I have previously explained that the vertical beam of a light clock remains
vertical in all horizontally moving frames. It merely moves sideways as
clearly evidenced by the fact that if the beam is projected up a vertical
pole the pole will not magically lean over. That being so, we will place a
45 degree mirror adjacent to the source in order to reflect the returning
beam horizontally to the moving observer.

|
|
|
|
-- |
\ ------------>Om ->
S ------------>

Rather than use a continuous light beam we will use a pulsed laser of
measurable frequency. It is obvious that the moving observer will measure
the same pulse frequency directly from the source and the returning beam
and that both frequencies will be Doppler shifted by the linear Newtonian
equation (c-v)/c.
The idea of the beam becoming diagonal and the clock frequency being slowed
by gamma in the moving frame is plainly laughable.

-- lover of truth

On Wednesday, May 31, 2023 at 2:59:05 PM UTC-7, Jane wrote:

> The idea of the beam becoming diagonal and the clock frequency being slowed
> by gamma in the moving frame is plainly laughable.
>
So, you never learned "light aberration". Fail.

On Monday, May 29, 2023 at 10:37:32 PM UTC-7, Jeremy wrote:
> > The clear and simple analysis gives the transit times for matched
> > wavecrests, which determines the amount of fringe shift. The
> > analysis does not involve "assuming different wavelengths". In
> > summary, your beliefs are both logically incoherent and
> > empirically falsified.
> >
> > In the nomenclature of the paper, the path lengths and speeds are
> > essentially equal except for the transit from one mirror to E and
> > back to the other mirror, and then back to the point where the mirror
> > would have been had it not moved by the distance d = (2D/c)v during
> > the transit to E and back.
> >
> > So, the relevant path lengths are 2(D+d) and 2(D-d), and Michelson
> > approximates by treating the speeds as c+rv and c-rv respectively
> > over these entire intervals, even though strictly speaking it is c
> > over the short back-tracked d, but the effect of that approximation
> > is negligible.
> >
> > Hence the difference between the arrival times of a given wave crest
> > (or any other phase) by the two paths is (to the lowest order) equal
> > to (4Dv/c^2) (2-r). Multiply this by c/L where L is the wavelength of
> > the light to give the fringe displacement (4D/L)(v/c)(2-r). Of course,
> > to avoid extinction, the experiment ought to be done in vacuum,
> > or with x-rays.
> >
> > Needless to say, Michelson was mistaken about the r=1 case, which
> > represents the case of absorption and re-emission at the characteristic
> > speed c in terms of the inertial coordinates in which the mirror is as
> > rest. Michelson overlooked the fact that for the case r=1 the light
> > reflecting off the distant mirror would have speed c, not speed c+rv,
> > in terms of the lab inertial coordinates. As a result, the actual time
> > delay for the case r=1 is 6Dv/c^2, compared with 8Dv/c^2 for the case
> > r=0, and 0 for the ballistic case r=2.
>
> c+rv is not c.

Well, if r=0 then c+rv = c, but if r=1 or 2, then c+rv does not equal c. Michelson examines all three alternatives, in a strictly Galilean setting.

> c/L is wave emission rate. It has nothing to do with anything
> that happens after emission.

The ratio c/L is the conversion between cycles and seconds, which is to say the phase at any location advances by that many cycles per second. So, given the difference in arrival times, we can calculate the corresponding phase difference. Don't be confused by the use of c rather than (say) c+rv, because when the light arrives at the receiver it has struck both the approaching and the receeding mirror, so it has speed c in each of the cases considered (since the last mirror each pulse struck was stationary in the lab. (Also, the effect on the result of using c versus c+-rv would be negligible anyway, differing by just the second order.)

> You know that the path lengths differ by 4D+d ...

No, the distances traveled by the two split pulses in the nominal configuration differ by 4d.

> how can here be the same number of waves in both paths?

As the mirrors move, near the nominal configuration, one optical path length is decreasing and the other is decreasing, and at any given instant of the lab time there are different configurations depending on which of the models (r=0,1, or 2) you are talking about. This isn't a situation in which there are two standing waves, it is a transient situation, and the objective is to determine how out-of-phase the pulses are when they reach the receiver, which is determined by computing the different in transit times and converting that to the difference in phase.

> Good old Newton got it right after all.

No, the Newtonian ballistic theory (r=2) predicts no fringe shift at all, and the emission theory (r=1) predicts (6D/L)(v/c), and the theory with r=0 predicts (8D/L)(v/c). The experimental result confirms the latter.

> Einstein on the other hand got it completely wrong....

Huh? Nothing in Michelson's calculation refers to special relativity. Remember, relativistic time dilation and length contraction are second-order effects, whereas this simple experiment, and the Doppler effect, and the Sagnac effect, etc., are all first order effects. Michelson's analysis of all three cases is exactly as Newton would have calculated it.

> [The explanation of rational adults] results in continuous fringe movement
> when it should be stationary.

Your absurd claim has been thoroughly debunked, remember? You seem to have a very strange mental block about all this.

> You cannot get into your head that using SR principles...

Again, the analysis of this experiment does not use any SR principles. It is carried out entirely in a Galilean context, with three difference naive premises as to how the speed of light is affected by reflection.

> The explanation of this experiment takes only a few lines in Newtonian
> logic to produce the right equation for fringe displacement, (4D/L)(v/c).

No, the correct answer, which is indeed produced by just a few lines of classical logic, results in predictions of 0, (6D/L)(v/c), and (8D/L)(v/c) for the premises r=2, 1, and 0 respectively. The experiment yields (8D/L)(v/c).

> 'r' is not significantly involved, as can be seen on casual inspection.

LOL. The whole point is to evaluate the effect of different possible speeds of light, as parameterized by r. Your utterly senseless and specious "analysis" does not involve r in a meaningful way, which is why your "analysis" is utterly senseless and specious.

> The vertical beam of a light clock remains vertical in all horizontally moving frames.

No, decompose a beam of light into individual pulses, and track each pulse as it moves from transmitter to receiver. If you are stationary relative to the apparatus, each pulse will follow a vertical path, but if you are moving to the left relative to the apparatus, then in terms of the inertial coordinate system in which you are at rest, the pulses are moving diagonally.

> A moving observer will measure the same pulse frequency directly from
> the source and the returning beam and both frequencies will be Doppler
> shifted by the linear Newtonian equation (c-v)/c.

The relativistic Doppler effect differs from the classical only in the second order, which you are neglecting. Look, this can be settled very easily:

Suppose you have a source of light with a known frequency f0 in terms of its own inertial rest coordinates, and this source is moving with speed v in terms of the lab frame, directly toward one lab receiver and directly away from another. The receiver in front of the moving source measures the frequency f1 (blue-shifted), and the receiver behind measures the frequency f2 (red-shifted). What is (f1+f2)/f0 - 2 ?

[Note to others: Please don't help; let him answer.]

On 2023-05-31 21:59:02 +0000, Jane said:

> Rather than use a continuous light beam we will use a pulsed laser of
> measurable frequency.

In the light clock there is no continuous beam nor a pulsed laser
with a measurable frequency. The purpose of the light clock is to
be a simple model of a clock that does not depend on any other clock.
A "masurable frequency" would involve another clock.

In the light clock there is a single pulse of light. The method of
creation of the pulse is not relevant. The pulse bounces between
two parallel mirrors as many times as needed.

If something else is desired that is not a light clock.

Mikko

On Wed, 31 May 2023 16:24:42 -0700, Dono. wrote:

> On Wednesday, May 31, 2023 at 2:59:05 PM UTC-7, Jane wrote:
>
>> The idea of the beam becoming diagonal and the clock frequency being
>> slowed by gamma in the moving frame is plainly laughable.
>>
> So, you never learned "light aberration". Fail.

Aberration has nothing to do with this, you idiot.

--
-- lover of truth

On Wed, 31 May 2023 17:14:33 -0700, Trevor Lange wrote:

> On Monday, May 29, 2023 at 10:37:32 PM UTC-7, Jeremy wrote:
>> > The clear and simple analysis gives the transit times for matched
>> > wavecrests, which determines the amount of fringe shift. The analysis
>> > does not involve "assuming different wavelengths". In summary, your
>> > beliefs are both logically incoherent and empirically falsified.
>> >
>> > In the nomenclature of the paper, the path lengths and speeds are
>> > essentially equal except for the transit from one mirror to E and
>> > back to the other mirror, and then back to the point where the mirror
>> > would have been had it not moved by the distance d = (2D/c)v during
>> > the transit to E and back.
>> >
>> > So, the relevant path lengths are 2(D+d) and 2(D-d), and Michelson
>> > approximates by treating the speeds as c+rv and c-rv respectively
>> > over these entire intervals, even though strictly speaking it is c
>> > over the short back-tracked d, but the effect of that approximation
>> > is negligible.
>> >
>> > Hence the difference between the arrival times of a given wave crest
>> > (or any other phase) by the two paths is (to the lowest order) equal
>> > to (4Dv/c^2) (2-r). Multiply this by c/L where L is the wavelength
>> > of the light to give the fringe displacement (4D/L)(v/c)(2-r). Of
>> > course,
>> > to avoid extinction, the experiment ought to be done in vacuum,
>> > or with x-rays.
>> >
>> > Needless to say, Michelson was mistaken about the r=1 case, which
>> > represents the case of absorption and re-emission at the
>> > characteristic speed c in terms of the inertial coordinates in which
>> > the mirror is as rest. Michelson overlooked the fact that for the
>> > case r=1 the light reflecting off the distant mirror would have speed
>> > c, not speed c+rv, in terms of the lab inertial coordinates. As a
>> > result, the actual time delay for the case r=1 is 6Dv/c^2, compared
>> > with 8Dv/c^2 for the case r=0, and 0 for the ballistic case r=2.
>>
>> c+rv is not c.

You have sent this to the wrong thread...however....
> Well, if r=0 then c+rv = c, but if r=1 or 2, then c+rv does not equal c.
> Michelson examines all three alternatives, in a strictly Galilean
> setting.

He did...but then he misinterpreted the observations.
>> c/L is wave emission rate. It has nothing to do with anything that
>> happens after emission.
>
> The ratio c/L is the conversion between cycles and seconds, which is to
> say the phase at any location advances by that many cycles per second.

c/L will tell how many cycles have been emitted in a certain time,
nothing else.

> So, given the difference in arrival times, we can calculate the
> corresponding phase difference. Don't be confused by the use of c
> rather than (say) c+rv, because when the light arrives at the receiver
> it has struck both the approaching and the receeding mirror, so it has
> speed c in each of the cases considered (since the last mirror each
> pulse struck was stationary in the lab. (Also, the effect on the result
> of using c versus c+-rv would be negligible anyway, differing by just
> the second order.)
>
>> You know that the path lengths differ by 4D+d ...
>
> No, the distances traveled by the two split pulses in the nominal
> configuration differ by 4d.
>
>> how can there be the same number of waves in both paths?
>
> As the mirrors move, near the nominal configuration, one optical path
> length is decreasing and the other is decreasing, and at any given
> instant of the lab time there are different configurations depending on
> which of the models (r=0,1, or 2) you are talking about. This isn't a
> situation in which there are two standing waves, it is a transient
> situation, and the objective is to determine how out-of-phase the pulses
> are when they reach the receiver, which is determined by computing the
> different in transit times and converting that to the difference in
> phase.
>
>> Good old Newton got it right after all.
>
> No, the Newtonian ballistic theory (r=2) predicts no fringe shift at
> all, and the emission theory (r=1) predicts (6D/L)(v/c), and the theory
> with r=0 predicts (8D/L)(v/c). The experimental result confirms the
> latter.

There is no point in talking to you. You obviously cannot understand
physics. You are just preaching the gospel of the E-religion.
>> Einstein on the other hand got it completely wrong....
>
> Huh? Nothing in Michelson's calculation refers to special relativity.
> Remember, relativistic time dilation and length contraction are
> second-order effects, whereas this simple experiment, and the Doppler
> effect, and the Sagnac effect, etc., are all first order effects.
> Michelson's analysis of all three cases is exactly as Newton would have
> calculated it.

Go away.... I am not going to waste my time trying to discuss science
with indoctrinated idiots.
>> [The explanation of rational adults] results in continuous fringe
>> movement when it should be stationary.
>
> Your absurd claim has been thoroughly debunked, remember? You seem to
> have a very strange mental block about all this.
>
>> You cannot get into your head that using SR principles...
>
> Again, the analysis of this experiment does not use any SR principles.
> It is carried out entirely in a Galilean context, with three difference
> naive premises as to how the speed of light is affected by reflection.

The experiment assumes that if dt = 0 there cannot be different numbers
of waves in the two paths.. THAT ASSUMES LIGHT SPEED IS C. If light speed
is NOT c then there can and will be different numbers of waves in the
two paths and the Newton predicted fringe displacement is exactly as
measured. I have given you all the details.
If you cannot read my derivation and understand that simple fact, you
should not be posting here.
>> The explanation of this experiment takes only a few lines in Newtonian
>> logic to produce the right equation for fringe displacement,
>> (4D/L)(v/c).
>
> No, the correct answer, which is indeed produced by just a few lines of
> classical logic, results in predictions of 0, (6D/L)(v/c), and
> (8D/L)(v/c) for the premises r=2, 1, and 0 respectively. The experiment
> yields (8D/L)(v/c).

Stop talking crap and study the facts I gave you.
>> 'r' is not significantly involved, as can be seen on casual inspection.
>
> LOL. The whole point is to evaluate the effect of different possible
> speeds of light, as parameterized by r. Your utterly senseless and
> specious "analysis" does not involve r in a meaningful way, which is why
> your "analysis" is utterly senseless and specious.

The path length difference is virtually independent of r. It is obviously
just ~4d for v<<c, ...where d/v=(2E+d)/(c+rv) ~ 2E
>> The vertical beam of a light clock remains vertical in all horizontally
>> moving frames.
>
> No, decompose a beam of light into individual pulses, and track each
> pulse as it moves from transmitter to receiver. If you are stationary
> relative to the apparatus, each pulse will follow a vertical path, but
> if you are moving to the left relative to the apparatus, then in terms
> of the inertial coordinate system in which you are at rest, the pulses
> are moving diagonally.
>
>> A moving observer will measure the same pulse frequency directly from
>> the source and the returning beam and both frequencies will be Doppler
>> shifted by the linear Newtonian equation (c-v)/c.
>
> The relativistic Doppler effect differs from the classical only in the
> second order, which you are neglecting. Look, this can be settled very
> easily:
>
> Suppose you have a source of light with a known frequency f0 in terms of
> its own inertial rest coordinates, and this source is moving with speed
> v in terms of the lab frame, directly toward one lab receiver and
> directly away from another. The receiver in front of the moving source
> measures the frequency f1 (blue-shifted), and the receiver behind
> measures the frequency f2 (red-shifted). What is (f1+f2)/f0 - 2 ?
>
> [Note to others: Please don't help; let him answer.]

You are completely out of your depth here. .. go away

--
-- lover of truth

On Thu, 01 Jun 2023 10:54:48 +0300, Mikko wrote:

> On 2023-05-31 21:59:02 +0000, Jane said:
>
>> Rather than use a continuous light beam we will use a pulsed laser of
>> measurable frequency.
>
> In the light clock there is no continuous beam nor a pulsed laser with a
> measurable frequency. The purpose of the light clock is to be a simple
> model of a clock that does not depend on any other clock.
> A "masurable frequency" would involve another clock.

It does....but in this modified light clock, that second 'clock' is used
only for comparison.
> In the light clock there is a single pulse of light. The method of
> creation of the pulse is not relevant. The pulse bounces between two
> parallel mirrors as many times as needed.
>
> If something else is desired that is not a light clock.

The idea of the light clock resulted from the successful TWLS
measurements by Fizeau and Foucault. It was reasoned that the measured
value should vary as the Earth moved through the supposed Ether, in which
case the beam used would have to be angled to ensure that it returned to
the observer. There was no evidence of that happening so Michelson and
Morely improved the accuracy by many orders of magnitude with their MMX.
The again null result led to Fitzgerald's and Lorentz's idea of length
and time contraction by the factor gamma, which would be the true change
in period of a light clock if an Ether existed...but of course there is
none.
> Mikko

--
-- lover of truth

On Thursday, June 1, 2023 at 4:22:15 PM UTC-7, Jane wrote:
> c+rv is not c.
Well, if r=0 then c+rv = c, but if r=1 or 2, then c+rv does not equal c. Michelson examines all three alternatives, in a strictly Galilean context.

> c/L is wave emission rate. It has nothing to do with anything that
> happens after emission.

The ratio c/L is the conversion between cycles and seconds, which is to say, the phase at any location advances by that many cycles per second.

> how can there be the same number of waves in both paths?
Ask someone who claims that. Again, as the mirrors move, near the nominal configuration, one optical path length is decreasing and the other is decreasing, and at any given instant of the lab time there are different configurations depending on which of the models (r=0,1, or 2) you are talking about. This isn't a situation in which there are two standing waves, it is a transient situation, and the objective is to determine how out-of-phase the pulses are when they reach the receiver, which is determined by computing the different in transit times and converting that to the difference in phase.

> > The Newtonian ballistic theory (r=2) predicts no fringe shift at
> > all, and the emission theory (r=1) predicts (6D/L)(v/c), and the theory
> > with r=0 predicts (8D/L)(v/c). The experimental result confirms the
> > latter.
>
> You are just preaching the gospel of the E-religion.

Huh? There is nothing involving special relativity in this discussion. Michelson's analysis is in a purely classical context, based on three different hypotheses for how light speed is affected by reflection. This is a purely first-order effect, not involving any relativistic effects at all.

> The experiment assumes that if dt = 0 there cannot be different numbers
> of waves in the two paths..

Nope, it "assumes" that if dt=0, meaning that the split wavecrests arrive at the same time, then the two wave crests arrive at the same time, i.e., in phase. Are you honestly unable to grasp this? How can the waves be out of phase, if they are in phase?

> THAT ASSUMES LIGHT SPEED IS C.

Goodness no (despite the impressive caps). The fact that if the phases arrive in phase implies that the phases are arriuving in phase does not depend upon or assume that the speed of light is c. Indeed, the analysis yielding zero phase shift is for the case when the speed of light reflect is c +- 2v, which is not equal to c. So, as always, your statements are completely incoherent and senseless.

> > The correct answer, which is indeed produced by just a few lines of
> > classical logic, results in predictions of 0, (6D/L)(v/c), and
> > (8D/L)(v/c) for the premises r=2, 1, and 0 respectively. The experiment
> > yields (8D/L)(v/c).
>
> [No substantive reply. duely noted.]

> > Suppose you have a source of light with a known frequency f0 in terms of
> > its own inertial rest coordinates, and this source is moving with speed
> > v in terms of the lab frame, directly toward one lab receiver and
> > directly away from another. The receiver in front of the moving source
> > measures the frequency f1 (blue-shifted), and the receiver behind
> > measures the frequency f2 (red-shifted). What is (f1+f2)/f0 - 2 ?
> >
> You are completely out of your depth here. ..

Your inability to answer the elementary question speaks for itself.

On Thursday, June 1, 2023 at 4:22:15 PM UTC-7, Jane wrote:
> c+rv is not c.

Well, if r=0 then c+rv = c, but if r=1 or 2, then c+rv doesn't equal c. Michelson examines all three alternatives, in a strictly Galilean context.

> c/L is wave emission rate. It has nothing to do with anything that
> happens after emission.

The ratio c/L is the conversion between cycles and seconds, i.e., the phase at any location advances by that many cycles per second.

> how can there be the same number of waves in both paths?

Ask someone who claims that. Again, as the mirrors move, near the nominal configuration, one optical path length is increasing and the other is decreasing, and at any given instant of the lab time there are different configurations depending on which of the models (r=0,1, or 2) you are talking about. This isn't a situation in which there are two standing waves, it is a transient situation, and the objective is to determine how out-of-phase the pulses are when they reach the receiver, which is determined by computing the different in transit times and converting that to the difference in phase.

> > The Newtonian ballistic theory (r=2) predicts no fringe shift at
> > all, and the emission theory (r=1) predicts (6D/L)(v/c), and the theory
> > with r=0 predicts (8D/L)(v/c). The experimental result confirms the
> > latter.
>
> You are just preaching the gospel of the E-religion.

Huh? There's nothing involving special relativity in this discussion. Michelson's analysis is in a purely classical context, based on three different hypotheses for how light speed is affected by reflection. Like Sagnac, this is a purely first-order effect, not involving any relativistic effects at all.

> The experiment assumes that if dt = 0 there cannot be different numbers
> of waves in the two paths..

Nope, it "assumes" that if dt=0, meaning that the split wavecrests arrive at the same time, then the two wave crests arrive at the same time, i.e., in phase. Are you honestly unable to grasp this? How could the waves be out of phase, if they are in phase?

> THAT ASSUMES LIGHT SPEED IS C.

Goodness no (despite the impressive caps). The fact that if the phases arrive in phase implies that the phases are arriving in phase does not depend upon or assume that the speed of light is c. Indeed, the analysis yielding zero phase shift is for the case when the speed of light reflect is c +- 2v, which is not equal to c. So, as always, your statements are completely incoherent and senseless.

> > The correct answer, which is indeed produced by just a few lines of
> > classical logic, results in predictions of 0, (6D/L)(v/c), and
> > (8D/L)(v/c) for the premises r=2, 1, and 0 respectively. The experiment
> > yields (8D/L)(v/c).

No substantive rely; duely noted.

> > Suppose you have a source of light with a known frequency f0 in terms of
> > its own inertial rest coordinates, and this source is moving with speed
> > v in terms of the lab frame, directly toward one lab receiver and
> > directly away from another. The receiver in front of the moving source
> > measures the frequency f1 (blue-shifted), and the receiver behind measures
> > the frequency f2 (red-shifted). To second order, what is (f1+f2)/f0 - 2 ?
>
> You are completely out of your depth here. ..

Your inability to answer the elementary question speaks for itself. Case closed.

On 2023-06-02 00:00:37 +0000, Jane said:

> On Thu, 01 Jun 2023 10:54:48 +0300, Mikko wrote:
>
>> On 2023-05-31 21:59:02 +0000, Jane said:
>>
>>> Rather than use a continuous light beam we will use a pulsed laser of
>>> measurable frequency.
>>
>> In the light clock there is no continuous beam nor a pulsed laser with a
>> measurable frequency. The purpose of the light clock is to be a simple
>> model of a clock that does not depend on any other clock.
>> A "masurable frequency" would involve another clock.
>
> It does....but in this modified light clock, that second 'clock' is used
> only for comparison.

The subject line is "The Correct Interpretation of the Light Clock". There
is no other clock in a light clock. No other clock is relevant to the
correctness of an interpretation of the light clock.

Mikko

On Friday, 2 June 2023 at 10:09:33 UTC+2, Mikko wrote:
> No other clock is relevant to the
> correctness of an interpretation of the light clock.

If a relativistic idiot is asserting - it
simply must be true. Anyway, no problem,
as light clocks exist only in his delusions.

On Fri, 02 Jun 2023 11:09:29 +0300, Mikko wrote:

> On 2023-06-02 00:00:37 +0000, Jane said:
>
>> On Thu, 01 Jun 2023 10:54:48 +0300, Mikko wrote:
>>
>>> On 2023-05-31 21:59:02 +0000, Jane said:
>>>
>>>> Rather than use a continuous light beam we will use a pulsed laser of
>>>> measurable frequency.
>>>
>>> In the light clock there is no continuous beam nor a pulsed laser with
>>> a measurable frequency. The purpose of the light clock is to be a
>>> simple model of a clock that does not depend on any other clock.
>>> A "masurable frequency" would involve another clock.
>>
>> It does....but in this modified light clock, that second 'clock' is
>> used only for comparison.
>
> The subject line is "The Correct Interpretation of the Light Clock".
> There is no other clock in a light clock. No other clock is relevant to
> the correctness of an interpretation of the light clock.

Ok, just use one pulse if you wish. You are then going to tell me that
the moving observer will not know the pulse rate he detects has slowed by
Newtonian Doppler because he is not allowed to have a clock and even if
he did it would have also slowed by gamma in the light clock frame and so
it would not detect any difference anyway. It that kind of circular logic
you are trying to inflict on us?
> Mikko

--
-- lover of truth

On 2023-06-02 09:49:55 +0000, Jane said:

> On Fri, 02 Jun 2023 11:09:29 +0300, Mikko wrote:

>> The subject line is "The Correct Interpretation of the Light Clock".
>> There is no other clock in a light clock. No other clock is relevant
>> to the correctness of an interpretation of the light clock.

> Ok, just use one pulse if you wish. You are then going to tell me that
> the moving observer will not know the pulse rate he detects has slowed by
> Newtonian Doppler because he is not allowed to have a clock and even if
> he did it would have also slowed by gamma in the light clock frame and so
> it would not detect any difference anyway.

No, I'm not. Whether the moving observer wants to use the light clock
for calibration of other instruments or to boil an egg is irrelevant
to the interpretation.

Mikko

On Friday, 2 June 2023 at 15:03:07 UTC+2, Mikko wrote:

> No, I'm not. Whether the moving observer wants to use the light clock
> for calibration of other instruments or to boil an egg is irrelevant
> to the interpretation.

I bet you have zilions of experiments confirming
it's irrelevant? A relativistic idiot always has them.
Still, no, it's not.

On Thursday, June 1, 2023 at 3:53:07 PM UTC-7, Jane wrote:
> On Wed, 31 May 2023 16:24:42 -0700, Dono. wrote:
>
> > On Wednesday, May 31, 2023 at 2:59:05 PM UTC-7, Jane wrote:
> >
> >> The idea of the beam becoming diagonal and the clock frequency being
> >> slowed by gamma in the moving frame is plainly laughable.
> >>
> > So, you never learned "light aberration". Fail.
> Aberration has nothing to do with this,

Actually, it DOES. This is why the light path is inclined. The irony is that this is a classical effect that you should have been familiar with.

>idiot.

Appropriate signature for you

On Fri, 02 Jun 2023 00:08:21 -0700, Trevor Lange wrote:

> On Thursday, June 1, 2023 at 4:22:15 PM UTC-7, Jane wrote:
>> c+rv is not c.
>
> Well, if r=0 then c+rv = c, but if r=1 or 2, then c+rv doesn't equal c.
> Michelson examines all three alternatives, in a strictly Galilean
> context.
>
>> c/L is wave emission rate. It has nothing to do with anything that
>> happens after emission.
>
> The ratio c/L is the conversion between cycles and seconds, i.e., the
> phase at any location advances by that many cycles per second.
>
>> how can there be the same number of waves in both paths?
>
> Ask someone who claims that. Again, as the mirrors move, near the
> nominal configuration, one optical path length is increasing and the
> other is decreasing, and at any given instant of the lab time there are
> different configurations depending on which of the models (r=0,1, or 2)
> you are talking about. This isn't a situation in which there are two
> standing waves, it is a transient situation, and the objective is to
> determine how out-of-phase the pulses are when they reach the receiver,
> which is determined by computing the different in transit times and
> converting that to the difference in phase.
>
>> > The Newtonian ballistic theory (r=2) predicts no fringe shift at all,
>> > and the emission theory (r=1) predicts (6D/L)(v/c), and the theory
>> > with r=0 predicts (8D/L)(v/c). The experimental result confirms the
>> > latter.
>>
>> You are just preaching the gospel of the E-religion.
>
> Huh? There's nothing involving special relativity in this discussion.
> Michelson's analysis is in a purely classical context, based on three
> different hypotheses for how light speed is affected by reflection. Like
> Sagnac, this is a purely first-order effect, not involving any
> relativistic effects at all.
>
>> The experiment assumes that if dt = 0 there cannot be different numbers
>> of waves in the two paths..
>
> Nope, it "assumes" that if dt=0, meaning that the split wavecrests
> arrive at the same time, then the two wave crests arrive at the same
> time, i.e., in phase. Are you honestly unable to grasp this? How could
> the waves be out of phase, if they are in phase?
>
>> THAT ASSUMES LIGHT SPEED IS C.
>
> Goodness no (despite the impressive caps). The fact that if the phases
> arrive in phase implies that the phases are arriving in phase does not
> depend upon or assume that the speed of light is c.

Don't 'goodness' me, you moron!

They arrive in phase because they leave in phase and travel for the same
time....BUT there are more waves in one path than the other, travelling
at different speeds.. Get it now?

> Indeed, the analysis
> yielding zero phase shift is for the case when the speed of light
> reflect is c +- 2v, which is not equal to c.

It is not the case for zero shift. It is the correct case of equal travel
times and more waves arriving from one path than the other.

So, as always, your statements are completely incoherent and senseless.

Like all relativists you are clueless. You cannot get into your head the
fact that light does not have an intrinsic frequency c/L and does not
necessarily move at c. When the mirrors are spinning, the path lengths
quite obviously differ by ~4d for v<<c. r plays no significant part in
the experiment, which is an absolute farce designed especially for
suckers like you who will do anything to prop up Einstein.
There are more 'wavelengths' in one path than the other but one is moving
at c+v and the other at c-v and the maths are such that the travel times
are exactly equal, WHICH THEY HAVE TO BE FOR THE FLASHES TO COINCIDE AND
FOR THE FRINGE PATTERN TO REMAIN STATIONARY. This happens because more
waves arrive at the receiver from one direction than the other. The
arrival frequencies are NOT c/L. One is (c+v)/L and the other is (c-v)/L.
The fringes move during an acceleration and remain displaced after is
ends. Get it now?

You obviously will not have the intelligence to understand any of this
even though it is not particularly difficult and quite obviously correct.
SR is proved wrong because it predicts the fringe pattern will move
during constant rotation.

--
-- lover of truth

On Friday, June 2, 2023 at 4:50:00 AM UTC-5, Jane wrote:
> On Fri, 02 Jun 2023 11:09:29 +0300, Mikko wrote:

> > The subject line is "The Correct Interpretation of the Light Clock".
> > There is no other clock in a light clock. No other clock is relevant to
> > the correctness of an interpretation of the light clock.
> Ok, just use one pulse if you wish. You are then going to tell me that
> the moving observer will not know the pulse rate he detects has slowed by
> Newtonian Doppler because he is not allowed to have a clock and even if
> he did it would have also slowed by gamma in the light clock frame and so
> it would not detect any difference anyway. It that kind of circular logic
> you are trying to inflict on us?

No circular logic at all.
1) Let the observer in S have a 1.5 meter light clock
aligned vertically along the y-axis
2) The observer in S will observe that his light clock
has a frequency of 10^8 pulses/second regardless
of the motions of anybody around him. The observer
in S sees the pulses go back and forth parallel to the
y-axis.
3) An observer in S1, in standard orientation relative
to S and motionless with respect to S, also sees
the pulses going back and forth relative to her
y-axis.
4) An observer in S2, in standard orientation relative
to S but moving along the x-axis at 0.1 c relative
to S, sees the pulses going along a zig-zag line
relative to its coordinates.

Do you have any objections to anything in (1)-(4)?

There is one and only one item that you object to,
and that is the FACT that over a century of experiment
and observation have always measured the speed of
light in vacuum to be c regardless of the relative
motions of source and receiver.

Over a century of experimental attempts to poke holes
in the second postulate or to find observational
evidence inconsistent with it have uniformly failed to
do so.

Do you have any reproducible experiments that you
can cite?

Experiment trumps all of your logic.

On 6/2/2023 8:48 AM, Jane wrote:
> On Fri, 02 Jun 2023 00:08:21 -0700, Trevor Lange wrote:
>
>> On Thursday, June 1, 2023 at 4:22:15 PM UTC-7, Jane wrote:
>>> c+rv is not c.
>>
>> Well, if r=0 then c+rv = c, but if r=1 or 2, then c+rv doesn't equal c.
>> Michelson examines all three alternatives, in a strictly Galilean
>> context.
>>
>>> c/L is wave emission rate. It has nothing to do with anything that
>>> happens after emission.
>>
>> The ratio c/L is the conversion between cycles and seconds, i.e., the
>> phase at any location advances by that many cycles per second.
>>
>>> how can there be the same number of waves in both paths?
>>
>> Ask someone who claims that. Again, as the mirrors move, near the
>> nominal configuration, one optical path length is increasing and the
>> other is decreasing, and at any given instant of the lab time there are
>> different configurations depending on which of the models (r=0,1, or 2)
>> you are talking about. This isn't a situation in which there are two
>> standing waves, it is a transient situation, and the objective is to
>> determine how out-of-phase the pulses are when they reach the receiver,
>> which is determined by computing the different in transit times and
>> converting that to the difference in phase.
>>
>>>> The Newtonian ballistic theory (r=2) predicts no fringe shift at all,
>>>> and the emission theory (r=1) predicts (6D/L)(v/c), and the theory
>>>> with r=0 predicts (8D/L)(v/c). The experimental result confirms the
>>>> latter.
>>>
>>> You are just preaching the gospel of the E-religion.
>>
>> Huh? There's nothing involving special relativity in this discussion.
>> Michelson's analysis is in a purely classical context, based on three
>> different hypotheses for how light speed is affected by reflection. Like
>> Sagnac, this is a purely first-order effect, not involving any
>> relativistic effects at all.
>>
>>> The experiment assumes that if dt = 0 there cannot be different numbers
>>> of waves in the two paths..
>>
>> Nope, it "assumes" that if dt=0, meaning that the split wavecrests
>> arrive at the same time, then the two wave crests arrive at the same
>> time, i.e., in phase. Are you honestly unable to grasp this? How could
>> the waves be out of phase, if they are in phase?
>>
>>> THAT ASSUMES LIGHT SPEED IS C.
>>
>> Goodness no (despite the impressive caps). The fact that if the phases
>> arrive in phase implies that the phases are arriving in phase does not
>> depend upon or assume that the speed of light is c.
>
> Don't 'goodness' me, you moron!
>
> They arrive in phase because they leave in phase and travel for the same
> time....BUT there are more waves in one path than the other, travelling
> at different speeds.. Get it now?
>
>> Indeed, the analysis
>> yielding zero phase shift is for the case when the speed of light
>> reflect is c +- 2v, which is not equal to c.
>
> It is not the case for zero shift. It is the correct case of equal travel
> times and more waves arriving from one path than the other.
>
> So, as always, your statements are completely incoherent and senseless.
>
> Like all relativists you are clueless. You cannot get into your head the
> fact that light does not have an intrinsic frequency c/L and does not
> necessarily move at c. When the mirrors are spinning, the path lengths
> quite obviously differ by ~4d for v<<c. r plays no significant part in
> the experiment, which is an absolute farce designed especially for
> suckers like you who will do anything to prop up Einstein.
> There are more 'wavelengths' in one path than the other but one is moving
> at c+v and the other at c-v and the maths are such that the travel times
> are exactly equal, WHICH THEY HAVE TO BE FOR THE FLASHES TO COINCIDE AND
> FOR THE FRINGE PATTERN TO REMAIN STATIONARY. This happens because more
> waves arrive at the receiver from one direction than the other. The
> arrival frequencies are NOT c/L. One is (c+v)/L and the other is (c-v)/L.
> The fringes move during an acceleration and remain displaced after is
> ends. Get it now?
>
> You obviously will not have the intelligence to understand any of this
> even though it is not particularly difficult and quite obviously correct.
> SR is proved wrong because it predicts the fringe pattern will move
> during constant rotation.

So when you bounced in here you made the claim that light in transit
has no frequency but then in this posting you are arguing about light
that "the phase at any location" and later "two wave crests arrive at
the same time."

You simply cannot have those statements be true without light having a
frequency. I'll save you the trouble of refuting this by saying you
simply don't understand what you are talking about. This what is called
painting oneself into a corner. All cranks do that eventually.

On Friday, 2 June 2023 at 18:40:24 UTC+2, Prokaryotic Capase Homolog wrote:

> There is one and only one item that you object to,
> and that is the FACT that over a century of experiment
> and observation have always measured the speed of
> light in vacuum to be c regardless of the relative
> motions of source and receiver.
>
> Over a century of experimental attempts to poke holes
> in the second postulate or to find observational
> evidence inconsistent with it have uniformly failed to
> do so.

And in the meantime in the real world, forbidden
by your bunch of idiots "improper" clocks keep
measuring improper t'=t in improper seconds.
Reproducibly.

On Fri, 02 Jun 2023 09:40:22 -0700, Prokaryotic Capase Homolog wrote:

> On Friday, June 2, 2023 at 4:50:00 AM UTC-5, Jane wrote:
>> On Fri, 02 Jun 2023 11:09:29 +0300, Mikko wrote:
>
>> > The subject line is "The Correct Interpretation of the Light Clock".
>> > There is no other clock in a light clock. No other clock is relevant
>> > to the correctness of an interpretation of the light clock.
>> Ok, just use one pulse if you wish. You are then going to tell me that
>> the moving observer will not know the pulse rate he detects has slowed
>> by Newtonian Doppler because he is not allowed to have a clock and even
>> if he did it would have also slowed by gamma in the light clock frame
>> and so it would not detect any difference anyway. It that kind of
>> circular logic you are trying to inflict on us?
>
> No circular logic at all.
> 1) Let the observer in S have a 1.5 meter light clock
> aligned vertically along the y-axis
> 2) The observer in S will observe that his light clock
> has a frequency of 10^8 pulses/second regardless of the motions of
> anybody around him.

Poor brainless Prokaryot, according to which 'clock' is that second
defined.

> The observer in S sees the pulses go back and
> forth parallel to the y-axis.
> 3) An observer in S1, in standard orientation relative
> to S and motionless with respect to S, also sees the pulses going
> back and forth relative to her y-axis.
> 4) An observer in S2, in standard orientation relative
> to S but moving along the x-axis at 0.1 c relative to S, sees the
> pulses going along a zig-zag line relative to its coordinates.
>
> Do you have any objections to anything in (1)-(4)?

Yes. Of course. The pulses do not move diagonally along one line.
The path of every infinitesimal element of the pulse can be represented
by a unique diagonal line on a graph. Each element moves at sqrt(c^2+v^2)
along its infinitesimally thin line. Nothing resembling a light beam
moves diagonally. The pulse of the clock, which must have finite length
and an axis, is not changed in any way. Its axis remains vertical in the
moving frame.

Understand now?
> There is one and only one item that you object to,
> and that is the FACT that over a century of experiment and observation
> have always measured the speed of light in vacuum to be c regardless of
> the relative motions of source and receiver.

There has never been a direct OWLS measurement from a moving source. The
fact that all TWLS measurements produce exactly the same answer is proof
that light is ballistic.
> Over a century of experimental attempts to poke holes in the second
> postulate or to find observational evidence inconsistent with it have
> uniformly failed to do so.

All claimed supporting evidence is full of much bigger holes.

I have stated one proof in another thread. Michelson's moving mirror
experiment. According to SR, the fringe pattern will continually move
during constant mirror rotation. That is the opposite of what is observed.
> Do you have any reproducible experiments that you can cite?

Just as Earth centrism can get things right occasionally, such as
predicting the tides, so will Einstein's SR...but for the wrong reasons.

Newtonian physics explains all observed phenomena.
> Experiment trumps all of your logic.

On Fri, 02 Jun 2023 11:57:03 -0500, whodat wrote:

> On 6/2/2023 8:48 AM, Jane wrote:
>> On Fri, 02 Jun 2023 00:08:21 -0700, Trevor Lange wrote:
>>

>>
>> It is not the case for zero shift. It is the correct case of equal
>> travel times and more waves arriving from one path than the other.
>>
>> So, as always, your statements are completely incoherent and senseless.
>>
>> Like all relativists you are clueless. You cannot get into your head
>> the fact that light does not have an intrinsic frequency c/L and does
>> not necessarily move at c. When the mirrors are spinning, the path
>> lengths quite obviously differ by ~4d for v<<c. r plays no significant
>> part in the experiment, which is an absolute farce designed especially
>> for suckers like you who will do anything to prop up Einstein.
>> There are more 'wavelengths' in one path than the other but one is
>> moving at c+v and the other at c-v and the maths are such that the
>> travel times are exactly equal, WHICH THEY HAVE TO BE FOR THE FLASHES
>> TO COINCIDE AND FOR THE FRINGE PATTERN TO REMAIN STATIONARY. This
>> happens because more waves arrive at the receiver from one direction
>> than the other. The arrival frequencies are NOT c/L. One is (c+v)/L and
>> the other is (c-v)/L.
>> The fringes move during an acceleration and remain displaced after is
>> ends. Get it now?
>>
>> You obviously will not have the intelligence to understand any of this
>> even though it is not particularly difficult and quite obviously
>> correct.
>> SR is proved wrong because it predicts the fringe pattern will move
>> during constant rotation.
>
> So when you bounced in here you made the claim that light in transit has
> no frequency but then in this posting you are arguing about light that
> "the phase at any location" and later "two wave crests arrive at the
> same time."

When you finish a physics degree I will explain how the moving wave
equation works...then you can explain it to Tommy.
> You simply cannot have those statements be true without light having a
> frequency.
Typical idiot crap. It has a wavelength.

> I'll save you the trouble of refuting this by saying you
> simply don't understand what you are talking about.

I know perfectly well what I am talking about and I know perfectly well
it is right. I also know you are not qualified to comment on whether it
is right or wrong because you know absolutely nothing about physics. You
are a typical obsequious, Einstein worshippping member of the E-church.

>This what is called
> painting oneself into a corner. All cranks do that eventually.

All the cranks are on your side...

On Friday, June 2, 2023 at 6:48:20 AM UTC-7, Jane wrote:
> > If dt=0, meaning that the split wavecrests arrive at the same time,
> > then the two wave crests arrive at the same time, i.e., in phase.
>
> They arrive in phase because they leave in phase and travel for the
> same time....

Yes! Bravo.

> BUT there are more waves in one path than the other, travelling
> at different speeds..

The point is that they arrive in phase, as you have acknowledged above. Of course, it goes without saying that if N wavecrests have been emitted (and split) by the instant when the first wavecrests arrive, then there are N crests in transit along each path. Surely you don't dispute this.

> > The analysis yielding zero phase shift is for the case when the speed of light
> > reflected is c +- 2v, which is not equal to c.
>
> It is not the case for zero shift.

Sure it is... the case r=2 is the ballistic Newtonian case where the particles of light are imagined to bounce elastically off the mirrors, and in this case the matched wavecrests arrive simultaneously, i.e., in phase, and hence there is no fringe shift.

> light does not have an intrinsic frequency c/L...

Well, the wavecrests arrive at a certain frequency f and have a certain phase speed V, and the quantity V/f is what's called the wavelength, for obvious reasons. Surely you don't dispute this.

> r plays no significant part in the experiment...

Sorry to be blunt, but that is insane, because the difference in speeds of the wavecrests along the two paths, together with the difference in path lengths, is what results in the differences in phase at the receiver. A correct statement would be: The parameter "r" plays no role in your analysis of the experiment, because your analysis is utterly brain-dead and specious.

> you who will do anything to prop up Einstein.

Huh? Again, Michelson's (no fan of relativity) analysis of this experiment has nothing to do with special relativity. There is no time dilation or length contraction or anything of the sort. Those are second-order effects, but this phenomenon (like Sagnac) is first-order. You keep being informed of this, and yet your brain fails to grasp it. Strange.

> The fringes move during an acceleration and remain displaced after is ends.

The fringe displacements are steady-state, not an artifact of some transient conditions. Again, in the simple case when the corresponding (split) wave crests along the two paths arrive at the same time, there is obviously the same number of wavecrests in transit. And, again, the point is that they are in phase. When they are out of phase (different transit times), there are the corresponding differences in the number (or fractional amount) of wave cycle differences, equal to the offset. Surely this is self-evident.

> SR is proved wrong because it predicts the fringe pattern will move
> during constant rotation.

That is an utterly absurd statement, first because special relativity is not even tested by this experiment -- falsifying ballistic and emission theories is not a positive "test of special relativity". Second, by no rational reasoning could any sentient being imagine that the r=0 case implies that the fringe pattern would behave differently than it does. The simple analysis of the cases is quite clear. Your beliefs on this subject are simply nuts, for the reasons explained above.

On Fri, 02 Jun 2023 06:35:56 -0700, Dono. wrote:

> On Thursday, June 1, 2023 at 3:53:07 PM UTC-7, Jane wrote:
>> On Wed, 31 May 2023 16:24:42 -0700, Dono. wrote:
>>
>> > On Wednesday, May 31, 2023 at 2:59:05 PM UTC-7, Jane wrote:
>> >
>> >> The idea of the beam becoming diagonal and the clock frequency being
>> >> slowed by gamma in the moving frame is plainly laughable.
>> >>
>> > So, you never learned "light aberration". Fail.
>> Aberration has nothing to do with this,
>
> Actually, it DOES. This is why the light path is inclined. The irony is
> that this is a classical effect that you should have been familiar with.

The light path is NOT inclined. Plot the bloody thing in the moving frame
and see for yourself.
>>idiot.
>

On Friday, June 2, 2023 at 5:01:05 PM UTC-7, Androcles' Ghost wrote:

> Yes. Of course. The pulses do not move diagonally along one line.
> The path of every infinitesimal element of the pulse can be represented
> by a unique diagonal line on a graph. Each element moves at sqrt(c^2+v^2)
> along its infinitesimally thin line.

Repeating the same imbecilities doesn't make them true

>Nothing resembling a light beam
> moves diagonally. The pulse of the clock, which must have finite length
> and an axis, is not changed in any way. Its axis remains vertical in the
> moving frame.

The above is false even in classical physics. "Aberration", remember, utter imbecile?

On Friday, June 2, 2023 at 5:15:34 PM UTC-7, Androcles' Ghost wrote:
> > If dt=0, meaning that the split wavecrests arrive at the same time,
> > then the two wave crests arrive at the same time, i.e., in phase.
>
> They arrive in phase because they leave in phase and travel for the
> same time....

Yes! Bravo!

> BUT there are more waves in one path than the other, travelling
> at different speeds..

The point is that they arrive in phase, as you've acknowledged above. Of course, it goes without saying that if N wavecrests have been emitted (and split) by the instant when the first wavecrests arrive, then there are N crests in transit along each path. Surely you don't dispute this.

> > The analysis yielding zero phase shift is for the case when the speed of light
> > reflected is c +- 2v, which is not equal to c.
>
> It is not the case for zero shift.

Sure it is... the case r=2 is the ballistic Newtonian case where the particles of light are imagined to bounce elastically off the mirrors, and in this case the matched wavecrests arrive simultaneously, i.e., in phase, so there would be no fringe shift.

> light does not have an intrinsic frequency c/L...

Well, the wavecrests arrive at a certain frequency f and have a certain phase speed V, and the quantity V/f is what's called the wavelength, for obvious reasons. Surely you do not dispute this.

> r plays no significant part in the experiment...

Sorry to be blunt, but that's insane, because the difference in speeds of the wavecrests along the two paths, together with the difference in path lengths, is what results in the differences in phase at the receiver. A correct statement would be: The parameter "r" plays no role in your analysis of the experiment, because your analysis is utterly brain-dead and specious.

> you who will do anything to prop up Einstein.

Say what? Again, Michelson's (no fan of relativity) analysis of this experiment has nothing to do with special relativity. There is no time dilation or length contraction or anything of the sort. Those are second-order effects, but this phenomenon (like Sagnac) is first-order. You keep being informed of this, and yet your brain fails to grasp it. Strange.

> The fringes move during an acceleration and remain displaced after is ends.

The fringe displacements are steady-state, not an artifact of some transient conditions. Again, in the simple case when the corresponding (split) wave crests along the two paths arrive at the same time, there is obviously the same number of wavecrests in transit. And, again, the point is that they are in phase. When they are out of phase (different transit times), there are the corresponding differences in the number (or fractional amount) of wave cycle differences, equal to the offset. This is self-evident.

> SR is proved wrong because it predicts the fringe pattern will move
> during constant rotation.

That's an utterly absurd statement, first because special relativity is not even tested by this experiment -- falsifying ballistic and emission theories is not a positive "test of special relativity". Second, by no rational reasoning could any sentient being imagine that the r=0 case implies that the fringe pattern would behave differently than it does. The simple analysis of the cases is quite clear. Your beliefs on this subject are simply nuts, for the reasons explained above.