On Monday, April 25, 2022 at 8:57:59 AM UTC-7, sep...@yahoo.com wrote:
The Lorentz transform gives the result that the clock at rest in the center of the polygon is running at a slower rate than the traveler's clock.

No, it doesn't. You need to stop lying, imbecile

On Monday, April 25, 2022 at 11:57:59 AM UTC-4, sep...@yahoo.com wrote:

> If a clock always shows a lesser value than another clock, and that difference in times
> between the two clocks keeps getting greater and greater with every reading of the clocks,
> doesn't that mean than one clock is running at a slower rate than the other clock?

No. It means " that difference in times between the two clocks keeps getting greater
and greater with every reading of the clocks".

There is no need to introduce the expression "running at a slower rate", especially that that expression is ambiguous and meaningless. 20 < 40 is far more clear, precise and succinct. You don't agree with that?

> So if the traveler is moving along one leg of a polygon (say a square)

Lets say, a circle.

> with velocity V

With *speed* v along this circle.

> relative to the inertial frame the polygon is in, and there is a clock at rest at the center of the polygon,

the circle...

> what does the traveler say the readings of his clock relative to the clock at the center of the polygon is?

He says nothing about it, since he is not at the center; he is not performing any measurements, any comparisons.
Don't forget, physics is about what is "displayed" on instruments, "observations", "comparisons".
If you are requiring no observations then there is nothing to say about it.

> The Lorentz transform gives the result that the clock at rest in the center of the polygon
> is running at a slower rate than the traveler's clock.

No, the LT's don't say that at all. The LT's say for instance a<b.
*people* say/call/express this as "slower" .
I rather be succinct and clear: a< b; t = 20 etc.

sepp623@yahoo.com <sepp623@yahoo.com> wrote:
> On Sunday, April 24, 2022 at 9:59:31 PM UTC-5, rotchm wrote:
>> On Sunday, April 24, 2022 at 4:47:13 PM UTC-4, sep...@yahoo.com wrote:
>>
>>> Clock At Rest Traveler's clock
>>> 0 0
>>> 40 20
>>> 80 40
>>> 120 60
>>> N*40 N*20
>>> Please explain how the traveler concludes that his clock is running at
>>> a faster rate than the clock at rest?
>> The traveler does not conclude that.
>>
>> What he says, or concludes is that as his clock coincides with the "at
>> rest clock(s)" his clock indicates a lesser value.
>> A lesser value does not mean "slower". In fact, "slower" here is an
>> unnecessary word; 20<40 suffices; 20N < 40N suffices and is clear. So he
>> notes experimentally that 20N < 40N. He also calculates that 20N < 40N.
>> He thus says, or concludes that 20N < 40N.
>>> Can you understand that?
>
> If a clock always shows a lesser value than another clock, and that
> difference in times between the two clocks keeps getting greater and
> greater with every reading of the clocks, doesn't that mean than one
> clock is running at a slower rate than the other clock?

No, not necessarily. And this is where things need to be stated carefully.
Whether a clock is running slow or fast is determined LOCALLY, by comparing
to some physical process that is happening in the vicinity of a clock. For
example, if 100 oscillations of a spring-mass system near the clock are
measured to have less total time than what the laws of physics would
suggest, then this is an indication that the clock is running slow.

However, it does NOT make sense to use distant clocks or clocks in
different states of motion to determine whether this clock can be said to
be running slow. To say it DOES make sense has buried in it the assumption
that time is absolute and global, when it is not.

>
> So if the traveler is moving along one leg of a polygon (say a square)
> with velocity V relative to the inertial frame the polygon is in, and
> there is a clock at rest at the center of the polygon, what does the
> traveler say the readings of his clock relative to the clock at the
> center of the polygon is? The Lorentz transform gives the result that the
> clock at rest in the center of the polygon is running at a slower rate
> than the traveler's clock.
> David Seppala
> Bastrop TX
>

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

On Monday, 25 April 2022 at 18:28:50 UTC+2, bodk...@gmail.com wrote:

> No, not necessarily. And this is where things need to be stated carefully.
> Whether a clock is running slow or fast is determined LOCALLY, by comparing
> to some physical process that is happening in the vicinity of a clock.

In the dreams of a brainwashed fanatic idiot. Not in the
real GPS, TAI, UTC.

On Monday, April 25, 2022 at 11:24:16 AM UTC-5, rotchm wrote:
> On Monday, April 25, 2022 at 11:57:59 AM UTC-4, sep...@yahoo.com wrote:
>
> > If a clock always shows a lesser value than another clock, and that difference in times
> > between the two clocks keeps getting greater and greater with every reading of the clocks,
> > doesn't that mean than one clock is running at a slower rate than the other clock?
> No. It means " that difference in times between the two clocks keeps getting greater
> and greater with every reading of the clocks".
>
> There is no need to introduce the expression "running at a slower rate", especially that that expression is ambiguous and meaningless. 20 < 40 is far more clear, precise and succinct. You don't agree with that?
> > So if the traveler is moving along one leg of a polygon (say a square)
> Lets say, a circle.
>
> > with velocity V
>
> With *speed* v along this circle.
> > relative to the inertial frame the polygon is in, and there is a clock at rest at the center of the polygon,
> the circle...
> > what does the traveler say the readings of his clock relative to the clock at the center of the polygon is?
> He says nothing about it, since he is not at the center; he is not performing any measurements, any comparisons.
> Don't forget, physics is about what is "displayed" on instruments, "observations", "comparisons".
> If you are requiring no observations then there is nothing to say about it.
> > The Lorentz transform gives the result that the clock at rest in the center of the polygon
> > is running at a slower rate than the traveler's clock.
> No, the LT's don't say that at all. The LT's say for instance a<b.
> *people* say/call/express this as "slower" .
> I rather be succinct and clear: a< b; t = 20 etc.

So if there is a light and a clock at rest at the center of the polygon or circle and the light flashes once each second shown on the clock, how many times per second does the traveler traveling around the polygon or circle say the light flashes as shown by his clock?
David Seppala
Bastrop TX

On Monday, April 25, 2022 at 12:40:33 PM UTC-4, sep...@yahoo.com wrote:

> So if there is a light and a clock at rest at the center of the polygon or circle and the light flashes
> once each second shown on the clock, how many times per second does the traveler
> traveling around the polygon or circle say the light flashes as shown by his clock?

For simplicity, the "polygon" will be a circle.
The traveler is moving at speed ß.
The time interval (difference in values) of two successive flashes (of the center source/clock)
is delta_t = 1.

Using the LT's, delta_t' = S g*dt = S √(1-ß²) dt = √(1-ß²) * 1 = √(1-ß²).
[I assume you have the brains to figure out what is meant by the S and g].

That is, the difference in values of the travelers clock as he coincides with two successive flashes
(along the circumference) is √(1-ß²) < 1.

Sloppy people would say that this traveling clock "ticks at a slower rate".
But better to say what the LT's actually say: √(1-ß²) < 1..

On Monday, April 25, 2022 at 8:57:59 AM UTC-7, sep...@yahoo.com wrote:
> If a clock always shows a lesser value than another clock, and that difference in
> times between the two clocks keeps getting greater and greater with every reading
> of the clocks, doesn't that mean than one clock is running at a slower rate than
> the other clock?

Again, a clock at rest in terms of F0 is running slow in terms of the inertial coordinates Fi in which the traveling clock is at rest at the moment when they pass, and likewise the traveling clock at rest in Fi at that moment is running slow in terms of F0. The system F0 and Fi are reciprocal. This is exactly like the two reciprocal rows of clocks that you ignored. These objective facts can be measured and verified by grids of clocks at rest and inertially synchronized in the respective frames.

Now suppose you ask: "For the circular loop, what is the rate of the clock at rest in F0 in terms of a system of accelerating coordinates C_accel, T,X, in which the traveler is continuously stationary (constant space coordinates) during his transit around the loop?" The answer to that is dtau/dT = 2, which arises from two countervailing effects, namely, the factor of 1/2 relating the instantaneous inertial coordinate systems, and the factor of 4 arising from the continual shift in the simultaneity mapping between the sequence of inertial systems.

Remember, your subject line asks about the Lorentz transformation (which you erroneously still call a "transform"), and as you know those transformations give the relationship between standard inertial coordinate systems. So, to answer the question in your subject line, yes, the Lorentz transformations are restricted to inertial coordinate systems. This is explained at the very beginning of Relativity 101.

If this is still unclear to you, could you describe what specifically is unclear? If you can't, it's difficult to know how to clarify it.

On Monday, April 25, 2022 at 1:39:01 PM UTC-5, Al Coe wrote:
> On Monday, April 25, 2022 at 8:57:59 AM UTC-7, sep...@yahoo.com wrote:
> > If a clock always shows a lesser value than another clock, and that difference in
> > times between the two clocks keeps getting greater and greater with every reading
> > of the clocks, doesn't that mean than one clock is running at a slower rate than
> > the other clock?
> Again, a clock at rest in terms of F0 is running slow in terms of the inertial coordinates Fi in which the traveling clock is at rest at the moment when they pass, and likewise the traveling clock at rest in Fi at that moment is running slow in terms of F0. The system F0 and Fi are reciprocal. This is exactly like the two reciprocal rows of clocks that you ignored. These objective facts can be measured and verified by grids of clocks at rest and inertially synchronized in the respective frames.
>
> Now suppose you ask: "For the circular loop, what is the rate of the clock at rest in F0 in terms of a system of accelerating coordinates C_accel, T,X, in which the traveler is continuously stationary (constant space coordinates) during his transit around the loop?" The answer to that is dtau/dT = 2, which arises from two countervailing effects, namely, the factor of 1/2 relating the instantaneous inertial coordinate systems, and the factor of 4 arising from the continual shift in the simultaneity mapping between the sequence of inertial systems.
>
> Remember, your subject line asks about the Lorentz transformation (which you erroneously still call a "transform"), and as you know those transformations give the relationship between standard inertial coordinate systems. So, to answer the question in your subject line, yes, the Lorentz transformations are restricted to inertial coordinate systems. This is explained at the very beginning of Relativity 101.
>
> If this is still unclear to you, could you describe what specifically is unclear? If you can't, it's difficult to know how to clarify it.
Clarify this: If there are two clocks at rest in an inertial reference frame at the same location, and Clock One reads 10 seconds and Clock Two reads 20 seconds, and then you look at the same two clocks 400 seconds later and Clock One reads 410 seconds and Clock 2 reads 820 seconds, and you look at them a third time and Clock One reads 600 seconds and Clock Two reads 1200 seconds, don't you say that Clock 2 is running at twice the clock rate of Clock 1?
If those same two clocks were side by side in an accelerating reference frame, and you got those same clock readings wouldn't you say that Clock 2 is running at twice the clock rate of Clock 1?
David Seppala
Bastrop TX

On 4/25/22 10:57 AM, sepp623@yahoo.com wrote:
> If a clock always shows a lesser value than another clock, and that
> difference in times between the two clocks keeps getting greater and
> greater with every reading of the clocks, doesn't that mean than one
> clock is running at a slower rate than the other clock?

If the clocks are co-located and co-moving, yes. In such a case the
clocks are clearly not identical. In physics nobody ever does that.

If the clocks are not co-located and co-moving, then you must describe
precisely how their readings are compared before any conclusion can be
reached.

[In gedankens like this we always consider identical,
ideal clocks, which implies their proper tick rates
are identical -- neither "runs at a slower rate" than
the other. This is independent of how they might be
compared.]

> So if the traveler is moving along one leg of a polygon (say a
> square) with velocity V relative to the inertial frame the polygon
> is in, and there is a clock at rest at the center of the polygon,
> what does the traveler say the readings of his clock relative to the
> clock at the center of the polygon is?

This depends on how the clocks' readings are compared/observed. Your
incomplete descriptions are unable to determine any useful conclusions.
But keep in mind that identical clocks always tick at identical proper
rates. Also keep in mind that in English, "a clock's tick rate" is
referring to its intrinsic (proper) tick rate, unless that phrase is
qualified with what tick rate is being discussed.

> The Lorentz transform gives the result that the clock at rest in the
> center of the polygon is running at a slower rate than the
> traveler's clock.

IF AND ONLY IF the traveler uses the usual method of measuring a
moving clock's tick rate -- that involves additional clocks at rest in
the traveler's inertial frame and synchronized in that frame. You don't
mention any of that....

If the traveler simply looks at the other clock (i.e. obtaining its
values via EM signals), then Doppler shift applies, and the traveler
could see the signals from the other clock ticking either faster or
slower than their own clock, depending on whether they are approaching
or receding from the other clock. This, of course, is NOT a comparison
of the other clock's tick rate to their own clock's rate, it is a
comparison of the SIGNALS' tick rate.

Tom Roberts

On Monday, April 25, 2022 at 3:46:25 PM UTC-5, tjrob137 wrote:
> On 4/25/22 10:57 AM, sep...@yahoo.com wrote:
> > If a clock always shows a lesser value than another clock, and that
> > difference in times between the two clocks keeps getting greater and
> > greater with every reading of the clocks, doesn't that mean than one
> > clock is running at a slower rate than the other clock?
> If the clocks are co-located and co-moving, yes. In such a case the
> clocks are clearly not identical. In physics nobody ever does that.
>
> If the clocks are not co-located and co-moving, then you must describe
> precisely how their readings are compared before any conclusion can be
> reached.
>
> [In gedankens like this we always consider identical,
> ideal clocks, which implies their proper tick rates
> are identical -- neither "runs at a slower rate" than
> the other. This is independent of how they might be
> compared.]
> > So if the traveler is moving along one leg of a polygon (say a
> > square) with velocity V relative to the inertial frame the polygon
> > is in, and there is a clock at rest at the center of the polygon,
> > what does the traveler say the readings of his clock relative to the
> > clock at the center of the polygon is?
> This depends on how the clocks' readings are compared/observed. Your
> incomplete descriptions are unable to determine any useful conclusions.
> But keep in mind that identical clocks always tick at identical proper
> rates. Also keep in mind that in English, "a clock's tick rate" is
> referring to its intrinsic (proper) tick rate, unless that phrase is
> qualified with what tick rate is being discussed.

In response to:
> > The Lorentz transform gives the result that the clock at rest in the
> > center of the polygon is running at a slower rate than the
> > traveler's clock.
Tom wrote:
> IF AND ONLY IF the traveler uses the usual method of measuring a
> moving clock's tick rate -- that involves additional clocks at rest in
> the traveler's inertial frame and synchronized in that frame. You don't
> mention any of that....
>
Let's say the traveler is given the information on the light source, knowing what the wavelength of the light is and the diameter of the beam when the light is at rest in F0, and observers in an inertial reference frame moving with velocity V=c*sqrt(3)/2 along the x-axis relative to the light know that the light is a distance of 5*sqrt(3)/2 light-seconds away from the x-axis and they had previously measured the flash rate of the light and made measurements of the size of the beam and the beams wavelength as it was received in previous experiments. Let the traveler have all this experimental data on hand so that as each flash of light is received by the traveler he immediately knows how far the flash of light traveled to meet him. He can then measure the received times of any two flashes, and knowing the distance each flash traveled he can compute the time between flashes. So he no longer needs an array of other observers and clocks.
What rate does the traveler say the beam of light is flashing during any leg of the polygon (square in this post)?
David Seppala
Bastrop TX

On Monday, April 25, 2022 at 12:41:18 PM UTC-7, sep...@yahoo.com wrote:
> Clarify this: If there are two clocks at rest in an inertial reference frame
> at the same location, and Clock One reads 10 seconds and Clock Two reads
> 20 seconds, and then you look at the same two clocks 400 seconds later
>[at the same location] and Clock One reads 410 seconds and Clock 2 reads
> 820 seconds... don't you say that Clock 2 is running at twice the clock rate
> of Clock 1?

In that case the *average* rate of clock 2 is twice the *average* rate of clock 1 during that interval, in terms of any specified system, but you cannot say that the instantaneous rate of clock 2 is always twice the rate of clock 1 in terms of arbitrary specified systems.

For example, in terms of the inertial coordinate system F1 in which the traveler is at rest while on edge 1, the source clock is running at half the rate of the "traveling clock" (which is at rest in that frame), but when the traveling clock is on edge 2 it is going much faster (in the opposite direction) in terms of F1, and if you figure out the average rates around the entire square, all in terms of F1, you find that the average rate of the traveling clock is half the average rate of the source clock (which is constant), even though it has twice the rate of the source clock while on edge 1. Similar results apply in terms of F2, and F3 and F4 and every other system of inertial coordinates. Understand?

On Monday, April 25, 2022 at 3:41:18 PM UTC-4, sep...@yahoo.com wrote:
> On Monday, April 25, 2022 at 1:39:01 PM UTC-5, Al Coe wrote:

> Clarify this: If there are two clocks at rest in an inertial reference frame at the same location,
> and Clock One reads 10 seconds and Clock Two reads 20 seconds, and then you look at the
> same two clocks 400 seconds later and Clock One reads 410 seconds and Clock 2 reads 820
> seconds, and you look at them a third time and Clock One reads 600 seconds and Clock
> Two reads 1200 seconds, don't you say that Clock 2 is running at twice the clock rate of Clock 1?

No *I* don't. I say " Clock One reads 600 seconds and Clock Two reads 1200 seconds"
isn't that more succinct?

> If those same two clocks were side by side in an accelerating reference frame, and you got
> those same clock readings wouldn't you say that Clock 2 is running at twice the clock rate of Clock 1?

As above, no.

On Monday, April 25, 2022 at 1:18:48 PM UTC-4, rotchm wrote:
> On Monday, April 25, 2022 at 12:40:33 PM UTC-4, sep...@yahoo.com wrote:
>
> > So if there is a light and a clock at rest at the center of the polygon or circle and the light flashes
> > once each second shown on the clock, how many times per second does the traveler
> > traveling around the polygon or circle say the light flashes as shown by his clock?
> For simplicity, the "polygon" will be a circle.
> The traveler is moving at speed ß.
> The time interval (difference in values) of two successive flashes (of the center source/clock)
> is delta_t = 1.
>
> Using the LT's, delta_t' = S g*dt = S √(1-ß²) dt = √(1-ß²) * 1 = √(1-ß²).
> [I assume you have the brains to figure out what is meant by the S and g]..
>
> That is, the difference in values of the travelers clock as he coincides with two successive flashes
> (along the circumference) is √(1-ß²) < 1.
>
> Sloppy people would say that this traveling clock "ticks at a slower rate".
> But better to say what the LT's actually say: √(1-ß²) < 1.

No comment/rebuttal?
So you agree with SR now?

On Monday, April 25, 2022 at 12:41:18 PM UTC-7, sep...@yahoo.com wrote:
> Clarify this: If there are two clocks at rest in an inertial reference frame
> at the same location, and Clock One reads 10 seconds and Clock Two reads
> 20 seconds, and then you look at the same two clocks 400 seconds later
>[at the same location] and Clock One reads 410 seconds and Clock 2 reads
> 820 seconds... don't you say that Clock 2 is running at twice the clock rate
> of Clock 1?

The *average* rate of clock 2 is twice the *average* rate of clock 1 during that interval, in terms of any specified system, but the instantaneous rate of clock 2 is not always twice the instantaneous rate of clock 1 in terms of arbitrary specified systems.
For example, in terms of the inertial coordinate system F1 in which the traveler is at rest while on edge 1, the source clock is running at half the rate of the "traveling clock" (which is at rest in that frame), but when the traveling clock is on edge 2 it is going much faster (in the opposite direction) in terms of F1, and faster still on edge 3, and if you figure out the average rates around the entire square, all in terms of F1, you find that the average rate of the traveling clock is half the average rate of the source clock (which is constant in terms of these coordinates), even though it has twice the rate of the source clock while on edge 1. Similar results apply in terms of F2, and F3 and F4 and every other system of inertial coordinates.

> Let's say the traveler is given the information on the light source, knowing what
> the wavelength of the light is and the diameter of the beam when the light is at
> rest in F0... so he no longer needs an array of other observers and clocks.

You're not making any sense. First, there is no "array of observers", we just have a grid of standard rulers and standard clocks at rest and inertially synchronized in any given inertial frame, and these can record all distances and angles and times in terms of that inertial coordinate system. All the Doppler frequency effects and spreading of the beam etc. are whatever they are in terms of these coordinates. All of this is fully specified and has been experimentally demonstrated. It's been shown that inertial coordinate systems are related by Lorentz transformations.

> What rate does the traveler say the beam of light is flashing during any leg of the polygon (square in this post)?

The answer to this question, along with fully detailed explanation, has been provided to you countless times. Can you explain what part of the explanation you think is wrong or unclear? (This calls for you to actually refer to the explanation, not simply repeat your question.) Also, where is your promised contradiction?

On Monday, 25 April 2022 at 22:46:25 UTC+2, tjrob137 wrote:

> [In gedankens like this we always consider identical,
> ideal clocks, which implies their proper tick rates
> are identical -- neither "runs at a slower rate" than
> the other. This is independent of how they might be
> compared.]

And in reality forbidden by your insane religion TAI and GPS
keep measuring t'=t, just like all serious clocks always
did.

On Tuesday, April 26, 2022 at 1:29:41 AM UTC-5, Al Coe wrote:
> On Monday, April 25, 2022 at 12:41:18 PM UTC-7, sep...@yahoo.com wrote:
> > Clarify this: If there are two clocks at rest in an inertial reference frame
> > at the same location, and Clock One reads 10 seconds and Clock Two reads
> > 20 seconds, and then you look at the same two clocks 400 seconds later
> >[at the same location] and Clock One reads 410 seconds and Clock 2 reads
> > 820 seconds... don't you say that Clock 2 is running at twice the clock rate
> > of Clock 1?
>
> The *average* rate of clock 2 is twice the *average* rate of clock 1 during that interval, in terms of any specified system, but the instantaneous rate of clock 2 is not always twice the instantaneous rate of clock 1 in terms of arbitrary specified systems.
> For example, in terms of the inertial coordinate system F1 in which the traveler is at rest while on edge 1, the source clock is running at half the rate of the "traveling clock" (which is at rest in that frame), but when the traveling clock is on edge 2 it is going much faster (in the opposite direction) in terms of F1, and faster still on edge 3, and if you figure out the average rates around the entire square, all in terms of F1, you find that the average rate of the traveling clock is half the average rate of the source clock (which is constant in terms of these coordinates), even though it has twice the rate of the source clock while on edge 1. Similar results apply in terms of F2, and F3 and F4 and every other system of inertial coordinates.
> > Let's say the traveler is given the information on the light source, knowing what
> > the wavelength of the light is and the diameter of the beam when the light is at
> > rest in F0... so he no longer needs an array of other observers and clocks.
>
> You're not making any sense. First, there is no "array of observers", we just have a grid of standard rulers and standard clocks at rest and inertially synchronized in any given inertial frame, and these can record all distances and angles and times in terms of that inertial coordinate system. All the Doppler frequency effects and spreading of the beam etc. are whatever they are in terms of these coordinates. All of this is fully specified and has been experimentally demonstrated. It's been shown that inertial coordinate systems are related by Lorentz transformations.
> > What rate does the traveler say the beam of light is flashing during any leg of the polygon (square in this post)?
> The answer to this question, along with fully detailed explanation, has been provided to you countless times. Can you explain what part of the explanation you think is wrong or unclear? (This calls for you to actually refer to the explanation, not simply repeat your question.) Also, where is your promised contradiction?

In your response you posted:
"> The *average* rate of clock 2 is twice the *average* rate of clock 1 during that interval, in terms of any specified system, but the instantaneous rate of clock 2 is not always twice the instantaneous rate of clock 1 in terms of arbitrary specified systems.
> For example, in terms of the inertial coordinate system F1 in which the traveler is at rest while on edge 1, the source clock is running at half the rate of the "traveling clock" (which is at rest in that frame), but when the traveling clock is on edge 2 it is going much faster (in the opposite direction) in terms of F1, and faster still on edge 3, and if you figure out the average rates around the entire square, all in terms of F1, you find that the average rate of the traveling clock is half the average rate of the source clock (which is constant in terms of these coordinates), even though it has twice the rate of the source clock while on edge 1. Similar results apply in terms of F2, and F3 and F4 and every other system of inertial coordinates."

Your post refers how each inertial frame F1, F2, F3, F4 measures the flash rate of the light that is in the inertial reference frame of the polygon(square). What does the traveler measure as the flash rate of the light during each leg of the journey around the polygon?
David Seppala
Bastrop TX

On Tuesday, April 26, 2022 at 12:42:40 PM UTC-4, sep...@yahoo.com wrote:
> On Tuesday, April 26, 2022 at 1:29:41 AM UTC-5, Al Coe wrote:

> Your post refers how
<snip>

Before you jump into a new topic, you should finish your previous ones instead of letting them dangling.
So, go finish your previous pending discussions, then we can continue with these newer ones.

On Tuesday, April 26, 2022 at 12:54:38 PM UTC-5, rotchm wrote:
> On Tuesday, April 26, 2022 at 12:42:40 PM UTC-4, sep...@yahoo.com wrote:
> > On Tuesday, April 26, 2022 at 1:29:41 AM UTC-5, Al Coe wrote:
> > Your post refers how
> <snip>
>
> Before you jump into a new topic, you should finish your previous ones instead of letting them dangling.
> So, go finish your previous pending discussions, then we can continue with these newer ones.
That isn't a new topic. That's the topic of my original post
David Seppala
Bastrop TX

On Tuesday, April 26, 2022 at 9:42:40 AM UTC-7, sep...@yahoo.com wrote:
> Your post refers how each inertial frame F1, F2, F3, F4 measures the flash rate of
> the light that is in the inertial reference frame of the polygon(square).

Frames don't measure things. Things are measured in terms of frames - or, more correctly, coordinate systems.

> What does the traveler measure as the flash rate of the light during each leg
> of the journey around the polygon?

How many times do you need to be given the answer to this question? Look, I encouraged you to not just repeat your question, but to actually point out what you think is wrong or unclear about the answer you've been given (a dozen times). In response, you just repeat your question, without pointing out anything wrong or unclear about the answer. This is why you never learn anything.

Again, everyone measures everything the same... relativity is not a subjectivist theory with different realities for people in different states of motion. If we want to measure the times and places of events in terms of F1, we use a grid of clocks and rulers at rest and inertially synchronized in terms of F1. To illustrate, consider a square path at rest in F0, and a clock is stationary at one corner, and another clock begins at that corner and moves at speed v around the square. The size of the square is such that it takes 1 year in terms of F0 for the moving clock to traverse each edge. When the traveling clock gets back to the original corner, 4 years of proper time have elapsed on that corner clock, but only 4*sqrt(1-v^2) years of proper time have elapsed on the traveling clock.

Now (try to concentrate), in terms of F1, the inertial coordinates in which the traveler is at rest while moving on edge 1, the elapsed proper time on the traveling clock from one corner to the next is sqrt(1-v^2), and the elapsed time for the "stationary" corner clock between the coordinate times of those turning events in terms of F1 is 1-v^2 year for the first leg, and for the second leg it is 1 year, and for the third leg it is 1+v^2 year, and for the forth leg it is 1 year, for a total of 4 years.

So, in terms of the traveler's rest frame while moving on edge 1, it has an elapsed proper time of sqrt(1-v^2) between the first two corners, during which time the stay-behind corner clock has an elapsed time of just 1-v^2 years. So, these measurements show that the stay-behind clock is slow by the factor sqrt(1-v^2) compared with the traveling clock for this leg. Understand?

On Tuesday, April 26, 2022 at 2:18:22 PM UTC-5, Al Coe wrote:
> On Tuesday, April 26, 2022 at 9:42:40 AM UTC-7, sep...@yahoo.com wrote:
> > Your post refers how each inertial frame F1, F2, F3, F4 measures the flash rate of
> > the light that is in the inertial reference frame of the polygon(square).
> Frames don't measure things. Things are measured in terms of frames - or, more correctly, coordinate systems.
> > What does the traveler measure as the flash rate of the light during each leg
> > of the journey around the polygon?
> How many times do you need to be given the answer to this question? Look, I encouraged you to not just repeat your question, but to actually point out what you think is wrong or unclear about the answer you've been given (a dozen times). In response, you just repeat your question, without pointing out anything wrong or unclear about the answer. This is why you never learn anything.
>
> Again, everyone measures everything the same... relativity is not a subjectivist theory with different realities for people in different states of motion. If we want to measure the times and places of events in terms of F1, we use a grid of clocks and rulers at rest and inertially synchronized in terms of F1. To illustrate, consider a square path at rest in F0, and a clock is stationary at one corner, and another clock begins at that corner and moves at speed v around the square. The size of the square is such that it takes 1 year in terms of F0 for the moving clock to traverse each edge. When the traveling clock gets back to the original corner, 4 years of proper time have elapsed on that corner clock, but only 4*sqrt(1-v^2) years of proper time have elapsed on the traveling clock.
>
> Now (try to concentrate), in terms of F1, the inertial coordinates in which the traveler is at rest while moving on edge 1, the elapsed proper time on the traveling clock from one corner to the next is sqrt(1-v^2), and the elapsed time for the "stationary" corner clock between the coordinate times of those turning events in terms of F1 is 1-v^2 year for the first leg, and for the second leg it is 1 year, and for the third leg it is 1+v^2 year, and for the forth leg it is 1 year, for a total of 4 years.
>
> So, in terms of the traveler's rest frame while moving on edge 1, it has an elapsed proper time of sqrt(1-v^2) between the first two corners, during which time the stay-behind corner clock has an elapsed time of just 1-v^2 years. So, these measurements show that the stay-behind clock is slow by the factor sqrt(1-v^2) compared with the traveling clock for this leg. Understand?

Using the numbers I've been using, where the length of each side is 10 * sqrt(3)/2 light-seconds, in F0 the time takes 40 seconds to complete one trip around the square, whereas the traveler shows that the time for a complete trip is 20 seconds - which agrees with your results when you use the length results in a year travel time.
If I use this equation:
t2' - t1' = ((t2-t1) - (v/c**2*(x2-x1))) / sqrt(1-v**2/c**2)
if V = c*sqrt(3)/2 and in F0 x2 = x1 and the light in F0 flashes one flash per second, I get that
the time between those two flashes leaving the light source at rest in F0 is 2 seconds in F1.
Do you agree? If not, what is the time between flashes leaving the light source as seen in F1?
David Seppala
Bastrop TX

On Tuesday, April 26, 2022 at 2:28:54 PM UTC-4, sep...@yahoo.com wrote:

> That isn't a new topic.

Yes it is.
This shows just how confused you are.

On Tuesday, April 26, 2022 at 1:41:11 PM UTC-7, sep...@yahoo.com wrote:
> the time between those two flashes leaving the light source at rest in F0 is
> 2 seconds in F1. Do you agree?

Of course. We've been over this a dozen times. In terms of F1 (and also in terms of F2, and in terms of F3, and in terms of F4), an object at rest in F0 is moving at speed v, and therefore has a rate of elapsed proper time of dtau/dt = sqrt(1-v^2), which with your numbers is 1/2, so if the pulses are emitted once per second in terms of F0 they are emitted once per 2 seconds in terms of F1 (and in terms of F2, and in terms of F3, and in terms of F4). Remember? And in each of these frames individually the totals around the loop work out perfectly, and in the case of stitching the four together we take account of the disjoint simultaneity mappings to show that, again, the totals work out perfectly. Remember?

I ask again, what do you think is wrong or unclear in the explanations and answers that have been given to all your questions? And where is your promised contradiction?

On Tuesday, April 26, 2022 at 5:51:29 PM UTC-5, Al Coe wrote:
> On Tuesday, April 26, 2022 at 1:41:11 PM UTC-7, sep...@yahoo.com wrote:
> > the time between those two flashes leaving the light source at rest in F0 is
> > 2 seconds in F1. Do you agree?
> Of course. We've been over this a dozen times. In terms of F1 (and also in terms of F2, and in terms of F3, and in terms of F4), an object at rest in F0 is moving at speed v, and therefore has a rate of elapsed proper time of dtau/dt = sqrt(1-v^2), which with your numbers is 1/2, so if the pulses are emitted once per second in terms of F0 they are emitted once per 2 seconds in terms of F1 (and in terms of F2, and in terms of F3, and in terms of F4). Remember? And in each of these frames individually the totals around the loop work out perfectly, and in the case of stitching the four together we take account of the disjoint simultaneity mappings to show that, again, the totals work out perfectly. Remember?
>
> I ask again, what do you think is wrong or unclear in the explanations and answers that have been given to all your questions? And where is your promised contradiction?

So if the length of each leg of the polygon (square) is 10*sqrt(3)/2 light-seconds and the relative velocity of the traveler along each leg of the polygon is V = c*sqrt(3)/2 with respect to the flashing light at rest at the center of the polygon in F0 flashing at a rate of one flash per second, the traveler as shown on his clock takes 5 seconds to pass along each leg or 20 seconds to complete the journey. If the traveler while in F1 says the light at the center of the polygon flashes at a rate of one flash every two seconds when he is in F1, and one flash every two seconds when he is in F2, and one flash every two seconds when he is in F3, and one flash every two seconds when he is in F4, he must conclude that the light flashes 10 times for every complete trip the traveler makes around the polygon. However, he actually receives 40 flashes of light.
Your posts seem to look at the entire trip around the polygon from say only the F1 perspective as the traveler makes his entire trip around the polygon, but you never explain how the traveler always receives 40 flashes each trip around the polygon while the traveler says the light in F0 only flashed 10 times as he made each completed trip around the polygon (square). So where does the traveler say these extra flashes that he received came from?
David Seppala
Bastrop TX

On Wednesday, April 27, 2022 at 6:35:30 AM UTC-7, sep...@yahoo.com wrote:
>> And in each of these frames individually the totals around the loop work out
>> perfectly, and in the case of stitching the four together we take account of the
>> disjoint simultaneity mappings to show that, again, the totals work out perfectly.
>> Remember?
>
> ... he must conclude that the light flashes 10 times for every complete trip the
> traveler makes around the polygon.

No. What is wrong with your brain? This was explicitly explained, with the disjoint
simultaneity mappings. Remember? Look at my very first reply to you in this thread.

> Your posts seem to look at the entire trip around the polygon from say only the F1
> perspective as the traveler makes his entire trip around the polygon, but you never
> explain how the traveler always receives 40 flashes each trip around the polygon...

That is a lie. Look at my very first reply in this thread (and repeated in several
subsequent posts). If there is something in the explanation that you think is wrong
or unclear, go ahead and point it out.

On Wednesday, April 27, 2022 at 10:07:36 AM UTC-4, Al Coe wrote:
> On Wednesday, April 27, 2022 at 6:35:30 AM UTC-7, sep...@yahoo.com wrote:

> No. What is wrong with your brain? This was explicitly explained,...

> That is a lie. Look at my very first reply in this thread (and repeated in several
> subsequent posts). If there is something in the explanation that you think is wrong
> or unclear, go ahead and point it out.

Either this person is severely mentally ill or is a troll.
When we explain something simply and clearly to him, he just runs away on changes the subject.
Then comes back later on and ask the same questions again.

May be best to ignore this idiot from now on, and to report him as spam.