How does a traveler moving along segments of a polygon reconcile his clock rate during each segment with the clock rate of a clock in an inertial reference frame? Here's the problem.

In inertial reference frame F0 there is a polygon at rest centered at (0,0) as measured in F0. The polygon has N segments each with a length of L = c*sqrt(3)/2 light-seconds as measured in F0. A traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0 along each segment. As measured in F0, the traveler takes 1 second to travel across each segment of the polygon. Using the Lorentz transform, the traveler's clock only changes by one half second as he travels across any segment of the polygon. At the center of the polygon, there is a light source that sends a pulse of light each second. That pulse arrives at the starting point of each segment of the polygon just as the traveler happens to start traveling across that segment and the next pulse arrives at the end point of that segment and the starting point of the next segment just as the traveler gets there.
So each complete trip around the polygon the traveler receives N light pulses, his clock shows that N/2 seconds have elapsed, and the clock at (0,0) where the light source is in F0 shows that N seconds have elapsed. So the traveler must conclude that his clock is running at half the rate of the clock in F0.
How does the traveler determine that during any segment, the clock in F0 is running at half the rate of his own clock, yet the data shows that the traveler's clock is running at the slower rate? We can make the number of segments (N) of the polygon very large so that the change of direction (accelerations) at the end of each segment do not significantly affect the clock rate of the traveler's clock.
Thanks,
David Seppala
Bastrop TX

On Thursday, February 23, 2023 at 4:34:58 PM UTC-8, sep...@yahoo.com wrote:
> How does the traveler determine that during any segment, the clock in F0 is
> running at half the rate of his own clock...

This is a standard quiz question in any introductory course on special relativity. In terms of the standard inertial coordinates in which the traveler is at rest as he traverses the first segment (in half a second), the elapsed time of the hub is 1/4 of a second. Likewise in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the second segment (in half a second), the elapsed time of the hub is 1/4 of a second. And so on. The sum of these elapsed times for the hub during the N segments is N/4, compared with N/2 for the traveler, but this does not account for all of the hub's elapsed time, because there is 3/4 second between the hub event that is simultaneous with the traveler at a given vertex in terms of the inertial coordinate system of the traveler on the previous segment and on the next segment. Over the whole N segments this contributes 3N/4 seconds to the hub, for a total of N seconds.

Each of the inertial coordinate systems can be constructed as a grid of standard inertial rulers and clocks (inertially synchronized), and each of the assertions in the above explanation is perfectly measurable in terms of those rulers and clocks.

On 2/23/23 6:34 PM, sepp623@yahoo.com wrote:
> How does a traveler moving along segments of a polygon reconcile his
> clock rate during each segment with the clock rate of a clock in an
> inertial reference frame? Here's the problem.
>
> In inertial reference frame F0 there is a polygon at rest centered
> at (0,0) as measured in F0. The polygon has N segments each with a
> length of L = c*sqrt(3)/2 light-seconds as measured in F0. A
> traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0
> along each segment. As measured in F0, the traveler takes 1 second
> to travel across each segment of the polygon. Using the Lorentz
> transform, the traveler's clock only changes by one half second as
> he travels across any segment of the polygon. At the center of the
> polygon, there is a light source that sends a pulse of light each
> second. That pulse arrives at the starting point of each segment of
> the polygon just as the traveler happens to start traveling across
> that segment and the next pulse arrives at the end point of that
> segment and the starting point of the next segment just as the
> traveler gets there. So each complete trip around the polygon the
> traveler receives N light pulses, his clock shows that N/2 seconds
> have elapsed, and the clock at (0,0) where the light source is in F0
> shows that N seconds have elapsed. So the traveler must conclude that
> his clock is running at half the rate of the clock in F0. How does
> the traveler determine that during any segment, the clock in F0 is
> running at half the rate of his own clock, yet the data shows that
> the traveler's clock is running at the slower rate? We can make the
> number of segments (N) of the polygon very large so that the change
> of direction (accelerations) at the end of each segment do not
> significantly affect the clock rate of the traveler's clock.

You keep confusing yourself by attempting to use non-inertial frames. In
the inertial frame of the center this is easy: the elapsed proper time
of the traveler is \integral sqrt(1-v^2/c^2) dt, where t is the time
coordinate of that inertial frame, and v is the traveler's speed
relative to that frame. So the elapsed proper time of the center clock
is N, and the elapsed proper time of the traveler around the polygon is
N/2. Since these are both invariant, the traveler must conclude the same.

In contrast, when the traveler attempts to measure the rate of the clock
at the center, they must have multiple assistants at rest in the
traveler's inertial frame that are pre-arranged so an assistant is
located at the center when the traveler begins a segment, and another
when the traveler ends a segment. This must be setup separately for each
segment. After completing the trip, when the traveler combines the
reports of all those assistants, they conclude the center clock ticks at
half the rate of their own clock. But where in your description are all
those assistants?

Bottom line: "time dilation" applies only to two inertial frames
measuring each other's clocks. Your scenario does not conform to the
inherent limitation, so you cannot simply apply the equation of "time
dilation", you must explicitly perform the calculation.

Tom Roberts

On Friday, February 24, 2023 at 1:34:58 AM UTC+1, sep...@yahoo.com wrote:
> How does a traveler moving along segments of a polygon reconcile his clock rate during each segment with the clock rate of a clock in an inertial reference frame? Here's the problem.
>
> In inertial reference frame F0 there is a polygon at rest centered at (0,0) as measured in F0. The polygon has N segments each with a length of L = c*sqrt(3)/2 light-seconds as measured in F0. A traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0 along each segment. As measured in F0, the traveler takes 1 second to travel across each segment of the polygon. Using the Lorentz transform, the traveler's clock only changes by one half second as he travels across any segment of the polygon. At the center of the polygon, there is a light source that sends a pulse of light each second. That pulse arrives at the starting point of each segment of the polygon just as the traveler happens to start traveling across that segment and the next pulse arrives at the end point of that segment and the starting point of the next segment just as the traveler gets there.
> So each complete trip around the polygon the traveler receives N light pulses, his clock shows that N/2 seconds have elapsed, and the clock at (0,0) where the light source is in F0 shows that N seconds have elapsed. So the traveler must conclude that his clock is running at half the rate of the clock in F0.
> How does the traveler determine that during any segment, the clock in F0 is running at half the rate of his own clock, yet the data shows that the traveler's clock is running at the slower rate? We can make the number of segments (N) of the polygon very large so that the change of direction (accelerations) at the end of each segment do not significantly affect the clock rate of the traveler's clock.
> Thanks,
> David Seppala
> Bastrop TX

You keep asking the same questions for more than a decade now.
Are you actually learning anything?

--
Jan

On Friday, 24 February 2023 at 07:12:14 UTC+1, Tom Roberts wrote:
> On 2/23/23 6:34 PM, sep...@yahoo.com wrote:
> > How does a traveler moving along segments of a polygon reconcile his
> > clock rate during each segment with the clock rate of a clock in an
> > inertial reference frame? Here's the problem.
> >
> > In inertial reference frame F0 there is a polygon at rest centered
> > at (0,0) as measured in F0. The polygon has N segments each with a
> > length of L = c*sqrt(3)/2 light-seconds as measured in F0. A
> > traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0
> > along each segment. As measured in F0, the traveler takes 1 second
> > to travel across each segment of the polygon. Using the Lorentz
> > transform, the traveler's clock only changes by one half second as
> > he travels across any segment of the polygon. At the center of the
> > polygon, there is a light source that sends a pulse of light each
> > second. That pulse arrives at the starting point of each segment of
> > the polygon just as the traveler happens to start traveling across
> > that segment and the next pulse arrives at the end point of that
> > segment and the starting point of the next segment just as the
> > traveler gets there. So each complete trip around the polygon the
> > traveler receives N light pulses, his clock shows that N/2 seconds
> > have elapsed, and the clock at (0,0) where the light source is in F0
> > shows that N seconds have elapsed. So the traveler must conclude that
> > his clock is running at half the rate of the clock in F0. How does
> > the traveler determine that during any segment, the clock in F0 is
> > running at half the rate of his own clock, yet the data shows that
> > the traveler's clock is running at the slower rate? We can make the
> > number of segments (N) of the polygon very large so that the change
> > of direction (accelerations) at the end of each segment do not
> > significantly affect the clock rate of the traveler's clock.
> You keep confusing yourself by attempting to use non-inertial frames. In
> the inertial frame of the center this is easy: the elapsed proper time
> of the traveler is \integral sqrt(1-v^2/c^2) dt, where t is the time

And in the meantime in the real world forbidden by
your insane cult GPS and TAI keep measuring t'=t.
Your screams "We're FORCED!!!!" somehow didn't
work.

On February 24, JanPB wrote:
> February 23, David Seppala wrote:
>> How does a traveler moving along segments of a polygon reconcile his clock
>> rate during each segment with the clock rate of a clock in an inertial
>> reference frame? Here's the problem.
>
> You keep asking the same questions for more than a decade now.
> Are you actually learning anything?

Learning isn't the motive.

An elementary result of probability theory says that,
in an endless sequence of repeated trials, for any non-zero
probability outcome, that outcome will eventually occur
with certainty.

So if any of David's refutations of relativity has non-zero
chance of validity, he will eventually be vindicated, hence
forever famous as the man who buried Einstein.

"If you bounce a tennis ball against a wall often enough,
eventually it will sail right through."
- Werner Heisenberg

--
Rich

On Friday, February 24, 2023 at 10:55:08 AM UTC-8, RichD wrote:
> So if any of David's refutations of relativity has non-zero
> chance of validity,

It doesn't. Besides, a theory can be invalidated by experiment only, Seppallotto is not capable of running any experiment.

On Friday, 24 February 2023 at 22:32:36 UTC+1, Dono. wrote:
> On Friday, February 24, 2023 at 10:55:08 AM UTC-8, RichD wrote:
>
> > So if any of David's refutations of relativity has non-zero
> > chance of validity,
> It doesn't. Besides, a theory can be invalidated by experiment only

Because a fanatic idiot is deeply believing that.

Den 24.02.2023 01:34, skrev sepp623@yahoo.com:
> How does a traveler moving along segments of a polygon reconcile his clock rate during each segment with the clock rate of a clock in an inertial reference frame? Here's the problem.
>
> In inertial reference frame F0 there is a polygon at rest centered at (0,0) as measured in F0. The polygon has N segments each with a length of L = c*sqrt(3)/2 light-seconds as measured in F0. A traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0 along each segment. As measured in F0, the traveler takes 1 second to travel across each segment of the polygon. Using the Lorentz transform, the traveler's clock only changes by one half second as he travels across any segment of the polygon. At the center of the polygon, there is a light source that sends a pulse of light each second. That pulse arrives at the starting point of each segment of the polygon just as the traveler happens to start traveling across that segment and the next pulse arrives at the end point of that segment and the starting point of the next segment just as the traveler gets there.
> So each complete trip around the polygon the traveler receives N light pulses, his clock shows that N/2 seconds have elapsed, and the clock at (0,0) where the light source is in F0 shows that N seconds have elapsed. So the traveler must conclude that his clock is running at half the rate of the clock in F0.

OK.

> How does the traveler determine that during any segment, the clock in F0 is running at half the rate of his own clock, yet the data shows that the traveler's clock is running at the slower rate? We can make the number of segments (N) of the polygon very large so that the change of direction (accelerations) at the end of each segment do not significantly affect the clock rate of the traveler's clock.

The answer is that the traveller can't with his single clock
measure that the coordinate clocks in F0 appear to run slow.
He can OTOH measure that the coordinate time in F0
appear to be running fast. When he compare his own clock to
the coordinate clocks at the beginning of each segment, he
will see that each consecutive clock show one second more
than the previous clock, while his own clock has advanced
only half a second.

See:
https://paulba.no/pdf/Mutual_time_dilation.pdf

--
Paul

https://paulba.no/

On Saturday, 25 February 2023 at 13:54:24 UTC+1, Paul B. Andersen wrote:
> Den 24.02.2023 01:34, skrev sep...@yahoo.com:
> > How does a traveler moving along segments of a polygon reconcile his clock rate during each segment with the clock rate of a clock in an inertial reference frame? Here's the problem.
> >
> > In inertial reference frame F0 there is a polygon at rest centered at (0,0) as measured in F0. The polygon has N segments each with a length of L = c*sqrt(3)/2 light-seconds as measured in F0. A traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0 along each segment. As measured in F0, the traveler takes 1 second to travel across each segment of the polygon. Using the Lorentz transform, the traveler's clock only changes by one half second as he travels across any segment of the polygon. At the center of the polygon, there is a light source that sends a pulse of light each second. That pulse arrives at the starting point of each segment of the polygon just as the traveler happens to start traveling across that segment and the next pulse arrives at the end point of that segment and the starting point of the next segment just as the traveler gets there.
> > So each complete trip around the polygon the traveler receives N light pulses, his clock shows that N/2 seconds have elapsed, and the clock at (0,0) where the light source is in F0 shows that N seconds have elapsed. So the traveler must conclude that his clock is running at half the rate of the clock in F0.
> OK.
> > How does the traveler determine that during any segment, the clock in F0 is running at half the rate of his own clock, yet the data shows that the traveler's clock is running at the slower rate? We can make the number of segments (N) of the polygon very large so that the change of direction (accelerations) at the end of each segment do not significantly affect the clock rate of the traveler's clock.
> The answer is that the traveller can't with his single clock
> measure that the coordinate clocks in F0 appear to run slow.

And in the meantime in the real world - forbidden
by your bunch of idiots GPS and TAI keep measuring
t'=t in forbidden by your bunch of idiots old seconds.

On Saturday, February 25, 2023 at 4:54:24 AM UTC-8, Paul B. Andersen wrote:
> The answer is that the traveller can't with his single clock
> measure that the coordinate clocks in F0 appear to run slow.

Measuring something about a distant and relatively moving clock "over there" with a single clock "over here" (and nothing else) is obviously never possible in any circumstance. There always needs to be some conveyance of information that specifies the situation of the distant object, and this can be done in infinitely many ways. In the example posed by Dave, we can stipulate that the traveler can ascertain (say, by looking at the readings on a grid of rulers as he passes them, or equivalently by a series of radar measurements, etc.) that he is moving along a polygonal path with a specified radius, and he can then use his clock to measure the time interval between the pulses received from the hub, and after accounting for aberration and Doppler, he can determine that the hub clock is running slow (half speed) in terms of his (the traveler's) co-moving inertial coordinate system. Also, don't say the clocks at rest in F0 "appear" to run slow in terms of the inertial coordinate system of the traveler on a given segment, say that they actually do run slow in terms of those coordinates.

Dave's question is a standard quiz question in any introductory course on special relativity, because students are often puzzled by how the rates can be related as they are will still getting the overall secular advance of the hub clock. The answer is that, in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the first segment (in half a second), the elapsed time of the hub is 1/4 of a second. Likewise in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the second segment (in half a second), the elapsed time of the hub is 1/4 of a second. And so on. The sum of these elapsed times for the hub during the N segments is N/4, compared with N/2 for the traveler, but this does not account for all of the hub's elapsed time, because there is 3/4 second between the hub event that is simultaneous with the traveler at a given vertex in terms of the inertial coordinate system of the traveler on the previous segment and on the next segment. Over the whole N segments this contributes 3N/4 seconds to the hub, for a total of N seconds.

Note that each of the inertial coordinate systems can be constructed as a grid of standard inertial rulers and clocks (inertially synchronized), and each of the assertions in the above explanation is perfectly measurable in terms of those rulers and clocks.

On Friday, February 24, 2023 at 12:12:14 AM UTC-6, Tom Roberts wrote:
> On 2/23/23 6:34 PM, sep...@yahoo.com wrote:
> > How does a traveler moving along segments of a polygon reconcile his
> > clock rate during each segment with the clock rate of a clock in an
> > inertial reference frame? Here's the problem.
> >
> > In inertial reference frame F0 there is a polygon at rest centered
> > at (0,0) as measured in F0. The polygon has N segments each with a
> > length of L = c*sqrt(3)/2 light-seconds as measured in F0. A
> > traveler travels at a speed of V = c*sqrt(3)/2 as measured in F0
> > along each segment. As measured in F0, the traveler takes 1 second
> > to travel across each segment of the polygon. Using the Lorentz
> > transform, the traveler's clock only changes by one half second as
> > he travels across any segment of the polygon. At the center of the
> > polygon, there is a light source that sends a pulse of light each
> > second. That pulse arrives at the starting point of each segment of
> > the polygon just as the traveler happens to start traveling across
> > that segment and the next pulse arrives at the end point of that
> > segment and the starting point of the next segment just as the
> > traveler gets there. So each complete trip around the polygon the
> > traveler receives N light pulses, his clock shows that N/2 seconds
> > have elapsed, and the clock at (0,0) where the light source is in F0
> > shows that N seconds have elapsed. So the traveler must conclude that
> > his clock is running at half the rate of the clock in F0. How does
> > the traveler determine that during any segment, the clock in F0 is
> > running at half the rate of his own clock, yet the data shows that
> > the traveler's clock is running at the slower rate? We can make the
> > number of segments (N) of the polygon very large so that the change
> > of direction (accelerations) at the end of each segment do not
> > significantly affect the clock rate of the traveler's clock.
> You keep confusing yourself by attempting to use non-inertial frames. In
> the inertial frame of the center this is easy: the elapsed proper time
> of the traveler is \integral sqrt(1-v^2/c^2) dt, where t is the time
> coordinate of that inertial frame, and v is the traveler's speed
> relative to that frame. So the elapsed proper time of the center clock
> is N, and the elapsed proper time of the traveler around the polygon is
> N/2. Since these are both invariant, the traveler must conclude the same.
>
> In contrast, when the traveler attempts to measure the rate of the clock
> at the center, they must have multiple assistants at rest in the
> traveler's inertial frame that are pre-arranged so an assistant is
> located at the center when the traveler begins a segment, and another
> when the traveler ends a segment. This must be setup separately for each
> segment. After completing the trip, when the traveler combines the
> reports of all those assistants, they conclude the center clock ticks at
> half the rate of their own clock. But where in your description are all
> those assistants?
>
> Bottom line: "time dilation" applies only to two inertial frames
> measuring each other's clocks. Your scenario does not conform to the
> inherent limitation, so you cannot simply apply the equation of "time
> dilation", you must explicitly perform the calculation.
>
> Tom Roberts
Tom,
If the traveler is traveling around a circle at rest in F0 instead of a polygon, can the traveler determine the distance from the center of the circle to his position or is it impossible for him to determine that length?
Thanks,
David Seppala

sepp623@yahoo.com wrote:

>> Tom Roberts
> Tom,
> If the traveler is traveling around a circle at rest in F0 instead of
> a polygon, can the traveler determine the distance from the center of
> the circle to his position or is it impossible for him to determine
> that length?
> Thanks,

at rest is NOT travelling.

On Saturday, February 25, 2023 at 11:13:44 AM UTC-8, sep...@yahoo.com wrote:
> If the traveler is traveling around a circle at rest in F0 instead of a polygon, can the
> traveler determine the distance from the center of the circle to his position...

Sure, he can determine any distance in terms of any specified system of coordinates, using any of the countless number of standard means. What is causing you to think that he could not?

On Thursday, February 23, 2023 at 4:34:58 PM UTC-8, sep...@yahoo.com wrote:
> How does the traveler determine that during any segment, the clock in F0 is
> running at half the rate of his own clock...

This is a standard question discussed in any introductory course on special relativity. In terms of the standard inertial coordinates in which the traveler is at rest as he traverses the first segment (in half a second), the elapsed time of the hub is 1/4 of a second. Likewise in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the second segment (in half a second), the elapsed time of the hub is 1/4 of a second. And so on. The sum of these elapsed times for the hub during the N segments is N/4, compared with N/2 for the traveler, but this does not account for all of the hub's elapsed time, because there is 3/4 second between the hub event that is simultaneous with the traveler at a given vertex in terms of the inertial coordinate system of the traveler on the previous segment and on the next segment. Over the whole N segments this contributes 3N/4 seconds to the hub, for a total of N seconds.

Each of the inertial coordinate systems can be constructed as a grid of standard inertial rulers and clocks (inertially synchronized), and each of the assertions in the above explanation is measurable in terms of those rulers and clocks.

On Saturday, February 25, 2023 at 4:54:24 AM UTC-8, Paul B. Andersen wrote:
> The answer is that the traveller can't with his single clock
> measure that the coordinate clocks in F0 appear to run slow.

Measuring something about a distant and relatively moving clock "over there" with a single clock "over here" (and nothing else) is obviously never possible in any circumstance. There always needs to be some conveyance of information that specifies the situation of the distant object, and this can be done in infinitely many different ways. In the example posed by Dave, we can stipulate that the traveler can ascertain (say, by looking at the readings on a grid of rulers as he passes them, or equivalently by a series of radar measurements, etc.) that he is moving along a polygonal path with a specified radius, and he can then use his clock to measure the time interval between the pulses received from the hub, and after accounting for aberration and Doppler, he can determine that the hub clock is running slow (half speed) in terms of his (the traveler's) co-moving inertial coordinate system. Also, don't say the clocks at rest in F0 "appear" to run slow in terms of the inertial coordinate system of the traveler on a given segment, say that they actually do run slow in terms of those coordinates.

On Saturday, 25 February 2023 at 18:25:02 UTC+1, Trevor Lange wrote:
> On Saturday, February 25, 2023 at 4:54:24 AM UTC-8, Paul B. Andersen wrote:
> > The answer is that the traveller can't with his single clock
> > measure that the coordinate clocks in F0 appear to run slow.
> Measuring something about a distant and relatively moving clock "over there" with a single clock "over here" (and nothing else) is obviously never possible in any circumstance.

Verty fortunate for your idiocies, that can't be
tested directly. Still, anyone can see in GPS,
the real clocks keep measuring t'=t, just like all
serious clock always did.

Den 25.02.2023 18:25, skrev Trevor Lange:
> On Saturday, February 25, 2023 at 4:54:24 AM UTC-8, Paul B. Andersen wrote:
>> The answer is that the traveller can't with his single clock
>> measure that the coordinate clocks in F0 appear to run slow.
>
> Measuring something about a distant and relatively moving clock "over there" with a single clock "over here" (and nothing else) is obviously never possible in any circumstance. There always needs to be some conveyance of information that specifies the situation of the distant object, and this can be done in infinitely many ways.
For a traveller with the only instrument "a clock",
the following is the only way:
> In the example posed by Dave, we can stipulate that the traveler can ascertain (say, by looking at the readings on a grid of rulers as he passes them,
Exactly!
As I said and you snipped:
>> He can OTOH measure that the coordinate time in F0
>> appear to be running fast. When he compare his own clock to
>> the coordinate clocks at the beginning of each segment, he
>> will see that each consecutive clock show one second more
>> than the previous clock, while his own clock has advanced
>> only half a second.
So he will see that the coordinate time in F0 passes
twice as fast as the proper time on his own clock.
> he can then use his clock to measure the time interval between the pulses received from the hub,
Exactly!
According to the definition of the scenario given by Dave,
the traveller will see the flash from the hub as he starts on
the next segment. According to his own clock, there is a half second
between the flashes, so he will observe (measure) the frequency of
the hub flashes to be 2Hz. If he knows that the proper frequency of
the flashes from the hub is 1Hz, he can conclude that the clock
at the hub in F0 appear to run twice as fast as his own clock.

> Also, don't say the clocks at rest in F0 "appear" to run slow in terms of the inertial coordinate system of the traveler on a given segment, say that they actually do run slow in terms of those coordinates.
Clocks always run at their proper rate by definition of a clock!
You can say: A clock in frame A runs slow as observed in frame B.
And you can say: A clock in frame A runs slow in terms of the inertial
coordinate system of frame B.
But only if the meaning of "in terms of the inertial coordinate
system of frame B" is "as observed/measured in frame B".
Because:
The speed of the observer relative to the observed object
can _not_ affect the observed object in any way.
But the speed of the observer relative to the observed object
_can_ affect the observer's observations of the observed object.
It is the _observation_ (measurement) of the clock that is affected.
The clock is _not_ affected, it runs at its proper rate, as always.
But if you say "the clock actually do run slow" it can hardly
be interpreted otherwise than that the clock is physically affected,
even if you add "in terms of those coordinates".
>
> Dave's question is a standard quiz question in any introductory course on special relativity,
Look.
Dave's question was:
"How does the traveler determine that during any segment,
the clock in F0 is running at half the rate of his own clock?"
Dave's question was NOT:
"Why is it that when two inertial frames are moving relative to
each other, then both frames will find that the clocks in
the other frame appear to run slow."
The answer to the former question is:
The traveller with a single clock can't.
The answer to the latter question can be found here:
https://paulba.no/pdf/Mutual_time_dilation.pdf
Note that all the four clocks always run at their proper
rate. None runs neither fast nor slow. EVER!

> because students are often puzzled by how the rates can be related as they are will still getting the overall secular advance of the hub clock. The answer is that, in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the first segment (in half a second), the elapsed time of the hub is 1/4 of a second. Likewise in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the second segment (in half a second), the elapsed time of the hub is 1/4 of a second. And so on. The sum of these elapsed times for the hub during the N segments is N/4, compared with N/2 for the traveler, but this does not account for all of the hub's elapsed time, because there is 3/4 second between the hub event that is simultaneous with the traveler at a given vertex in terms of the inertial coordinate system of the traveler on the previous segment and on the next segment. Over the whole N segments this contributes 3N/4 seconds to the hub, for a total of N seconds.
I don't find it very obvious what you are trying to explain,
but it probably is the same as this:
It is true that, measured in an inertial frame co-moving with
the traveller, the hub clock in F0 will advance only 1/4 second
during the half second it takes the travelling clock to traverse
a section. At the end of the section the traveller abruptly
changes the direction of his velocity. He is 'jumping frame',
that is changing co-moving inertial frame. Due to the fact that
simultaneity is different in different frames, this means that
the hub clock will appear to jump 3/4 seconds forward, so at
the beginning of the next section the hub clock will have
advanced 1 second compared to the beginning of the previous section.
(Remember that we _know_ that the traveller will see a flash from
the hub at the beginning of each section.)
So after a complete circle, the travellers clock will have advanced
N/2 seconds while the hub clock will have advanced N seconds.
>
> Note that each of the inertial coordinate systems can be constructed as a grid of standard inertial rulers and clocks (inertially synchronized), and each of the assertions in the above explanation is perfectly measurable in terms of those rulers and clocks.
--
Paul
https://paulba.no/

On Sunday, 26 February 2023 at 15:12:16 UTC+1, Paul B. Andersen wrote:
> Den 25.02.2023 18:25, skrev Trevor Lange:
> > On Saturday, February 25, 2023 at 4:54:24 AM UTC-8, Paul B. Andersen wrote:
> >> The answer is that the traveller can't with his single clock
> >> measure that the coordinate clocks in F0 appear to run slow.
>
> >
> > Measuring something about a distant and relatively moving clock "over there" with a single clock "over here" (and nothing else) is obviously never possible in any circumstance. There always needs to be some conveyance of information that specifies the situation of the distant object, and this can be done in infinitely many ways.
> For a traveller with the only instrument "a clock",
> the following is the only way:

Fortunately, we have GPS now, so everyone can
test your moronic bullshit directly and see
it's just some moronic bullshit, having nothing
in common with real clocks, real observations,
real measurements or real anything.

On 2/24/23 12:55 PM, RichD wrote:
> An elementary result of probability theory says that, in an endless
> sequence of repeated trials, for any non-zero probability outcome,
> that outcome will eventually occur with certainty.

How silly. The math underlying SR has been proven to be as
self-consistent as that of Euclidean geometry, and as that of real
analysis. So there is zero "probability" that a gedanken can refute SR [#].

An experiment could certainly refute SR, but to date none has come close
to doing so.

[#] Unless you can also show an inconsistency in
Euclidean geometry, and one in real analysis.
Good luck with that....

Tom Roberts

On Sunday, February 26, 2023 at 6:12:16 AM UTC-8, Paul B. Andersen wrote:
> As I said and you snipped: He can OTOH measure that the coordinate
> time in F0 appear to be running fast...

The question was, how can the traveler determine that the hub clock is running slow in terms of his co-moving inertial coordinates. (He obviously can't determine it is running fast in terms of those coordinates, because it is not.) The answer is as given in the previous message, i.e., examining the signals from the hub, taking aberration and Doppler into account (which Dave always fails to do).

> > Don't say the clocks at rest in F0 "appear" to run slow in terms of the inertial
> > coordinate system of the traveler on a given segment, say that they actually do
> > run slow in terms of those coordinates.
>
> Clocks always run at their proper rate by definition of a clock!

Again, the hub clock runs slow in terms of the inertial coordinates in which the traveler is at rest on any given edge of the polygon. Likewise, the traveler's clock runs slow in terms of the inertial coordinates in which the hub is at rest.

> You can say: A clock in frame A runs slow as observed in frame B.

You can certainly say that, but it's wrong and conceptually garbled, because the phrase "as observed in frame B" is hopelessly ambiguous, to the point of being meaningless, noting, for example, that every object is "in" every "frame", and moreover that relativity does not provide a subjectivist account of phenomena in which (as in quantum theory) an "observation" is (arguably) an efficacious element. Further, any observer at rest in any frame can make measurements in terms of any system of coordinates, e.g., we can and often do make measurements in terms of the ECI coordinate system, even when (on the earth's rotating surface) we are not at rest in terms of that system. And so on. Observations "in" a frame involve Doppler, aberration, etc., and statements about such observations (sense impressions) are not to be confused with statements about the objective descriptions of phenomena in terms of well-defined systems of coordinates.

Talking in terms of "as observed in frame B" is precisely what leads people like Dave and you into confusion and contradictions. Just state the objective facts: The hub clock runs slow in terms of the inertial coordinate system in which the traveler is at rest on any given segment. Likewise, the traveler's clock runs slow in terms of the inertial coordinates in which the hub is at rest.

>But only if the meaning of "in terms of the inertial coordinate
>system of frame B" is "as observed/measured in frame B".

You have that backwards: The former is the clear statement of the objective fact, whereas the latter is the hopelessly vague and ambiguous (not to say meaningless) phrase that can only be considered correct if it is simply defined to mean the former. The usage of such misguided phrases (i.e., referring to systems of inertial coordinates as observers) in popularizations contributes to the creations of anti-relativity cranks.

> The traveller with a single clock can't.

Your misconception about this was explained in the previous message. Again, without some conveyance of information about a distant objects, we obviously can't know anything about it, barring acausal telepathy.

> > because students are often puzzled by how the rates can be related as they are will still getting the overall secular advance of the hub clock. The answer is that, in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the first segment (in half a second), the elapsed time of the hub is 1/4 of a second. Likewise in terms of the standard inertial coordinates in which the traveler is at rest as he traverses the second segment (in half a second), the elapsed time of the hub is 1/4 of a second. And so on. The sum of these elapsed times for the hub during the N segments is N/4, compared with N/2 for the traveler, but this does not account for all of the hub's elapsed time, because there is 3/4 second between the hub event that is simultaneous with the traveler at a given vertex in terms of the inertial coordinate system of the traveler on the previous segment and on the next segment. Over the whole N segments this contributes 3N/4 seconds to the hub, for a total of N seconds.

> I don't find it very obvious what you are trying to explain...

If something is unclear, point it out. What has been explained there is precisely how the mutual time dilation between hub and traveler along each segment is reconciled with the secular advance of the hub clock's time versus the traveler's elapsed time in terms of the aggregate cycle, which is another aspect of the situation that Dave doesn't understand.

> the hub clock will appear to jump 3/4 seconds forward, so at

The hub clock never "appears" to jump, i.e., if it was transmitting it's time readings to the traveler they will not contain any jumps, obviously. Every time you type the word "appear" or "appears", you should backspace through it, and try to type the actual objective facts. For example, at each vertex, the traveler is simultaneous with a hub event in terms of the incoming inertial coordinate system, and it is simultaneous with a different hub event in terms of the outgoing inertial coordinate system, and those two hub events are 3/4 sec apart. When you succeed in describing things correctly, without referring to imaginary "apparent jumps", you will have reproduced the above.

> > Note that each of the inertial coordinate systems can be constructed as a grid of
> > standard inertial rulers and clocks (inertially synchronized), and each of the assertions
> > in the above explanation is perfectly measurable in terms of those rulers and clocks.

On Sunday, 26 February 2023 at 18:58:34 UTC+1, Tom Roberts wrote:
> On 2/24/23 12:55 PM, RichD wrote:
> > An elementary result of probability theory says that, in an endless
> > sequence of repeated trials, for any non-zero probability outcome,
> > that outcome will eventually occur with certainty.
> How silly. The math underlying SR has been proven to be as

Go fuck yourself, even with consistent math underlying
your SR is still some inconsistent mumble of an insane
crazie.

> SR - which is still some inconsiste
> self-consistent as that of Euclidean geometry

And speaking of EG it's always good to
remind that your idiot guru has announced
it false as well in his next step.

On Sunday, 26 February 2023 at 19:28:06 UTC+1, Trevor Lange wrote:
> On Sunday, February 26, 2023 at 6:12:16 AM UTC-8, Paul B. Andersen wrote:
> > As I said and you snipped: He can OTOH measure that the coordinate
> > time in F0 appear to be running fast...
>
> The question was, how can the traveler determine that the hub clock is running slow in terms of his co-moving inertial coordinates.

He doesn't. Your moronic tales had never anything
in common with real clocks, real travellers
or real anything.

Maciej Wozniak wrote:
> On Sunday, 26 February 2023 at 18:58:34 UTC+1, Tom Roberts wrote:
>> On 2/24/23 12:55 PM, RichD wrote:
>>> An elementary result of probability theory says that, in an endless
>>> sequence of repeated trials, for any non-zero probability outcome,
>>> that outcome will eventually occur with certainty.
>> How silly. The math underlying SR has been proven to be as
>
> Go fuck yourself, even with consistent math underlying
> your SR is still some inconsistent mumble of an insane
> crazie.
>
>> SR - which is still some inconsiste
>> self-consistent as that of Euclidean geometry
>
> And speaking of EG it's always good to
> remind that your idiot guru has announced
> it false as well in his next step.

The way you are both an asshole and an idiot is quite
fascinating, Wozniak. I wonder how such a degraded kind of
human can come to existence.

Le 26/02/2023 à 21:29, Python a écrit :
> The way you are both an asshole and an idiot is quite
> fascinating, Wozniak. I wonder how such a degraded kind of
> human can come to existence.

That's for diagnosis. It remains to propose a therapy.

Adolf Hitler had planned the extermination of the handicapped and the
"degraded kins of human".

You feel like you want to put this back on the agenda.

R.H.

On Sunday, 26 February 2023 at 21:29:38 UTC+1, Python wrote:
> Maciej Wozniak wrote:
> > On Sunday, 26 February 2023 at 18:58:34 UTC+1, Tom Roberts wrote:
> >> On 2/24/23 12:55 PM, RichD wrote:
> >>> An elementary result of probability theory says that, in an endless
> >>> sequence of repeated trials, for any non-zero probability outcome,
> >>> that outcome will eventually occur with certainty.
> >> How silly. The math underlying SR has been proven to be as
> >
> > Go fuck yourself, even with consistent math underlying
> > your SR is still some inconsistent mumble of an insane
> > crazie.
> >
> >> SR - which is still some inconsiste
> >> self-consistent as that of Euclidean geometry
> >
> > And speaking of EG it's always good to
> > remind that your idiot guru has announced
> > it false as well in his next step.
> The way you are both an asshole and an idiot is quite

Oh, stinker Python is opening its muzzle again,
and trying to pretend he knows something.
Tell me, poor stinker, what is your definition of
a "theory" in the terms of Peano arithmetic?
See: if a theorem is going to be a part of a theory,
it has to be formulable in the language of the
theory. Do you get it? Or are you too stupid even for
that, poor stinker?

Den 26.02.2023 19:28, skrev Trevor Lange:
> On Sunday, February 26, 2023 at 6:12:16 AM UTC-8, Paul B. Andersen wrote:
>> As I said and you snipped: He can OTOH measure that the coordinate
>
>> the hub clock will appear to jump 3/4 seconds forward, so at
>
> The hub clock never "appears" to jump, i.e., if it was transmitting it's time readings to the traveler they will not contain any jumps, obviously. Every time you type the word "appear" or "appears", you should backspace through it, and try to type the actual objective facts.

And the objective facts are:

According to the definition of the scenario:
Observed in the traveller's co-moving inertial frame
will the hub clock advance 1 second between each time
the traveller is at the beginning of a section.

You know of course that "to observe the hub clock"
means to compare the proper time of the hub clock
to the coordinate times of the co-moving frame.

(Hard to do in reality, but well defined in theory.)

When I in the following say "is observed to", I mean
"is observed in the traveller's co-moving inertial frame."

If we for simplicity say that the hub clock is
observed to show 0 when the traveller is at the beginning
of a section, we know that it will be observed to show
1/4 second when the traveller is at the end of the section.
When the traveller abruptly changes the direction of his
velocity, and the traveller is at the beginning of
the next section, the hub clock is observed to show 1 second.
The reason is obviously that it is a different co-moving
inertial frame.

The _observed_ reading of the hub clock will really change
abruptly from 1 second to one second.

But the reading of the clock never changes abruptly. Obviously.

So the the reading of the clock appear to change abruptly.

> For example, at each vertex, the traveler is simultaneous with a hub event in terms of the incoming inertial coordinate system, and it is simultaneous with a different hub event in terms of the outgoing inertial coordinate system, and those two hub events are 3/4 sec apart.

Quite.
At one instant the hub clock simultaneously shows 1/4 second,
at the next instant it simultaneously shows 1 second.

No apparent jump there, is it?

Must be a real jump, then.

--
Paul

https://paulba.no/