On Mon, 19 Jun 2023 22:42:52 -0700, Trevor Lange wrote:

> On Monday, June 19, 2023 at 9:07:17 PM UTC-7, Jane wrote:
>> >> > Acceleration does not (and can not) change the fact that the
>> >> > wavecrests enter in pairs, and given that a pair of wavecrests
>> >> > enter at time t1 and they reach the exit simultaneously at time
>> >> > t2, and given that N additional matched pairs enter between t1 and
>> >> > t2, with one component of each pair taking each path, there are
>> >> > N-1 wavecrests in transit on each path (not counting the pairs at
>> >> > the entrance and exit) at that instant.
>> >>
>> >> During an acceleration...
>> >
>> > Again, the only condition -- which you fully stipulated -- is that
>> > one particular matched pair of wavecrests, which emerge from the
>> > entrance splitter at the same time t1, take the same time from from
>> > entrance to exit, and arrive at the exit at the same time, t2, which
>> > is when another matched pair is emerging from the splitter at the
>> > entrance. In between t1 and t2, exactly the same number of wavecrests
>> > have been sent down each path.
>>
>> Light does not always move at c.
>
> The speeds of the wavecrests (or the men and women walking down the
> aisles), and whether they are speeding up or slowing down or traveling
> shorter or longer distances, etc, etc.,... none of that matters, given
> that the transit times are equal, they arrive in phase with the same
> number of wavecrests in transit along each path. There is no ambiguity
> about this, it is not a debatable or controversial point. It is
> self-evident to any sentient being.

Look that might be true in general but in this experiment, there is a
very subtle difference due to the way light reflects from moving mirrors.
It changes speed but not wavelength. Try to follow the next bit.
The fringe displacement occurs during the passage of the light flashes
between C and D, in opposite directions. During rotation, their paths are
not equal and nor are their travel times...and their speeds are also
different. Waves that leave together do not arrive together and in this
actual experiment there were 3.7 move waves in one path than the other.
However when the time taken to travel from those mirrors to the observing
device is added, the overall travel times are the same and the arrival
rate is quite mysteriously the same as the emission rate. However, one
path has very cleverly acquired 3.7 more waves that the other.
>> > Whether those wavecrests are speeding up or slowing down or the path
>> > lengths are changing, or some of the matched pairs are not going to
>> > arrive simultaneously (meaning they have different travel times) make
>> > absolutely no difference. Provided only that the travel times for
>> > this particular pair are the same, there are exactly N-1 wavecrests
>> > in both paths at the moment this pair arrives at the exit. This is
>> > self-evident, and refutes your fundamental claim, i.e., your absurd
>> > denial of the obvious fact that equal travel times implies arrival in
>> > phase. When matched phases arrive at the same time, they are in phase
>> > by definition.
>>
>> They are in phase.
>
> Right! Bravo.

>>,,,but that does not determine the fringe displacement.
>
> Sure it does. The fringe displacement is proportional to the phase
> difference. This is elementary Optics 101.

Elementally wrong. Light is not an oscillator.
In both Sagnac and the 1913X, the beams reunite is phase as they must.
Phase difference is a sole consequence of different wavelengths in the
two paths.

I really can't imagine what
> mental block could be preventing you from grasping this elementary and
> self-evident fact.

No mental block, just way ahead of you.
I will start a new thread about this soon.

--
-- lover of truth

On Tuesday, June 20, 2023 at 7:42:14 PM UTC-7, Jane wrote:
> > The speeds of the wavecrests (or the men and women walking down the
> > aisles), and whether they are speeding up or slowing down or traveling
> > shorter or longer distances, etc, etc.,... none of that matters, given
> > that the transit times are equal, they arrive in phase with the same
> > number of wavecrests in transit along each path. There is no ambiguity
> > about this.
>
> In this experiment, there is a very subtle difference...

There is no kind of difference, subtle or otherwise, that can change the simple elementary fact that if the split wavecrests take the same transit time and arrive in phase, there are equal numbers of wavecrests along the respective paths.

> > Whether those wavecrests are speeding up or slowing down or the path
> > lengths are changing, or some of the matched pairs are not going to
> > arrive simultaneously (meaning they have different travel times) make
> > absolutely no difference. Provided only that the travel times for
> > this particular pair are the same, there are exactly N-1 wavecrests
> > in both paths at the moment this pair arrives at the exit. This is
> > self-evident, and refutes your fundamental claim, i.e., your absurd
> > denial of the obvious fact that equal travel times implies arrival in
> > phase. When matched phases arrive at the same time, they are in phase
> > by definition.
>
> In both Sagnac and the 1913X, the beams reunite is phase as they must.

The transit times for a rotating Sagnac device (or fiber optic gyro) are different due to the different optical path lengths in the two directions, and this results in the observed fringe shift. When the phases take equal times they arrive in phase (by definition), and there is no fringe shift. This is elementary optics.

> I will start a new thread about this soon.

Starting a new thread is not going to change the facts explained above.

On Tue, 20 Jun 2023 22:14:37 -0700, Trevor Lange wrote:

> On Tuesday, June 20, 2023 at 7:42:14 PM UTC-7, Jane wrote:
>> > The speeds of the wavecrests (or the men and women walking down the
>> > aisles), and whether they are speeding up or slowing down or
>> > traveling shorter or longer distances, etc, etc.,... none of that
>> > matters, given that the transit times are equal, they arrive in phase
>> > with the same number of wavecrests in transit along each path. There
>> > is no ambiguity about this.
>>
>> In this experiment, there is a very subtle difference...
>
> There is no kind of difference, subtle or otherwise, that can change the
> simple elementary fact that if the split wavecrests take the same
> transit time and arrive in phase, there are equal numbers of wavecrests
> along the respective paths.

You obviously did not even try to understand what I just explained to
you. The difference in waves numbers is quite obviously the difference in
path lengths divided by lambda. The difference in path lengths is quite
obviously 2(d1+d2) or ~4d for v<<c.
>> > Whether those wavecrests are speeding up or slowing down or the path
>> > lengths are changing, or some of the matched pairs are not going to
>> > arrive simultaneously (meaning they have different travel times) make
>> > absolutely no difference. Provided only that the travel times for
>> > this particular pair are the same, there are exactly N-1 wavecrests
>> > in both paths at the moment this pair arrives at the exit. This is
>> > self-evident, and refutes your fundamental claim, i.e., your absurd
>> > denial of the obvious fact that equal travel times implies arrival in
>> > phase. When matched phases arrive at the same time, they are in phase
>> > by definition.
>>
>> In both Sagnac and the 1913X, the beams reunite is phase as they must.
>
> The transit times for a rotating Sagnac device (or fiber optic gyro) are
> different due to the different optical path lengths in the two
> directions, and this results in the observed fringe shift. When the
> phases take equal times they arrive in phase (by definition), and there
> is no fringe shift. This is elementary optics.

No. The transit times are definitely the same. That is easy to prove.
>> I will start a new thread about this soon.
>
> Starting a new thread is not going to change the facts explained above.

You are only preaching the Einstein gospel. You are not interested in
learning. This is an important topic and worth a new thread. I will not
reply to this one again.

--
-- lover of truth

On Wednesday, June 21, 2023 at 4:44:42 PM UTC-7, Jane wrote:
> > There is no kind of difference, subtle or otherwise, that can change the
> > simple elementary fact that if the split wavecrests take the same
> > transit time and arrive in phase, there are equal numbers of wavecrests
> > along the respective paths.
>
> The difference in waves numbers...

The number of wavecrests along each path at any given time t2 equals the number at time t1 plus how many have entered minus how many have exited. Hence (barring infinite or negative frequencies) if the matched split crests arrive simultaneously, the number along each path is the same. This is tautological, it is not a debatable or controversial fact.

> No. The transit times [in a device where they are self-evidently different]
> are definitely the same.

Nope, we clearly observe a fringe shift, which is proportional to the difference in transit times for pulses in opposite directions.

> > Starting a new thread is not going to change the facts explained above.
>
> You are only preaching the Einstein gospel.

Huh? There is nothing in our discussion about special relativity. This is basic logic and rationality. Remember, special relativistic effects like time dilation and length contraction are second-order, whereas what are discussing (things like Sagnac) are first-order phenomena.