On a very distant planet, aliens who know nothing about the theory of
relativity manage to send their first rocket into space.
The goal, simply fun for this mission, is, among other things, to place
two large clocks in space.
The clocks are placed very far from their original planet, and are
separated by 900,000 kms.
Suddenly Commander Cromar notices that something bizarre has happened, and
he begins to gently scratch the green scales with bluish reflections,
which cover his skull. From his spaceship placed near the first clock, he
notices that the two clocks do not mark the same time. Immediately, he
begins to think that something must have happened when disembarking the
clocks, and that one of them must have gone out of tune during the
manipulation.
This is what a normal human being would have thought first. Aliens are not
dumber, nor smarter than us (otherwise, it would be known).

R.H.

Richard Hachel <r.hachel@tiscali.fr> wrote:
> On a very distant planet, aliens who know nothing about the theory of
> relativity manage to send their first rocket into space.
> The goal, simply fun for this mission, is, among other things, to place
> two large clocks in space.
> The clocks are placed very far from their original planet, and are
> separated by 900,000 kms.
> Suddenly Commander Cromar notices that something bizarre has happened, and
> he begins to gently scratch the green scales with bluish reflections,
> which cover his skull. From his spaceship placed near the first clock, he
> notices that the two clocks do not mark the same time. Immediately, he
> begins to think that something must have happened when disembarking the
> clocks, and that one of them must have gone out of tune during the
> manipulation.

The commander’s first instinct might be to check whether it is a
propagation delay issue. Fortunately, it is a simple matter to check. He
arranges to send a pulse to the distant clock when the near clock marks
noon on the 11th day of Urg, after recoding the far clock to send a
timestamped signal back the instant it receives the pulse. This is enough
information to separate the propagation delay from the clock readings and
to see if the two clocks are indeed synced. The method should be obvious to
you.

> This is what a normal human being would have thought first. Aliens are not
> dumber, nor smarter than us (otherwise, it would be known).
>
> R.H.
>
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

Le 14/12/2021 à 16:06, Odd Bodkin a écrit :
> Richard Hachel <r.hachel@tiscali.fr> wrote:
>> On a very distant planet, aliens who know nothing about the theory of
>> relativity manage to send their first rocket into space.
>> The goal, simply fun for this mission, is, among other things, to place
>> two large clocks in space.
>> The clocks are placed very far from their original planet, and are
>> separated by 900,000 kms.
>> Suddenly Commander Cromar notices that something bizarre has happened, and
>> he begins to gently scratch the green scales with bluish reflections,
>> which cover his skull. From his spaceship placed near the first clock, he
>> notices that the two clocks do not mark the same time. Immediately, he
>> begins to think that something must have happened when disembarking the
>> clocks, and that one of them must have gone out of tune during the
>> manipulation.
>
> The commander’s first instinct might be to check whether it is a
> propagation delay issue. Fortunately, it is a simple matter to check. He
> arranges to send a pulse to the distant clock when the near clock marks
> noon on the 11th day of Urg, after recoding the far clock to send a
> timestamped signal back the instant it receives the pulse. This is enough
> information to separate the propagation delay from the clock readings and
> to see if the two clocks are indeed synced. The method should be obvious to
> you.
>
>> This is what a normal human being would have thought first. Aliens are not
>> dumber, nor smarter than us (otherwise, it would be known).
>>
>> R.H.
>>
>>
>>

"and several hypotheses are now being imagined by our scientists"

You will notice the deep intelligence of the people of this planet, who
immediately make tests, and seek the advice of the scientists of their
planet.

several hypotheses.... LOL.

Not only one, like us.

R.H.

Richard Hachel <r.hachel@tiscali.fr> wrote:
> Le 14/12/2021 à 16:06, Odd Bodkin a écrit :
>> Richard Hachel <r.hachel@tiscali.fr> wrote:
>>> On a very distant planet, aliens who know nothing about the theory of
>>> relativity manage to send their first rocket into space.
>>> The goal, simply fun for this mission, is, among other things, to place
>>> two large clocks in space.
>>> The clocks are placed very far from their original planet, and are
>>> separated by 900,000 kms.
>>> Suddenly Commander Cromar notices that something bizarre has happened, and
>>> he begins to gently scratch the green scales with bluish reflections,
>>> which cover his skull. From his spaceship placed near the first clock, he
>>> notices that the two clocks do not mark the same time. Immediately, he
>>> begins to think that something must have happened when disembarking the
>>> clocks, and that one of them must have gone out of tune during the
>>> manipulation.
>>
>> The commander’s first instinct might be to check whether it is a
>> propagation delay issue. Fortunately, it is a simple matter to check. He
>> arranges to send a pulse to the distant clock when the near clock marks
>> noon on the 11th day of Urg, after recoding the far clock to send a
>> timestamped signal back the instant it receives the pulse. This is enough
>> information to separate the propagation delay from the clock readings and
>> to see if the two clocks are indeed synced. The method should be obvious to
>> you.
>>
>>> This is what a normal human being would have thought first. Aliens are not
>>> dumber, nor smarter than us (otherwise, it would be known).
>>>
>>> R.H.
>>>
>>>
>>>
>
> "and several hypotheses are now being imagined by our scientists"
>
> You will notice the deep intelligence of the people of this planet, who
> immediately make tests, and seek the advice of the scientists of their
> planet.
>
> several hypotheses.... LOL.

Prioritized. Do one first.

Do you understand the method I described and how it tells you the answer?

>
> Not only one, like us.
>
> R.H.
>
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Tuesday, December 14, 2021 at 9:43:49 AM UTC-5, Richard Hachel wrote:
<off topics snipped>

You once again started a new thread without finishing a previous one. that's bad netiquette and very impolite.
So I ask you again,

" ...Did you not notice that it was the troll? "

Den 14.12.2021 15:43, skrev Richard Hachel:
> On a very distant planet, aliens who know nothing about the theory of
> relativity manage to send their first rocket into space.
> The goal, simply fun for this mission, is, among other things, to place
> two large clocks in space.

A good idea.
But instead of inventing several variation of nonsense,
let's see what we can conclude from visually observation
of distant clocks.

-------------------------------

It is an empirical fact that the longitudinal Doppler shift is:
fᵣ = fₜ⋅√((1−v/c)/(1+v/c)) (1)
where fₜ is the transmitted frequency, fᵣ is the received frequency
and v is the speed of the transmitter relative to the receiver.

I will use this empirical fact to calculate how much the twins
will age in the following 'twin scenario':

We will use the units:
time : year [y]
distance: light years [ly]
speed : light years per year [ly/y]
c = 1 ly/y

The "travelling twin" leaves the Earth at the speed v = 0.8 ly/y.
When he is a distance L = 12 ly away from the Earth as
measured in Earth's rest frame, he turns abruptly around
and travels back to the Earth at the speed 0.8 ly/y.

Both the Earth clock and the travelling twin's clock are
set to zero when the travelling twin starts.

Each twin has a transmitter which transmits a video signal
showing his clock, and each twin has a receiver which receives
the video signal and display the other twin's clock on a monitor.

When the twins are receding from each other, they will both
receive a signal Doppler shifted by:
D₁ = √((1−0.8)/(1+0.8)) = 1/3

When the twins are approaching each other, they will both
receive a signal Doppler shifted by:
D₂ = √((1+0.8)/(1−0.8)) = 3

Observations done by the travelling twin:
=========================================
When the travelling twin is on his way out,
he will see the Earth twin's clock on the monitor.
Since this picture is Doppler shifted, the clock on
the monitor must run slow by the factor D₁.
So when his clock shows 9 years, the clock on the monitor
will show 9⋅D₁ y = 3 y.

At this time the travelling twin turns abruptly around and
travels towards the Earth at speed 0.8 ly/y relative to the Earth.
Immediately after the turnaround, the clock on the monitor will
still show 3 years, but now this clock will run fast by
the factor D₂. Symmetry makes it obvious that the travelling twin
will use 9 years to get back to the Earth.
So when his clock shows 18 years, the clock on the monitor,
which now will show the same as the clock on Earth,
will show 3+9⋅D₂ y = 30 y.

Conclusion:
-----------
This means that when the travelling twin is back, he MUST
observe that the Earth clock shows 30 years, while his own
clock shows 18 years.

This is an inevitable consequence of the Doppler shift in equation (1).

Observations made by the Earth twin:
=====================================
The Earth twin will observe the Doppler shift D₁ during
the time the travelling twin uses to reach the turning
point, PLUS the time the signal uses to reach the Earth.
This time is:
t₁ = L/v + L/c = 12/0.8+12/1 y = 27 y

So when the Earth twin's clock shows 27 y, the clock on
the monitor will show 27⋅D₁ y = 9 y.

Note that when the Earth twin sees the clock on the monitor
showing 9 y, this signal has been on its way for the time
L/c = 12 y. That means that the he will see the signal
Doppler shifted by D₂ for the time:
t₂ = L/v − L/c = 12/0.8 − 12/1 y = 3 y

So when the travelling twin is back on the Earth
the Earth clock will show: 27+3 y = 30 y and
the travelling twin's clock will show 9+3⋅D₂= 18 years.

Conclusion:
-----------
This means that when the travelling twin is back, the Earth twin MUST
observe that the travelling twin's clock shows 18 years, while his own
clock shows 30 years.

This is an inevitable consequence of the Doppler shift in equation (1).

https://paulba.no/pdf/TwinsByDoppler.pdf

--
Paul

https://paulba.no/

--
Paul

https://paulba.no/

On Tuesday, 14 December 2021 at 19:46:22 UTC+1, Paul B. Andersen wrote:
> Den 14.12.2021 15:43, skrev Richard Hachel:
> > On a very distant planet, aliens who know nothing about the theory of
> > relativity manage to send their first rocket into space.
> > The goal, simply fun for this mission, is, among other things, to place
> > two large clocks in space.
> A good idea.
> But instead of inventing several variation of nonsense,
> let's see what we can conclude from visually observation
> of distant clocks.
>
> -------------------------------
>
> It is an empirical fact that the longitudinal Doppler shift is:
> fᵣ = fₜ⋅√((1−v/c)/(1+v/c)) (1)

In the meantime in the real world, however, forbidden by your
moronic religion GPS clocks keep measuring t'=t, just
like all serious clocks always did.

Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :
>
>
> Den 14.12.2021 15:43, skrev Richard Hachel:
>> On a very distant planet, aliens who know nothing about the theory of
>> relativity manage to send their first rocket into space.
>> The goal, simply fun for this mission, is, among other things, to place
>> two large clocks in space.
>
> A good idea.
> But instead of inventing several variation of nonsense,
> let's see what we can conclude from visually observation
> of distant clocks.
>
> -------------------------------
>
> It is an empirical fact that the longitudinal Doppler shift is:
> fᵣ = fₜ⋅√((1−v/c)/(1+v/c)) (1)
> where fₜ is the transmitted frequency, fᵣ is the received frequency
> and v is the speed of the transmitter relative to the receiver.
>
> I will use this empirical fact to calculate how much the twins
> will age in the following 'twin scenario':
>
> We will use the units:
> time : year [y]
> distance: light years [ly]
> speed : light years per year [ly/y]
> c = 1 ly/y
>
> The "travelling twin" leaves the Earth at the speed v = 0.8 ly/y.
> When he is a distance L = 12 ly away from the Earth as
> measured in Earth's rest frame, he turns abruptly around
> and travels back to the Earth at the speed 0.8 ly/y.
>
> Both the Earth clock and the travelling twin's clock are
> set to zero when the travelling twin starts.
>
> Each twin has a transmitter which transmits a video signal
> showing his clock, and each twin has a receiver which receives
> the video signal and display the other twin's clock on a monitor.
>
> When the twins are receding from each other, they will both
> receive a signal Doppler shifted by:
> D₁ = √((1−0.8)/(1+0.8)) = 1/3
>
> When the twins are approaching each other, they will both
> receive a signal Doppler shifted by:
> D₂ = √((1+0.8)/(1−0.8)) = 3

> Paul

Fantastic!

It's exactly that.

You should post more often.

R.H.

Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :

> Conclusion:
> -----------
> This means that when the travelling twin is back, the Earth twin MUST
> observe that the travelling twin's clock shows 18 years, while his own
> clock shows 30 years.
>
> This is an inevitable consequence of the Doppler shift in equation (1).

> Paul

OKAY.

You answered it really well, and I can say that, frankly, I have rarely
read anything so interesting on the forums in the almost 30 years I have
been posting there.

But now you have to go further, because I think you have the capacity to
understand what I am saying.

The rest is just human behavior.

I refer you to my little diagrams.

You will obviously agree with the eight diagrams I made for the earth
traveler.

I think everyone is okay with that on all relativity forums.

BUT watch how I deal with the evolution of the star traveler, this is
quite new, and quite amazing.

And yet it is the reality of things.

This is how this traveler would see it if he had the opportunity to make
this trip for real.

When a grandmaster plays a brilliant chess move, it is called a "point".

Here, the point of the reasoning, it is to think that the situation is
symmetrical on the length and the distances.

This is called covariance.

Just as I see a rocket that turns and returns three times as long, the
subject placed in the rocket sees an earth three times as stretched in the
direction of displacement.

BUT ABOVE ALL, IT IS ALL THE SPACE THAT IS STRETCHED.

See how, in the example, the extraordinary spatial zoom effect occurs.

Once this is understood, we howl with laughter at a hundred years of
incomprehension, because we can only see that.

We then ask ourselves the question: for nine years, the traveler sees a
land coming back to him 36 light years?

OBVIOUSLY.

It is of fantastic mathematical beauty and infinitely obvious.

What is the apparent speed of an object (the earth or a rocket) coming
back towards you at 0.8c?

Vapp = 4c.

And that is exactly what is happening.

Vapp = x / T

Vapp = 36/9 = 4c

I beg any readers to understand this, which I have been trying to explain
for decades on the French forums, without achieving anything other than to
make idiots laugh who do not understand it.

R.H.

Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :

> https://paulba.no/

J'apprécie votre énorme travail.

Bravo.

R.H.

Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :

> It is an empirical fact that the longitudinal Doppler shift is:
> fᵣ = fₜ⋅√((1−v/c)/(1+v/c)) (1)
> where fₜ is the transmitted frequency, fᵣ is the received frequency
> and v is the speed of the transmitter relative to the receiver.

> Paul

This is entirely correct.

But we have to explain why.

This is correct because there are two things to consider:

1. Spatial anichrony, which is of the first degree, according to the
direction of the object and which evolves according to (1+cosµ.v/c)

2. The relativistic chronotropy which means that each watch sees
constantly, whatever its direction, the other watch systematically beat
less quickly. According to the sqrt factor (1-v²/c²)

We will therefore have, in total, the following effect, which is not a
simple longitudinal Doppler effect to be added to the relativistic
transverse effect, but a real effect.

We then obtain the equation of
t '(seen on the other watch) compared to t (seen on my own watch).

Let here t'= √(1-v²/c²)/(1+cosµ.v/c)

If the object approaches, we therefore have µ = 180 °.

cosµ = -1

let :
t'=t.√[(1-v/c)/(1+v/c)]

This is what your equation gives, which is obviously correct.

fᵣ = fₜ⋅√((1−v/c)/(1+v/c))

R.H.

Richard Hachel wrote:

> Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :
>
>> https://paulba.no/
>
> J'apprécie votre énorme travail.

you guys are frogs.

Richard Hachel wrote:

> Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :
>> It is an empirical fact that the longitudinal Doppler shift is:
>> fᵣ = fₜ⋅√((1−v/c)/(1+v/c)) (1)
>> where fₜ is the transmitted frequency, fᵣ is the received frequency
>> and v is the speed of the transmitter relative to the receiver.
>
> This is entirely correct.
>
> But we have to explain why.

It has already been done:

<https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Derivation>
> This is correct because there are two things to consider:
>
> 1. Spatial anichrony, which is of the first degree, according to the
> direction of the object and which evolves according to (1+cosµ.v/c)
>
> 2. The relativistic chronotropy which means that each watch sees
> constantly, whatever its direction, the other watch systematically beat
> less quickly. According to the sqrt factor (1-v²/c²)

Pure fantasy.
> We will therefore have, in total, the following effect, which is not a
> simple longitudinal Doppler effect to be added to the relativistic
> transverse effect, but a real effect.
>
> We then obtain the equation of
> t '(seen on the other watch) compared to t (seen on my own watch).
>
> Let here t'= √(1-v²/c²)/(1+cosµ.v/c)
[^1]

Still wrong, see below.

And unless you learn the difference between the transformation of *a*
coordinat*e*, and the transformation of a difference between coordinate*s*,
you have no hope of ever understanding this correctly, let alone obtaining
a correct, observable result.

I recommend that you start by writing coordinates with letters (and maybe
extra marks), e.g. “t'”, and, as is customary, differences between
coordinates by prepending “Δ” (Delta), i.e. “Δt'”:

For two events

e₁ ≔ (t₁', x₁', y₁', z₁')
e₂ ≔ (t₂', x₂', y₂', z₂')

one typically defines

Δt' ≔ t₂' − t₁'.

If the events are simultaneous in that frame of reference, then t₂' = t₁',
and

Δt' = t₂' − t₁' = t₁' − t₁' = 0.

Another convention is to use lowercase letters for coordinates and uppercase
letters for differences between coordinates, but as uppercase letters are
often taken for physical quantities (e.g. “T” usually stands for kinetic
energy in special relativity), this can interfere or lead to confusion.
> If the object approaches, we therefore have µ = 180 °.

At least you have defined “μ” once. This will make it easier (for me) to
show how your equations are wrong.

[The unit symbol “°” ought to be written without preceding space
according to the Système international d’unités (SI). This is
different e.g. from the unit symbol “°C” which ought to be written
with preceding space (but often, falsely, is not).]
> cosµ = -1

That is trivially true.¹ But it means that according to you[^1], the
temporal coordinate or the difference between them (you are still unclear
about that) would transform as follows:

t' = √(1 − v²/c²)/(1 + cos(µ) · v/c)
= √(1 − v²/c²)/(1 + (−1) · v/c)
= √(1 − v²/c²)/(1 − v/c)
= 1/γ · 1/(1 − v/c)
= 1/[γ (1 − v/c)].

That is not what is observed.² Instead, ALL observations are consistent
with the Lorentz transformation of time:

t' = γ (t − v/c² x)

and for time intervals

Δt' = γ (Δt − v/c² Δx)

[as I derived in detail earlier], where x is the spatial coordinate of an
event on the line-of-sight.

[And that means that moving clocks not only appear to run slower than
stationary ones: it means that they would be observed to not be
synchronous if they are spatially separated (Δx ≠ 0), even though an
observer moving with them (v = 0) would disagree – the relativity of
simultaneity that you deny.]

PointedEars
___________
¹ Why you would use such an impractical coordinate system is beyond me.
Does it make you feel smart perhaps?

² I sincerely hope that you do NOT think that

√(1 − v²/c²)/(1 − v/c) = √[(1 + v/c)/(1 − v/c)]

or some ridiculous, non-mathematical nonsense like that.

--
I heard that entropy isn't what it used to be.

(from: WolframAlpha)

On 15-Dec-21 1:43 am, Richard Hachel wrote:
> On a very distant planet, aliens who know nothing about the theory of
> relativity manage to send their first rocket into space.
> The goal, simply fun for this mission, is, among other things, to place
> two large clocks in space.
> The clocks are placed very far from their original planet, and are
> separated by 900,000 kms.
> Suddenly Commander Cromar notices that something bizarre has happened,
> and he begins to gently scratch the green scales with bluish
> reflections, which cover his skull. From his spaceship placed near the
> first clock, he notices that the two clocks do not mark the same time.
> Immediately, he begins to think that something must have happened when
> disembarking the clocks, and that one of them must have gone out of tune
> during the manipulation.
> This is what a normal human being would have thought first. Aliens are
> not dumber, nor smarter than us (otherwise, it would be known).
>
> R.H.
>

And after they've tried this repeatedly, and with the same deviation
each time, what would they conclude then?

Of would they just persist with the first thought, despite mounting
evidence the contrary?

Sylvia.

On Tuesday, December 14, 2021 at 3:46:22 PM UTC-3, Paul B. Andersen wrote:

<snip>

> The "travelling twin" leaves the Earth at the speed v = 0.8 ly/y.
> When he is a distance L = 12 ly away from the Earth as
> measured in Earth's rest frame, he turns abruptly around
> and travels back to the Earth at the speed 0.8 ly/y.

Inertia, Paul, inertia.

Turning abruptly around while traveling at 0.8 c would left any atom of the "travelling twin"
splashed all over the inner front of the ship.

And start traveling back due to an hyperdrive to instantly get v = 0.8 c would left their subatomic
components all over the rear end of the ship.

And that's why this kind of games are stupid when played by grown ups. Let it to a child being 8 or less.

On Tuesday, December 14, 2021 at 8:33:36 PM UTC-5, Richard Hertz wrote:
> On Tuesday, December 14, 2021 at 3:46:22 PM UTC-3, Paul B. Andersen wrote:
>
> <snip>
> > The "travelling twin" leaves the Earth at the speed v = 0.8 ly/y.
> > When he is a distance L = 12 ly away from the Earth as
> > measured in Earth's rest frame, he turns abruptly around
> > and travels back to the Earth at the speed 0.8 ly/y.
> Inertia, Paul, inertia.
>
> Turning abruptly around while traveling at 0.8 c would left any atom of the "travelling twin"
> splashed all over the inner front of the ship....

> And that's why this kind of games are stupid when played by grown ups.

So why are YOU initiating that stupid game?
We all know about the "instantaneous turnaround" is not "instantaneous". We just don't mention these details since its so obvious, irrelevant to the scenario, and addressed in many ways (deceleration/acceleration, or a 3rd twin). We all know this, and you do too (or, should know).

So it is childish of you to invoke your above complaint. You are being childish.

On Tuesday, December 14, 2021 at 10:52:05 PM UTC-3, rotchm wrote:
> On Tuesday, December 14, 2021 at 8:33:36 PM UTC-5, Richard Hertz wrote:

<snip>

> > Inertia, Paul, inertia.
> >
> > Turning abruptly around while traveling at 0.8 c would left any atom of the "travelling twin"
> > splashed all over the inner front of the ship....
> > And that's why this kind of games are stupid when played by grown ups.

> So why are YOU initiating that stupid game?
> We all know about the "instantaneous turnaround" is not "instantaneous". We just don't mention these details since its so obvious, irrelevant to the scenario, and addressed in many ways (deceleration/acceleration, or a 3rd twin). We all know this, and you do too (or, should know).
>
> So it is childish of you to invoke your above complaint. You are being childish.

All right, all right. Let's follow the game and assume that decelerates at a = -10 g, to avoid their eyeballs to pop out.

It implies t(v = 0) = 0.8 c/10g ≈ 35 days to a full stop and another equal time to reach - 0.8 c.

Don't forget to include almost 1/6 year (proper time) in the calculations.

By the way, how come the ship knows reached v = 0 so it can turn back? I bet that the ship is using non linear acceleration in both cases.
But it takes much more time to reach to a full stop, maybe 1 year or so.

See? Incorporating some FICTIONAL reality ruins the THOUGHT experiment. Now you have TWO EXTRA years to deal with.

On Wednesday, December 15, 2021 at 1:23:57 AM UTC-3, Richard Hertz wrote:
> On Tuesday, December 14, 2021 at 10:52:05 PM UTC-3, rotchm wrote:
> > On Tuesday, December 14, 2021 at 8:33:36 PM UTC-5, Richard Hertz wrote:
> <snip>
> > > Inertia, Paul, inertia.
> > >
> > > Turning abruptly around while traveling at 0.8 c would left any atom of the "travelling twin"
> > > splashed all over the inner front of the ship....
> > > And that's why this kind of games are stupid when played by grown ups..
>
> > So why are YOU initiating that stupid game?
> > We all know about the "instantaneous turnaround" is not "instantaneous".. We just don't mention these details since its so obvious, irrelevant to the scenario, and addressed in many ways (deceleration/acceleration, or a 3rd twin). We all know this, and you do too (or, should know).
> >
> > So it is childish of you to invoke your above complaint. You are being childish.
> All right, all right. Let's follow the game and assume that decelerates at a = -10 g, to avoid their eyeballs to pop out.
>
> It implies t(v = 0) = 0.8 c/10g ≈ 35 days to a full stop and another equal time to reach - 0.8 c.
>
> Don't forget to include almost 1/6 year (proper time) in the calculations..
>
> By the way, how come the ship knows reached v = 0 so it can turn back? I bet that the ship is using non linear acceleration in both cases.
> But it takes much more time to reach to a full stop, maybe 1 year or so.
>
> See? Incorporating some FICTIONAL reality ruins the THOUGHT experiment. Now you have TWO EXTRA years to deal with.

Silly me! I forgot to add that the TWO EXTRA years are the SAME for both twins, because SR MATH is invalid if acceleration is present.

Den 14.12.2021 22:54, skrev Richard Hachel:
> Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :
>
>> It is an empirical fact that the longitudinal Doppler shift is:
>>   fᵣ = fₜ⋅√((1−v/c)/(1+v/c))    (1)
>> where fₜ is the transmitted frequency, fᵣ is the received frequency
>> and v is the speed of the transmitter relative to the receiver.
>
>> Paul
>
> This is entirely correct.
>
> But we have to explain why.

Einstein explained it 116 years ago.
https://paulba.no/paper/Electrodynamics.pdf
see §7

And it is repeated several times:
https://paulba.no/pdf/AberrationDoppler.pdf

>
> This is correct because there are two things to consider:

We know it is correct because it is solidly
experimentally confirmed.

>
> 1. Spatial anichrony, which is of the first degree, according to the
> direction of the object and which evolves according to (1+cosµ.v/c)
>
> 2. The relativistic chronotropy which means that each watch sees
> constantly, whatever its direction, the other watch systematically beat
> less quickly. According to the sqrt factor (1-v²/c²)
>
> We will therefore have, in total, the following effect, which is not a
> simple longitudinal Doppler effect to be added to the relativistic
> transverse effect, but a real effect.

I won't bother to try to figure out what you are talking about.

>
> We then obtain the equation of
> t '(seen on the other watch) compared to t (seen on my own watch).

Let's look at an example.

A clock C is showing the time tC.
Two spaceships A and B are travelling towards this clock
with the speeds (relative to C) 0.6c and 0.8c respectively.
At one instant, the ships are adjacent 1 light year from C.
At that instant, the ships clocks are showing tA and tB
respectively.

A->0.6c C
B->0.8c
|------------ 1 ly -------|

Let tC be what clock C shows when the light from C that
hits the ships when they are adjacent, was emitted.

Let tA' and tB' be the the visually observed reading of clock C.

So we have at this instant:
tA' = tC
tB' = tC

The point is that the light front emitted
from the clock when it showed t will hit both
space ships at the same time, so the spaceships
will at that instant see the same time even
if their speeds relative to C are different.

But the observed rate of the clocks is
depend on their speed.

dtA'/dtA = √((1+0.6)/(1-0.6)) = 2
dtB'/dtB = √((1+0.8)/(1-0.8)) = 3

What A and B can say, is:

If A, when his clock shows tA1, observes tA1' = tC,
then he will, when his clock shows tA2 = tA1 + Δt,
observe tA2' = tC + 2⋅Δt

If B, when his clock shows tB1, observes tB1' = tC,
then he will, when his clock shows tB2 = tB1 + Δt,
observe tB2' = tC + 3⋅Δt

>
> Let here t'= √(1-v²/c²)/(1+cosµ.v/c)
>
> If the object approaches, we therefore have µ = 180 °.
>
> cosµ = -1
>
> let :
> t'=t.√[(1-v/c)/(1+v/c)]

This equation is meaningless.

The function t' = f(t) will obviously depend on:
- How the observer's clock and the observed clock
are synchronised.
- The reading t of the observer's clock.
- The distance between the observer and the observed clock
- The speed of the observer relative to the observed clock

> This is what your equation gives, which is obviously correct.
>
> fᵣ = fₜ⋅√((1−v/c)/(1+v/c))
> R.H.

--
Paul

https://paulba.no/

On Wednesday, 15 December 2021 at 13:43:56 UTC+1, Paul B. Andersen wrote:
> Den 14.12.2021 22:54, skrev Richard Hachel:
> > Le 14/12/2021 à 19:46, "Paul B. Andersen" a écrit :
> >
> >> It is an empirical fact that the longitudinal Doppler shift is:
> >> fᵣ = fₜ⋅√((1−v/c)/(1+v/c)) (1)
> >> where fₜ is the transmitted frequency, fᵣ is the received frequency
> >> and v is the speed of the transmitter relative to the receiver.
> >
> >> Paul
> >
> > This is entirely correct.
> >
> > But we have to explain why.
> Einstein explained it 116 years ago.
> https://paulba.no/paper/Electrodynamics.pdf
> see §7
>
> And it is repeated several times:
> https://paulba.no/pdf/AberrationDoppler.pdf
> >
> > This is correct because there are two things to consider:
> We know it is correct because it is solidly
> experimentally confirmed.

In the meantime in the real world, however, forbidden
by your moronic religion GPS clocks keep measuring
t'=t, just like all serious clocks always did.

Paul B. Andersen wrote:

>>> It is an empirical fact that the longitudinal Doppler shift is:
>>>   fᵣ = fₜ⋅√((1−v/c)/(1+v/c))    (1)
>>> where fₜ is the transmitted frequency, fᵣ is the received frequency
>>> and v is the speed of the transmitter relative to the receiver. Paul
>>
>> This is entirely correct. But we have to explain why.
>
> Einstein explained it 116 years ago.
> https://paulba.no/paper/Electrodynamics.pdf see §7

you must be meaning Poincare, something you should correct in whatever
you write in that "paper.pdf".

Le 15/12/2021 à 00:40, Thomas 'PointedEars' Lahn a écrit :

> ² I sincerely hope that you do NOT think that
>
> √(1 − v²/c²)/(1 − v/c) = √[(1 + v/c)/(1 − v/c)]

J'en arrive à me demander s'il ne se trouve pas de réels malades sur
les news.

R.H.

On Tuesday, December 14, 2021 at 11:23:57 PM UTC-5, Richard Hertz wrote:

> All right, all right. Let's follow the game and assume that decelerates at a = -10 g, to avoid their eyeballs to pop out.

Unnecessary. Why not just have a 3rd twin already coming back at constant speed?
No accelerations involved. Take it from there...

Richard Hertz <hertz778@gmail.com> wrote:
> On Tuesday, December 14, 2021 at 3:46:22 PM UTC-3, Paul B. Andersen wrote:
>
> <snip>
>
>> The "travelling twin" leaves the Earth at the speed v = 0.8 ly/y.
>> When he is a distance L = 12 ly away from the Earth as
>> measured in Earth's rest frame, he turns abruptly around
>> and travels back to the Earth at the speed 0.8 ly/y.
>
> Inertia, Paul, inertia.
>
> Turning abruptly around while traveling at 0.8 c would left any atom of
> the "travelling twin"
> splashed all over the inner front of the ship.
>
> And start traveling back due to an hyperdrive to instantly get v = 0.8 c
> would left their subatomic
> components all over the rear end of the ship.
>
> And that's why this kind of games are stupid when played by grown ups.
> Let it to a child being 8 or less.
>

Idealization is core to scientific modeling. Any system of any complexity
is not going to be completely accurately described by any simple model, SR
included. What other factors need to be brought into consideration depends
on whether the intent is to compare to an actual experimental result, what
the desired precision of the prediction and the measurement need to be, and
so on.

Your boondoggle complaint is pointless. You might as well go to any
freshman text where projectile motion examples are discussed in an early
chapter with the disclaimer that air friction will be neglected, and then
you splutter that air friction is ALWAYS present in terrestrial projectile
motion. All that spluttering would demonstrate is your profound tendency to
miss the point entirely.

--
Odd Bodkin -- maker of fine toys, tools, tables

Le 15/12/2021 à 13:43, "Paul B. Andersen" a écrit :

> Let's look at an example.

I beg you (and any other reader who truly seeks to understand SR) to have
a cup of coffee, and to think calmly about my description of the traveler
from Langevin.
I have never seen such a precise description in the world without any
paradox.
I find it unfortunate that many fight against people (insults, threats,
denunciations) and not against ideas, with the desire to improve them.
I refer you, if you can display them on your screen, to the drawings that
I made and edited in the posts of December 13 at 15:51 and 16:26
I think you should take these designs very seriously, and possibly take
them as an example to place them on your web pages. There is no better one
in the world on the subject of Langevin's travelor.

R.H.