[Kryptid]

(ie zero fringeshift)

Why do you keep talking about fringe shift? The experiment in the link looks for changes in frequency using a frequency counter.

(1)(2)(3) cancel exactly. 

Please demonstrate this mathematically. Not that it matters even if they do cancel out. The device isn't looking for differences in travel time between the two arms anyway. It's looking for changes in the frequency of the laser beams using a frequency counter.

LIGO itself should have actually detected an aether if it was there. Kip Thorne explains in "Black Holes and Time Warps" the principle behind a gravitational wave interferometer like LIGO. A laser beam is sent through a beam splitter that sends half of the beam off in one direction and half off at a 90 degree angle to the first. They are then reflected by mirrors back towards the beam splitter where they form interference patterns with each other. In one direction, this interference is destructive and there is no longer a beam. In the other direction (at 90 degrees to the first), the interference is constructive and you now have a full-strength beam.

By placing a photodetector in the path of the self-destructive half of the beam, you can guarantee that any frequency shifts experienced by one laser beam can be detected because the two interfering beams would no longer perfectly cancel. Thus, some of the light gets to leak through. Travel time and fringe shift are irrelevant here. In the case of LIGO passing through the aether, one arm traveling parallel to the aether wind would experience a frequency shift whereas the arm perpendicular to the aether wind would not. When these two beams are recombined, their differences in frequency would prevent perfect cancellation and thus resulting in a detected signal. It isn't something that could have been ignored. It would have been an anomaly that would call for an explanation.

[mad aetherist (OP)]

(ie zero fringeshift)

Why do you keep talking about fringe shift? The experiment in the link looks for changes in frequency using a frequency counter.

I know, i was surprised that your wordage was directed at the VMMX (not the VRX).

(1)(2)(3) cancel exactly. 

Please demonstrate this mathematically. Not that it matters even if they do cancel out. The device isn't looking for differences in travel time between the two arms anyway. It's looking for changes in the frequency of the laser beams using a frequency counter.

Yes, i will do an Excel for (1)(2)(3) tomorrow.

LIGO itself should have actually detected an aether if it was there. Kip Thorne explains in "Black Holes and Time Warps" the principle behind a gravitational wave interferometer like LIGO. A laser beam is sent through a beam splitter that sends half of the beam off in one direction and half off at a 90 degree angle to the first. They are then reflected by mirrors back towards the beam splitter where they form interference patterns with each other. In one direction, this interference is destructive and there is no longer a beam. In the other direction (at 90 degrees to the first), the interference is constructive and you now have a full-strength beam.
By placing a photodetector in the path of the self-destructive half of the beam, you can guarantee that any frequency shifts experienced by one laser beam can be detected because the two interfering beams would no longer perfectly cancel. Thus, some of the light gets to leak through. Travel time and fringe shift are irrelevant here. In the case of LIGO passing through the aether, one arm traveling parallel to the aether wind would experience a frequency shift whereas the arm perpendicular to the aether wind would not. When these two beams are recombined, their differences in frequency would prevent perfect cancellation and thus resulting in a detected signal. It isn't something that could have been ignored. It would have been an anomaly that would call for an explanation.

Yes & no. I dont understand that there description of how LIGO works (or would work), but i will comment by assuming that its a standard MMX, & assuming that where u say frequency shift u mean fringeshift.
Firstly an MMX in vacuum gives zero fringeshift.
However i  think that when Demjanov & Cahill say that a VMMX gives a null rezult they actually understand but dont mention that actually there would be a weak 3rd order fringeshift (even in vacuum)(i seem to remember that i have calculated it in Excel).
LIGO would presumably detect this weak fringeshift which when using 4 km arms would i suppose not be so weak.  But all fringeshifts big or small in a fixed MMX or fixed VMMX would occur very gradually & very smoothly over 6 hrs which is 0.000023 hertz.  LIGO must automatically filter that out.  They probly dont even notice.  Actually they probly do notice, but aint saying.

[Kryptid]

assuming that where u say frequency shift u mean fringeshift.

I don't. When I say a frequency shift, I literally mean a change in frequency. No more, no less.

LIGO must automatically filter that out. They probly dont even notice.

It isn't looking for fringe shifts at all, so that's irrelevant. LIGO detects a signal when something causes the crests and troughs of the lasers to mismatch and it can detect such a mismatch down to one ten-thousandth the diameter of a proton. If the local speed of light changes for one beam more than it changes for the other beam, then the frequencies of the beams are no longer identical at the point of recombination. This means that you would get a sequence where some of the troughs perfectly cancel out the crests, then they partially cancel out the crests, then the crests perfectly add up to create a full-strength signal, then the crests and troughs partially cancel again and so on.

This anomalous signal would vary in a predictable manner where it becomes strongest at one time of the day (where the Earth's rotation adds to its orbital velocity) and weakest at the opposite time of the day (where the rotation subtracts from the orbital velocity). You'd also get a predictable change of signal throughout the year, where it is strongest when the Earth's velocity adds to the Sun's velocity through the Milky Way and weakest when it subtracts from it. This would rule out the possibility of the signal being caused by random noise. Since it would be detectable by both LIGO installations as well as VIRGO, they would also know that it wasn't caused by local environmental effects or glitches.

LIGO also has sensors which can detect such things as vibrations and temperature. This would allow for any signal to be back-checked against their readings to see if it was caused by such noise. Any aether signal would not be capable of being correlated to such readings and as such could not be explained by them.

Actually they probly do notice, but aint saying.

And now you need to demonstrate that this is what is actually happening.

[Bored chemist]

I havnt looked it up, but absorption & re-emission smells fishy.  How does the electron know the correct angle for emission (ie reflexion)?  My bounce theory makes more sense.

"I havnt looked it up"
You can stop advertising your ignorance directly  like that. We already know you haven't done any research
"but absorption & re-emission smells fishy."
Only to you, and that's because you haven't looked it up.
"How does the electron know the correct angle for emission (ie reflexion)?"
Conservation of momentum and energy.
You would know that it you weren't too stubborn/ lazy to look stuff up.

"My bounce theory makes more sense."
Only to you, and that's because you haven't looked it up.

Are you starting to see some sort of connection here?

[mad aetherist (OP)]

assuming that where u say frequency shift u mean fringeshift.

I don't. When I say a frequency shift, I literally mean a change in frequency. No more, no less.

LIGO must automatically filter that out. They probly dont even notice.

It isn't looking for fringe shifts at all, so that's irrelevant. LIGO detects a signal when something causes the crests and troughs of the lasers to mismatch and it can detect such a mismatch down to one ten-thousandth the diameter of a proton. If the local speed of light changes for one beam more than it changes for the other beam, then the frequencies of the beams are no longer identical at the point of recombination. This means that you would get a sequence where some of the troughs perfectly cancel out the crests, then they partially cancel out the crests, then the crests perfectly add up to create a full-strength signal, then the crests and troughs partially cancel again and so on.

This anomalous signal would vary in a predictable manner where it becomes strongest at one time of the day (where the Earth's rotation adds to its orbital velocity) and weakest at the opposite time of the day (where the rotation subtracts from the orbital velocity). You'd also get a predictable change of signal throughout the year, where it is strongest when the Earth's velocity adds to the Sun's velocity through the Milky Way and weakest when it subtracts from it. This would rule out the possibility of the signal being caused by random noise. Since it would be detectable by both LIGO installations as well as VIRGO, they would also know that it wasn't caused by local environmental effects or glitches.

LIGO also has sensors which can detect such things as vibrations and temperature. This would allow for any signal to be back-checked against their readings to see if it was caused by such noise. Any aether signal would not be capable of being correlated to such readings and as such could not be explained by them.

Actually they probly do notice, but aint saying.

And now you need to demonstrate that this is what is actually happening.

No i see the opposite. If LIGO could detect & isolate an aetherwind signal & if none then they would have already done a paper trumpeting a null result for aetherwind (praps at the 19th decimal).  And if they could & indeed did find an aetherwind signal then they would not have done any paper trumpeting a non-null result. But as i said the truth must be that they cant detect & isolate any such very gradual fringeshift. 
They might be able to in a  technical sense, but i think that it would mean a different set up or running of their whole contraption especially their auto corrections & filterings. And they would havtahav a good idea of what they were looking for.  An aetherwind chirp is predictable, if u know the primary vector of the background aetherwind.  For example some of your remarks re the effect of Earth's spin & orbit would not apply at some latitudes, because the primary vector is 20 deg off Earth's axis RA 4:30.

[mad aetherist (OP)]

I havnt looked it up, but absorption & re-emission smells fishy.  How does the electron know the correct angle for emission (ie reflexion)?  My bounce theory makes more sense.

"I havnt looked it up"
You can stop advertising your ignorance directly  like that. We already know you haven't done any research
"but absorption & re-emission smells fishy."
Only to you, and that's because you haven't looked it up.
"How does the electron know the correct angle for emission (ie reflexion)?"
Conservation of momentum and energy.
You would know that it you weren't too stubborn/ lazy to look stuff up.
"My bounce theory makes more sense."
Only to you, and that's because you haven't looked it up.
Are you starting to see some sort of connection here?

I dont understand.
(1) Every time an electron absorbs a photon there must be conservation of momentum & energy no matter which direction the photon came from.  Am i correct or wrong?
(2) And every time an electron emits a photon there must be conservation of momentum & energy no matter which direction the photon is sent.  Am i correct or wrong?

(3) But u are telling me that if an electron absorbs & then almost instantly emits a photon then there can only be conservation of  momentum & energy if the photon is emitted at the mirror-image angle of it's absorption.

Well, i'll be a monkey's uncle. Apparently (1) & (2) no longer apply at mirrors.  Who would have thort?  Live & learn! 
Electrons are amazing little buggers. 

[jeffreyH]

Dark quarks? You can't just join up disparate words and phrases and hope something scientific pops out. There are some decent books on particle physics if you really want to know what you are talking about. I doubt if you do. That means applying yourself and actually doing some study. Don't you have any gardening to do?

I think that i invented dark quarks, ie quarks made from neutrinos (which are dark photons). I am not in a hurry to do more reading re the standard atomic or sub-atomic model, not enough time, & far too many rubbish particles (about half of them), & u have to wade throo all of that krapp re virtual particles etc, but of course there is good stuff in there too.

You post long, rambling, multi-coloured text with no apparent substance to them. Why would I expect you to study. You could be doing something useful like weeding. Probably more suitable for you than reading.

[mad aetherist (OP)]

Dark quarks? You can't just join up disparate words and phrases and hope something scientific pops out. There are some decent books on particle physics if you really want to know what you are talking about. I doubt if you do. That means applying yourself and actually doing some study. Don't you have any gardening to do?

I think that i invented dark quarks, ie quarks made from neutrinos (which are dark photons). I am not in a hurry to do more reading re the standard atomic or sub-atomic model, not enough time, & far too many rubbish particles (about half of them), & u have to wade throo all of that krapp re virtual particles etc, but of course there is good stuff in there too.

You post long, rambling, multi-coloured text with no apparent substance to them. Why would I expect you to study. You could be doing something useful like weeding. Probably more suitable for you than reading.

My words are a model of sensible new theories. More facts & good ideas per square inch than any others i have read here. Mostly i refer to my heroes, mostly re aether & aetherwind.
But i am especially happy re my own contributions, eg that ........
(1) photaenos are em radiation & em radiation is photaenos.......
(2) photaenos explain slowing of light in air water glass...
(3) photaenos explain slowing of light near mass (ie Einsteinian slowing)(Einstein was right but for wrong reasons).....
(4) photaenos explain bending of light near mass....
(5) photaenos explain refraction at interfaces & diffraction at edges ...
(6) photaenos explain reflexion of light.......
(7) centrifuging of aether due to macro spinning & macro orbiting creates pseudo-macro-gravity & affects g & big G....
(8 ) coe due to micro spinning & micro orbiting affects pseudo-micro-gravity & gives the strong force...
(9) confined neutrinos (confined dark photons) give us dark particles & dark matter.....
(10) EZ water causes blurring of eyesight due to EZ zones at each dot (haze) in the eyeball....

(11) plus my espousing of the ideas of geniuses like Pollack Catt Cahill Crothers Demjanov Michelson & Co..
(12) plus my weeding of noxious standard science (gravity waves)(SR & GR)......
All in color.  My new theories (& others') are amazingly simple & shockingly lethal to the standard dogma & canon. I think that i an amateur have already achieved more than all u professionals combined. And just by spending 4 or 5 years of detailed reading of good science.  I too believed that Einstein was a genius, the bigbang was true, etc etc.  I forget what started me off.  And i am still learning.
 

[Kryptid]

No i see the opposite. If LIGO could detect & isolate an aetherwind signal & if none then they would have already done a paper trumpeting a null result for aetherwind (praps at the 19th decimal).

That isn't what LIGO was built for, so they have no motivation to write such a paper (especially when the results would be unsurprising to the majority of scientists). That would be like scientists who did an experiment investigating the molecular dynamics of combustion writing a paper showing how their results refute the existence of phlogiston.

And if they could & indeed did find an aetherwind signal then they would not have done any paper trumpeting a non-null result.

Given that this would be an important, unexpected result, why would they keep such a discovery under wraps? No mafia or conspiracy arguments, please.

But as i said the truth must be that they cant detect & isolate any such very gradual fringeshift. 

Quit it with the "fringe shift" talk already. That has nothing to do with the mechanism by which LIGO would find an aether wind signal. Being gradual also has nothing to do with it. If it takes six hours for the signal to build to full strength, then at the end of that six hours you have a full strength signal that is detectable by the photodetector. It would be recorded as data. It takes millions of years to get a mountain but that doesn't mean you can't see the mountain once it is finally there.

They might be able to in a  technical sense, but i think that it would mean a different set up or running of their whole contraption especially their auto corrections & filterings.

The signal from a frequency shift would allow a full strength signal to get through to the photodetector at times because there would be times when the peaks of both waves add together instead of canceling out. That would represent a signal many, many times stronger than a gravitational wave detection. So they can't just filter it out. Filtering out a time-varying signal like the aether wind would require them to know what the pattern of the signal was in advance. Since they wouldn't be expecting such a signal, they wouldn't know to filter it out.

And they would havtahav a good idea of what they were looking for.

In order to properly interpret the signal, perhaps. But that wouldn't keep them from recognizing that they were getting a predictable signal of some unknown origin that couldn't be attributed to the thermal or vibrational noise from their other sensors.

For example some of your remarks re the effect of Earth's spin & orbit would not apply at some latitudes, because the primary vector is 20 deg off Earth's axis RA 4:30.

I don't understand how that would keep any detections from occurring. The arms of LIGO would be changing their orientations and directions into the aether throughout the day and year. The speed and direction of the aether relative to one arm should therefore always be different relative to the other arm.

[mad aetherist (OP)]

But as i said the truth must be that they cant detect & isolate any such very gradual fringeshift. 

Quit it with the "fringe shift" talk already. That has nothing to do with the mechanism by which LIGO would find an aether wind signal. Being gradual also has nothing to do with it. If it takes six hours for the signal to build to full strength, then at the end of that six hours you have a full strength signal that is detectable by the photodetector. It would be recorded as data. It takes millions of years to get a mountain but that doesn't mean you can't see the mountain once it is finally there.

Do LIGO record intensity or flux of the combined light from the two arms? And versus time?  And if so then isnt that in effect a measurement of fringeshift?

[Kryptid]

Do LIGO record intensity or flux of the combined light from the two arms? And versus time?

Yes, because because that's what you would need to do in order to see the stress-strain GW pattern of a black hole or neutron star inspiral.

And if so then isnt that in effect a measurement of fringeshift?

Not as I understand it. Not in the same way as the MMX did, anyway. This doesn't involve looking at fringe patterns on a viewing surface: https://en.wikipedia.org/wiki/Fringe_shift

There would technically be two beams upon recombination for LIGO, as the original laser beams are sent through the beam splitter a second time, resulting in the two beams becoming four beams. However, two of the beams are headed back towards the laser while the other two are headed towards the detector. The device is set up such that the two beams towards the laser contructively interfere while those towards the detector destructively interfere. A change in the frequency of any of the beams results in imperfect construction/destruction, creating a signal.

[mad aetherist (OP)]

Do LIGO record intensity or flux of the combined light from the two arms? And versus time?

Yes, because because that's what you would need to do in order to see the stress-strain GW pattern of a black hole or neutron star inspiral.

And if so then isnt that in effect a measurement of fringeshift?

Not as I understand it. Not in the same way as the MMX did, anyway. This doesn't involve looking at fringe patterns on a viewing surface: https://en.wikipedia.org/wiki/Fringe_shift

There would technically be two beams upon recombination for LIGO, as the original laser beams are sent through the beam splitter a second time, resulting in the two beams becoming four beams. However, two of the beams are headed back towards the laser while the other two are headed towards the detector. The device is set up such that the two beams towards the laser contructively interfere while those towards the detector destructively interfere. A change in the frequency of any of the beams results in imperfect construction/destruction, creating a signal.

Its a while since i read LIGOs stuff, i might have a closer look.
But re auto recording of light, thats what both Demjanov did in 1968-72 & Cahill in 2002-15, to give their modern MMX fringeshifts. They didnt have any oldfashioned dark/bright fringes anywhere, nor did they measure the movement of any such fringes.
All of which reminds me that LIGO has a noise or false signal to GW signal ratio of 1000 to 1. 

Whereas Bored Chemist is insisting that the oldendays MMXs had giant error bars much bigger than the signals. He is getting good at that.
Firstly insist that they were null. 
Then insist that even if not null that their MMX error bars if properly done would have been off the page.
Then insist that (even if the raw results were consistent & concentrated) that their signals were due to temp.
Then when shown that the results accord with the sidereal day not solar day he responds with hey everyone look over there a black hole.
Then when shown that lots of MMXs done by different teams using different gizmos gave the same consistent concentrated results he tags the A-team who say that these are all trumped by the new modern vacuum pseudo MMXs that  give a null result. Here it is like the building inspector who insists that a wall has no studs because he forgot to put batteries in his newfangled laser stud finder, & he wont listen to the guys who located the studs by drilling.   

[Bored chemist]

Whereas Bored Chemist is insisting that the oldendays MMXs had giant error bars much bigger than the signals. He is getting good at that.

I am in good company.
M + M pointed it out.

[mad aetherist (OP)]

Whereas Bored Chemist is insisting that the oldendays MMXs had giant error bars much bigger than the signals. He is getting good at that.

I am in good company. M + M pointed it out.

I think that Michelson & Morley mentioned possible/probable error & it was a small fraction of the nett signal, but didnt draw error bars. And likewise i think Miller & Morley mentioned error, but likewise smallish, & didnt draw error bars.
But as i said smallish. But what did u read?

Shankland (1953?) mentioned temperature effects re Miller & Morley (which is rubbish)(they had tested temp very well).
And Roberts (2005?) mentioned a systematic error re Miller & Morley (which i think was merely the evergrowing  linear nonperiodic fringeshift), plus he mentioned that their error bars would be off the page (which is rubbish).

Reg Cahill (2003?) drew error bars for at least one of thems old MMXs, but i dont know whether these were his own estimates or whether he was using old numbers.

Anyhow for sure over the years Michelson & Co did lots of MMXs & MGXs & FresnelXs etc & often did a detailed error analysis, so they were on top of all of that, but of course drawing error bars was not the standard practice back then.

But the error bars on modern Xs make me laugh.  How do u draw error bars for your krapp theories & stupid Xs.

[mad aetherist (OP)]

(1)(2)(3) cancel exactly. 

Please demonstrate this mathematically. Not that it matters even if they do cancel out. The device isn't looking for differences in travel time between the two arms anyway. It's looking for changes in the frequency of the laser beams using a frequency counter.

In terms of your argument that a laser beam traveling upwind and then downwind will take the same time to cover the same distance as that from a stationary laser would, let's see about it. We'll assume that the laser cavity is 1 meter in length. So the time it takes for beam from the stationary laser to travel from one end of the cavity to the other (1 meter there and 1 meter back) is simply 2 meters divided by the speed of light, which is 6.6712819039630409915115342894984 x 10^-9 seconds.

In the case of a 500,000 m/s tailwind, the laser beam will take (1 meter/(299,792,458 m/s + 500,000 m/s)) = 1/300,292,458 = 3.330086964754872398427002785398 x 10^-9 seconds to travel from one end of the cavity to the other. On the return journey, it experiences a headwind instead of a tailwind, causing it to take (1 meter/(299,792,458 m/s - 500,000 m/s)) = 1/299,292,458 = 3.3412134962652483545041418985573 x 10^-9 seconds to come back. Add these two values together and you get a total trip time of 6.6713004610201207529311446839553 x 10^-9.

Subtract the two times and you get a difference of about 1.8557057 x 10^-14 seconds. The laser beam in a stationary device takes about 99.99972% as long to take its trip as the one in the moving device. So the tailwind-headwind combination does not compensate and make the travel times equal for each laser beam.

Yes i agree, i think u misquoted me or i wasnt clear.  I will reply to this part first & the rest later today.

I said that in a VMMX (a michelson morley experiment done in vacuum) the light takes the same time in both arms (& hencely u get a null result)(ie zero fringeshift) because of (1) LLC in the arm that is parallel to the aetherwind (this arm has a tailwind or headwind)(of say 500 kmps), & this shortening results in photons taking less time to go up & back.  The full explanation for the zero fringeshift etc involves (2) the photons in that arm take longer to go up & back because the headwind hurts more than the tailwind helps, & (3) the photons in the arm with a sidewind take longer to go up & back because they have to crab into the wind (like a plane on a windy day) & hencely have to travel throo more aether.  (1)(2)(3) cancel exactly. 

In the northern hemisphere the horizontal wind varies over a sidereal day from a tailwind of 140 kmps to a tailwind of 480 kmps measured at Obninsk (latitude 53 deg i think).  The full vector is say 500 kmps south to north say 20 deg off Earth's axis RA 4:30.  Hencely (4)(5) photons in both arms of the VMMX will have to crab downwards & thusly take longer in both arms, but (4)&(5) cancel exactly i suppose (anyhow they are allways ignored). 

In a non-vacuum MMX (ie in air usually) the photons are (6)(7) slowed to c/n, which affects both arms equally, however the slowing in (2) is magnified moreso than the slowing in (3).  And because of the air we now have to take into account (8 ) Fresnel Drag slowing the photons during tailwind, (9) Fresnel Drag fasting the photons during headwind, here (8 ) has a greater effect than (9) i think.  And (10) Fresnel Drag magnifies slowing in (3).  And (11)(12) Fresnel Drag also affects (4)&(5), equally i think.

So in an air MMX u can get a fringeshift.  I say can because at latitudes above 70 deg the aetherwind can be vertical in which case a horizontal MMX will measure zero fringeshift.  Except that a continuously rotating MMX will always give a systematic linear non-periodic evergrowing fringeshift that can be very large for some designs especially if the rotation is more rapid.  Demjanov explains the (geometric) reasons for this parasitic SLNPEGFS, & he designed his twin-media MMX such that his SLNPEGFS was nearly zero.  Of course anyone can reduce it to zero simply by stopping the rotation for each reading, but apparently this starting-stopping upsets most MMXs & Michelson Miller Morley & Co preferred to accept the need for the arithmetic corrections.
Demjanov (1968) & Cahill (2002) formulated the calibration equation needed for converting fringeshift to kmps. They used (1)(2)(3)(6)(7)(8 )(9) & ignored (4)(5)(10)(11)(12).

Re Fresnel Drag, Fresnel's equation appears ok for water at 6 m/s, but Demjanov's & Cahill's use of that equation for air at say 480,000 m/s might be suspect.

I used Excel to calculate the time in each of the 2 arms of an MMX with arms one light second long.  For a 500 kmps aetherwind (v).  For three cases (1) in vacuum, (2) in air but ignoring Fresnel Drag, & (3) in air including Fresnel Drag.
fresnel           c + v                   time to go   c - v               time to             average               reduced c           time   
drag d            c/n + v         one lightsec         c/n - v          return                  time taken          using           taken                   nett time taken
kmps          c/n + v + d             tailwind           c/n - v - d        headwind              sec            Pythagoras              sec                     sec
              300292.45800   0.99833357   299292.45800   1.00166921   1.0000013908   299792.0410   1.0000013908   0.0000000000000
             300211.12449   0.99833312   299211.12449   1.00166967   1.0000013923   299710.7074   1.0000013916   0.0000000007550
0.2713   300211.39576   0.99833221   299210.85323   1.00167058   1.0000013953   299710.7074   1.0000013916   0.0000000037756
                       tailwind        tailwind            headwind     headwind               headw+tailw      crosswind crosswind
For all three cases the arm length of one light second was contracted by gamma in the arm with a tailwind-headwind, but the arm with the crosswind was not contracted. 
As can be seen, in vacuum the delay in the arm with the tailwind/headwind exactly equaled the delay in the arm with the crosswind (ie the nett time difference was  0.00000000000000 sec).
In air there was a non-zero time difference even if Fresnel Drag is ignored, &
that difference is 5 times greater (5.0011 actually) when Fresnel Drag is not ignored.
The Fresnel Drag in the arm with a crosswind was ignored, it is too small to bother with.
I used c = 299,792.458 kmps & n for air = 1.000271373.
[edit 10march2019]  The Fresnel equation might be ok for water at 6 m/s but that doesnt mean that Fresnel is ok at 600,000 m/s.  E Falkner -- Classical Ether Theory Explains the Fizeau Experiment.
https://www.scribd.com/document/78584947/Emil-Falkner-Classical-Ether-Theory-Explains-the-Fizeau-Experiment
Galina Weinstein -- Albert Einstein and the Fizeau 1851 Water Tube Experiment.

[Kryptid]

I used Excel to calculate the time in each of the 2 arms of an MMX with arms one light second long.  For a 500 kmps aetherwind (v).  For three cases (1) in vacuum, (2) in air but ignoring Fresnel Drag, & (3) in air including Fresnel Drag.
fresnel           c + v                   time to go   c - v               time to             average               reduced c           time   
drag d            c/n + v         one lightsec         c/n - v          return                  time taken          using           taken                   nett time taken
kmps          c/n + v + d             tailwind           c/n - v - d        headwind              sec            Pythagoras              sec                     sec
              300292.45800   0.99833357   299292.45800   1.00166921   1.0000013908   299792.0410   1.0000013908   0.0000000000000
             300211.12449   0.99833312   299211.12449   1.00166967   1.0000013923   299710.7074   1.0000013916   0.0000000007550
0.2713   300211.39576   0.99833221   299210.85323   1.00167058   1.0000013953   299710.7074   1.0000013916   0.0000000037756
                       tailwind        tailwind            headwind     headwind               headw+tailw      crosswind crosswind
For all three cases the arm length of one light second was contracted by gamma in the arm with a tailwind-headwind, but the arm with the crosswind was not contracted. 
As can be seen, in vacuum the delay in the arm with the tailwind/headwind exactly equaled the delay in the arm with the crosswind (ie the nett time difference was  0.00000000000000 sec).
In air there was a non-zero time difference even if Fresnel Drag is ignored, &
that difference is 5 times greater (5.0011 actually) when Fresnel Drag is not ignored.
The Fresnel Drag in the arm with a crosswind was ignored, it is too small to bother with.
I used c = 299,792.458 kmps & n for air = 1.000271373.

I think I understand your data with the exception of "reduced c using Pythagoras". What does that represent?

[mad aetherist (OP)]

I used Excel to calculate the time in each of the 2 arms of an MMX with arms one light second long.  For a 500 kmps aetherwind (v).  For three cases (1) in vacuum, (2) in air but ignoring Fresnel Drag, & (3) in air including Fresnel Drag.
fresnel           c + v                   time to go   c - v               time to             average               reduced c           time   
drag d            c/n + v         one lightsec         c/n - v          return                  time taken          using           taken                   nett time taken
kmps          c/n + v + d             tailwind           c/n - v - d        headwind              sec            Pythagoras              sec                     sec
              300292.45800   0.99833357   299292.45800   1.00166921   1.0000013908   299792.0410   1.0000013908   0.0000000000000
             300211.12449   0.99833312   299211.12449   1.00166967   1.0000013923   299710.7074   1.0000013916   0.0000000007550
0.2713   300211.39576   0.99833221   299210.85323   1.00167058   1.0000013953   299710.7074   1.0000013916   0.0000000037756
                       tailwind        tailwind            headwind     headwind               headw+tailw      crosswind crosswind
For all three cases the arm length of one light second was contracted by gamma in the arm with a tailwind-headwind, but the arm with the crosswind was not contracted. 
As can be seen, in vacuum the delay in the arm with the tailwind/headwind exactly equaled the delay in the arm with the crosswind (ie the nett time difference was  0.00000000000000 sec).
In air there was a non-zero time difference even if Fresnel Drag is ignored, &
that difference is 5 times greater (5.0011 actually) when Fresnel Drag is not ignored.
The Fresnel Drag in the arm with a crosswind was ignored, it is too small to bother with.
I used c = 299,792.458 kmps & n for air = 1.000271373.

I think I understand your data with the exception of "reduced c using Pythagoras". What does that represent?

In the arm with crosswind the photon goes at c but it has to crab/sidle into the wind to get to the nearest bit of the end mirror in that arm, & then likewise coming back, & so we have a triangle with hypotenuse = c (or = c/n) & opposite is 500 kmps (or c/600) so the effective speed of the photon is the adjacent side which is (c*c - c*c/600*600)^0.5 (or) (c/n*c/n - c*c/600*600)^0.5.

I wasnt sure whether the Fresnel Drag is handled by having ++ & then - - (which is what i did)(& which is what i show in my calcs above), or whether + - & - + (which i did too)(& for which the delay was negative & the size of the delay was only -2.9989 times instead of the +5.0011 times shown in my calcs).

I did some more Excel calcs & i found that these ratios -3 & +5 change by less than 1% for a large range of kmps (eg 100 kmps) & a large range of n (eg n=1.001000).

[Bored chemist]

I think that Michelson & Morley mentioned possible/probable error & it was a small fraction of the nett signal, but didnt draw error bars. And likewise i think Miller & Morley mentioned error, but likewise smallish, & didnt draw error bars.
But as i said smallish. But what did u read?

Rather unoriginally, I read their paper on their experiment.
https://history.aip.org/exhibits/gap/PDF/michelson.pdf
There's a transcription here
https://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether


What they found was "Considering the motion of the Earth in its orbit the displacement should be about 2 D (v^2 /V^2)...
hence the displacement to be expected was about 0.4 fringes. The actual displacement was certainly less than one twentieth part of this and probably less than one fortieth"

So, even though they only considered the orbital speed of the Earth (about 30 km/s)  (and not the much larger  movement of the Sun rond the galaxy about 200 km/s) they expected to see  a fringe shift, but didn't.

They did not see any ether effect.
And, because their experiment would have observed the effect from the Sun's orbit the experiment was much more sensitive than they thought.

The reason they did not see the effect of the ether wind is simple.
It does not exist.

[Kryptid]

In the arm with crosswind the photon goes at c but it has to crab/sidle into the wind to get to the nearest bit of the end mirror in that arm, & then likewise coming back, & so we have a triangle with hypotenuse = c (or = c/n) & opposite is 500 kmps (or c/600) so the effective speed of the photon is the adjacent side which is (c*c - c*c/600*600)^0.5 (or) (c/n*c/n - c*c/600*600)^0.5.

I wasnt sure whether the Fresnel Drag is handled by having ++ & then - - (which is what i did)(& which is what i show in my calcs above), or whether + - & - + (which i did too)(& for which the delay was negative & the size of the delay was only -2.9989 times instead of the +5.0011 times shown in my calcs).

I did some more Excel calcs & i found that these ratios -3 & +5 change by less than 1% for a large range of kmps (eg 100 kmps) & a large range of n (eg n=1.001000).

Alright. I'm going to see if I can replicate these values using a spreadsheet myself. A cursory look at it does seem to indicate that you might actually be right about the travel times being equal in a vacuum.

[mad aetherist (OP)]

In a VMMX the light takes the same time in both arms because the arm with the tailwind-headwind is Lorentz length contracted due to the aetherwind V kmps in accordance with V/c*V/c in the Lorentz gamma.

Let's find out if the numbers pan out. Let's say we start with a laser that is 1 meter long emitting a laser beam with a wavelength of 0.00001 meters. One arm is heading parallel to the aether at a velocity of 500,000 m/s while the other is perpendicular to it.  The length contraction due to the Lorentz factor is L = L₀ √(1-(v²/c²)). Inserting 500,000 m/s, we get: = 0.9999986.  So the laser apparatus would contract to be 99.99986% of its original length. The wavelength and frequency emitted by the laser would presumably change by the same amount.

Yes i think that most of us would agree that starting with zero kmps aetherwind & then introducing a 500 kmps say headwind would give a LLC of the laser & hencely a shorter wavelength & higher frequency.  But i have a few thorts, & i think i might change my mind. 
Yes i see a problem.  The LLC is 0.999 998 609 (ie gamma), which has the effect of shortening the time taken for a photon to go up & back by that same factor (actually the headwind-tailwind effect adds to the time but the size of this addition is smaller than what Excel can handle)(Excel can only handle about 16 decimals i think)(thats why elsewhere i call it a 3rd order effect), but the headwind-tailwind effect re the time taken to go up & back is twice as powerfull as the LLC effect (ie it equals 2*gamma) & the resulting time actually increases to  1.000 001 391 sec  (ie 1/gamma) despite the LLC.  U can see this in the Excel numbers that i posted yesterday. 
Hencely the laser frequency drops in a 500 kmps headwind (likewise tailwind) by the inverse of the LLC (ie by 1/gamma).  And the wavelength must increase by the same factor.
 
But wait.  When the laser has a 500 kmps crosswind the time taken for a photon to go up & back is also slowed, due to the photon having to crab/sidle into the wind a little, in vacuum the factor being  1.000001391 (as shown yesterday).  In vacuum the two slowings in effect cancel, & the laser's frequency doesnt change if rotated in an aetherwind. 
However lasers have some gas inside, & hencely the frequency must change a little during rotation.  For a gas with a density of  1% of air i think the  n would be  1.00000271373, & the numbers would be as follows.

1.000 001 390 860 76   slowing due to headwind-tailwind effect if  1% air in laser.
1.000 001 390 823 02   slowing due to crosswind, if 1% air in laser.
0.000 000 000 037 74   nett slowing, giving a lower frequency, & longer wavelength.
0.000 000 000 003 774   nett slowing if 0.1% air in laser.

Hmmmm. One little problem. Photons do not suffer LLC.  During a crosswind the laser has its full length (ie no LLC), & the photons are crabbing (& the wavelengths are in effect shorter measured axially along the laser), hencely there are a max of photons (waves) standing in the laser (the photons/waves going both ways), an equal number in each direction. 
Now if we rotate the laser to get a headwind or tailwind the laser contracts, plus the photons/waves are no longer crabbing, hencely we must have ejected a lot of waves (because photons dont contract).  Where did they go?   What does this do to the laser's frequency?  The (sticky) waves wont go without a fight. 
And then later in the rotation it all happens in reverse. Two maximums & two minimums per rev. 
And during a headwind or tailwind there must be an equal number of photons/waves in each direction, because as i said earlier photons do not suffer LLC (but a laser can i guess emit a changing wavelength).
Its all very complicated.  Still thinking. I must read up on lasers.

But does this match the frequency shift for the laser caused by the changing relative speed of light in the aether? Since frequency is velocity divided by wavelength, a laser beam traveling into an aether headwind at 500,000 m/s would have a frequency of (299,792,458 - 500,000)/0.00001 = 299,292,458/0.00001 = 29,929,245,800,000 hertz. Without the wind, the frequency would simply be 299,792,458/0.00001 = 29,979,245,800,000 hertz.  Divide these two frequencies and you find that the laser beam in the headwind has 99.833217% the frequency of a stationary laser.

This number does not match the frequency shift caused by length contract and therefore could not be masked by it.

Yes, but i will comment later today. Still thinking.