Ricardo Jimenez wrote:

> The meaning is that if all clocks in the same inirtial systen are
> synched to the clock at a fixed point, say the origin, comparing the
> time coordinates of events can be done by comparing the times on the
> clock at the origin that signals from those events reach the origin.

No; instead, the idea is that the synchronization signal originates at the
master clock (in the middle).

> Well at least I think so.

You are mistaken about that, too.
> In the inaugural paper, Einstein doesn't mention the procedure for
> synching two clocks by putting an observer in the middle.

He does not mention *any* synchronization *procedure* there, but describes
what would be the case *if* two clocks were _synchronous_. Thereby he
*defines* what he means when he says that they are _synchronous_.

> I got the definition of synching with the clock at a fixed point by
> simultaneously solving the two linear equations he gives.

He simply does not do that there. Get that /idée fixe/ out of your mind.

> Defined your way, simultaneity might lead to an equivalence relation among
> a finite number of clocks which might be enough clocks to do relativity
> theory.

“Might”? *facepalm*

If clock A is synchronous with clock B, then B is synchronous with A. If B
is synchronous with A, and C is synchronous with B, then C is of course
synchronous with A, too: A and B show the *same* time. How can you even
begin to think that would not be so?

> However, this seems to require some extra assumptions given
> say, in the Wikipedia article
> https://en.wikipedia.org/wiki/Einstein_synchronisation where
> Macdonald's paper is mentioned.

Einstein synchronization is NOT what Einstein describes in the paper that
you are referring to.

You should take this online course:

<https://online.stanford.edu/courses/som-y0009-understanding-einstein-special-theory-relativity>

I can highly recommend it from first-hand experience: I took it (and passed
it early with full points) before I started studying Astrophysics at the
(another, real-life) university. It was one of the things that showed me
that I would be able to understand this, that I wanted to study and
understand it more deeply, and that I wanted to go to university again in
order to do that.

You are lucky: The next round (it is scheduled for 8 weeks, then it repeats)
is starting *today*.

PointedEars
--
Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)

You cannot measure the velocity of light using electrons that are propagating slower than the velocity of light

carl eto <carleto4157990662@gmail.com> wrote:
> You cannot measure the velocity of light using electrons that are
> propagating slower than the velocity of light
>

Of course you can.

You can also measure the speed of a race car going 200 mph with just people
holding stopwatches, who are going no faster than walking pace.

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

On Wednesday, October 13, 2021 at 11:03:09 AM UTC-7, bodk...@gmail.com wrote:
> carl eto <carleto4...@gmail.com> wrote:
> > You cannot measure the velocity of light using electrons that are
> > propagating slower than the velocity of light
> >
> Of course you can.
>
> You can also measure the speed of a race car going 200 mph with just people
> holding stopwatches, who are going no faster than walking pace.
>
> --
> Odd Bodkin -- maker of fine toys, tools, tables

How?

On Wednesday, October 13, 2021 at 2:03:09 PM UTC-4, bodk...@gmail.com wrote:
> carl eto <carleto4...@gmail.com> wrote:

You are stroking the troll....

carl eto <carleto4157990662@gmail.com> wrote:
> On Wednesday, October 13, 2021 at 11:03:09 AM UTC-7, bodk...@gmail.com wrote:
>> carl eto <carleto4...@gmail.com> wrote:
>>> You cannot measure the velocity of light using electrons that are
>>> propagating slower than the velocity of light
>>>
>> Of course you can.
>>
>> You can also measure the speed of a race car going 200 mph with just people
>> holding stopwatches, who are going no faster than walking pace.
>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>
> How?
>

Oh, come on. There are two people, Bob and Tom, standing at two locations
on the track. They both started their stopwatches when the race started.
The race car passes Bob and Bob clicks his stopwatch. The race car then
passes Tom and Tom clicks his stopwatch. Bob and Tom then casually walk to
a tent a few hundred feet away and compare their stopwatch times. The
difference in times goes in the denominator, the distance between where Bob
and Tom were standing goes in the numerator, and the quotient is the speed
of the race car.

This eludes you? What’s wrong with you? ….. Oh never mind, I remember now
what’s wrong with you.

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

rotchm <rotchm@gmail.com> wrote:
> On Wednesday, October 13, 2021 at 2:03:09 PM UTC-4, bodk...@gmail.com wrote:
>> carl eto <carleto4...@gmail.com> wrote:
>
> You are stroking the troll....
>

I’m certainly paying a visit to the psych wing at the hospital, yes.

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

You cannot use a stop watch to measure the velocity of light. You have to use an electrical circuit that signal is produced by propagating electrons. So when you attempt to measure the velocity of a radio wave the radio wave has already passed by. Plus, the satellite signal is continuous and GPS is formed by the intensity difference.

carl eto <carleto4157990662@gmail.com> wrote:
> You cannot use a stop watch to measure the velocity of light.

That’s true. But you can use a slow device to measure a fast signal. I just
gave you an example. Slow humans, fast car.

> You have to use an electrical circuit that signal is produced by
> propagating electrons. So when you attempt to measure the velocity of a
> radio wave the radio wave has already passed by. Plus, the satellite
> signal is continuous and GPS is formed by the intensity difference.
>

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

Odd Bodkin wrote:

> rotchm <rotchm@gmail.com> wrote:
>> On Wednesday, October 13, 2021 at 2:03:09 PM UTC-4, bodk...@gmail.com
>> wrote:
>>> carl eto <carleto4...@gmail.com> wrote:
>>
>> You are stroking the troll....
>>
> I’m certainly paying a visit to the psych wing at the hospital, yes.

certainly you just did, your "rotchm" is a known imbecile.

On Monday, October 11, 2021 at 10:09:31 AM UTC-7, Ricardo Jimenez wrote:
> AFAIK, the standard procedure to sync all clocks in a frame with the
> one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
> point in space. If you perform a translation of space, how do you
> show all clocks are synchonized with the one at the new origin? If
> you try to evaluate t(p + q) where q is the translation vector you get
> t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
> trick. What am I missing?

This is actually not Einstein's sync process exactly.

The sync process is tB - tA = tA' - tB, where tA = clock reading at the
light pulse emission site, tB = clock reading at the mirror, tA' = clock
reading again at the emission site after the light ray returns.

It is not immediately obvious that this process is free of contradiction.
Einstein in his 1905 paper simply says we assume that it is - that's
because it's easy to demonstrate and he was writing for professional
physicists.

Anyway, to prove that there is no contradiction there, it suffices to
demonstrate that the "sync relation" between clocks is an equivalence
relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
requires an extra consideration to take care of (the other two are
trivial from the definition): one has to assume that given a single
clock at some location A and two mirrors at B and C (so that
a triangular light path from A to A is formed), the elapsed time
recorded by that single clock for light going A->B->C->A is equal
to the time with the light going the other way: A->C->B->A.

This follows from the presumed (silently) homogeneity of space and
it had been in fact verified experimentally by Fizeau (to acceptable
limits).

What you describe seems to be taken from Taylor & Wheeler's book
(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
and your question is valid, and it has the resolution I've described,
I think.

--
Jan

On Thu, 14 Oct 2021 07:59:20 -0700 (PDT), JanPB <filmart@gmail.com>
wrote:

>On Monday, October 11, 2021 at 10:09:31 AM UTC-7, Ricardo Jimenez wrote:
>> AFAIK, the standard procedure to sync all clocks in a frame with the
>> one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
>> point in space. If you perform a translation of space, how do you
>> show all clocks are synchonized with the one at the new origin? If
>> you try to evaluate t(p + q) where q is the translation vector you get
>> t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
>> trick. What am I missing?
>
>This is actually not Einstein's sync process exactly.
>
>The sync process is tB - tA = tA' - tB, where tA = clock reading at the
>light pulse emission site, tB = clock reading at the mirror, tA' = clock
>reading again at the emission site after the light ray returns.
>
>It is not immediately obvious that this process is free of contradiction.
>Einstein in his 1905 paper simply says we assume that it is - that's
>because it's easy to demonstrate and he was writing for professional
>physicists.
>
>Anyway, to prove that there is no contradiction there, it suffices to
>demonstrate that the "sync relation" between clocks is an equivalence
>relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
>requires an extra consideration to take care of (the other two are
>trivial from the definition): one has to assume that given a single
>clock at some location A and two mirrors at B and C (so that
>a triangular light path from A to A is formed), the elapsed time
>recorded by that single clock for light going A->B->C->A is equal
>to the time with the light going the other way: A->C->B->A.
>
>This follows from the presumed (silently) homogeneity of space and
>it had been in fact verified experimentally by Fizeau (to acceptable
>limits).
>
>What you describe seems to be taken from Taylor & Wheeler's book
>(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
>and your question is valid, and it has the resolution I've described,
>I think.

I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things. It is fairly terse and I am still trying to absorb it.
I am at first, taken aback, by the use of Einstein's definition of two
clocks being synchronized since it allows one to add any amount of
time to either clock's setting and they will still be synchronized
according to it. But I suppose you might want to call two clocks,
using the official time in two different time zones, synchronized.

On Wednesday, October 13, 2021 at 2:51:55 PM UTC-7, bodk...@gmail.com wrote:
> carl eto <carleto4...@gmail.com> wrote:
> > You cannot use a stop watch to measure the velocity of light.
> That’s true. But you can use a slow device to measure a fast signal. I just
> gave you an example. Slow humans, fast car.
> > You have to use an electrical circuit that signal is produced by
> > propagating electrons. So when you attempt to measure the velocity of a
> > radio wave the radio wave has already passed by. Plus, the satellite
> > signal is continuous and GPS is formed by the intensity difference.
> >
> --
> Odd Bodkin -- maker of fine toys, tools, tables

On Wednesday, October 13, 2021 at 2:51:55 PM UTC-7, bodk...@gmail.com wrote:
> carl eto <carleto4...@gmail.com> wrote:
> > You cannot use a stop watch to measure the velocity of light.
> That’s true. But you can use a slow device to measure a fast signal. I just
> gave you an example. Slow humans, fast car.
> > You have to use an electrical circuit that signal is produced by
> > propagating electrons. So when you attempt to measure the velocity of a
> > radio wave the radio wave has already passed by. Plus, the satellite
> > signal is continuous and GPS is formed by the intensity difference.
> >
> --
> Odd Bodkin -- maker of fine toys, tools, tables
The question is how does the GPS function ---- by the time difference of the radio waves or the intensity of the radio waves since the time difference implies that the velocity of light is being measured.

On Thursday, 14 October 2021 at 16:59:22 UTC+2, JanPB wrote:
> On Monday, October 11, 2021 at 10:09:31 AM UTC-7, Ricardo Jimenez wrote:
> > AFAIK, the standard procedure to sync all clocks in a frame with the
> > one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
> > point in space. If you perform a translation of space, how do you
> > show all clocks are synchonized with the one at the new origin? If
> > you try to evaluate t(p + q) where q is the translation vector you get
> > t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
> > trick. What am I missing?
>
> This is actually not Einstein's sync process exactly.
>
> The sync process is tB - tA = tA' - tB, where tA = clock reading at the
> light pulse emission site, tB = clock reading at the mirror, tA' = clock
> reading again at the emission site after the light ray returns.
>
> It is not immediately obvious that this process is free of contradiction.
> Einstein in his 1905 paper simply says we assume that it is - that's
> because it's easy to demonstrate and he was writing for professional
> physicists.
>
> Anyway, to prove that there is no contradiction there, it suffices to
> demonstrate that the "sync relation" between clocks is an equivalence
> relation, i.e., reflexive, symmetric, and transitive

No, it doesn't.

carl eto <carleto4157990662@gmail.com> wrote:
> On Wednesday, October 13, 2021 at 2:51:55 PM UTC-7, bodk...@gmail.com wrote:
>> carl eto <carleto4...@gmail.com> wrote:
>>> You cannot use a stop watch to measure the velocity of light.
>> That’s true. But you can use a slow device to measure a fast signal. I just
>> gave you an example. Slow humans, fast car.
>>> You have to use an electrical circuit that signal is produced by
>>> propagating electrons. So when you attempt to measure the velocity of a
>>> radio wave the radio wave has already passed by. Plus, the satellite
>>> signal is continuous and GPS is formed by the intensity difference.
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>
> The question is how does the GPS function ---- by the time difference of
> the radio waves or the intensity of the radio waves since the time
> difference implies that the velocity of light is being measured.
>

It implies nothing of the kind. Nor does measuring the speed of something
require using something else capable of traveling faster.

You do not think well. You know this already.

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

Time across the universe has gone on since the initial
conditions of the BB. Gravity would be in the way
of beginning singularity expanding.

How did the singularity form first Hydrogen atoms
if its gravity didn't allow it to expand?

Mitchell Raemsch

On 10/14/21 10:42 AM, Ricardo Jimenez wrote:
> I am at first, taken aback, by the use of Einstein's definition of
> two clocks being synchronized since it allows one to add any amount
> of time to either clock's setting and they will still be synchronized
> according to it.

I have no idea where you got this notion. It is WRONG.

Note that Einstein described several different techniques for
synchronizing two clocks, in different papers over many years.

Here are three techniques he described to synchronize clock C2 to clock
C1. In all cases both clocks must be at rest in a single inertial frame
F, and all light pulses are in vacuum.

A) Repeatedly send a light pulse from C1 to C2, where it is
reflected back to C1. Set C2 so at the reflection it
displays a value that is midway between the values on C1 when
it was emitted and received.

B) Place a pulsed light source at rest in F at the midpoint
between C1 and C2, able to send light pulses to both
clocks. Set C2 so it displays the same value as C1 when
they receive a given pulse.

C) Measure the distance in F between C1 and C2, call it D. Let
C1 emit light pulses to C2, and set C2 so when it receives a
pulse it displays a value that is the emission value on C1
plus D/c.

It is easy to show that:
* a pair of clocks synchronized with any method will be
synchronized according to all of them.
* Such synchronization is reflexive, symmetric, and
transitive.
So synchronization can be extended to any number of clocks at rest in F.

> But I suppose you might want to call two clocks, using the official
> time in two different time zones, synchronized.

Nobody in their right mind would do so.

Tom Roberts

On Thursday, October 14, 2021 at 8:42:38 AM UTC-7, Ricardo Jimenez wrote:
> On Thu, 14 Oct 2021 07:59:20 -0700 (PDT), JanPB <fil...@gmail.com>
> wrote:
> >On Monday, October 11, 2021 at 10:09:31 AM UTC-7, Ricardo Jimenez wrote:
> >> AFAIK, the standard procedure to sync all clocks in a frame with the
> >> one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
> >> point in space. If you perform a translation of space, how do you
> >> show all clocks are synchonized with the one at the new origin? If
> >> you try to evaluate t(p + q) where q is the translation vector you get
> >> t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
> >> trick. What am I missing?
> >
> >This is actually not Einstein's sync process exactly.
> >
> >The sync process is tB - tA = tA' - tB, where tA = clock reading at the
> >light pulse emission site, tB = clock reading at the mirror, tA' = clock
> >reading again at the emission site after the light ray returns.
> >
> >It is not immediately obvious that this process is free of contradiction.
> >Einstein in his 1905 paper simply says we assume that it is - that's
> >because it's easy to demonstrate and he was writing for professional
> >physicists.
> >
> >Anyway, to prove that there is no contradiction there, it suffices to
> >demonstrate that the "sync relation" between clocks is an equivalence
> >relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
> >requires an extra consideration to take care of (the other two are
> >trivial from the definition): one has to assume that given a single
> >clock at some location A and two mirrors at B and C (so that
> >a triangular light path from A to A is formed), the elapsed time
> >recorded by that single clock for light going A->B->C->A is equal
> >to the time with the light going the other way: A->C->B->A.
> >
> >This follows from the presumed (silently) homogeneity of space and
> >it had been in fact verified experimentally by Fizeau (to acceptable
> >limits).
> >
> >What you describe seems to be taken from Taylor & Wheeler's book
> >(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
> >and your question is valid, and it has the resolution I've described,
> >I think.
> I you go to this webpage http://www.faculty.luther.edu/~macdonal/
> there is a short paper on Clock Synchonization which it is claimed
> fixes things.

He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).

> It is fairly terse and I am still trying to absorb it.
> I am at first, taken aback, by the use of Einstein's definition of two
> clocks being synchronized since it allows one to add any amount of
> time to either clock's setting and they will still be synchronized
> according to it.

Not sure what you are saying here. This is just a property of any
reasonable definition of "time". Just like "space" relationship are
not altered by changing the position of some reference point
like the origin of coordinates.

> But I suppose you might want to call two clocks,
> using the official time in two different time zones, synchronized.

Again, not sure what you are saying. The 1905 paper uses
a *definition* as a means to establish whether two spatially
separated clocks are "at the same time".

--
Jan

On Thursday, October 14, 2021 at 8:42:38 AM UTC-7, Ricardo Jimenez wrote:
> On Thu, 14 Oct 2021 07:59:20 -0700 (PDT), JanPB <fil...@gmail.com>
> wrote:
> >On Monday, October 11, 2021 at 10:09:31 AM UTC-7, Ricardo Jimenez wrote:
> >> AFAIK, the standard procedure to sync all clocks in a frame with the
> >> one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
> >> point in space. If you perform a translation of space, how do you
> >> show all clocks are synchonized with the one at the new origin? If
> >> you try to evaluate t(p + q) where q is the translation vector you get
> >> t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
> >> trick. What am I missing?
> >
> >This is actually not Einstein's sync process exactly.
> >
> >The sync process is tB - tA = tA' - tB, where tA = clock reading at the
> >light pulse emission site, tB = clock reading at the mirror, tA' = clock
> >reading again at the emission site after the light ray returns.
> >
> >It is not immediately obvious that this process is free of contradiction.
> >Einstein in his 1905 paper simply says we assume that it is - that's
> >because it's easy to demonstrate and he was writing for professional
> >physicists.
> >
> >Anyway, to prove that there is no contradiction there, it suffices to
> >demonstrate that the "sync relation" between clocks is an equivalence
> >relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
> >requires an extra consideration to take care of (the other two are
> >trivial from the definition): one has to assume that given a single
> >clock at some location A and two mirrors at B and C (so that
> >a triangular light path from A to A is formed), the elapsed time
> >recorded by that single clock for light going A->B->C->A is equal
> >to the time with the light going the other way: A->C->B->A.
> >
> >This follows from the presumed (silently) homogeneity of space and
> >it had been in fact verified experimentally by Fizeau (to acceptable
> >limits).
> >
> >What you describe seems to be taken from Taylor & Wheeler's book
> >(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
> >and your question is valid, and it has the resolution I've described,
> >I think.
> I you go to this webpage http://www.faculty.luther.edu/~macdonal/
> there is a short paper on Clock Synchonization which it is claimed
> fixes things.

He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).

> I am at first, taken aback, by the use of Einstein's definition of two
> clocks being synchronized since it allows one to add any amount of
> time to either clock's setting and they will still be synchronized
> according to it.

No, just look at the sync criterion again: tB - tA = tA' - tB.

If you add some nonzero amount of time, T, to clock B, the above
equation becomes:

(tB + T) - tA = tA' - (tB + T)

i.e.

tB - tA + T = tA' - tB - T

i.e.

T = -T

....which is plainly false. Adding T to clock A results in a similar
contradiction.

--
Jan

carl eto wrote:

> The question is how does the GPS function ----

The question has already been answered (by me, and many other people)
multiple times before.

> by the time difference of the radio waves

Yes.

> or the intensity of the radio waves

No, that is not feasible. Consider that there is not just vacuum between
sender and receiver: The signal will be partially reflected, absorbed, or
scattered on its way from a satellite to the receiver on the ground.

> since the time difference implies that the velocity of light is being
> measured.

No; instead, the _speed_ of light¹ in vacuum (c) is *known*², and from that
and the light-travel delay, T, the distance to a satellite, S, can be
calculated.

You should have learned in secondary school:

(mean) speed := (distance traveled)/(time taken).

We can write this formally as

v := S/T.

If the speed is the speed of light in vacuum, v := c, then

c = S/T.

So the distance traveled by light in vacuum in the duration T is

S = c * T.

Do that calculation for 4 satellites and 4 light-travel delays, and you can
obtain the observer’s position given by the coordinates x_r, y_r, and z_r.

Because, notice: If we know the position of a satellite at a certain time³
as given by the coordinates x_s, y_s, and z_s, and define the differences in
the coordinates between satellite and receiver as

X := x_s - x_r
Y := y_s - y_r
Z := z_s - z_r,

then

sqrt(X^2 + Y^2 + Z^2) = S = c T [3D-Pythagorean theorem]
==> X^2 + Y^2 + Z^2 = S^2 = c^2 * T^2
<=> 0 = S^2 = c^2 * T^2 - X^2 - Y^2 - Z^2. ⁴)
<=> 0 = c^2 * T^2 - (x_s - x_r)^2 - (y_s - y_r)^2 - (z_s - z_r)^2.

This gives at least 4 such equations (since there are at least 4
satellites), and the values of x_r, y_r and z_r must be found that solve all
of them. Then we know the receiver’s position as its coordinates are
precisely those values.

*That* is how GPS⁵ *basically* works⁶:

<https://en.wikipedia.org/wiki/Global_Positioning_System#Navigation_equations>

PointedEars
___________
¹ (from observation) to be constant and the same in all reference frames
² I am using here, as is customary in physics, the term “light” as a
shorthand for electromagnetic radiation; that includes radio waves.
If we only want to talk about electromagnetic radiation that one
(humans) can perceive directly (visual perception), we call it
“(human-)visible light” instead.
³ and we do; it is in the ephemerides stored in the receiver firmware,
updateable from the signal payload
⁴ That is nothing else than a discrete version of the Minkowski metric
of *special relativity*, with S^2 = 0. This is NOT a coincidence :)
⁵ and several other GNSSs
⁶ For technical and physical reasons, the actual equations are more complex;
there is also another way, for scientific purposes.

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

(from: WolframAlpha)

On Thu, 14 Oct 2021 12:22:12 -0700 (PDT), JanPB <filmart@gmail.com>
wrote:

>> I you go to this webpage http://www.faculty.luther.edu/~macdonal/
>> there is a short paper on Clock Synchonization which it is claimed
>> fixes things.
>
>He just states Einstein's definition I quoted above. He does not address
>the transitivity requirement which Einstein left unsaid but is necessary
>for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
>never leads to a situation in which one clock is assigned two *different*
>times).
>
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
>> I am at first, taken aback, by the use of Einstein's definition of two
>> clocks being synchronized since it allows one to add any amount of
>> time to either clock's setting and they will still be synchronized
>> according to it.
>
>No, just look at the sync criterion again: tB - tA = tA' - tB.
>
>If you add some nonzero amount of time, T, to clock B, the above
>equation becomes:
>
> (tB + T) - tA = tA' - (tB + T)
>
>i.e.
>
> tB - tA + T = tA' - tB - T
>
>i.e.
>
> T = -T
>
>...which is plainly false. Adding T to clock A results in a similar
>contradiction.
>
>--
>Jan

Oops, I should checked it before writing. :-)

[Subject repaired]

Ricardo Jimenez wrote:

> On Thu, 14 Oct 2021 12:22:12 -0700 (PDT), JanPB <filmart@gmail.com>
> wrote:

If only your references would be that precise. *facepalm*

Remove the *unnecessary* attribution novel (it is all in the header fields),
and add the *necessary* precision to your references instead.

>>> I you go to this webpage http://www.faculty.luther.edu/~macdonal/
>>> there is a short paper on Clock Synchonization which it is claimed
>>> fixes things.
>>
>>He just states Einstein's definition I quoted above. He does not address
>>the transitivity requirement which Einstein left unsaid but is necessary
>>for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
>>never leads to a situation in which one clock is assigned two *different*
>>times).
>
> He claims to address transitivity on pgs 2 and 3

Who is “he”, and the pages of which paper are you referring to?

You *really* need to learn how to cite your sources *properly* (including
*unambiguously*). “Einstein, p.115” or “there is a short paper on …” on
$WEBSITE simply does not cut it.

PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven

On Thursday, 14 October 2021 at 21:12:37 UTC+2, tjrob137 wrote:
> On 10/14/21 10:42 AM, Ricardo Jimenez wrote:
> > I am at first, taken aback, by the use of Einstein's definition of
> > two clocks being synchronized since it allows one to add any amount
> > of time to either clock's setting and they will still be synchronized
> > according to it.
> I have no idea where you got this notion. It is WRONG.
>
> Note that Einstein described several different techniques for
> synchronizing two clocks, in different papers over many years.

And note that in the meantime in the real world,
"impossibly" synchronized GPS clocks keep
measuring t'=t, just like all serious clocks always
did.

On Thursday, October 14, 2021 at 1:16:14 PM UTC-7, Ricardo Jimenez wrote:
> On Thu, 14 Oct 2021 12:22:12 -0700 (PDT), JanPB <fil...@gmail.com>
> wrote:
> >> I you go to this webpage http://www.faculty.luther.edu/~macdonal/
> >> there is a short paper on Clock Synchonization which it is claimed
> >> fixes things.
> >
> >He just states Einstein's definition I quoted above. He does not address
> >the transitivity requirement which Einstein left unsaid but is necessary
> >for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
> >never leads to a situation in which one clock is assigned two *different*
> >times).
> >
> He claims to address transitivity on pgs 2 and 3 with the assumption
> going around the triangle in either direction yields the same result.
> And he references a laser experiment for justification of that
> assumption. Is he wrong? Einstein does state transitivity holds but
> doesn't offer a proof.

I don't see any mention of transitivity in either Einstein or Alan Macdonald's
paper you've referenced above.

--
Jan