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)?
Luigi Fortunati <fortuna...@gmail.com> writes:
The Earth takes 24 hours of the Earth's twin time for one complete(I take the rotating Earth to be a clock that flashes every
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)?
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.)
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?
Il giorno domenica 16 luglio 2023 alle 09:58:06 UTC+2 Stefan Ram ha scritto:
Luigi Fortunati <fortuna...@gmail.com> writes:It is not (yet) the paradox of the twins, because the return of the traveling twin is missing.
The Earth takes 24 hours of the Earth's twin time for one complete(I take the rotating Earth to be a clock that flashes every
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)?
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.)
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?
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?
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.
There is an additional time of aging on Earth that you have not
taken into account.
You are failing to take into account the acceleration when the travelingbrother turns around. There is a sudden shift in "now" on earth for the traveling brother during that acceleration.
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?
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...
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).
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.
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).
What I want to point out is that in (1) there is the=20This is true from the point of view of the space traveler if=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).
he considers himself to be at rest. (But which traveler would=20
seriously consider himself to be at rest?)=20
However, if we don't look at this from a special coordinate system...
[[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 timeI think one issue that causes confusion is the difference between
(which runs slower in the spaceship frame) and in (3) there is the
*contraction
* of terrestrial time (which runs faster in the spaceship frame).
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.
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.
[[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]]
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
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* ofI think one issue that causes confusion is the difference between
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).
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]]
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 *big* change visible on earth,
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!
Luigi Fortunati <fortuna...@gmail.com> writes:
What I want to point out is that in (1) there is theThis is true from the point of view of the space traveler if
*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).
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 theThis is true from the point of view of the space traveler if
*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).
he considers himself to be at rest. (But which traveler would
seriously consider himself to be at rest?)
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