Most relativity gurus teach that if there are two inertial reference frames, F0 and F1 with relative velocity V, per relativity, each frame where all the clocks are initialized and synchronized measure that clocks in the other frame run slower.
For example, let there be two inertial reference frames, F0 and F1, that have a relative velocity of V=c*sqrt(3)/2. If clocks in the other frame were running slower, each frame would say that the other frame's clocks are running at half the rate of clocks in their own frame. So to test this concept, we do the following gedanken experiment. In frame F1 there is an ideal clock. Frame F0 says that clocks in F1 run at half the rate of F0 clocks, so observers in F0 make a clock that runs twice as fast as a standard clock, and place that clock in F1 next to the standard F1 clock. Now as these two clocks pass every synchronized clock in F0, the clock in F1 that is running at twice the normal rate as a standard clock matches the time shown on every F0 clock that it passes. The F1 observers now must consider that the difference in times has something to do with the initialization of clocks, not merely that the clocks in F0 are running at half the rate.
David Seppala
Bastrop TX

On Thursday, September 30, 2021 at 1:26:28 PM UTC-7, sep...@yahoo.com wrote:
> Most relativity gurus teach that if there are two inertial reference frames, F0 and
> F1 with relative velocity V, each frame where all the clocks are initialized and
> synchronized measure that clocks in the other frame run slower.

Most? Every competent scientist knows that, given two relatively moving clocks, each clock runs slow in terms of the inertial coordinates in which the other clock is at rest. [Note "coordinate system", not "frame".]

> Let there be two inertial [coordinate systems], F0 and F1, that have a relative velocity
> of V=c*sqrt(3)/2. ...each frame would say that the other frame's clocks are running at
> half the rate of clocks in their own frame.

Frames don't talk, and what you really need to refer to is coordinate systems, which don't talk either. Correct statement: Each clock runs at half the rate of the time coordinate of the inertial coordinate system in which the other clock is at rest.

> So to test this concept, we do the following gedanken experiment.

"Thought experiments" do not constitute experimental tests, they are just thoughts.

> In frame F1 there is an ideal clock.

Translation: There is a clock at rest in inertial coordinate system F1.

> Frame F0 says that clocks in F1 run at half the rate of F0 clocks...

Frames don't talk, and you need to refer to coordinate systems, which don't talk either, and no, it is not correct to say that some clocks run slower than other clocks... such a statement does not express the facts. The facts are that if we have two rows of clocks, one row at rest and inertially synchronized in F0 and the other at rest and inertially synchronized in F1, the elapsed time on each F0 clock between passing successive F1 clocks is half the difference in the readings of the F1 clocks as it passes them... and vice versa. That's the meaning of "each clock runs slow in terms of the inertial coordinates in which the other clock is at rest".

> so observers in F0 make a clock that runs twice as fast as a standard clock,
> and place that clock in F1 next to the standard F1 clock.

A clocks at rest in F0 cannot be "placed next to" a clock in F1, except for one instant, because they are moving relative to each other. Your statement is ambiguous. Do you mean a clock that runs at twice the standard proper rate is placed at rest in F1?

> Now as these two clocks pass every synchronized clock in F0, the clock in F1
> that is running at twice the normal rate as a standard clock matches the time
> shown on every F0 clock that it passes.

Right. Yawn.

> The F1 observers now must consider that the difference in times has something
> to do with the initialization of clocks, not merely that the clocks in F0 are running
> at half the rate.

Of course it does, nitwit, if by initialization you mean the initial synchronization. That's why I keep telling you that the clocks in the grid are at rest and inertially synchronized in a given frame. Yes, simultaneity is relative. Duh. This is what people have been telling you for decades. Now suddenly you've realized that simultaneity is relative? Congratulations. Now that's you've realized this, you can see that special relativity is perfectly simple and correct, and you've been utterly wrong all these years, right?

Gravity strength and speed slows down the retarded clocks parts in space.
Rotation time dilates the clock and gives it more energy.

Mitchell Raemsch

On Thursday, September 30, 2021 at 1:26:28 PM UTC-7, sep...@yahoo.com wrote:
> Most relativity gurus teach that if there are two inertial reference frames, F0 and
> F1 with relative velocity V, each frame where all the clocks are initialized and
> synchronized measure that clocks in the other frame run slower.

Most? Every competent scientist knows that, given two relatively moving clocks, each clock runs slow in terms of the inertial coordinates in which the other clock is at rest. [Note "coordinate system", not "frame".]

> Let there be two inertial [coordinate systems], F0 and F1, that have a relative velocity
> of V=c*sqrt(3)/2. ...each frame would say that the other frame's clocks are running at
> half the rate of clocks in their own frame.

Frames don't talk, and what you really need to refer to is coordinate systems, which don't talk either. A correct statement is: Each clock runs at half the rate of the time coordinate of the inertial coordinate system in which the other clock is at rest.

> So to test this concept, we do the following gedanken experiment.

"Thought experiments" don't constitute experimental tests, they are just thoughts.

> In frame F1 there is an ideal clock.

Translation: There's an ideal clock at rest in inertial coordinate system F1.

> Frame F0 says that clocks in F1 run at half the rate of F0 clocks...

Frames don't talk, and you need to refer to coordinate systems, which don't talk either, and no, it's not correct to say that some clocks run slower than other clocks... such a statement does not express the facts. The facts are that if we have two rows of clocks, one row at rest and inertially synchronized in F0 and the other at rest and inertially synchronized in F1, the elapsed time on each F0 clock between passing successive F1 clocks is half the difference in the readings of the F1 clocks as it passes them... and vice versa. That's the meaning of "each clock runs slow in terms of the inertial coordinates in which the other clock is at rest".

> so observers in F0 make a clock that runs twice as fast as a standard clock,
> and place that clock in F1 next to the standard F1 clock.

A clocks at rest in F0 cannot be "placed next to" a clock in F1, except for one instant. Were you perhaps groping to say a clock that runs at twice the standard proper rate is placed at rest in F1?

> Now as these two clocks pass every synchronized clock in F0, the clock in F1
> that is running at twice the normal rate as a standard clock matches the time
> shown on every F0 clock that it passes.

Right... and trivial.

> The F1 observers now must consider that the difference in times has something
> to do with the initialization of clocks, not merely that the clocks in F0 are running
> at half the rate.

Of *course* it does, if by initialization you mean the initial synchronization. That's why I keep telling you that the clocks in the grid are at rest and inertially synchronized in a given frame. Yes, inertial simultaneity is relative, which is an unavoidable consequence of the inertia of energy (E=mc^2). This is what people have been telling you for decades. Now that this has finally sunk into your thick skull, you should be able to see that special relativity is perfectly simple and correct, just as you've been told all along. You're welcome.

Den 30.09.2021 22:26, skrev sepp623@yahoo.com:
> Most relativity gurus teach that if there are two inertial reference frames, F0 and F1 with relative velocity V, per relativity, each frame where all the clocks are initialized and synchronized measure that clocks in the other frame run slower.
> For example, let there be two inertial reference frames, F0 and F1, that have a relative velocity of V=c*sqrt(3)/2. If clocks in the other frame were running slower, each frame would say that the other frame's clocks are running at half the rate of clocks in their own frame. So to test this concept, we do the following gedanken experiment. In frame F1 there is an ideal clock. Frame F0 says that clocks in F1 run at half the rate of F0 clocks, so observers in F0 make a clock that runs twice as fast as a standard clock, and place that clock in F1 next to the standard F1 clock. Now as these two clocks pass every synchronized clock in F0, the clock in F1 that is running at twice the normal rate as a standard clock matches the time shown on every F0 clock that it passes. The F1 observers now must consider that the difference in times has something to do with the initialization of clocks, not merely that the clocks in F0 are running at half the rate.
> David Seppala
> Bastrop TX
>
https://paulba.no/pdf/Mutual_time_dilation.pdf

--
Paul

https://paulba.no/

On Friday, October 1, 2021 at 2:15:43 AM UTC-5, Al Coe wrote:
> On Thursday, September 30, 2021 at 1:26:28 PM UTC-7, sep...@yahoo.com wrote:
> > Most relativity gurus teach that if there are two inertial reference frames, F0 and
> > F1 with relative velocity V, each frame where all the clocks are initialized and
> > synchronized measure that clocks in the other frame run slower.
> Most? Every competent scientist knows that, given two relatively moving clocks, each clock runs slow in terms of the inertial coordinates in which the other clock is at rest. [Note "coordinate system", not "frame".]
>
> > Let there be two inertial [coordinate systems], F0 and F1, that have a relative velocity
> > of V=c*sqrt(3)/2. ...each frame would say that the other frame's clocks are running at
> > half the rate of clocks in their own frame.
> Frames don't talk, and what you really need to refer to is coordinate systems, which don't talk either. A correct statement is: Each clock runs at half the rate of the time coordinate of the inertial coordinate system in which the other clock is at rest.
> > So to test this concept, we do the following gedanken experiment.
> "Thought experiments" don't constitute experimental tests, they are just thoughts.
> > In frame F1 there is an ideal clock.
> Translation: There's an ideal clock at rest in inertial coordinate system F1.
>
> > Frame F0 says that clocks in F1 run at half the rate of F0 clocks...
>
> Frames don't talk, and you need to refer to coordinate systems, which don't talk either, and no, it's not correct to say that some clocks run slower than other clocks... such a statement does not express the facts. The facts are that if we have two rows of clocks, one row at rest and inertially synchronized in F0 and the other at rest and inertially synchronized in F1, the elapsed time on each F0 clock between passing successive F1 clocks is half the difference in the readings of the F1 clocks as it passes them... and vice versa. That's the meaning of "each clock runs slow in terms of the inertial coordinates in which the other clock is at rest".
> > so observers in F0 make a clock that runs twice as fast as a standard clock,
> > and place that clock in F1 next to the standard F1 clock.
> A clocks at rest in F0 cannot be "placed next to" a clock in F1, except for one instant. Were you perhaps groping to say a clock that runs at twice the standard proper rate is placed at rest in F1?
> > Now as these two clocks pass every synchronized clock in F0, the clock in F1
> > that is running at twice the normal rate as a standard clock matches the time
> > shown on every F0 clock that it passes.
> Right... and trivial.
> > The F1 observers now must consider that the difference in times has something
> > to do with the initialization of clocks, not merely that the clocks in F0 are running
> > at half the rate.
> Of *course* it does, if by initialization you mean the initial synchronization. That's why I keep telling you that the clocks in the grid are at rest and inertially synchronized in a given frame. Yes, inertial simultaneity is relative, which is an unavoidable consequence of the inertia of energy (E=mc^2). This is what people have been telling you for decades. Now that this has finally sunk into your thick skull, you should be able to see that special relativity is perfectly simple and correct, just as you've been told all along. You're welcome.

If the clock that is designed to run twice as fast as a standard clock is at rest side by side to a standard clock in inertial reference frame F1, and they pass synchronized clocks in a frame that is moving at V=c*sqrt(3)/2, inertial reference frame F0, the clock that is running twice as fast as the standard clock matches the time difference of all the F0 clocks it passes. Why would someone at rest in frame F1 say that the clocks in F0 are running at 1/4 the rate of the clock that always shows the same elapsed time? Wouldn't it be more logical to say the difference in times shown on the standard clocks includes how the clocks were initially synchronized, not just the rate at which they are running?
David Seppala
Bastrop TX

On Friday, October 1, 2021 at 5:58:58 AM UTC-7, sep...@yahoo.com wrote:
> If the clock that is designed to run twice as fast as a standard clock is at rest side
> by side to a standard clock in inertial reference frame F1, and they pass synchronized
> clocks in a frame that is moving at V=c*sqrt(3)/2, inertial reference frame F0, the clock
> that is running twice as fast as the standard clock matches the time difference of all
> the F0 clocks it passes.

Right, you are just repeating from the previous message.

> Why would someone at rest in frame F1 say that the clocks in F0 are running at 1/4 the
> rate of the clock that always shows the same elapsed time?

The elapsed time on a standard clock at rest in F0 between encountering successive double-time clocks at rest in F1 is just 1/4 the difference between the readings of those double-time clocks at those encounters. Do you understand this?

You see, this is no different than the original scenario with the two rows of standard clocks, with each clock showing half the elapsed time between encounters as the difference between the readings of the encountered clocks in the other row. Once you understand this, all other variants are trivial..

> Wouldn't it be more logical to say the difference in times shown on the standard
> clocks includes how the clocks were initially synchronized, not just the rate at
> which they are running?

But that is precisely what everyone has been telling you for decades. The relationship between inertia-based coordinates does not just exhibit time dilation and length contraction, it exhibits the skew of simultaneity due to the inertia of energy, and this skew is perfectly reciprocal. You see, the inertia-based coordinate systems are the unique systems in terms of which the equations of physics take their simple homogeneous and isotropic form, and these system are empirically related by Lorentz transformations, meaning they have skewed simultaneity. Now do you finally understand?

On 9/30/21 3:26 PM, sepp623@yahoo.com wrote:
> Most relativity gurus teach that if there are two inertial reference
> frames, F0 and F1 with relative velocity V, per relativity, each
> frame where all the clocks are initialized and synchronized measure
> that clocks in the other frame run slower.

Not "most", ALL. Note that this does not mean the "clocks run slower",
it means instruments at rest in F0 measure clocks at rest in F1
to tick slower than identical clocks at rest in F0, and vice-versa.

> [...] each frame would say that the other frame's clocks are running
> at half the rate of clocks in their own frame.

Only if they are as incompetent as you are. Competent people would say
that their MEASUREMENTS of the clocks at rest in the other frame yield a
rate half that of identical clocks at rest in their own frame. This is
an artifact of how the clocks' tick rates are measured, not of the
clocks themselves (which are, of course, completely unaffected by
observers at rest in other frames).

> [... long-winded claim that synchronization also matters]

Yes, synchronization matters, whenever more than one clock is involved. DUH!

Tom Roberts

On 2021-10-01, Tom Roberts <tjroberts137@sbcglobal.net> wrote:
> it means instruments at rest in F0 measure clocks at rest in F1
> to tick slower than identical clocks at rest in F0, and vice-versa.
>
Clocks measure speed of events flow....
so more faster events happen, faster clock ticks...

--

7-77-777
Evil Sinner!

Branimir Maksimovic wrote:

> On 2021-10-01, Tom Roberts <tjroberts137@sbcglobal.net> wrote:
>> it means instruments at rest in F0 measure clocks at rest in F1
>> to tick slower than identical clocks at rest in F0, and vice-versa.
>>
> Clocks measure speed of events flow....
> so more faster events happen, faster clock ticks...

You have not thought this through. It is a fallacy to conclude from “time
is what can be measured with a clock” that “if a clock ticks faster, more
time has passed”.

To give a simple example why your reasoning is flawed: Suppose that is
midnight at your location, as you can read from your wristwatch and other
clocks. Now your wristwatch starts malfunctioning: it is ticking too fast.
After what you think would have been 24 hours, you compare your watch with
other, but perfect clocks (relative to which you had been at rest all the
time, and they are also at the same altitude above sea level). Your watch
is necessarily ahead of all the other clocks: it shows a later time than
they do. If your logic would be sound, then more time would have passed for
you, and Earth would have rotated more than (approximately) one time about
its axis just because of your broken watch. This would mean that because of
your broken watch the planet would have rotated faster. Obviously that is a
nonsensical idea, and simply not so.

However, if you consider the “light clock” thought experiment that I
described earlier, you can see that for an observer who would see the
light clock moving (relative to them), "events"¹ actually "happen" *slower*
than for an observer who is moving with the light clock. In particular,
more time elapses for the first observer to see that light pulse bouncing
between the mirrors of the light clock.

By your logic (which is somewhat correct there) this means that time is
observed to pass slower where the (moving) clock is at rest. In a way (but
only *in a way*) this is correct, and called "time *dilation*" (lengthening
of time), and put simplified as “moving clocks run slow(er)”.

In reality, though, simply *less* proper time has elapsed instead in the
frame where the clock is at rest than in the frame where the external
observer is at rest. So it only *seems* as if time had passed slower in
the clock’s moving rest frame than in the external observer’s.

PointedEars
___________
¹ An event in physics is a *point* in spacetime; it happens/occurs at a
single place at a single point in time. It does not happen in the sense
that time could elapse while it is going on. There is no “going on”.
Instead, time elapses *between* events; it is a component of the distance
*between* them. One must separate the everyday meaning of “event” from
the scientific meaning.
--
“Science is empirical: knowing the answer means nothing;
testing your knowledge means everything.”
—Dr. Lawrence M. Krauss, theoretical physicist,
in “A Universe from Nothing” (2009)

On 2021-10-01, Thomas 'PointedEars' Lahn <PointedEars@web.de> wrote:
>
> To give a simple example why your reasoning is flawed: Suppose that is
> midnight at your location, as you can read from your wristwatch and other
> clocks.
Wristwatch just counts mechanism, nothing real.

--

7-77-777
Evil Sinner!

Branimir Maksimovic wrote:

> On 2021-10-01, Thomas 'PointedEars' Lahn <PointedEars@web.de> wrote:
>> To give a simple example why your reasoning is flawed: Suppose that is
>> midnight at your location, as you can read from your wristwatch and other
>> clocks.
> Wristwatch just counts mechanism, nothing real.

You are wrong. Every (working) clock is working based on real, physical
processes.

A wristwatch, for example, may be working based on the oscillation of a
spring or a quartz crystal – caused by the tendency of the spring to return
to its equilibrium position, or the property of a quartz crystal to
oscillate only at a specific frequency.

This is very basic physics (introductory mechanics, electromagnetism). If
you do not know it, you should study it before you venture to understand,
and discuss about, the theories of relativity:

<https://openstax.org/books/university-physics-volume-1/pages/15-introduction>

<https://www.feynmanlectures.caltech.edu/I_21.html>

PointedEars
--
Q: Why is electricity so dangerous?
A: It doesn't conduct itself.

(from: WolframAlpha)

On Friday, October 1, 2021 at 6:44:45 PM UTC-4, Thomas 'PointedEars' Lahn wrote:
> Branimir Maksimovic wrote:
<snip>

You got got by the troll.

DO NOT STROKE THE TROLLS

On Friday, October 1, 2021 at 7:30:44 PM UTC-4, Thomas 'PointedEars' Lahn wrote:
> Branimir Maksimovic wrote:

Again, you got got by the troll.
Not too perceptive, aren't ya!

Be careful!

On 10/1/21 5:44 PM, Thomas 'PointedEars' Lahn wrote:
> [...] It is a fallacy to conclude from “time
> is what can be measured with a clock” that “if a clock ticks faster, more
> time has passed”.

That's not merely a fallacy, it is a counterfactual -- no clock ever
"ticks faster". Clocks always tick at their usual rate.

N.B. a clock that is broken is not a clock.

Tom Roberts

On Friday, 1 October 2021 at 23:59:49 UTC+2, tjrob137 wrote:

> Only if they are as incompetent as you are. Competent people would say
> that their MEASUREMENTS of the clocks at rest in the other frame yield a
> rate half that of identical clocks at rest in their own frame.

Only if they are as incompetent as you are. GPS clocks
keep measurng t'=t, just like all serious clocks always
did. Youir moronic screams "WE ARE FORCED!!!!"
didn't work.

On Saturday, 2 October 2021 at 04:33:32 UTC+2, tjrob137 wrote:
> On 10/1/21 5:44 PM, Thomas 'PointedEars' Lahn wrote:
> > [...] It is a fallacy to conclude from “time
> > is what can be measured with a clock” that “if a clock ticks faster, more
> > time has passed”.
> That's not merely a fallacy, it is a counterfactual -- no clock ever
> "ticks faster". Clocks always tick at their usual rate.

Sorry, poor halfbrain, anyone can check GPS:
some clocks run at 9 192 631 770, other at
9 192 631 774. Whatever you imagine, we're
not FORCED at all.

> N.B. a clock that is broken is not a clock.

And it's Tom Roberts deciding, which one is
the broken one! Because he is a part of THE
COMMUNITY OF PHYSICISTS!!!!!!!!!!!!!!!!!!
Keep dreaming, poor trash.

A clock can be retarded by sharing rotation.
Such as on a neutron stars rotating surface...
There is rotation time dilation.