I posted Circular logic with clocks. Here's one add-on to that I would like a relativity guru to explain.
The original problem is: There is a clock, C0, at rest in inertial reference frame F0 placed at (x,y) = (0,0). There is a second clock, C1, traveling with velocity V around a circle of diameter D that is centered at (0,0). The question is which clock runs slower or do both clocks run at the same rate?
Let a light source in F0 at (0,0) emit one pulse per second. Let the velocity magnitude of the traveling clock C1 be slightly greater than |V| = c*sqrt(3)/2. Make the diameter the circle very large such that a third clock,C2, is traveling in a straight line path with velocity V = c*sqrt(3)/2 relative to F0 at the circumference of the circle. Per Einstein's theory, clocks in the inertial reference frame that C2 is in measure the time between the emitted pulses in F0 as occurring one pulse every two seconds. Since the circling clock C1 is traveling almost in a straight line with clock C2, and with virtually the same velocity, during this arc segment where two of these pulses hit the circumference of the circle, clock C1 must also show that the time between the two pulses is about 2 seconds.
Now let there be N of these arc segments that complete one round trip around the circumference of the circle. If a light source moving with clock C1 emits one pulse for each second of elapsed time shown on clock C1, then that source will emit about 2N + K pulses during each trip around the circumference of the circle, whereas the light source with clock C0 in F0 only emits N pulses during C1's trip around the circumference.
So observers in F0 must conclude that the C1 clock is running at twice the rate of the C0 clock since a pulse every half second is received at (0,0) in F0. But when the C1 clock and the C2 are virtually side by side with virtually the same velocity, per Einstein, F0 observers measure that the C2 and therefore the C1 clock are running at a slower rate than the C0 clock. Furthermore F0 observers measure that the C1 clock that is circling (0,0) always runs at a constant rate. Can anyone explain that contradiction?
David Seppala
Bastrop TX

sepp623@yahoo.com <sepp623@yahoo.com> wrote:

> I posted Circular logic with clocks. Here's one add-on to that I would
> like a relativity guru to explain.

[Clocks going round in circles]
See under 'Transverse Doppler effect',
<https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect>

Jan

Le 20/02/2022 à 20:00, "sepp623@yahoo.com" a écrit :
> I posted Circular logic with clocks. Here's one add-on to that I would like a
> relativity guru to explain.
> The original problem is: There is a clock, C0, at rest in inertial
> reference frame F0 placed at (x,y) = (0,0). There is a second clock, C1,
> traveling with velocity V around a circle of diameter D that is centered at (0,0).
> The question is which clock runs slower or do both clocks run at the same rate?
> Let a light source in F0 at (0,0) emit one pulse per second. Let the
> velocity magnitude of the traveling clock C1 be slightly greater than |V| =
> c*sqrt(3)/2. Make the diameter the circle very large such that a third clock,C2,
> is traveling in a straight line path with velocity V = c*sqrt(3)/2 relative to F0
> at the circumference of the circle. Per Einstein's theory, clocks in the inertial
> reference frame that C2 is in measure the time between the emitted pulses in F0 as
> occurring one pulse every two seconds. Since the circling clock C1 is traveling
> almost in a straight line with clock C2, and with virtually the same velocity,
> during this arc segment where two of these pulses hit the circumference of the
> circle, clock C1 must also show that the time between the two pulses is about 2
> seconds.
> Now let there be N of these arc segments that complete one round trip
> around the circumference of the circle. If a light source moving with clock C1
> emits one pulse for each second of elapsed time shown on clock C1, then that
> source will emit about 2N + K pulses during each trip around the circumference of
> the circle, whereas the light source with clock C0 in F0 only emits N pulses
> during C1's trip around the circumference.
> So observers in F0 must conclude that the C1 clock is running at twice the
> rate of the C0 clock since a pulse every half second is received at (0,0) in F0.
> But when the C1 clock and the C2 are virtually side by side with virtually the
> same velocity, per Einstein, F0 observers measure that the C2 and therefore the C1
> clock are running at a slower rate than the C0 clock. Furthermore F0 observers
> measure that the C1 clock that is circling (0,0) always runs at a constant rate.
> Can anyone explain that contradiction?
> David Seppala

There is a Chinese proverb that says: "It is better to look stupid for
five minutes than to remain so all your life".
"Zhuāng shǎ wǔ fēnzhōng zǒng bǐ shǎ shǎ yībèizi hǎo".
This means that when you don't know something, or don't understand
something, you shouldn't hesitate to ask, even if it means looking stupid.

R.H.

R.H.

Le 20/02/2022 à 23:12, Richard Hachel a écrit :
> Le 20/02/2022 à 20:00, "sepp623@yahoo.com" a écrit :
>> I posted Circular logic with clocks. Here's one add-on to that I would like a
>> relativity guru to explain.
>> The original problem is: There is a clock, C0, at rest in inertial
>> reference frame F0 placed at (x,y) = (0,0). There is a second clock, C1, traveling
>> with velocity V around a circle of diameter D that is centered at (0,0). The
>> question is which clock runs slower or do both clocks run at the same rate?
>> Let a light source in F0 at (0,0) emit one pulse per second. Let the
>> velocity magnitude of the traveling clock C1 be slightly greater than |V| =
>> c*sqrt(3)/2. Make the diameter the circle very large such that a third clock,C2,
>> is traveling in a straight line path with velocity V = c*sqrt(3)/2 relative to F0
>> at the circumference of the circle. Per Einstein's theory, clocks in the inertial
>> reference frame that C2 is in measure the time between the emitted pulses in F0 as
>> occurring one pulse every two seconds. Since the circling clock C1 is traveling
>> almost in a straight line with clock C2, and with virtually the same velocity,
>> during this arc segment where two of these pulses hit the circumference of the
>> circle, clock C1 must also show that the time between the two pulses is about 2
>> seconds.
>> Now let there be N of these arc segments that complete one round trip
>> around the circumference of the circle. If a light source moving with clock C1
>> emits one pulse for each second of elapsed time shown on clock C1, then that source
>> will emit about 2N + K pulses during each trip around the circumference of the
>> circle, whereas the light source with clock C0 in F0 only emits N pulses during
>> C1's trip around the circumference.
>> So observers in F0 must conclude that the C1 clock is running at twice the
>> rate of the C0 clock since a pulse every half second is received at (0,0) in F0.
>> But when the C1 clock and the C2 are virtually side by side with virtually the same
>> velocity, per Einstein, F0 observers measure that the C2 and therefore the C1 clock
>> are running at a slower rate than the C0 clock. Furthermore F0 observers measure
>> that the C1 clock that is circling (0,0) always runs at a constant rate. Can
>> anyone explain that contradiction?
>> David Seppala
>
>
> There is a Chinese proverb that says: "It is better to look stupid for five
> minutes than to remain so all your life".
> "Zhuāng shǎ wǔ fēnzhōng zǒng bǐ shǎ shǎ yībèizi hǎo".
> This means that when you don't know something, or don't understand something,
> you shouldn't hesitate to ask, even if it means looking stupid.
>
> R.H.

My translator is not very efficient, so I don't quite understand your
question.
I'll try to answer on what I understand.
We place a clock at rest in a reference frame R, and we make gravitate
around it, very far away, and at a tangential speed of 0.8c, a second
clock.
You ask, if I understand correctly, which clock runs the fastest?
According to the theory of relativity, the watch which moves around the
other will beat less quickly, according to Tr=To.sqrt(1-v²/c²).
That is to say that when a second elapses for the watch placed in the
center of the system, only 0.6 seconds elapse on the other.
At this moment, a third clock passes, in uniform Galilean motion with a
speed of v=0.8c as well. For a few moments, they are interchangeable, and
an observer placed there would make no difference between the two watches.
Both beat the same way, and slower than the central watch.

R.H.

Le 20/02/2022 à 21:32, nospam@de-ster.demon.nl (J. J. Lodder) a écrit :
> sepp623@yahoo.com <sepp623@yahoo.com> wrote:
>
>> I posted Circular logic with clocks. Here's one add-on to that I would
>> like a relativity guru to explain.
>
> [Clocks going round in circles]
> See under 'Transverse Doppler effect',
>
> <https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect>
>
> Jan

Le 20/02/2022 à 21:32, nospam@de-ster.demon.nl (J. J. Lodder) a écrit :
> sepp623@yahoo.com <sepp623@yahoo.com> wrote:
>
>> I posted Circular logic with clocks. Here's one add-on to that I would
>> like a relativity guru to explain.
>
> [Clocks going round in circles]
> See under 'Transverse Doppler effect',
>
> <https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect>
>
> Jan

There are some interesting things in the link you gave.
I would like us to come back to something, because it is very important.
Let's say that the watch which rotates at 0.8c around the other, at a
distance of nine light-minutes, ie a little further than the Earth's
orbit.
The central watch will observe a watch which beats less quickly, as I said
in the previous post, and which revolves around it at a distance of NINE
light-minutes.
BUT, it is necessary to correctly apply the transformations of Lorentz for
the other watch, and one understands directly that the other watch, for
it, sees the central watch which revolves around it at FIFTEEN
light-minutes.
Concept of reference mollusk.
D=sqrt(x'²+y²)
with x'=g.(x-vTo)
So FIFTEEN light-minutes and no longer nine as for the first watch.

If the two watches transmit every revolution, one transmits every 70.686
minutes, the other, the control unit transmits every 117.809 minutes.
gamma=0.6

Apply the Lorentz transformations.

It's very logical.

You can also apply Hachel's transformations (that's me) by writing the
same thing said differently:
x'=g.[x+sqrt(x²+y²+z²).v/c]

But it's the same thing.

R.H.

On Sunday, February 20, 2022 at 11:01:03 AM UTC-8, sep...@yahoo.com wrote:
> There is a clock, C0, at rest in inertial reference frame F0 placed at (x,y) = (0,0). Let a light source [at rest] in F0 at (0,0) emit one pulse per second... Make the diameter the circle very large such that a clock, C2, is traveling in a straight line path with velocity V = c*sqrt(3)/2 relative to F0 [tangent to] the circumference of the circle. Per [special relativity], clocks in the inertial reference frame F2 that C2 is in measure the time between the emitted pulses in F0 as occurring one pulse every two seconds... where two of these pulses hit the circumference of the circle... the time [on C2] between the two pulses is about 2 seconds.

No, the emitter is moving in terms of F2, so the times of emission are necessarily read on a sequence of clocks, all at rest and inertially synchronized in F2, and the difference between the readings of this sequence of clocks (as the emitter passes them) between consecutive emissions is indeed 2 seconds, but the difference between the readings of the individual clock C2 at rest in F2 between consecutive arrivals is 0.5 seconds, after accounting for the Doppler effect due to the fact that this clock is approaching the source at just a 30 deg angle from straight ahead (due to aberration). Naturally the circling clock (C1) sees this same blue shift (and aberration) continually.

Special Relativity: 850 ..... Barnpole Dave: 0

Richard Hachel <r.hachel@tiscali.fr> wrote:

> Le 20/02/2022 à 21:32, nospam@de-ster.demon.nl (J. J. Lodder) a écrit :
> > sepp623@yahoo.com <sepp623@yahoo.com> wrote:
> >
> >> I posted Circular logic with clocks. Here's one add-on to that I would
> >> like a relativity guru to explain.
> >
> > [Clocks going round in circles]
> > See under 'Transverse Doppler effect',
> >
> > <https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Dopple
r_effect>
> >
> > Jan
>
> There are some interesting things in the link you gave.
> I would like us to come back to something, because it is very important.
> Let's say that the watch which rotates at 0.8c around the other, at a
> distance of nine light-minutes, ie a little further than the Earth's
> orbit.

There is no need to invent complicated examples.
A GPS sat does just what you want, it is almost in a circular orbit,
and the relativistic clock shift is observed just as predicted.
(after taking the general relativistic shift also into account)

Jan

On Monday, 21 February 2022 at 15:03:46 UTC+1, J. J. Lodder wrote:

> A GPS sat does just what you want, it is almost in a circular orbit,
> and the relativistic clock shift is observed just as predicted.

Apart of that the clocks of GPS are synchronized, i.e.
indicating t'=t (with the precision of an acceptable error),
just like all serious clocks always did.
As anyone can easily check, poor halfbrain. That's where
the impudent lies of your insane Shit end.