Dono dunno much about MMX

[Prokaryotic Caspase Homolog]

Reference: Thread Michelson got his mirrors wrong

The discussion with Dono begins from approximately HERE:
https://groups.google.com/d/msg/sci.physics.relativity/Pyw8wLVN-Qk/XJBl_R3aBQAJ

Dono's post to which I am replying is HERE:
https://groups.google.com/d/msg/sci.physics.relativity/Pyw8wLVN-Qk/ZoM-1OooBgAJ

On Friday, October 28, 2016 at 11:18:54 PM UTC-5, Dono, wrote:
> On Friday, October 28, 2016 at 7:56:18 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >
> > Note that in these final adjustments, Joos was NOT concerned with trying to get
> > the path lengths equal;
>
> Of course he is, you are just too stupid to understand it. Let me see if I can dumb it down even further for you:
>
> You have an interferometer with an adjustable arm. Somehow, you start with the arms of equal length. You observe the interference pattern and, like in the days of Michelson, you scratch the position of the center fringe.
> You start lengthening the adjustable arms-the center fringe moves one way.
> You shorten the adjustable arm, the center fringe moves the OPPOSITE way.
> The above is indistiguishable with starting with unequal arm lengths and rotating the interferometer 90 degrees. If you do not get this, you will never understand the basis of MMX.

What is the relative orientation of the telescope and the arms of
the interferometer if you rotate the interferometer 45 degrees?

What is the relative orientation of the telescope and the arms of
the interferometer if you rotate the interferometer 22.5 degrees?

Remember that Michelson and Morley took measurements every 1/16 of
a turn. What does this imply about the relative orientation of the
telescope and the arms of the interferometer?

Joos photographed the fringes continuously as the apparatus was
rotated at a rate of one rotation every 10 minutes. What does this
imply about the relative orientation of the detector and the arms
of the interferometer?

[Dono,]

On Saturday, October 29, 2016 at 7:06:46 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> Reference: Thread Michelson got his mirrors wrong
>

> snip imbecilities due to your lack of understanding <

No point in wasting any more time with you. Continue to bask in your ignorance.

[Prokaryotic Caspase Homolog]

You can't answer the question, can you?

Because the answer will demonstrate that your understanding of
how MMX is set up was COMPLETELY OFF.

Coward.

[Dono,]

On Saturday, October 29, 2016 at 7:17:27 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Saturday, October 29, 2016 at 9:12:38 AM UTC-5, Dono, wrote:
> > On Saturday, October 29, 2016 at 7:06:46 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > Reference: Thread Michelson got his mirrors wrong
> > >
> > > snip imbecilities due to your lack of understanding <
> >
> > No point in wasting any more time with you. Continue to bask in your ignorance.
>
> You can't answer the question, can you?
>

Imbecile,

Your questions have nothing to do with the subject being discussed. The fact that you fail to see the disconnect proves that you have no clue.

[Prokaryotic Caspase Homolog]

I think that everybody who has bothered to follow the original
thread understands that you have been a fool.

[Dono,]

I thought that I had no way of dumbing down things any further for you. Turns out that I can:

1. If the arms start unequal, that induces a fringe difference.
2. The larger the difference in arms lengths, the larger the fringe difference:

[tex]n \lambda = \Delta d[/tex]

3. When the interferometer is rotated 90 degrees the fringe difference DOUBLES (by the very clever design of the experiment).

I do not expect an autistic imbecile like you to ever admit that you do not understand any of the above. I do expect you to continue to bask in your arrogant ignorance.

[Prokaryotic Caspase Homolog]

What happens when the interferometer rotates 45 degrees?
What happens when the interferometer rotates 22.5 degrees?

Remember the following exchange?

Me: "The eyepiece rotates along with the arms.
Didn't you know that?"
You: "Not in all setups. Apparently it is you who doesn't know. Look at all the embarrassing mistakes you have posted on the subject ("aether detection". "violation of Noether theorem", etc.). Stop puffing yourself."

https://groups.google.com/d/msg/sci.physics.relativity/Pyw8wLVN-Qk/QOpXLPr_BQAJ

Now. ***HOW IN THE WORLD*** can you have a fixed eyepiece in the MMX while
rotating the arms, say, 45 degrees?

Can you draw me a picture? Pretty please?

*** SHEESH!!! ***

Original Reference: Thread Michelson got his mirrors wrong
The discussion with Dono began from approximately HERE:
https://groups.google.com/d/msg/sci.physics.relativity/Pyw8wLVN-Qk/XJBl_R3aBQAJ

Google maintains a record, Dono.

Your inability to properly visualize the setup orders on the miraculous.
Unfortunately, I've seen worse on these groups...

[Dono,]

On Sunday, October 30, 2016 at 1:21:58 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> Me: "The eyepiece rotates along with the arms.

Not in the setup that I uploaded for you autistic imbecile. The OTHER arm lines yp with the eyepiece when the interferometer is rotated 90 degrees.

> Google maintains a record, Dono.
>

Yes,it does. Of your stubborn cretinism.

[Dono,]

In Ray Skinner "relativity for Scientists and Engineers" there is a solved example that I would recommend for you. It is example 1/2 on page 23.

It gives you a glimpse as to what happens if the arms of the interferometer were not adjusted to be equal.
Let's say that the arms are of length [tex]SM_1[/tex] and [tex]SM_2[/tex] using Skinner's notation.
Let's say that [tex]SM_1=SM_2+\epsilon[/tex]
Then, if you do the calculations, you will get that SR predicts a fringe shift of [tex]m=\frac{2 \epsilon}{\lambda}[/tex]

The "aether theory" predicts a shift of

[tex]n=2 \frac{SM_1+\epsilon}{\lambda} \frac{v^2}{c^2}[/tex]

Michelson would have not liked the presence of the "+\epsilon" in his formula, I can assure you.

Get the book, do the exercise. STFU for awhile.

[Paul B. Andersen]

On 30.10.2016 20:57, Dono, wrote:
>
> 1. If the arms start unequal, that induces a fringe difference.
> 2. The larger the difference in arms lengths, the larger the fringe difference:
>
> [tex]n \lambda = \Delta d[/tex]
>
> 3. When the interferometer is rotated 90 degrees the fringe difference DOUBLES (by the very clever design of the experiment).

So if the arms are unequal, there is a fringe difference
(whatever that may mean) which will double when the arms are
rotated 90 degrees?

What happened when the KTX interferometer was rotated 90 degrees?
https://paulba.no/paper/Kennedy_Thorndike.pdf

--
Paul

https://paulba.no/

[Dono,]

On Sunday, October 30, 2016 at 2:00:40 PM UTC-7, Paul B. Andersen wrote:
> On 30.10.2016 20:57, Dono, wrote:
> >
> > 1. If the arms start unequal, that induces a fringe difference.
> > 2. The larger the difference in arms lengths, the larger the fringe difference:
> >
> > [tex]n \lambda = \Delta d[/tex]
> >
> > 3. When the interferometer is rotated 90 degrees the fringe difference DOUBLES (by the very clever design of the experiment).
>
> So if the arms are unequal, there is a fringe difference
> (whatever that may mean) which will double when the arms are
> rotated 90 degrees?
>

Correct. Get the ray Skinner book, rework example 1.2 on page 23 for the case of unequal arms.

> What happened when the KTX interferometer was rotated 90 degrees?
> https://paulba.no/paper/Kennedy_Thorndike.pdf
>

One thing at the time, start by understanding MMX firs, we will get to KTX AFTER you really understand MMX. Based on your post above, you don't.

[Paul B. Andersen]

On 30.10.2016 21:21, Prokaryotic Caspase Homolog wrote:
>
> https://groups.google.com/d/msg/sci.physics.relativity/Pyw8wLVN-Qk/QOpXLPr_BQAJ

An interesting dialogue:
| On Friday, October 28, 2016 at 11:11:24 AM UTC-5, Dono, wrote:
|> On Friday, October 28, 2016 at 8:56:27 AM UTC-7, Prokaryotic Caspase
Homolog wrote:
|>> On Friday, October 28, 2016 at 10:48:40 AM UTC-5, Dono, wrote:
|>>> On Friday, October 28, 2016 at 8:44:53 AM UTC-7, Prokaryotic
Caspase Homolog wrote:

|>>>> On Friday, October 28, 2016 at 10:34:38 AM UTC-5, Dono, wrote:
|>>>>
|>>>> The eyepiece rotates along with the arms.
|>>>> Didn't you know that?
|>>>

|>>> Not in all setups.
|>>> Apparently it is you who doesn't know.
|>>

|>> Oh???
|>> Show me one, single, solitary MMX setup where the eyepiece, screen,
camera or
|>> other such detector does not rotate with the apparatus.
|>
|> You are getting really desperate. You could have made a much better
use of your
|> time if you did the simple calculations of fringe motion I showed you
in the initial post.
|> Your problem is not that you are incompetent, it is that you are too
arrogant to realize the magnitude of your incompetence.

Who is desperate and incompetent? :-D

--
Paul, amused

https://paulba.no/

[Paul B. Andersen]

So you don't know what happened when an interferometer
with unequal arms was rotated 90 degrees?

--
Paul

https://paulba.no/

[Dono,]

Yu tend to be a good guy but when you align yourself with the jerk you become as pathetic as he is.

[Dono,]

I know exactly what happens. Why don't you STFU and do the exercise from the Skinner book. When you align yourself with the jerk you become as much as a jerk as he is.

[Paul B. Andersen]

On 30.10.2016 21:45, Dono, wrote:
> On Sunday, October 30, 2016 at 1:21:58 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>>
>> Me: "The eyepiece rotates along with the arms.
>
> Not in the setup that I uploaded for you autistic imbecile. The OTHER arm lines yp with the eyepiece when the interferometer is rotated 90 degrees.
>

Wasn't the discussion about the MMX?
Was the eyepiece rotating along with the arms in the MMX interferometer?
https://paulba.no/paper/Michelson_1887.pdf

--
Paul

https://paulba.no/

[Dono,]

On Sunday, October 30, 2016 at 2:53:50 PM UTC-7, Paul B. Andersen wrote:
> On 30.10.2016 21:45, Dono, wrote:
> > On Sunday, October 30, 2016 at 1:21:58 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >>
> >> Me: "The eyepiece rotates along with the arms.
> >
> > Not in the setup that I uploaded for you autistic imbecile. The OTHER arm lines yp with the eyepiece when the interferometer is rotated 90 degrees.
> >
>
> Wasn't the discussion about the MMX?

Yes, it is about why it is so important that the arms in MMX are equal. Now, you can continue to troll or you can do the suggested exercise from the Ray Skinner book and learn something. Which one is it going to be?

[Paul B. Andersen]

On 30.10.2016 22:49, Dono, wrote:
> On Sunday, October 30, 2016 at 2:33:40 PM UTC-7, Paul B. Andersen wrote:
>> On 30.10.2016 22:14, Dono, wrote:
>>> On Sunday, October 30, 2016 at 2:00:40 PM UTC-7, Paul B. Andersen wrote:
>>>> On 30.10.2016 20:57, Dono, wrote:
>>>>>
>>>>> 1. If the arms start unequal, that induces a fringe difference.
>>>>> 2. The larger the difference in arms lengths, the larger the fringe difference:
>>>>>
>>>>> [tex]n \lambda = \Delta d[/tex]
>>>>>
>>>>> 3. When the interferometer is rotated 90 degrees the fringe difference DOUBLES (by the very clever design of the experiment).
>>>>
>>>> So if the arms are unequal, there is a fringe difference
>>>> (whatever that may mean) which will double when the arms are
>>>> rotated 90 degrees?
>>>>
>>>
>>> Correct. Get the ray Skinner book, rework example 1.2 on page 23 for the case of unequal arms.
>>>
>>>
>>>> What happened when the KTX interferometer was rotated 90 degrees?
>>>> https://paulba.no/paper/Kennedy_Thorndike.pdf
>>>>
>>>
>>> One thing at the time, start by understanding MMX firs, we will get to KTX AFTER you really understand MMX. Based on your post above, you don't.

And in the MMX the eyepiece is not rotating along with the arms,
and that's why there are fringe shifts if the arms are unequal
and are rotated 90 degrees ? :-D

>> So you don't know what happened when an interferometer
>> with unequal arms was rotated 90 degrees?
>>
>

> I know exactly what happens.

So you know that there was no fringe shifts when
the interferometer with very unequal arms was
rotated 90 degrees.

And now you are desperate to divert the attention
from your blunder:

> Why don't you STFU and do the exercise from the Skinner book.
> When you align yourself with the jerk you become as much as a jerk as he is.

:-D

--
Paul, having fun

https://paulba.no/

[Paul B. Andersen]

On 30.10.2016 23:06, Dono, wrote:
> On Sunday, October 30, 2016 at 2:53:50 PM UTC-7, Paul B. Andersen wrote:
>> On 30.10.2016 21:45, Dono, wrote:
>>> On Sunday, October 30, 2016 at 1:21:58 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>>>>
>>>> Me: "The eyepiece rotates along with the arms.
>>>
>>> Not in the setup that I uploaded for you autistic imbecile.
>>> The OTHER arm lines yp with the eyepiece when the interferometer is rotated 90 degrees.

And in this interferometer which never was used in
the MMX the fringes will obviously change.

BUT:

>> Wasn't the discussion about the MMX?
>
> Yes, it is about why it is so important that the arms in MMX are equal.
> Now, you can continue to troll or you can do the suggested exercise from
> the Ray Skinner book and learn something. Which one is it going to be?
>

Any particular reason why you snipped this?

>> Was the eyepiece rotating along with the arms in the MMX interferometer?
>> https://paulba.no/paper/Michelson_1887.pdf

A rhetoric question, of course.
The reason is obvious to all.

Bedtime. Good night.

--
Paul, yawn

https://paulba.no/

[al...@interia.pl]

Another stupidity from the pseudoscientists labs.

they all showed and used the fact:
r' = c't is automatically replaced by:
ct' under the c = inv fallacy,
due to the fact: r' = r', or in other way: 1 = 1.

[Prokaryotic Caspase Homolog]

On Sunday, October 30, 2016 at 3:45:32 PM UTC-5, Dono, wrote:
> On Sunday, October 30, 2016 at 1:21:58 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >
> > Me: "The eyepiece rotates along with the arms.
>
> Not in the setup that I uploaded for you autistic imbecile. The OTHER arm lines yp with the eyepiece when the interferometer is rotated 90 degrees.

It is very annoying that you prohibit downloads. It makes me have to work a
little bit before I can bypass your prohibitions.

Having downloaded the file despite your efforts to prevent me from doing so,
I've uploaded it here so that other people can easily get their own copy:
https://drive.google.com/open?id=0B8XIf0XcrpOcY0RraEVoUDZHSTA

What you have is a lab manual for students to learn some basics about fringe
formation in the Michelson interferometer, along with an exercise to learn
a practical application of the interferometer for precise measurement of
distance.

The interferometer described in this lab manual does not rotate, and nowhere
is there any swapping of beams.

The Michelson interferometer in this lab manual is not set up for the MMX
experiment.

[Prokaryotic Caspase Homolog]

On Sunday, October 30, 2016 at 5:06:46 PM UTC-5, Dono, wrote:

> Yes, it is about why it is so important that the arms in MMX are equal.

It was important to Michelson and Morley because they observed fringes in
white light. White light has a very short coherence length.

Experimenters using monochromatic light didn't need to match the arms with
the same sort of precision that Michelson and Morley required. Kennedy didn't
need precisely matched arms, Illingworth didn't need precisely matched arms,
Joos didn't need precisely matched arms, etc.

I have elsewhere gone into detail as to why the early experimenters preferred
performing the MMX experiment using white light. You've ignored my explanations.

[Dono,]

On Sunday, October 30, 2016 at 5:00:30 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Sunday, October 30, 2016 at 5:06:46 PM UTC-5, Dono, wrote:
>
> > Yes, it is about why it is so important that the arms in MMX are equal.
>
> It was important to Michelson and Morley because they observed fringes in
> white light. White light has a very short coherence length.
>

This is the closest that you will come to admitting that you were wrong all along.

[Dono,]

On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
> snip imbecilities<

It is sad to see that you turned into an imbecile. You could have done the exercise, yet you elected to troll. Must be that you are getting old and dementia is setting in. sad to see how you have deteriorated.

[Dono,]

On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
>
> So you know that there was no fringe shifts when
> the interferometer with very unequal arms was
> rotated 90 degrees.
>

Posting the above shows that you do not understand KTX (in addition to your proving earlier that you do not understand MMX).

KTX is not, contrary to your naive beliefs, "an MMX with unequal arms". The two experiments are VERY different in concept, there is no such thing as "rotating the interferometer 90 degrees" in KTX. KTX relies on a slow revolution, over a long period of time such that the wavelength has the chance to change when transformed from the frame of the lab into the "aether" frame.
MMX relies on an instantaneous rotation such that the wavelength DOES NOT CHANGE when transformed between the same frames as above.
I could explain a lot of things to you, you miss a lot of the basics. But that would require that you behave less as the other jerk and with more respect.

[Prokaryotic Caspase Homolog]

(sigh)
Michelson and Morley's original experimental plan was to perform comparative
checks of fringe shifts (1) with the apparatus oriented at different angles
with respect to the aether, by rotating the apparatus, and (2) with the
apparatus traveling at different velocities through the aether, by repeating
the experiment at multiple times throughout the year.

Their first set of experimental runs were so utterly discouraging and
unexpected that they did not carry out their second program of repeating the
experiment throughout the year. Only Miller was to fulfill this second part of
the experimental plan.

The speed of rotation was unimportant. Michelson and Moreley might have adopted
Kennedy and Thorndike's strategy of letting the Earth perform the rotation
except for fundamental

***TECHNOLOGICAL LIMITATIONS***

which were
1) Air conditioning hadn't been invented yet. There was no way of shielding
the apparatus from temperature change, and temperature changes of as little
as 0.001°C would make perceptible shifts in fringe position.
2) The apparatus was too large to shield from changes in atmospheric pressure,
nor could they perform the experiment in vacuum.

Michelson and Morley rotated the apparatus at a rate of one rotation every
several minutes because they were in a *race* to perform each set of fringe
shift measurements before the environment conditions had changed too much.

Kennedy and Thorndike, in addition to testing for the *isotropy* of light in
different orientations with respect to the hypothetical aether, also wanted to
test for time dilation effects by checking for fringe shifts with the apparatus
in various states of movement with respect to the aether. This they did by
using an apparatus with arms as unequal in length as possible, given the
coherence length of their mercury source.

Kennedy and Thorndike performed their experiment in a constant temperature
room that maintained the environmental temperature to 0.1°C. The apparatus
itself was much scaled down in size from Michelson and Morley's experiment, so
that it was possible to use a temperature controlled water jacket to maintain
the critical optical assembly at a temperature constant to 0.001°C.
Furthermore, the optical assembly was in vacuum.
https://en.wikipedia.org/wiki/File:Kennedy-Thorndike_experiment_DE.svg

It was possible for Kennedy and Thorndike to scale down the dimensions of
their apparatus from that used in Michelson and Morley's experiment because of
several other technological innovations.
1) They recorded the fringes photographically. This enabled post-experimental
measurements of fringe position to thousandths of fringe.
2) They maximized fringe contrast by using polarized light with the beam
splitter set at Brewster's angle, thus eliminating back surface reflections.
The greater contrast greatly improved their ability to measure fringe
position.

Later in life, Michelson was to propose construction of a giant interferometer,
too large to rotate manually, which would depend on Earth's rotation rather than
manual rotation to orient the apparatus at different directions with respect to
the aether. Nothing ever came of this later proposal, but it does show that
"instantaneous rotation", as you put it, was not an *essential* part of the
experimental design of the MMX. It was adopted because of the technological
limitations of the time, not because of theoretical necessity.

[Paul B. Andersen]

The KTX interferometer was a modified Michelson interferometer with
very unequal arms. It was rotated by the Earth, and the fringe
shifts were consistent with null.
You will know this, so you know that there are no fringe shifts
when an interferometer with unequal arms are rotated any angle.


Case closed for my part.

--
Paul

https://paulba.no/

[Paul B. Andersen]

On 31.10.2016 11:38, Prokaryotic Caspase Homolog wrote:
>
> Kennedy and Thorndike, in addition to testing for the *isotropy* of light in
> different orientations with respect to the hypothetical aether, also wanted to
> test for time dilation effects by checking for fringe shifts with the apparatus
> in various states of movement with respect to the aether. This they did by
> using an apparatus with arms as unequal in length as possible, given the
> coherence length of their mercury source.
>

Right.
Just a little supplementing comment to this.
The MMX showed that the speed of light was isotropic in
the lab frame. If it had been repeated with null result
several times during the year, it would have shown
that the speed of light was isotropic in all frames,
but it would _not_ have shown that the speed of light
was the _same_ in all frames of reference.

If you have an interferometer with unequal arms,
and change nothing but the speed of light, there
will be fringe shifts by obvious reasons.
So when the KTX interferometer was rotated by
the Earth and repeated at several times during the year
(January, April, May, July, August and October)
and the result always was consistent with null,
it showed that the speed of light was isotropic
and the same in all frames of reference.
In other words, the speed of light is invariant.
The "Relativity of Time" (in the title) is an inevitable
consequence of the invariance of the speed of light.

[Dono,]

On Monday, October 31, 2016 at 3:38:46 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Sunday, October 30, 2016 at 11:44:05 PM UTC-5, Dono, wrote:
> > On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
> > >
> > > So you know that there was no fringe shifts when
> > > the interferometer with very unequal arms was
> > > rotated 90 degrees.
> > >
> >
> > Posting the above shows that you do not understand KTX (in addition to your proving earlier that you do not understand MMX).
> >
> > KTX is not, contrary to your naive beliefs, "an MMX with unequal arms". The two experiments are VERY different in concept, there is no such thing as "rotating the interferometer 90 degrees" in KTX. KTX relies on a slow revolution, over a long period of time such that the wavelength has the chance to change when transformed from the frame of the lab into the "aether" frame.
> > MMX relies on an instantaneous rotation such that the wavelength DOES NOT CHANGE when transformed between the same frames as above.
> > I could explain a lot of things to you, you miss a lot of the basics. But that would require that you behave less as the other jerk and with more respect.
>

>snip stuff copied off the internet<
While what you copied is true, it is not relevant to our debate. Your time would have been spent much better doing the exercise from Skinner. You would have learned why the arms NEED to be equal.

[Dono,]

On Monday, October 31, 2016 at 6:20:26 AM UTC-7, Paul B. Andersen wrote:
> On 31.10.2016 05:44, Dono, wrote:
> > On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
> >>
> >> So you know that there was no fringe shifts when
> >> the interferometer with very unequal arms was
> >> rotated 90 degrees.
> >>
> >
> > Posting the above shows that you do not understand KTX (in addition to your proving earlier that you do not understand MMX).
> >
> > KTX is not, contrary to your naive beliefs, "an MMX with unequal arms". The two experiments are VERY different in concept, there is no such thing as "rotating the interferometer 90 degrees" in KTX. KTX relies on a slow revolution, over a long period of time such that the wavelength has the chance to change when transformed from the frame of the lab into the "aether" frame.
> > MMX relies on an instantaneous rotation such that the wavelength DOES NOT CHANGE when transformed between the same frames as above.
> > I could explain a lot of things to you, you miss a lot of the basics. But that would require that you behave less as the other jerk and with more respect.
> >
>
> The KTX interferometer was a modified Michelson interferometer with
> very unequal arms. It was rotated by the Earth, and the fringe
> shifts were consistent with null.

While this is all true it doesn't change the fact that you have some very serious holes in your understanding.

> You will know this, so you know that there are no fringe shifts
> when an interferometer with unequal arms are rotated any angle.
>
>

You did not understand a word of what I explained to you. You did not even try.

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 10:57:48 AM UTC-5, Dono, wrote:
> On Monday, October 31, 2016 at 3:38:46 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > On Sunday, October 30, 2016 at 11:44:05 PM UTC-5, Dono, wrote:
> > > On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
> > > >
> > > > So you know that there was no fringe shifts when
> > > > the interferometer with very unequal arms was
> > > > rotated 90 degrees.
> > > >
> > >
> > > Posting the above shows that you do not understand KTX (in addition to your proving earlier that you do not understand MMX).
> > >
> > > KTX is not, contrary to your naive beliefs, "an MMX with unequal arms". The two experiments are VERY different in concept, there is no such thing as "rotating the interferometer 90 degrees" in KTX. KTX relies on a slow revolution, over a long period of time such that the wavelength has the chance to change when transformed from the frame of the lab into the "aether" frame.
> > > MMX relies on an instantaneous rotation such that the wavelength DOES NOT CHANGE when transformed between the same frames as above.
> > > I could explain a lot of things to you, you miss a lot of the basics. But that would require that you behave less as the other jerk and with more respect.
> >
> >snip stuff copied off the internet<

Oh? Where did you find this stuff on the network that you claim that I copied?
Please. Find a url and put my text through a plagiarism detector. I extremely doubt that you'll find any site on the network that will echo my particular
"takes" on MMX and KTX, which include, so far as I can tell, a significant
amount of analysis not found in any other secondary source (although obvious
from primary source readings).

(restoring snipped EXTREMELY RELEVANT stuff)

> While what you copied is true, it is not relevant to our debate. Your time would have been spent much better doing the exercise from Skinner. You would have learned why the arms NEED to be equal.

But they don't need to be, not if you are using monochromatic light.

Otherwise KTX would have given false positives.

[al...@interia.pl]

W dniu poniedziałek, 31 października 2016 11:38:46 UTC+1 użytkownik Prokaryotic Caspase Homolog

> Kennedy and Thorndike, in addition to testing for the *isotropy* of light in
> different orientations with respect to the hypothetical aether, also wanted to
> test for time dilation effects by checking for fringe shifts with the apparatus
> in various states of movement with respect to the aether. This they did by
> using an apparatus with arms as unequal in length as possible, given the
> coherence length of their mercury source.
>
> Kennedy and Thorndike performed their experiment in a constant temperature
> room that maintained the environmental temperature to 0.1°C. The apparatus
> itself was much scaled down in size from Michelson and Morley's experiment, so
> that it was possible to use a temperature controlled water jacket to maintain
> the critical optical assembly at a temperature constant to 0.001°C.
> Furthermore, the optical assembly was in vacuum.
> https://en.wikipedia.org/wiki/File:Kennedy-Thorndike_experiment_DE.svg

Unfortunately in the vaccum the contraction alone perfectly
compensates the lightspeed anisotropy.

c'(f) = c/(1 + v/c cosf)

thus the measured two-way time (and of a phase shift)
are independent of the agnle f,
because a mean speed of light is now:

<c'> = 2/[1/c'(f) + 1/(c'(f+180)] =
2c/[(1 + v/c cosf) + (1 - v/c cosf)] = 2c /2 = c = inv

because: cos(x+180) = -cosx;

> It was possible for Kennedy and Thorndike to scale down the dimensions of
> their apparatus from that used in Michelson and Morley's experiment because of
> several other technological innovations.
> 1) They recorded the fringes photographically. This enabled post-experimental
> measurements of fringe position to thousandths of fringe.
> 2) They maximized fringe contrast by using polarized light with the beam
> splitter set at Brewster's angle, thus eliminating back surface reflections.
> The greater contrast greatly improved their ability to measure fringe
> position.

Only the experiment with n <> 1
can show the light anisotropy, because the shift depends additionaly on this:

sqrt(n^2-1), what is perfectly null for vaccum: n = 1!

for the air: n = 1.00025 ->
sqrt(n^2-1) = 1/45

thus the measured speeds in the air conditions are underestimated 40-50 times!

for example: Michelson 'measured' about 8 km/s,
thus the correct value, which he measured in fact, was just equal to:
8 x 45 km/s =~ 300-400 km/s !!!

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 10:57:48 AM UTC-5, Dono, wrote:

> While what you copied is true, it is not relevant to our debate. Your time would have been spent much better doing the exercise from Skinner. You would have learned why the arms NEED to be equal.

Dono, every year in high school science fairs across the country, you see dozens
of repetitions of the easier-to-do classic experiments. I know. As a former
judge, I've seen MMX, Cavendish, RFLP analyses, simple genetic engineering
projects and so forth. I've seen a demonstration of the Woodward–Hoffmann
rules in organic chemistry, and a demonstration of gravitational time
dilation using cheap rubidium clocks purchased from eBay.

Extrapolating from what I've seen, I would imagine the number of re-enactments
of MMX across the nation number probably in the dozens each year. Almost always
the students use cheap pen lasers, and they *never* bother to get the arm
lengths exactly equal. (How could they? One dark fringe looks just like
another!)

Yet these high school kids consistently report null results. I twirled the
MMX apparatus that I saw that was floating in a tub of water, arm length
maybe half a meter. No fringe shifts, despite the arm lengths probably being
mismatched by a few millimeters. The complete setup wasn't there, because the
student mounted his cell phone to take pictures of the fringes.

[al...@interia.pl]

W dniu poniedziałek, 31 października 2016 18:25:46 UTC+1 użytkownik Prokaryotic Caspase Homolog:

> Yet these high school kids consistently report null results.

Indeed. the kids permanently measures just the nothing.

But that is nothing yet:
they are invariable very proud of this... his own great stupidity, like you are.

[Dono,]

On Monday, October 31, 2016 at 9:43:32 AM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> > While what you copied is true, it is not relevant to our debate. Your time would have been spent much better doing the exercise from Skinner. You would have learned why the arms NEED to be equal.
>
> But they don't need to be, not if you are using monochromatic light.

Actually, they DO NEED to be equal. Obviously you are incapable of doing the simple exercise from the Skinner book.

>
> Otherwise KTX would have given false positives.

Nope, KTX relies on the wavelength change with time (time dilation). MMX relies on the fact that wavelength DOESN'T change. You do not understand the basics of the experiments. Keep basking in your ignorance.

[Dono,]

On Monday, October 31, 2016 at 10:25:46 AM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> Yet these high school kids consistently report null results. I twirled the
> MMX apparatus that I saw that was floating in a tub of water, arm length
> maybe half a meter. No fringe shifts, despite the arm lengths probably being
> mismatched by a few millimeters. The complete setup wasn't there, because the
> student mounted his cell phone to take pictures of the fringes.

This is what happens when you are incapable of doing the simple exercise from the Skinner book: you continue to bask in your ignorance.

[Prokaryotic Caspase Homolog]

YOUR problem is that you seemingly are *only* capable of doing math. You
have no understanding of how the math relates to reality.

[Tom Roberts]

On 10/31/16 10/31/16 - 1:28 PM, Dono, wrote:
> Actually, [the arms of the MMX] DO NEED to be equal.

Yes and no -- yes for the instrumentation they used a century ago; no for the
principle of the experiment and for modern instrumentation. The need for equal
arms depends ONLY on the coherence length of the light source. The white light
source used by Michelson and Morley had a coherence length of a few microns, and
their arms had to be equal in length to that accuracy in order to get visible
fringes. Today there are lasers that can achieve coherence lengths of several
kilometers; we have two in our lab and routinely use a Michelson interferometer
with arms ~10 cm and ~60 cm -- DELIBERATELY not equal because that reduces the
free spectral range of the instrument (we are NOT looking for aether drift!).

BTW this is purely an instrumentation effect; there's nothing in the underlying
physics that requires equal arms.

My library does not have the Skinner book you reference. But
I doubt very much that the exercise you mention does what you
claim: you claim that it implies that SR predicts a "fringe
shift" of "2 \epsilon / \lambda", where \epsilon is the
difference in length of the two arms. This is just plain
wrong. I do not know if the error is Skinner's or yours, but
I strongly suspect the latter.

In fact, for a Michelson interferometer with arms of different
lengths (within the coherence length of the source), SR
predicts ZERO fringe shift as the interferometer is rotated,
independent of the lengths of the arms. But changing the length
of one arm (i.e. changing \epsilon) will shift the fringes,
and I suspect that is what you calculated (the value you give
"just happens" to be equal to this, including the factor of 2).

In particular, the Michelson interferometer in our lab has 2 \epsilon / \lambda
~ 600,000. But as the earth rotates beneath it, it displays no fringe shift at
all (to an accuracy of ~ 0.000015 fringe) -- billions of times smaller than your
claim.

Tom Roberts

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 8:41:07 AM UTC-5, Paul B. Andersen wrote:
> On 31.10.2016 11:38, Prokaryotic Caspase Homolog wrote:
> >
> > Kennedy and Thorndike, in addition to testing for the *isotropy* of light in
> > different orientations with respect to the hypothetical aether, also wanted to
> > test for time dilation effects by checking for fringe shifts with the apparatus
> > in various states of movement with respect to the aether. This they did by
> > using an apparatus with arms as unequal in length as possible, given the
> > coherence length of their mercury source.
> >
>
> Right.
> Just a little supplementing comment to this.
> The MMX showed that the speed of light was isotropic in
> the lab frame. If it had been repeated with null result
> several times during the year, it would have shown
> that the speed of light was isotropic in all frames,
> but it would _not_ have shown that the speed of light
> was the _same_ in all frames of reference.
>
> If you have an interferometer with unequal arms,
> and change nothing but the speed of light, there
> will be fringe shifts by obvious reasons.

Good point that I had neglected to explain properly!

[Dono,]

On Monday, October 31, 2016 at 1:14:19 PM UTC-7, tjrob137 wrote:
> On 10/31/16 10/31/16 - 1:28 PM, Dono, wrote:
> > Actually, [the arms of the MMX] DO NEED to be equal.
>
> Yes and no -- yes for the instrumentation they used a century ago; no for the
> principle of the experiment and for modern instrumentation. The need for equal
> arms depends ONLY on the coherence length of the light source.

The coherence length of white light is only ONE factor, the weaker one.
The stronger reason is that unequal arms will create a fringe shift. I am going to recommend to you example 1.2 on page 23 on the Ray Skinner book" "Relativity: for Scientists and Engineers".
The other two are incapable of doing the simple exercise, maybe you will be able to.

> My library does not have the Skinner book you reference. But
> I doubt very much that the exercise you mention does what you
> claim: you claim that it implies that SR predicts a "fringe
> shift" of "2 \epsilon / \lambda", where \epsilon is the
> difference in length of the two arms. This is just plain
> wrong.

Actually, it is correct. Go to the library and get the book.

> In fact, for a Michelson interferometer with arms of different
> lengths (within the coherence length of the source), SR
> predicts ZERO fringe shift as the interferometer is rotated,
> independent of the lengths of the arms.

Let's check this claim:

In the frame of the lab, the time difference is [tex]t_0=\frac{2L_1}{c}-\frac{2L_2}{c}[/tex]

[tex]L_1=L_2+\epsilon[/tex]

So

[tex]t_0=\frac{2 \epsilon}{c}[/tex]

When you rotate the interferometer 90 degrees, the difference doubles and the fringes travel about the zero point by a total excursion of [tex]4 \epsilon[/tex]

[Dono,]

Your problem is that you have no clue, either physics or math.

[Dono,]

On Monday, October 31, 2016 at 1:14:19 PM UTC-7, tjrob137 wrote:
>

> In particular, the Michelson interferometer in our lab has 2 \epsilon / \lambda
> ~ 600,000. But as the earth rotates beneath it, it displays no fringe shift at
> all (to an accuracy of ~ 0.000015 fringe) -- billions of times smaller than your
> claim.
>

This is because you are not truly exchanging the roles of the two arms, you rotate the light source (and the intereference detector) WITH the arms. Duh!

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 4:21:01 PM UTC-5, Dono, wrote:
> On Monday, October 31, 2016 at 1:14:19 PM UTC-7, tjrob137 wrote:

> > In fact, for a Michelson interferometer with arms of different
> > lengths (within the coherence length of the source), SR
> > predicts ZERO fringe shift as the interferometer is rotated,
> > independent of the lengths of the arms.
>
> Let's check this claim:

<snip>

Experience trumps your faulty interpretation of the theory, Dono.

I have personally twirled a student-built MMX apparatus floating in a tub of
water with approximately 0.5 m arms and I don't remember how many back-and-forth
reflections without seeing any fringe shift.

Tom has in his lab unequal arm MMX setups that would be *zillions* of times more
sensitive to unequal arm length effects than the student-built MMX that I played
with.

[Prokaryotic Caspase Homolog]

*** BUT THAT IS HOW THE MICHELSON-MORLEY EXPERIMENT WAS SET UP!!! ***

[Dono,]

On Monday, October 31, 2016 at 2:34:18 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Monday, October 31, 2016 at 4:21:01 PM UTC-5, Dono, wrote:
> > On Monday, October 31, 2016 at 1:14:19 PM UTC-7, tjrob137 wrote:
>
> > > In fact, for a Michelson interferometer with arms of different
> > > lengths (within the coherence length of the source), SR
> > > predicts ZERO fringe shift as the interferometer is rotated,
> > > independent of the lengths of the arms.
> >
> > Let's check this claim:
>
> <snip>
>
> Experience trumps your faulty interpretation of the theory, Dono.
>

Precisely

1. If you rotate the whole assembly (interferometer+light source+detector) you will not get a fringe shift because you are NOT exchanging the roles of the arms. I pointed that out to you several times but you are too dumb to get the importance. The setup in our lab rotates the arms ONLY.

2. Michelson did not know about SR, his theory of the experiment predicts fringe shift if the whole setup is rotated (see the Skinner example that you are incapable of solving)

3. If you rotate the arms only such you truly exchange the roles of the arms, you will get a fringe shift because of the exchange in the arms role. SR predicts such a fringe shift (see the simple arithmetic that you snipped like the idiot you are).

> I have personally twirled a student-built MMX apparatus floating in a tub of
> water with approximately 0.5 m arms and I don't remember how many back-and-forth
> reflections without seeing any fringe shift.
>

BECAUSE you "twirled" the whole enchilada. You still don't get it. You had a chance to learn but you are too set in your ways to even TRY to comprehend.

> Tom has in his lab unequal arm MMX setups that would be *zillions* of times more
> sensitive to unequal arm length effects than the student-built MMX that I played
> with.

Tom is as dumb as you, he is unwilling to engage brain and learn that there are setups DIFFERENT from his. Setups that are a lot more interesting than his.
Nor does he understand that Michelson applied a DIFFERENT theory to his experiment. In THAT framework, the theory predicts fringe shift for unequal arms. Dumb and dumber.

[Dono,]

[HGW]

On 01/11/16 00:41, Paul B. Andersen wrote:
> On 31.10.2016 11:38, Prokaryotic Caspase Homolog wrote:

> If you have an interferometer with unequal arms,
> and change nothing but the speed of light, there
> will be fringe shifts by obvious reasons.
> So when the KTX interferometer was rotated by
> the Earth and repeated at several times during the year
> (January, April, May, July, August and October)
> and the result always was consistent with null,
> it showed that the speed of light was isotropic
> and the same in all frames of reference.

Rubbish. It merely showed that the light used moved at the same speed
relative to the source no matter what the orientation.

> In other words, the speed of light is invariant.
> The "Relativity of Time" (in the title) is an inevitable
> consequence of the invariance of the speed of light.
> https://paulba.no/paper/Kennedy_Thorndike.pdf

When the MM interferometer was rotated, they hoped to see fringe
MOVEMENT rather the displacement. The pattern would remain the same.
They had to look through the eyepiece the whole time it was being
rotated and hopefully count the number of fringes that passed a
crosshair in the eyepiece. Of course there was no movement because light
is ballistic.

It doesn't matter if the arm lengths are different. It merely affects
the space between the lines or circles. In the interferometer I made, I
just used flourescent light passed through a collimator.


--

[Prokaryotic Caspase Homolog]

[facepalm]

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 4:47:26 PM UTC-5, Dono, wrote:

> 1. If you rotate the whole assembly (interferometer+light source+detector) you will not get a fringe shift because you are NOT exchanging the roles of the arms. I pointed that out to you several times but you are too dumb to get the importance. The setup in our lab rotates the arms ONLY.

You can't even read your own lab handout.
You aren't rotating the arms.
You are laterally translating a mirror.

That is not MMX.
The lab experiment is using a Michelson interferometer for *length measurement*

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 4:47:26 PM UTC-5, Dono, wrote:

> 1. If you rotate the whole assembly (interferometer+light source+detector) you will not get a fringe shift because you are NOT exchanging the roles of the arms. I pointed that out to you several times but you are too dumb to get the importance. The setup in our lab rotates the arms ONLY.

Bullshit. You can't even read your own lab manual.

> 2. Michelson did not know about SR, his theory of the experiment predicts fringe shift if the whole setup is rotated (see the Skinner example that you are incapable of solving)
>
> 3. If you rotate the arms only such you truly exchange the roles of the arms, you will get a fringe shift because of the exchange in the arms role. SR predicts such a fringe shift (see the simple arithmetic that you snipped like the idiot you are).

Have you considered whether it is physically possible to perform such an action
with any degree of precision?

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 4:47:26 PM UTC-5, Dono, wrote:

> Tom is as dumb as you, he is unwilling to engage brain and learn that there are setups DIFFERENT from his. Setups that are a lot more interesting than his.

Your fantasized setup is impossible to realize.

Draw me
1) a picture of how your setup looks initially, and
2) how it looks after rotating the arms.

How would you manage the switcheroo?

[Dono,]

On Monday, October 31, 2016 at 3:22:16 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Monday, October 31, 2016 at 4:47:26 PM UTC-5, Dono, wrote:
>
> > 1. If you rotate the whole assembly (interferometer+light source+detector) you will not get a fringe shift because you are NOT exchanging the roles of the arms. I pointed that out to you several times but you are too dumb to get the importance. The setup in our lab rotates the arms ONLY.
>
> You can't even read your own lab handout.
> You aren't rotating the arms.
> You are laterally translating a mirror.
>

The translation is just for equalizing the length , imbecile.
You will continue to hang by your fingernails rather than trying to learn something new. It is a trademark of your old age,

> That is not MMX.
> The lab experiment is using a Michelson interferometer for *length measurement*

It is ALSO the pre-adjustment that equalizes the arms lengths, stubborn cretin.

[Dono,]

On Monday, October 31, 2016 at 3:31:43 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> Have you considered whether it is physically possible to perform such an action
> with any degree of precision?

Yes, we are doing it routinely. It is trivial. But you are too much of a stubborn idiot to want to make the effort to learn something new, you would rather cling to the old.

[Tom Roberts]

But that's what Michelson did! -- he rotated the entire instrument, including
the light source and telescope and observer (who walked around as it rotated).

Neither library available to me has Skinner's book, so I have
requested it via interlibrary loan. That will take a while.

Tom Roberts

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 6:19:53 PM UTC-5, Dono, wrote:
> On Monday, October 31, 2016 at 3:31:43 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >
> > Have you considered whether it is physically possible to perform such an action
> > with any degree of precision?
>
> Yes, we are doing it routinely. It is trivial.

Bullshit.

Draw me
1) an illustration of how your setup looks initially, and

2) how it looks after "rotating" the arms.

Post your before/after drawings on your Goggle Drive account.

I'm really interested in how you manage to interchange the roles
of the longitudinal and transverse arms via a "rotation" of the
arms, keeping the eyepiece/screen/detector fixed.

[Prokaryotic Caspase Homolog]

The lab experiment as written does not require equal length arms.
You set up equal thickness fringes and a photodiode, translating
the movable mirror until you've counted about 10,000 fringes.
From that, you can either calculate the wavelength assuming that
your distance traveled measurement is accurate, or you can
calculate the distance traveled assuming that your figure for the
wavelength is accurate. The lab experiment presumes the first.

Question: How do you "pre-adjust" the arms for equal length using
the lab setup? One fringe of He-Ne laser light looks pretty much
like another.

[Dono,]

On Monday, October 31, 2016 at 5:55:21 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Monday, October 31, 2016 at 6:18:28 PM UTC-5, Dono, wrote:
> > On Monday, October 31, 2016 at 3:22:16 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> > > That is not MMX.
> > > The lab experiment is using a Michelson interferometer for *length measurement*
> >
> > It is ALSO the pre-adjustment that equalizes the arms lengths, stubborn cretin.
>
> The lab experiment as written does not require equal length arms.

You refuse to THINK:

1. One does not have to re-create the experiment as originally designed
2. As originally designed, the experiment has the flaw that it is not symmetric, the arms do not truly exchange roles. The only thing that exchanges is the direction of the "aether wind"/ Since there is no "aether wind". the experiment, as conceived. detects nothing
3. As a consequence of 1+2 we decided to MODIFY the experiment such that it is truly symmetric, the arms DO exchange roles. The one requirement is that the arms are adjusted to be equal prior to running. The adjustment is described in the document I posted.

[Dono,]

On Monday, October 31, 2016 at 5:43:42 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Monday, October 31, 2016 at 6:19:53 PM UTC-5, Dono, wrote:
> > On Monday, October 31, 2016 at 3:31:43 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > >
> > > Have you considered whether it is physically possible to perform such an action
> > > with any degree of precision?
> >
> > Yes, we are doing it routinely. It is trivial.
>
> Bullshit.
>
> Draw me

No, no "draw me". I don't have to follow the orders of an utter imbecile. Go order your servants.

>
> I'm really interested in how you manage to interchange the roles
> of the longitudinal and transverse arms via a "rotation" of the
> arms, keeping the eyepiece/screen/detector fixed.

Engage brain, it is trivial.

[Prokaryotic Caspase Homolog]

If it is so trivial, make a drawing and post it.

You need to show:
1) how your setup looks initially, and

2) how it looks after "rotating" the arms.

There is a rotation that will do it, but it sure ain't trivial,
and it sure ain't something that can be done accurately.

[Dono,]

On Monday, October 31, 2016 at 6:18:00 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Monday, October 31, 2016 at 8:10:34 PM UTC-5, Dono, wrote:
> > On Monday, October 31, 2016 at 5:43:42 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > On Monday, October 31, 2016 at 6:19:53 PM UTC-5, Dono, wrote:
> > > > On Monday, October 31, 2016 at 3:31:43 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > > >
> > > > > Have you considered whether it is physically possible to perform such an action
> > > > > with any degree of precision?
> > > >
> > > > Yes, we are doing it routinely. It is trivial.
> > >
> > > Bullshit.
> > >
> > > Draw me
> >
> > No, no "draw me". I don't have to follow the orders of an utter imbecile. Go order your servants.
> > >
> > > I'm really interested in how you manage to interchange the roles
> > > of the longitudinal and transverse arms via a "rotation" of the
> > > arms, keeping the eyepiece/screen/detector fixed.
> >
> > Engage brain, it is trivial.
>
> If it is so trivial, make a drawing and post it.
>

Engage brain: what do you have to do to in order for the arms to full exchange roles.

>
> There is a rotation that will do it, but it sure ain't trivial,
> and it sure ain't something that can be done accurately.

Because you are so dim witted

[Prokaryotic Caspase Homolog]

Why are you unable to keep on subject? We were discussing the Michelson
interferometer lab handout as posted here:
https://drive.google.com/open?id=0B8XIf0XcrpOcY0RraEVoUDZHSTA

We were NOT discussing experiments intended to detect aether wind.

The lab experiment demonstrates how to use a Michelson interferometer for
length measurement, and now you go off in a completely different tangent.

When you state that the lateral mirror translation "is ALSO the pre-adjustment
that equalizes the arms lengths", please describe the criteria that you use to
determine that the arm lengths are equal.

The problem is, one dark fringe from a He-Ne laser looks just like any other
dark fringe. How do you tell which is the central fringe representing equal path
length?

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 8:32:19 PM UTC-5, Dono, wrote:
> On Monday, October 31, 2016 at 6:18:00 PM UTC-7, Prokaryotic Caspase Homolog wrote:

> > If it is so trivial, make a drawing and post it.
>
> Engage brain: what do you have to do to in order for the arms to full exchange roles.

That's for *you* to tell *me*.

After all, it's just a trivial 90 degree rotation, right?

[Dono,]

On Monday, October 31, 2016 at 7:55:54 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> Why are you unable to keep on subject?

I keep on subject lying POS. The subject has always been the effect of using unequal arms in MMX. I proposed a simple exercise that explains what happens when one uses unequal arms. Instead of solving it, you have been trolling the whole day. Because you are incapable of doing a simple calculation even AFTER I gave you the results. Here it is, all done for you:

https://faculty.etsu.edu/gardnerr/5310/5310pdf/dg2-2.pdf

> We were discussing the Michelson
> interferometer lab handout as posted here:
> https://drive.google.com/open?id=0B8XIf0XcrpOcY0RraEVoUDZHSTA
>

This is the setup used by our lab to equalize the arms. You are not only an imbecile, you are an autistic one.

> When you state that the lateral mirror translation "is ALSO the pre-adjustment
> that equalizes the arms lengths", please describe the criteria that you use to
> determine that the arm lengths are equal.
>

I did. in one of the earlier posts. You are too much of an imbecile to have paid attention.

[Dono,]

Yep, with a few trivial changes. (hint: the half silvered mirror is replaced by something else).

[Dono,]

Here is the answer, I suggest you study it. Once you understand it, feel free to explain it to the idiot you've been dutifully supporting.
It is interesting to note that even Feynman, in his lectures, missed this fine point (section 15-3, book 1)

[Dono,]

Hi harder, it might jar your brain. Look, you are an idiot, and a stubborn one to boot. This doesn't bother me, you were born this way. Your arrogance and your being a jerk is what bothers me.

[Prokaryotic Caspase Homolog]

On Monday, October 31, 2016 at 10:31:08 PM UTC-5, Dono, wrote:
> On Monday, October 31, 2016 at 7:55:54 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >
> > Why are you unable to keep on subject?
>
> I keep on subject lying POS. The subject has always been the effect of using unequal arms in MMX. I proposed a simple exercise that explains what happens when one uses unequal arms. Instead of solving it, you have been trolling the whole day. Because you are incapable of doing a simple calculation even AFTER I gave you the results. Here it is, all done for you:
>
> https://faculty.etsu.edu/gardnerr/5310/5310pdf/dg2-2.pdf

Yes. And 2.2.2 to 2.2.4 provides a *partial* description of the theory behind
Kennedy-Thorndike. Length contraction alone cannot explain Kennedy-Thorndike's
null result. What this section doesn't explain is why the addition of time
dilation does.

*** Paul, Tom and I all know this stuff! Or haven't you been reading? ***

> > We were discussing the Michelson
> > interferometer lab handout as posted here:
> > https://drive.google.com/open?id=0B8XIf0XcrpOcY0RraEVoUDZHSTA
> >
>
> This is the setup used by our lab to equalize the arms. You are not only an imbecile, you are an autistic one.
>
> > When you state that the lateral mirror translation "is ALSO the pre-adjustment
> > that equalizes the arms lengths", please describe the criteria that you use to
> > determine that the arm lengths are equal.
>
> I did. in one of the earlier posts. You are too much of an imbecile to have paid attention.

Listen, Dono. I do know how to equalize the arms, using techniques not explained
in the lab handout that you posted.
https://drive.google.com/open?id=0B8XIf0XcrpOcY0RraEVoUDZHSTA

NOTHING in the experiments described in the above lab handout requires equality
of arm length, nor can equality of arm length be achieved using *only* the
materials and methods described in the lab handout.

What are the criteria that ***YOU*** use to achieve equal arm length?

[Prokaryotic Caspase Homolog]

Oh boy. You're going to completely revamp the optics and then tell me that
what you have is still MMX.

[Prokaryotic Caspase Homolog]

On Sunday, October 30, 2016 at 11:44:05 PM UTC-5, Dono, wrote:
> On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
> >
> > So you know that there was no fringe shifts when
> > the interferometer with very unequal arms was
> > rotated 90 degrees.
> >
>
> Posting the above shows that you do not understand KTX (in addition to your proving earlier that you do not understand MMX).
>
> KTX is not, contrary to your naive beliefs, "an MMX with unequal arms". The two experiments are VERY different in concept, there is no such thing as "rotating the interferometer 90 degrees" in KTX.
> KTX relies on a slow revolution, over a long period of time such that the wavelength has the chance to change when transformed from the frame of the lab into the "aether" frame.

Whatever that means!

> MMX relies on an instantaneous rotation such that the wavelength DOES NOT CHANGE when transformed between the same frames as above.
> I could explain a lot of things to you, you miss a lot of the basics. But that would require that you behave less as the other jerk and with more respect.

All right. I see that you have just provided a link to a "partial" explanation
of KTX. (Most definitely NOT your own work!!!)

So it is time to address your totally *gobbledegook* explanation of the difference
between MMX and KTX. Let us consider an unequal arm Michelson apparatus with arm
lengths L_1 and L_2. Assume the validity of the FitzGerald-Lorentz contraction.

Let L_1 be aligned in parallel with the flow of the hypothetical aether drift.
The time it takes for light to go back and forth along the Lorentz-contracted
arm is 2Ɣ(v)L_1/c. The time it takes to traverse the perpendicular arm is
2Ɣ(v)L_2/c, and the difference in time that it takes for light to traverse the
two paths is 2Ɣ(v)(L_1-L_2)/c.

Rotate the apparatus 90 degrees so that L_2 is parallel with the aether flow.
The time it takes for light to go back and forth along the Lorentz-contracted
arm is 2Ɣ(v)L_2/c. The time it takes to traverse the perpendicular arm is
2Ɣ(v)L_1/c, and the difference in time that it takes for light to traverse the
two paths is 2Ɣ(v)(L_1-L_2)/c, which is exactly the same as the difference in
time between the two beams with the apparatus in the first orientation.

Given the validity of Fitzgerald-Lorentz contraction, rotating an unequal arm
Michelson interferometer will yield a null result.

On the other hand, if we wait a few months for the Earth to travel at a different
velocity with respect to the aether, the magnitude of 2Ɣ(v)(L_1-L_2)/c *will*
change, and the fringes will show a differential displacement.

A null result would require a modification in the frequency of light (i.e. time
dilation) that will equalize the number of cycles difference between the arms
between measurements taken at different times of the year.

[Dono,]

On Monday, October 31, 2016 at 11:26:00 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Monday, October 31, 2016 at 10:31:08 PM UTC-5, Dono, wrote:
> > On Monday, October 31, 2016 at 7:55:54 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > >
> > > Why are you unable to keep on subject?
> >
> > I keep on subject lying POS. The subject has always been the effect of using unequal arms in MMX. I proposed a simple exercise that explains what happens when one uses unequal arms. Instead of solving it, you have been trolling the whole day. Because you are incapable of doing a simple calculation even AFTER I gave you the results. Here it is, all done for you:
> >
> > https://faculty.etsu.edu/gardnerr/5310/5310pdf/dg2-2.pdf
>
> Yes. And 2.2.2 to 2.2.4 provides a *partial* description of the theory behind
> Kennedy-Thorndike. Length contraction alone cannot explain Kennedy-Thorndike's
> null result. What this section doesn't explain is why the addition of time
> dilation does.
>

While the above is true, you are not addressing your own ineptitude in figuring out why the unequal arms result into an additional fringe shift even AFTER I gave you the reference and the answer.

> *** Paul, Tom and I all know this stuff!

Actually, you demonstrated that you DON'T. You are not only an idiot, you are also a pathological liar.

[Dono,]

Hint, cretin : MMX uses DIFFERENT setups. Not all of them are half silvered mirrors. The fact that you are ignorant about that is your problem.

[Dono,]

On Tuesday, November 1, 2016 at 6:40:59 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Sunday, October 30, 2016 at 11:44:05 PM UTC-5, Dono, wrote:
> > On Sunday, October 30, 2016 at 3:17:25 PM UTC-7, Paul B. Andersen wrote:
> > >
> > > So you know that there was no fringe shifts when
> > > the interferometer with very unequal arms was
> > > rotated 90 degrees.
> > >
> >
> > Posting the above shows that you do not understand KTX (in addition to your proving earlier that you do not understand MMX).
> >
> > KTX is not, contrary to your naive beliefs, "an MMX with unequal arms". The two experiments are VERY different in concept, there is no such thing as "rotating the interferometer 90 degrees" in KTX.
> > KTX relies on a slow revolution, over a long period of time such that the wavelength has the chance to change when transformed from the frame of the lab into the "aether" frame.
>
> Whatever that means!
>

It means that you are a puffed up imbecile.

> > MMX relies on an instantaneous rotation such that the wavelength DOES NOT CHANGE when transformed between the same frames as above.
> > I could explain a lot of things to you, you miss a lot of the basics. But that would require that you behave less as the other jerk and with more respect.
>
> All right. I see that you have just provided a link to a "partial" explanation
> of KTX. (Most definitely NOT your own work!!!)
>
> So it is time to address your totally *gobbledegook* explanation of the difference
> between MMX and KTX. Let us consider an unequal arm Michelson apparatus with arm
> lengths L_1 and L_2. Assume the validity of the FitzGerald-Lorentz contraction.
>
> Let L_1 be aligned in parallel with the flow of the hypothetical aether drift.
> The time it takes for light to go back and forth along the Lorentz-contracted
> arm is 2Ɣ(v)L_1/c. The time it takes to traverse the perpendicular arm is
> 2Ɣ(v)L_2/c, and the difference in time that it takes for light to traverse the
> two paths is 2Ɣ(v)(L_1-L_2)/c.
>
> Rotate the apparatus 90 degrees so that L_2 is parallel with the aether flow.
> The time it takes for light to go back and forth along the Lorentz-contracted
> arm is 2Ɣ(v)L_2/c. The time it takes to traverse the perpendicular arm is
> 2Ɣ(v)L_1/c, and the difference in time that it takes for light to traverse the
> two paths is 2Ɣ(v)(L_1-L_2)/c, which is exactly the same as the difference in
> time between the two beams with the apparatus in the first orientation.
>

Err, wrong. I already gave you the correct calculations. It is a sign of your utter imbecility to produce the incorrect ones above. What sort of imbecile comes up with the wrong answer AFTER he has been provided the correct one? Answer: you!.

[Dono,]

On Tuesday, November 1, 2016 at 6:40:59 AM UTC-7, Prokaryotic Caspase Homolog wrote:
>

> Let L_1 be aligned in parallel with the flow of the hypothetical aether drift.
> The time it takes for light to go back and forth along the Lorentz-contracted
> arm is 2Ɣ(v)L_1/c. The time it takes to traverse the perpendicular arm is
> 2Ɣ(v)L_2/c, and the difference in time that it takes for light to traverse the
> two paths is 2Ɣ(v)(L_1-L_2)/c.
>

Correct. You don't even need to do the calculations in the "aether frame". You could have done it the way I showed you earlier, in the lab frame:

2(L1-L2)/c

> Rotate the apparatus 90 degrees so that L_2 is parallel with the aether flow.
> The time it takes for light to go back and forth along the Lorentz-contracted
> arm is 2Ɣ(v)L_2/c. The time it takes to traverse the perpendicular arm is
> 2Ɣ(v)L_1/c, and the difference in time that it takes for light to traverse the
> two paths is 2Ɣ(v)(L_1-L_2)/c, which is exactly the same as the difference in
> time between the two beams with the apparatus in the first orientation.
>

Incorrect. When you exchange the arms roles by the 90 degree rotation, you need to be careful. The writeup that I sent you (and the exercise in the Skinner book) point out a key thing, the difference CHANGES sign:

2(L2-L1)/c

So, the total fringe displacement is DOUBLE the single one, not zero.

This is the basic calculation for MMX. Since you have repeatedly demonstrated that you are incapable of doing it right for the simpler case of MMX, there is no point in you trying to progress to the more complicated case of KTX.

[Tom Roberts]

On 10/31/16 10/31/16 - 4:21 PM, Dono, wrote:
> On Monday, October 31, 2016 at 1:14:19 PM UTC-7, tjrob137 wrote:

[... Skinner's example, which I'll not discuss until I get the book]

>> In fact,
>> for a Michelson interferometer with arms of different lengths (within the
>> coherence length of the source), SR predicts ZERO fringe shift as the
>> interferometer is rotated, independent of the lengths of the arms.
>
> Let's check this claim:

To do that, you must explain your notation and actually USE SR.


> In the frame of the lab, the time difference is
> [tex]t_0=\frac{2L_1}{c}-\frac{2L_2}{c}[/tex]

WHAT "time difference"????? I am not clairvoyant. You must describe the physical
situation and the meanings of symbols.

Your use of TeX is obfuscatory. Please use the usual ASCII
notation everyone else uses around here:
t_0 = 2 L_1/c - 2 L_2/c
We can all read this MUCH more easily than what you wrote.

> [tex]L_1=L_2+\epsilon[/tex]
> So
> [tex]t_0=\frac{2 \epsilon}{c}[/tex]
> When you rotate the interferometer 90 degrees, the difference doubles and the
> fringes travel about the zero point by a total excursion of [tex]4
> \epsilon[/tex]

More undefined nonsense. When you don't explain your physical situation, your
terms, and your notation, all you can write is NONSENSE.


Using SR the analysis of the MMX is simple and straightforward:

The interferometer is at rest in an inertial frame [#]. Light propagates
isotropically in this frame, and with equal-length arms the optical axis has
equal-length paths, and thus has reinforcing interference and a white fringe
(the light source is white). A fringe pattern extends left-and-right from the
central fringe, with "rainbows" at the edges that uniquely identify the central
fringe.

[#] Not strictly true, but as I have explained several times
before, the non-inertial motions of the lab have negligible
effects on this measurement. So I neglect them.

Note I did not need to specify the orientation of the instrument -- this is a
GREAT BIG HINT that the result is independent of orientation: when the
instrument is rotated [@], NOTHING in the previous paragraph changes, and the
fringe pattern visible in the eyepiece remains in the same place -- NO FRINGE SHIFT.

[@] This is the MMX, and the entire instrument is rotated rigidly,
including both arms, the beam splitter, the light source, the
telescope, and the observer (who walks around the rotating
instrument). Dono seems to have some other notions about what
rotates.

Note there is no "aether frame", because THIS IS SR. In SR I can choose any
inertial frame for the analysis, and the inertial frame of the lab and
instrument makes everything simple. As light propagates isotropically in this
inertial frame, the fringe position is completely independent of orientation. NO
FRINGE SHIFT.

Now consider changing the length of one arm [%]. The fringes move as its length
is changed, but once its length is fixed they stop moving. As before, rotating
the instrument changes NOTHING and the fringe pattern visible in the eyepiece
remains in the same place -- NO FRINGE SHIFT.

[%] Keeping the path-length difference within the
coherence length of the source. For a white source this
is not possible, so we must switch to a different source
with longer coherence length. But nothing else changes.

Yes, as one changes the length of one arm, the fringes move. This is the basis
of using a Michelson interferometer to measure distance with an accuracy of a
fraction of a wavelength. But this is NOT the MMX (in particular, no rotation is
involved -- one arm is kept aligned with the distance to be measured).

Tom Roberts

[Prokaryotic Caspase Homolog]

Slight clarification in wording from deleted post.
Correction of factor-of-two error.

On Tuesday, November 1, 2016 at 10:50:32 AM UTC-5, Dono, wrote:

> Incorrect. When you exchange the arms roles by the 90 degree rotation, you need to be careful. The writeup that I sent you (and the exercise in the Skinner book) point out a key thing, the difference CHANGES sign:
>
> 2(L2-L1)/c
>
> So, the total fringe displacement is DOUBLE the single one, not zero.
>
> This is the basic calculation for MMX. Since you have repeatedly demonstrated that you are incapable of doing it right for the simpler case of MMX, there is no point in you trying to progress to the more complicated case of KTX.

Are you sure about the sign change, now?

Let us mount the unequal arm MMX on the back of a very, very slow snail. This
snail creeps forth at a rate of 1 mm/hr into the static aether. L_1 is initially
parallel to the 1 mm/hr aether "breeze", L_2 is perpendicular.

Let's give numbers to this: Let L_1 equal 300,000 km, and L_2 equal 1 mm. The
light from L_1 arrives 2 seconds after the light from L_2.

Now let's rotate the apparatus by 90 degrees. You claim that that because of
this sign change, light from L_1 will arrive 2 seconds BEFORE the light from
L_2, and that there will consequently be a massive fringe shift upon rotation?

[Dono,]

On Tuesday, November 1, 2016 at 9:09:45 AM UTC-7, tjrob137 wrote:
> On 10/31/16 10/31/16 - 4:21 PM, Dono, wrote:
> > On Monday, October 31, 2016 at 1:14:19 PM UTC-7, tjrob137 wrote:
>
> [... Skinner's example, which I'll not discuss until I get the book]
>
> >> In fact,
> >> for a Michelson interferometer with arms of different lengths (within the
> >> coherence length of the source), SR predicts ZERO fringe shift as the
> >> interferometer is rotated, independent of the lengths of the arms.
> >
> > Let's check this claim:
>
> To do that, you must explain your notation and actually USE SR.
>
>
> > In the frame of the lab, the time difference is
> > [tex]t_0=\frac{2L_1}{c}-\frac{2L_2}{c}[/tex]
>
> WHAT "time difference"????? I am not clairvoyant. You must describe the physical
> situation and the meanings of symbols.
>
> Your use of TeX is obfuscatory.

You come up with the same objection as the crackpot Stephen Baune (:rotchm:). You need to learn Tex. Anyways, I linked in a detailed explanation, before you post more of your incorrect formulas, I suggest that you read it.

Please use the usual ASCII
> notation everyone else uses around here:
> t_0 = 2 L_1/c - 2 L_2/c
> We can all read this MUCH more easily than what you wrote.
>
> > [tex]L_1=L_2+\epsilon[/tex]
> > So
> > [tex]t_0=\frac{2 \epsilon}{c}[/tex]
> > When you rotate the interferometer 90 degrees, the difference doubles and the
> > fringes travel about the zero point by a total excursion of [tex]4
> > \epsilon[/tex]
>
> More undefined nonsense. When you don't explain your physical situation, your
> terms, and your notation, all you can write is NONSENSE.
>

Read the explanation I linked in, utter jerk.

> Dono seems to have some other notions about what
>rotates.

I explained that long ago, utter cretin. I also explained the rationale of decoupling the detector+light source from the interferometer arms.

> Yes, as one changes the length of one arm, the fringes move. This is the basis
> of using a Michelson interferometer to measure distance with an accuracy of a
> fraction of a wavelength. But this is NOT the MMX (in particular, no rotation is
> involved -- one arm is kept aligned with the distance to be measured).
>

This is what happens when you butt in without reading. This is what you keep doing and this is what makes you look like an puffed up idiot.

[Dono,]

On Tuesday, November 1, 2016 at 9:29:35 AM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> Are you sure about the sign change, now?
>

Yes, cretin. Have you read the solution I linked for you? Obviously not.

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 11:09:45 AM UTC-5, tjrob137 wrote:

> The interferometer is at rest in an inertial frame [#]. Light propagates
> isotropically in this frame, and with equal-length arms the optical axis has
> equal-length paths, and thus has reinforcing interference and a white fringe
> (the light source is white).

Also, it appears that you are using a cube beam splitter with a reflective
surface that introduces an equal phase change to the light regardless of whether
it impinges on the diagonal surface in the forwards or reverse direction.
(Not all cube beam splitters will do so.) Otherwise the central fringe could be
dark.

[Prokaryotic Caspase Homolog]

Think very carefully about the parable of the MMX snail.

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 11:36:06 AM UTC-5, Dono, wrote:

> I explained that long ago, utter cretin. I also explained the rationale of decoupling the detector+light source from the interferometer arms.

No MMX-like experiment has EVER adopted such a decoupling strategy, and for
good reason.

You don't even know how to draw me an illustration of how you propose to
accomplish the "rotation" which is to exchange the parallel and transverse arms.

[Dono,]

You can't even read and comprehend a freshman relativity exercise, that how dumb and stubborn you are.

The time shift at initial orientation is:

[tex]\Delta t=\frac{2}{c}(L_{parallel}-L_{perpendicular})[/tex]
[tex]L_{parallel=L_1[/tex]
[tex]L_{perpendicular}=L_2[/tex]
So:
[tex]\Delta t=\frac{2}{c}(L_1-L_2)[/tex]

After 90 degree rotation of the arms:
[tex]L'_{parallel=L_2[/tex]
[tex]L'_{perpendicular}=L_1[/tex]

So:
[tex]\Delta t'=\frac{2}{c}(L'_{parallel}-L'_{perpendicular})[/tex]

meaning that:
[tex]\Delta t'=\frac{2}{c}(L_2-L_1)[/tex]

If there is a difference in arms length, there is going to be an "excursion" in fringe shift. BTW, this is the principle used in order to equalize the arms length. I told you this twice before but your skull is too thick and your ego is too big.

[Dono,]

On Tuesday, November 1, 2016 at 9:54:23 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Tuesday, November 1, 2016 at 11:36:06 AM UTC-5, Dono, wrote:
>
> > I explained that long ago, utter cretin. I also explained the rationale of decoupling the detector+light source from the interferometer arms.
>
> No MMX-like experiment has EVER adopted such a decoupling strategy, and for
> good reason.
>

But CRETIN

I explained to you REPEATEDLY that THIS is what I am talking about and THIS is what we are running in the lab.
How cretin are you? Really.

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 11:56:55 AM UTC-5, Dono, wrote:
> On Tuesday, November 1, 2016 at 9:41:24 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > On Tuesday, November 1, 2016 at 11:38:32 AM UTC-5, Dono, wrote:
> > > On Tuesday, November 1, 2016 at 9:29:35 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > >
> > > > Are you sure about the sign change, now?
> > > >
> > >
> > > Yes, cretin. Have you read the solution I linked for you? Obviously not.
> >
> > Think very carefully about the parable of the MMX snail.
>
>
> You can't even read and comprehend a freshman relativity exercise, that how dumb and stubborn you are.
>
> The time shift at initial orientation is:
>
> [tex]\Delta t=\frac{2}{c}(L_{parallel}-L_{perpendicular})[/tex]
> [tex]L_{parallel=L_1[/tex]
> [tex]L_{perpendicular}=L_2[/tex]
> So:
> [tex]\Delta t=\frac{2}{c}(L_1-L_2)[/tex]
>
> After 90 degree rotation of the arms:
> [tex]L'_{parallel=L_2[/tex]
> [tex]L'_{perpendicular}=L_1[/tex]
>
> So:
> [tex]\Delta t'=\frac{2}{c}(L'_{parallel}-L'_{perpendicular})[/tex]
>
> meaning that:
> [tex]\Delta t'=\frac{2}{c}(L_2-L_1)[/tex]
>
> If there is a difference in arms length, there is going to be an "excursion" in fringe shift.

Note that v does not appear in your math.

According to your math, you would see a sign change in Δt upon 90 degrees
rotation even in the limit as v approaches 0.

Is this reasonable?

> BTW, this is the principle used in order to equalize the arms length. I told you this twice before but your skull is too thick and your ego is too big.

What do you ***see*** that will allow you to equalize the arms length.

With a He-Ne laser, you see a monotonous set of red and black fringes.
As you adjust the mirror distance, you see a monotonous set of fringes.
When you get exact equality of arm length, you see a monotonous set of fringes.
If you go too far, you see a monotonous set of red and black fringes.

What is it about these fringes that enables you to find the point of equal
path length?

[Dono,]

On Tuesday, November 1, 2016 at 10:38:17 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Tuesday, November 1, 2016 at 11:56:55 AM UTC-5, Dono, wrote:
> > On Tuesday, November 1, 2016 at 9:41:24 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > On Tuesday, November 1, 2016 at 11:38:32 AM UTC-5, Dono, wrote:
> > > > On Tuesday, November 1, 2016 at 9:29:35 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > > >
> > > > > Are you sure about the sign change, now?
> > > > >
> > > >
> > > > Yes, cretin. Have you read the solution I linked for you? Obviously not.
> > >
> > > Think very carefully about the parable of the MMX snail.
> >
> >
> > You can't even read and comprehend a freshman relativity exercise, that how dumb and stubborn you are.
> >
> > The time shift at initial orientation is:
> >
> > [tex]\Delta t=\frac{2}{c}(L_{parallel}-L_{perpendicular})[/tex]
> > [tex]L_{parallel=L_1[/tex]
> > [tex]L_{perpendicular}=L_2[/tex]
> > So:
> > [tex]\Delta t=\frac{2}{c}(L_1-L_2)[/tex]
> >
> > After 90 degree rotation of the arms:
> > [tex]L'_{parallel=L_2[/tex]
> > [tex]L'_{perpendicular}=L_1[/tex]
> >
> > So:
> > [tex]\Delta t'=\frac{2}{c}(L'_{parallel}-L'_{perpendicular})[/tex]
> >
> > meaning that:
> > [tex]\Delta t'=\frac{2}{c}(L_2-L_1)[/tex]
> >
> > If there is a difference in arms length, there is going to be an "excursion" in fringe shift.
>
> Note that v does not appear in your math.
>

The calculations are in the frame of the lab. Imbecile.Incurable.

[Prokaryotic Caspase Homolog]

So without a velocity vector, how do you tell which arm is parallel to the wind
and which arm is transverse to the wind? How do you know what magnitude of
fringe shift to apply?

[Dono,]

On Tuesday, November 1, 2016 at 12:23:32 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Tuesday, November 1, 2016 at 1:12:59 PM UTC-5, Dono, wrote:
> > On Tuesday, November 1, 2016 at 10:38:17 AM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> > > Note that v does not appear in your math.
> >
> > The calculations are in the frame of the lab. Imbecile.Incurable.
>
> So without a velocity vector, how do you tell which arm is parallel to the wind

There is NO wind, you are stuck in your paradigms and cannot get out. "Parallel" means parallel with the direction of observation.

[Prokaryotic Caspase Homolog]

We are working within the context of a ***TEST THEORY***. The articles that
you downloaded from East Tennessee State University analyzed MMX and KTX within
the context of (1) the classical aether, and (2) a classical aether modified
by the introduction of FitzGerald-Lorentz contraction.

In the parable of the MMX snail, a pulse emitted from the source traveling
along L_1 will arrive at the detector two seconds after the pulse that travels
along L_2, and the sign of 2Ɣ(v)(L_1-L_2)/c will remain *unchanged* no matter
*how* the MMX apparatus is aligned with the direction of the snail's crawl.

Your claim is that the sign of 2Ɣ(v)(L_1-L_2)/c should reverse when the MMX
apparatus is rotated 90 degrees. That can't work unless you've got a mighty
fast snail. :-)

[Dono,]

On Tuesday, November 1, 2016 at 12:51:00 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Tuesday, November 1, 2016 at 2:32:54 PM UTC-5, Dono, wrote:
> > On Tuesday, November 1, 2016 at 12:23:32 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > On Tuesday, November 1, 2016 at 1:12:59 PM UTC-5, Dono, wrote:
> > > > On Tuesday, November 1, 2016 at 10:38:17 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> > >
> > > > > Note that v does not appear in your math.
> > > >
> > > > The calculations are in the frame of the lab. Imbecile.Incurable.
> > >
> > > So without a velocity vector, how do you tell which arm is parallel to the wind
> >
> > There is NO wind, you are stuck in your paradigms and cannot get out. "Parallel" means parallel with the direction of observation.
>
> We are working within the context of a ***TEST THEORY***.

First of all, you are not working, you are trolling and posting imbecilities (look at the title of your thread).
Second of all, a test theory can be worked through ANY experimental setup one chooses. Your problem is that you are unable to understand the small modification that we made to the original MMX setup.
Third, you cannot do the math for the basic SR, there is no way that you will EVER be capable of doing the math for a test theory.
You need to stop puffing yourself, take a break from trolling and stalking and use your time to learn.

> In the parable of the MMX snail, a pulse emitted from the source traveling
> along L_1 will arrive at the detector two seconds after the pulse that travels
> along L_2, and the sign of 2Ɣ(v)(L_1-L_2)/c will remain *unchanged* no matter
> *how* the MMX apparatus is aligned with the direction of the snail's crawl.
>

No, imbecile. If the arms rotate independently of the source+detector, what you are saying is false.

> Your claim is that the sign of 2Ɣ(v)(L_1-L_2)/c should reverse when the MMX
> apparatus is rotated 90 degrees. That can't work unless you've got a mighty
> fast snail. :-)

Once an imbecile, always an imbecile.

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 4:40:16 PM UTC-5, Dono, wrote:
> On Tuesday, November 1, 2016 at 12:51:00 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > On Tuesday, November 1, 2016 at 2:32:54 PM UTC-5, Dono, wrote:

> > > There is NO wind, you are stuck in your paradigms and cannot get out. "Parallel" means parallel with the direction of observation.
> >
> > We are working within the context of a ***TEST THEORY***.
>
> First of all, you are not working, you are trolling and posting imbecilities (look at the title of your thread).
> Second of all, a test theory can be worked through ANY experimental setup one chooses. Your problem is that you are unable to understand the small modification that we made to the original MMX setup.
> Third, you cannot do the math for the basic SR, there is no way that you will EVER be capable of doing the math for a test theory.
> You need to stop puffing yourself, take a break from trolling and stalking and use your time to learn.

From whom? YOU????

> > In the parable of the MMX snail, a pulse emitted from the source traveling
> > along L_1 will arrive at the detector two seconds after the pulse that travels
> > along L_2, and the sign of 2Ɣ(v)(L_1-L_2)/c will remain *unchanged* no matter
> > *how* the MMX apparatus is aligned with the direction of the snail's crawl.
>
> No, imbecile. If the arms rotate independently of the source+detector, what you are saying is false.

No MMX apparatus has ever been built that acts the way that you want it to work,
and you have ***NO IDEA*** how to implement your concept. I have repeatedly
challenged you to provide a simple drawing of your experimental design, and your
only response has been to run away shouting insults.

You're obviously clueless. You want people to believe that you have invented
something great and wonderful, but you have nothing to show. That sort of
behavior doesn't work anywhere.

> > Your claim is that the sign of 2Ɣ(v)(L_1-L_2)/c should reverse when the MMX
> > apparatus is rotated 90 degrees. That can't work unless you've got a mighty
> > fast snail. :-)
>
> Once an imbecile, always an imbecile.

You don't get the joke? There isn't *ANY* speed of snail that will work!

[Dono,]

On Tuesday, November 1, 2016 at 4:21:18 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> No MMX apparatus has ever been built that acts the way that you want it to work,

Therein lies the nobelty , old fart.

> and you have ***NO IDEA*** how to implement your concept.

We have had it working in the lab for over a year. Coming from an idiot like you, this is hilarious.

>I have repeatedly
> challenged you to provide a simple drawing of your experimental design, and your
> only response has been to run away shouting insults.
>

First things, first, you do not even understand the basics.

> You don't get the joke? There isn't *ANY* speed of snail that will work!

This is because you are an utter cretin who cannot follow even the simplest basics.

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 6:27:19 PM UTC-5, Dono, wrote:
> On Tuesday, November 1, 2016 at 4:21:18 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >
> > No MMX apparatus has ever been built that acts the way that you want it to work,
>
> Therein lies the nobelty , old fart.

Oh. You think it's worth a Nobel Prize?

> > and you have ***NO IDEA*** how to implement your concept.
>
> We have had it working in the lab for over a year. Coming from an idiot like you, this is hilarious.

Then why won't you share your design? If you've using it for over a year, then
you have obviously been collecting data. You've given everybody strong hints
that your results have not been null, that you have found a novel effect, still
consistent with SR, that has never before been reported, based on a fundamental
departure from legacy MMX implementations.

Am I in the ball park?

> >I have repeatedly
> > challenged you to provide a simple drawing of your experimental design, and your
> > only response has been to run away shouting insults.
> >
>
> First things, first, you do not even understand the basics.

There are others here that do. Educate them.

[Dono,]

On Tuesday, November 1, 2016 at 4:45:19 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Tuesday, November 1, 2016 at 6:27:19 PM UTC-5, Dono, wrote:
> > On Tuesday, November 1, 2016 at 4:21:18 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > >
> > > No MMX apparatus has ever been built that acts the way that you want it to work,
> >
> > Therein lies the nobelty , old fart.
>
> Oh. You think it's worth a Nobel Prize?
>

No, it is just a different way of looking at an old problem. You could benefit from learning how to look at old issues in a novel way.

> > > and you have ***NO IDEA*** how to implement your concept.
> >
> > We have had it working in the lab for over a year. Coming from an idiot like you, this is hilarious.
>
> Then why won't you share your design?

Because you aren't even at the point of learning the theoretical foundations, yet you have been arguing and pinballing from contention to contention, from misunderstanding to misunderstanding, from error to error. Because you have been behaving like a jerk.

> If you've using it for over a year, then
> you have obviously been collecting data. You've given everybody strong hints
> that your results have not been null, that you have found a novel effect,

No , imbecile, what I have been telling you is that IF you do not start with equal arms you get a non-null , erroneous result. If you calibrate the arms to be equal prior to running, you get a null result, as expected. You keep talking without making even the smallest effort to understand.

> still
> consistent with SR, that has never before been reported, based on a fundamental
> departure from legacy MMX implementations.
>
> Am I in the ball park?
>

Yes. It was never reported because no one decoupled the arms from the source+detector. It is not a big deal but it is a fun thing to try. It is simply an exercise of thinking a little out of the box. Michelson setup is not perfectly symmetric, we simply played with what happens if we make it fully symmetric. Not a revolutionary discovery, just a fun thing to experiment with.

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 7:02:38 PM UTC-5, Dono, wrote:
> On Tuesday, November 1, 2016 at 4:45:19 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > On Tuesday, November 1, 2016 at 6:27:19 PM UTC-5, Dono, wrote:
> > > On Tuesday, November 1, 2016 at 4:21:18 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > > >
> > > > No MMX apparatus has ever been built that acts the way that you want it to work,
> > >
> > > Therein lies the nobelty , old fart.
> >
> > Oh. You think it's worth a Nobel Prize?
> >
>
> No, it is just a different way of looking at an old problem. You could benefit from learning how to look at old issues in a novel way.
>
>
> > > > and you have ***NO IDEA*** how to implement your concept.
> > >
> > > We have had it working in the lab for over a year. Coming from an idiot like you, this is hilarious.
> >
> > Then why won't you share your design?
>
> Because you aren't even at the point of learning the theoretical foundations, yet you have been arguing and pinballing from contention to contention, from misunderstanding to misunderstanding, from error to error. Because you have been behaving like a jerk.

Uh uh. We all know what your *real* reason is.

> > If you've using it for over a year, then
> > you have obviously been collecting data. You've given everybody strong hints
> > that your results have not been null, that you have found a novel effect,
>
> No , imbecile, what I have been telling you is that IF you do not start with equal arms you get a non-null , erroneous result. If you calibrate the arms to be equal prior to running, you get a null result, as expected. You keep talking without making even the smallest effort to understand.

Then KTX should have given a non-null result, since the arms were deliberately
made non-equal. It was rotated on a daily basis as well as being taken around
the Sun at different speeds relative to the hypothetical aether.

> > still
> > consistent with SR, that has never before been reported, based on a fundamental
> > departure from legacy MMX implementations.
> >
> > Am I in the ball park?
> >
>
> Yes. It was never reported because no one decoupled the arms from the source+detector. It is not a big deal but it is a fun thing to try. It is simply an exercise of thinking a little out of the box. Michelson setup is not perfectly symmetric, we simply played with what happens if we make it fully symmetric. Not a revolutionary discovery, just a fun thing to experiment with.

So share your design.

I've dreamed up several means of uncoupling the source+detector from the arms,
absolutely NONE of which are worth much!!!
1) Pass light from the laser source through single mode optical fiber and an
optical isolator. Pass the laser signal through a 2x2 coupler, which splits
the beam into two outputs. Feed each output into a fiber terminated by a
retroreflector. Merge the two reflected return beams through the same 2x2
coupler that was used to split the beam, and split the merged beam so that
one return split is blocked by the optical isolator from returning to the
laser, while the other split goes to a photodetector. The fibers may be moved
around without moving the laser source or detector.

laser
\ Retroreflector
isolator /
\ /
>--< 2x2 coupler
/ \
/ \
Detector Retroreflector

Pretty horrible design, no?

I think Dono's design could be even worse!

[Dono,]

On Tuesday, November 1, 2016 at 5:50:20 PM UTC-7, Prokaryotic Caspase Homolog wrote:
>
> Then KTX should have given a non-null result, since the arms were deliberately
> made non-equal.

No, imbecile, we have already been over this.

> I've dreamed up several means of uncoupling the source+detector from the arms,
> absolutely NONE of which are worth much!!!

LOL. This is what happens when you are a narrow minded idiot.

>
> I think Dono's design could be even worse!

Actually it is quite straightforward and uncomplicated. :-)

[Prokaryotic Caspase Homolog]

On Tuesday, November 1, 2016 at 8:01:51 PM UTC-5, Dono, wrote:
> On Tuesday, November 1, 2016 at 5:50:20 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> >
> > Then KTX should have given a non-null result, since the arms were deliberately
> > made non-equal.
>
> No, imbecile, we have already been over this.

And you have repeatedly failed to come up with a method whereby Kennedy,
Illingworth, Joos etc. could have equalized path lengths in their Michelson
interferometers using monochromatic light.

Equalization of path lengths was not important to them.

> > I've dreamed up several means of uncoupling the source+detector from the arms,
> > absolutely NONE of which are worth much!!!
>
> LOL. This is what happens when you are a narrow minded idiot.
>
> > I think Dono's design could be even worse!
>
> Actually it is quite straightforward and uncomplicated. :-)

So how come you don't show me how it is done so as to embarrass the heck out
of me?

The evidence is that you can't come up with the goods, so stop trying.

[Nick]

Tom Roberts wrote:

> On 10/31/16 10/31/16 - 1:28 PM, Dono, wrote:
>> Actually, [the arms of the MMX] DO NEED to be equal.
>
> Yes and no -- yes for the instrumentation they used a century ago; no
> for the principle of the experiment and for modern instrumentation. The
> need for equal arms depends ONLY on the coherence length of the light
> source.

However there are two kind of coherence in Lasers. Length Oriented
coherence, then Time Oriented Coherence. You need to make sure we both are
talking about same kind of coherence.

> The white light source used by Michelson and Morley had a
> coherence length of a few microns, and their arms had to be equal in
> length to that accuracy in order to get visible fringes. Today there are
> lasers that can achieve coherence lengths of several kilometers; we have
> two in our lab and routinely use a Michelson interferometer with arms
> ~10 cm and ~60 cm -- DELIBERATELY not equal because that reduces the
> free spectral range of the instrument (we are NOT looking for aether
> drift!).

I strongly suspect that you mix the two types of coherence together in
here. By doing it you convert the Time Oriented into Length Oriented. Not
necessary directly convertible.

> In particular, the Michelson interferometer in our lab has 2 \epsilon /
> \lambda ~ 600,000. But as the earth rotates beneath it, it displays no
> fringe shift at all (to an accuracy of ~ 0.000015 fringe) -- billions of
> times smaller than your claim. Tom Roberts

Focusing on Fringe Shits is Length Oriented Coherence dominated only.
Hence, you mostly have a Half of the Answer.

[Dono,]

On Wednesday, November 2, 2016 at 1:05:31 AM UTC-7, Prokaryotic Caspase Homolog wrote:
> On Tuesday, November 1, 2016 at 8:01:51 PM UTC-5, Dono, wrote:
> > On Tuesday, November 1, 2016 at 5:50:20 PM UTC-7, Prokaryotic Caspase Homolog wrote:
> > >
> > > Then KTX should have given a non-null result, since the arms were deliberately
> > > made non-equal.
> >
> > No, imbecile, we have already been over this.
>
> And you have repeatedly failed to come up with a method whereby Kennedy,
> Illingworth, Joos etc. could have equalized path lengths in their Michelson
> interferometers using monochromatic light.
>

Actually is is quite trivial but you are too stupid to figure it out and too much of a jerk for me to show you.

> Equalization of path lengths was not important to them.
>

Because the equalization is necessary ONLY when one dcoples the motion of the arms from the source+detector.

> > Actually it is quite straightforward and uncomplicated. :-)
>
> So how come you don't show me how it is done so as to embarrass the heck out
> of me?
>

Because you are too much of a jerk and of an ignorant. I like to keep you that way.

[Nick]

Dono, wrote:

>> And you have repeatedly failed to come up with a method whereby
>> Kennedy, Illingworth, Joos etc. could have equalized path lengths in
>> their Michelson interferometers using monochromatic light.
>
> Actually is is quite trivial but you are too stupid to figure it out and
> too much of a jerk for me to show you.

Not to omit mentioning that "monochromatic light" simply does NOT exists.
Reveals they never had their ass into a laboratory. They take everything
theoretically, so to speech.