigi Fortunati <fortuna...@gmail.com> writes:

What I want to point out is that in (1) there is the
*dilation* of terrestrial time (which runs slower in the
spaceship frame) and in (3) there is the *contraction* of
terrestrial time (which runs faster in the spaceship frame).

This is true from the point of view of the space traveler if
he considers himself to be at rest. (But which traveler would
seriously consider himself to be at rest?)

I read you write this "But which traveler would seriously consider
himself to be at rest?"
Your question is very strange.
The other day on the train I heard the conversation of two academic
professors of relativity. The first saying to the second: "But what is
the station platform waiting for to move??!! The second: "You're right.
The departure time is over by gamma times three minutes" The first -
outbidding - "With my relativist colleagues we refuse to pay the carbon
tax for aircraft travels as long as the terminals which move while we
are at rest refuse to pay it first"

The second "Why only the terminals? All these poors moving in relation
to us should pay too. Their relative movements in relation to us, the intellectual elite, contribute to global warming"

I then stopped listening to them to reread "The Dialectic of Nature"

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