• #### Synchronization

From Luigi Fortunati@21:1/5 to All on Sat Aug 12 07:58:03 2023
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1
corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Richard Livingston@21:1/5 to Luigi Fortunati on Sun Aug 13 07:56:51 2023
On Saturday, August 12, 2023 at 2:58:08 AM UTC-5, Luigi Fortunati wrote:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?

No, not as viewed in telescopes (i.e. via light from each earth) nor as calculated by K' as the time "now".

Rich L.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Sylvia Else@21:1/5 to Luigi Fortunati on Sun Aug 13 07:56:51 2023
On 12-Aug-23 5:58 pm, Luigi Fortunati wrote:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

To be clear, for this to mean anything, it has to be a statement in the
frame of reference of the two Earths. It will not be the case that when
the travelling twin reaches Q, they will determine that both Earth
clocks show 24 hours, since, in answer to your later question, the Earth
clocks are not synchronized in the travelling twin's frame.

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?

Sylvia.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Sun Aug 13 15:43:08 2023
Il giorno domenica 13 agosto 2023 alle 09:56:56 UTC+2 Richard Livingston ha scritto:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each other in the reference K' of the spaceship?
No, not as viewed in telescopes (i.e. via light from each earth) nor as calculated by K' as the time "now".

My animation really serves to make the time of distant objects current.

Indeed, the traveling twin of my animation does not need to watch from afar how Earth 1 and Earth 2 move.

He just needs to look at the belt that runs in front of his eyes to know how much the two Earths have rotated.

The advancement of the belt measures the time of the rotation of *both* Earths and not only one: how could the belt move regularly (as indeed it does) if the two Earths rotated at different times?

Luigi.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Mon Aug 14 06:55:56 2023
Il giorno domenica 13 agosto 2023 alle 09:56:55 UTC+2 Sylvia Else ha scritto:
On 12-Aug-23 5:58 pm, Luigi Fortunati wrote:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).
To be clear, for this to mean anything, it has to be a statement in the
frame of reference of the two Earths.

Certain! My animation is (exactly!) in the K frame of reference of the two Earths.

It will not be the case that when
the travelling twin reaches Q, they will determine that both Earth
clocks show 24 hours, since, in answer to your later question, the Earth clocks are not synchronized in the travelling twin's frame.

If the Earth's clocks are not synchronized, how can the belt move regularly (as indeed it does) if the two Earths rotate at different times?

Luigi.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Tue Aug 15 07:14:29 2023
Il giorno luned=C3=AC 14 agosto 2023 alle 08:56:01 UTC+2 Luigi Fortunati ha=
scritto:
In my animation=20
https://www.geogebra.org/m/ez4jk4qm=20
...

In terrestrial reference frame, the belt moves at the speed of rotation of = the two Earths: if the Earths rotate fast, the belt moves fast, if they rot= ate slowly, the belt also moves slowly.

In the spaceship reference frame, if Earth-1 rotates at a different angular=
velocity than Earth-2, does the belt adjusts its speed at Earth-1's rotati=
on or Earth-2's rotation?

Luigi.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Richard Livingston@21:1/5 to Luigi Fortunati on Tue Aug 15 16:55:14 2023
On Sunday, August 13, 2023 at 10:43:13=E2=80=AFAM UTC-5, Luigi Fortunati wrote:
Il giorno domenica 13 agosto 2023 alle 09:56:56 UTC+2 Richard Livingston ha scritto:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q when the time of the two Earths is exactly equal to 24 hours (one complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each other in the reference K' of the spaceship?
No, not as viewed in telescopes (i.e. via light from each earth) nor as calculated by K' as the time "now".
My animation really serves to make the time of distant objects current.

Indeed, the traveling twin of my animation does not need to watch from afar how Earth 1 and Earth 2 move.

He just needs to look at the belt that runs in front of his eyes to know how much the two Earths have rotated.

The advancement of the belt measures the time of the rotation of *both* Earths and not only one: how could the belt move regularly (as indeed it does) if the two Earths rotated at different times?

Luigi.

In light of special relativity, starting up that belt is problematic.
First of all,
simultaneous is not a well defined condition for starting because different observers can have different ideas about when is the same time on both
earths. Second, starting such a long belt will take a long time due to
finite stiffness of the belt. When one earth starts rotating (or engages
with the belt) the motion of the belt will propagate at the speed of sound
in the belt towards the other earth.

Now, in principle, you could start up the belt in such a way that eventually the entire belt is moving and at constant tension and the two sides of the
belt are synchronized as you are assuming. But that does not establish
a unique synchronization between the two earths. Moving observers,
including the distant observer looking at the belts, will still see different synchronizations between the two earths depending on the state of motion
of the observer. You can't get around special relativity this way.

Rich L.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Sylvia Else@21:1/5 to Luigi Fortunati on Tue Aug 15 16:54:04 2023
On 14-Aug-23 1:43 am, Luigi Fortunati wrote:
Il giorno domenica 13 agosto 2023 alle 09:56:56 UTC+2 Richard Livingston ha scritto:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1
corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?
No, not as viewed in telescopes (i.e. via light from each earth) nor as
calculated by K' as the time "now".

My animation really serves to make the time of distant objects current.

Indeed, the traveling twin of my animation does not need to watch from afar how Earth 1 and Earth 2 move.

He just needs to look at the belt that runs in front of his eyes to know how much the two Earths have rotated.

The advancement of the belt measures the time of the rotation of *both* Earths and not only one: how could the belt move regularly (as indeed it does) if the two Earths rotated at different times?

Luigi.

People often think of special relativity as being a complete description
of events. Popular science descriptions do nothing to dispel this
illusion. But it is not. It is part of a model that includes other
aspects of physics, such as the propagation of light. The complete model describes the results of measurements, including things that are seen by
way of light that has been emitted by objects at a distance and later
received by a person's eye.

Attempting to construe special relativity alone as describing events
leads to the kind of confusion illustrated by your example. The
travelling twin looks at the belt and infers a remote time on Earth from
it. But that remote time has no physical meaning, and cannot be measured directly.

Instead, the travelling twin can look at the remote Earth's clock
through a telescope. If the twin observes the belt and infers a time
"now" for Earth, and then waits a time for light to travel the distance
from Earth, the twin would expect to see the previously identified "now"
time through the telescope. But that's not what they would see, because
they've used the wrong model to make their prediction.

Sylvia.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Tue Aug 15 16:55:34 2023
Op 12/08/2023 om 9:58 schreef Luigi Fortunati:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?

No of course. To both "earths" the traveler will be at the same location
X at a given, common (synchro), moment T.

But at that event X,T, traveler's space axis will point toward a past
event of Earth1, and toward a future event of Earth2. So no, they're not synchronised in a moving system like traveler's.

Their time is running at the same pace though in traveler's system, as co-members of the same inertial system, obviously.

--
guido wugi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Fri Aug 18 07:44:31 2023
Il giorno mercoledì 16 agosto 2023 alle 01:54:10 UTC+2 Sylvia Else ha scritto: >> In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.
...
The travelling twin looks at the belt and infers a remote time on Earth from it. But that remote time has no physical meaning, and cannot be measured directly.

Instead it is a direct measurement because the length of the belt that
passes in front of the traveling twin's eyes is *directly* connected to
the rotation of the clock hand highlighted in red in my animation.

Each section of the belt corresponds to a precise rotation of the red hand, no more and no less.

Instead, the travelling twin can look at the remote Earth's clock
through a telescope...

Why should the traveling twin prefer the complications relating to the observation of a distant watch (with all the related problems) when he
can know the terrestrial time by directly observing the belt passing in
front of his eyes?

Luigi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Sylvia Else@21:1/5 to Luigi Fortunati on Sat Aug 19 06:32:55 2023
On 18/08/2023 5:44 pm, Luigi Fortunati wrote:
Il giorno mercoledì 16 agosto 2023 alle 01:54:10 UTC+2 Sylvia Else ha scritto:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.
...
The travelling twin looks at the belt and infers a remote time on Earth from >> it. But that remote time has no physical meaning, and cannot be measured
directly.

Instead it is a direct measurement because the length of the belt that
passes in front of the traveling twin's eyes is *directly* connected to
the rotation of the clock hand highlighted in red in my animation.

Each section of the belt corresponds to a precise rotation of the red hand, no more and no less.

Instead, the travelling twin can look at the remote Earth's clock
through a telescope...

Why should the traveling twin prefer the complications relating to the observation of a distant watch (with all the related problems) when he
can know the terrestrial time by directly observing the belt passing in
front of his eyes?

Luigi

Determining the time on Earth based on observations of the belt involves
an inference. Such an inference unavoidably requires a model of space
time. The model you're trying to use says that the time on Earth is just
the amount of belt that has reached the observer divided by the belt's velocity. It's a perfectly good model - it just happens not to be the
correct one for our universe.

Sylvia

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Sat Aug 19 22:50:22 2023
Il giorno mercoled=C3=AC 16 agosto 2023 alle 01:55:39 UTC+2 wugi ha scritto:
Op 12/08/2023 om 9:58 schreef Luigi Fortunati:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that moves like a transmission belt, so that each revolution of Earth 1 corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each other in the reference K' of the spaceship?
No of course. To both "earths" the traveler will be at the same location
X at a given, common (synchro), moment T.

But at that event X,T, traveler's space axis will point toward a past
event of Earth1, and toward a future event of Earth2. So no, they're not synchronised in a moving system like traveler's.

Their time is running at the same pace though in traveler's system, as co-members of the same inertial system, obviously.

Velocity v=0.866c, gamma=2.

Spaceship reference frame: t=Spaceship time, t1=Earth-1 time,
t2=Earth-2 time.

Start: t=0, t1=0, t2=0.

After a spaceship time 2 is t=2: what are the values of t1 and t2
in the spaceship reference frame?

Luigi.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Sun Aug 20 09:59:30 2023
Il giorno mercoledì 16 agosto 2023 alle 01:55:39 UTC+2 wugi ha scritto:
Op 12/08/2023 om 9:58 schreef Luigi Fortunati:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1
corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?
No of course. To both "earths" the traveler will be at the same location
X at a given, common (synchro), moment T.

But at that event X,T, traveler's space axis will point toward a past
event of Earth1, and toward a future event of Earth2. So no, they're not synchronised in a moving system like traveler's.

Their time is running at the same pace though in traveler's system, as co-members of the same inertial system, obviously.

Spaceship reference frame: t=Spaceship time, t1=Earth-1 time, t2=Earth-2 time.

Start: t=0, t1=0, t2=0.

After a spaceship time 2 is t=2: what are the values of t1 and t2 in the spaceship reference frame?

Luigi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Sun Aug 20 11:37:35 2023
Il giorno mercoled=C3=AC 16 agosto 2023 alle 01:55:20 UTC+2 Richard Livingston ha scritto:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.
...
In light of special relativity, starting up that belt is problematic.
First of all,
simultaneous is not a well defined condition for starting because different observers can have different ideas about when is the same time on both earths. Second, starting such a long belt will take a long time due to
finite stiffness of the belt. When one earth starts rotating (or engages
with the belt) the motion of the belt will propagate at the speed of sound
in the belt towards the other earth.

Now, in principle, you could start up the belt in such a way that eventually the entire belt is moving and at constant tension and the two sides of the belt are synchronized as you are assuming.

That's exactly the idea.

But that does not establish
a unique synchronization between the two earths. Moving observers,
including the distant observer looking at the belts, will still see different synchronizations between the two earths depending on the state of motion
of the observer. You can't get around special relativity this way.

The connection between the traveler twin and the 2 Terre can take place
in two different ways: either through light (with all the complications
of the case) or through the belt (with all the simplicity of the case).

The second case is clearly more reliable than the first.

Luigi

[[Mod. note --
As Sylvia Else, Richard Livingston (who wrote the passage you quoted
that begins "But that does not establish a unique synchronization"),
and Guido Wugi have pointed out in this thread, using a belt doesn't
provide a simple or unique synchronization. You have *asserted* that
the synchronization is unique, but you haven't *proven* it.

What does it mean for a synchronization convention to be "reliable"?
I truly have no idea what the criteria are for "reliability".
-- jt]]

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Mon Aug 21 07:11:23 2023
Op 20/08/2023 om 7:50 schreef Luigi Fortunati:
Il giorno mercoled=C3=AC 16 agosto 2023 alle 01:55:39 UTC+2 wugi ha scritto:
Op 12/08/2023 om 9:58 schreef Luigi Fortunati:
In my animation
https://www.geogebra.org/m/ez4jk4qm
there is Earth 1 (ours) and there is Earth 2 at some fixed distance.

The 2 Earths (which rotate at the same speed) are joined by a belt that
moves like a transmission belt, so that each revolution of Earth 1
corresponds to a revolution of Earth 2 and an advancement of points P
and Q equal to a circumference of the earth .

The traveling twin V starts with a certain speed v and reaches point Q
when the time of the two Earths is exactly equal to 24 hours (one
complete rotation around its own axis).

These are the times in the terrestrial reference K, where the clocks of
the 2 Terre are always synchronized with each other.

Question: do the clocks of the two Earths remain synchronized with each
other in the reference K' of the spaceship?
No of course. To both "earths" the traveler will be at the same location
X at a given, common (synchro), moment T.

But at that event X,T, traveler's space axis will point toward a past
event of Earth1, and toward a future event of Earth2. So no, they're not
synchronised in a moving system like traveler's.

Their time is running at the same pace though in traveler's system, as
co-members of the same inertial system, obviously.

Velocity v=0.866c, gamma=2.
Spaceship reference frame: t=Spaceship time, t1=Earth-1 time,
t2=Earth-2 time.

Start: t=0, t1=0, t2=0.

After a spaceship time 2 is t=2: what are the values of t1 and t2
in the spaceship reference frame?

Homework?

Here you can toy with parameters: https://www.desmos.com/calculator/kxi1hft38c?lang=nl

T = the two terras, Earths
V = voyager position at terra events T
T' = terra events at voyager's V event

b = velocity, g = gamma value, L = scale param;
s = Voyager position

Redundant data:
B = synchro belt control events (belt linking two rotating terras)
a = Earth diameter (has to be small to correspond with time unit 1)...
d = belt velocity (following the terra rotations)

--
guido wugi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Mon Aug 21 13:43:11 2023
wugi il 21/08/2023 09:11:23 ha scritto:
...
Here you can toy with parameters: https://www.desmos.com/calculator/kxi1hft38c?lang=nl

T = the two terras, Earths
V = voyager position at terra events T
T' = terra events at voyager's V event

b = velocity, g = gamma value, L = scale param;
s = Voyager position

Redundant data:
B = synchro belt control events (belt linking two rotating terras)
a = Earth diameter (has to be small to correspond with time unit 1)...
d = belt velocity (following the terra rotations)

In your animation where there is nothing spinning:
- are the two red circles marked by T' the 2 Earths in the reference
frame of the spaceship?

- have these small circles (in the reference frame of the spaceship)
carried out the same number of rotations since the moment of departure
or has one rotated more than the other?

Luigi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Wed Aug 23 07:40:19 2023
Op 21/08/2023 om 22:43 schreef Luigi Fortunati:
wugi il 21/08/2023 09:11:23 ha scritto:
...
Here you can toy with parameters:
https://www.desmos.com/calculator/kxi1hft38c?lang=nl

T = the two terras, Earths
V = voyager position at terra events T
T' = terra events at voyager's V event

b = velocity, g = gamma value, L = scale param;
s = Voyager position

Redundant data:
B = synchro belt control events (belt linking two rotating terras)
a = Earth diameter (has to be small to correspond with time unit 1)...
d = belt velocity (following the terra rotations)

In your animation where there is nothing spinning:
- are the two red circles marked by T' the 2 Earths in the reference
frame of the spaceship?

- have these small circles (in the reference frame of the spaceship)
carried out the same number of rotations since the moment of departure
or has one rotated more than the other?

Sorry, in my answer I forgot to put the link to the new desmos version: https://www.desmos.com/calculator/wreelz925x?lang=nl
(I don't understand how and why Desmos saves updated but un-renamed
versions everytime under a new, different link):

--
guido wugi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Luigi Fortunati@21:1/5 to All on Thu Aug 24 22:41:21 2023
Luigi Fortunati il 20/08/2023 06:37:35 ha scritto:
In my animation
https://www.geogebra.org/m/ez4jk4qm
...
[[Mod. note --
As Sylvia Else, Richard Livingston (who wrote the passage you quoted
that begins "But that does not establish a unique synchronization"),
and Guido Wugi have pointed out in this thread, using a belt doesn't
provide a simple or unique synchronization. You have *asserted* that
the synchronization is unique, but you haven't *proven* it.

What does it mean for a synchronization convention to be "reliable"?
I truly have no idea what the criteria are for "reliability".
-- jt]]

The graduated belt of my animation is directly *bound* to the Earth and
its rotation.

This makes its measurement (direct and immediate) more reliable than
the measurement at distance with light.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Thu Aug 24 22:41:43 2023
Op 21/08/2023 om 22:43 schreef Luigi Fortunati:
wugi il 21/08/2023 09:11:23 ha scritto:
...
Here you can toy with parameters:
https://www.desmos.com/calculator/kxi1hft38c?lang=nl

T = the two terras, Earths
V = voyager position at terra events T
T' = terra events at voyager's V event

b = velocity, g = gamma value, L = scale param;
s = Voyager position

Redundant data:
B = synchro belt control events (belt linking two rotating terras)
a = Earth diameter (has to be small to correspond with time unit 1)...
d = belt velocity (following the terra rotations)

In your animation where there is nothing spinning:
- are the two red circles marked by T' the 2 Earths in the reference
frame of the spaceship?

Yes.

- have these small circles (in the reference frame of the spaceship)
carried out the same number of rotations since the moment of departure
or has one rotated more than the other?

The time axis shows their different times in the Terra system, so they
do 'carry' a different number of rotations. That's precisely why Voyager
has a 'relativistic' POV!
The time unit cT=1 is defined as a 'day', corresponding to one Terra
rotation. So you can count the time for the various points/events.
I've made a few corrections for some point constraints, and clarified
the graph a bit more.

The animation omits the belt time during a half Earth turn, so the belt
is supposed running around a small axis, but at equator rotation speed.
The Earths size is shown though at y=-2.

Another caveat: the belt is also running at 'relativistic' velocities,
so changing its velocity d would impact the relativistic length of the
belt, yet this one is for convenience always equalled to twice the
Earths' proper distance.

--
guido wugi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Richard Livingston@21:1/5 to Luigi Fortunati on Sat Aug 26 11:39:13 2023
On Friday, August 25, 2023 at 12:41:27=E2=80=AFAM UTC-5, Luigi Fortunati wrote: ...
The graduated belt of my animation is directly *bound* to the Earth and
its rotation.

This makes its measurement (direct and immediate) more reliable than
the measurement at distance with light.

I'll try one last time to make this point. The time "now" for events outside the light cone, e.g. events that are spatially separated, are not uniquely defined for different observers.

I must ask what meaning you attach to a time "now" back on earth? It
is not the time you can see, via light, from earth. In fact there is nothing you can do to interact with earth "now" in any way. The only way to
know what is happening "now" is to wait for that event to enter your
past light cone, at which time you can see it. At that later time you can
know what happened back when you identified a time as "now".

If you want to define some time as "now" by some mechanism like you
describe, that is OK. But the conventional way to define "now" is to synchronize clocks as described in most any book on special relativity.
That process allows you to define a "now" that conforms with what
we all normally think of as "now". It gives a different result for different inertial observers, but it gives a result that each observer can make
sense of as "now". That is, if they do some experiment such as send
a time message via light, the result is consistent with special relativity.

There is another factor that you are not taking into account: The
Lorentz contraction of each side of the belt will be different for
moving observers. This may seem as it should be negligible for
such a slow moving belt, but note that magnetic fields are the result
of just such a difference in Lorentz contraction even though the
electrons in a wire are moving at less than 1 mm/second. In the
case of your belt model, a rapidly moving observer will see that
one side of the belt has shrunk compared to the other, and thus
the earth is rotated relative to what the "stationary" observer
calculates.

Rich L.

[[Mod. note -- That's a very clear (and completely correct) exposition.
Thank you!
-- jt]]

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Sun Aug 27 13:18:38 2023
Op 26/08/2023 om 20:39 schreef Richard Livingston:
On Friday, August 25, 2023 at 12:41:27=E2=80=AFAM UTC-5, Luigi Fortunati wrote:
...
The graduated belt of my animation is directly *bound* to the Earth and
its rotation.

This makes its measurement (direct and immediate) more reliable than
the measurement at distance with light.

I'll try one last time to make this point. The time "now" for events outside the light cone, e.g. events that are spatially separated, are not uniquely defined for different observers.

I must ask what meaning you attach to a time "now" back on earth? It
is not the time you can see, via light, from earth. In fact there is nothing you can do to interact with earth "now" in any way. The only way to
know what is happening "now" is to wait for that event to enter your
past light cone, at which time you can see it. At that later time you can know what happened back when you identified a time as "now".

If you want to define some time as "now" by some mechanism like you
describe, that is OK. But the conventional way to define "now" is to synchronize clocks as described in most any book on special relativity.
That process allows you to define a "now" that conforms with what
we all normally think of as "now". It gives a different result for different inertial observers, but it gives a result that each observer can make
sense of as "now". That is, if they do some experiment such as send
a time message via light, the result is consistent with special relativity.

There is another factor that you are not taking into account: The
Lorentz contraction of each side of the belt will be different for
moving observers. This may seem as it should be negligible for
such a slow moving belt, but note that magnetic fields are the result
of just such a difference in Lorentz contraction even though the
electrons in a wire are moving at less than 1 mm/second. In the
case of your belt model, a rapidly moving observer will see that
one side of the belt has shrunk compared to the other, and thus
the earth is rotated relative to what the "stationary" observer
calculates.

Rich L.

[[Mod. note -- That's a very clear (and completely correct) exposition.
Thank you!
-- jt]]

It sounds nice, but it is yet wrong. There is no 'rotation' of the two
Earths relative to a traveller along their axis of symmetry.

What *is* true and correct, is that the belts are observed (Lorentz-wise
as well as 'looking'-wise) to advance at different speeds 'back' and
'fro', by the traveller.

Take a look at the graph I proposed before: https://www.desmos.com/calculator/wreelz925x?lang=nl

For an example, choose b=0.43 (traveller velocity) and d=0.75 (belt
velocity). Choose "Show voyager's POV" (and unchoose "Show Earths'
POV"). Now let the traveller go from Earth1 (left) to Earth2 (right)
with parameter s. Look at the following events:

O. s=0 :
traveller leaves Earth1. Belt check points B (in Earths' POV) leave
their respective Earth, say, B1 and B2. But in traveller's POV, B'1
coincides with B1, but B'2 is already underway, as traveller's x-axis
points to Earths' future.

A. s=0.24 :
Checkpoints B'1 and B'2 cross each other midways (same event for both
traveller and Earths)

B. s=0.364 :
Traveller crosses checkpoint B'2 (heading to the left).

C. s=0.47 :
Checkpoint B'1 reaches Earth2 at the right.
! Immediately a "new" checkpoint B'2bis starts heading to the left! [not
shown]
Checkpoint B'2 is still on its way left.
! In the upper part there is no checkpoint at present!

D. s=0.7 :
Checkpoint B'2 reaches Earth1.
! Immediately, but only just now, a new checkpoint B'1bis starts heading
to the right! [not shown]

E. s=1 :
Voyager reaches Earth2.

From this we can conclude:
The belt running in voyager's direction seems shorter and running at
higher velocity (at times, without checkpoint on it).
The belt running in voyager's opposite direction seems longer and
running at lower velocity (at times, with two checkpoints on it).

--
guido wugi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Wed Aug 30 07:11:48 2023
Op 27/08/2023 om 22:18 schreef wugi:

From this we can conclude:
The belt running in voyager's direction seems shorter and running at
higher velocity (at times, without checkpoint on it).
The belt running in voyager's opposite direction seems longer and
running at lower velocity (at times, with two checkpoints on it).

(Corrections)
The three distances:
between Eart1 and Earth2,
of belt moving 'along' with traveller, and
of belt moving 'opposite to' traveller,
are of course equal! (And just for a reminder, length-contracted in
traveller's POV WRT Earths' POV.)

But whereas the belt 'at rest' would be *symmetrical* in both halves,
the 'running' belt in traveller's POV will display different velocities
'to and fro', and appear
- "elongated" in his own direction of motion, so that "less than half"
is visible, running at a higher speed; and
- "shrunk" in the opposite direction, so that "more than half" is
visible, running at a lower speed.

New simulation:
The Earths and the belt running around and between them, according to Traveller's POV, in this new simulation: https://www.desmos.com/calculator/izfxoxinku?lang=nl

--
guido wugi

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From wugi@21:1/5 to All on Fri Sep 1 07:34:46 2023
Op 30/08/2023 om 9:11 schreef wugi:

New simulation:
The Earths and the belt running around and between them, according to Traveller's POV, in this new simulation: https://www.desmos.com/calculator/izfxoxinku?lang=nl

A last (hopefully:) version, showing traveller's POV in both the Earths'
and traveller's rest systems: https://www.desmos.com/calculator/tddxwyq5mf?lang=nl

And now some homework: do the same for Earths' POV (that is, in both
Earths' and traveller's rest system;)

--
guido wugi

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