Roberts' explanation of Miller's MMMX.

[crawler]

Ok here is what i reckon re the cause of Miller's incline.
Firstly the main problem is that the cause needs to be ever-growing. How? Good luck with that. How can an interferometer store fringes, ie accumulate them? Where are they stored?

Temperature can be ever-growing, in fact Miller calls his method of averaging "compensation", & on his worksheets he calls it a "Temp" thing (which it can be)(but usually it aint). Cahill comes straight out & calls it a correction for temperature (no it aint), & ignores it thereafter (he aint silly).

Anyhow, its no good coming up with causes that are obviously periodic (either in a halfturn or in a fullturn), it has to be non-periodic (which automatically means ever-growing & linear), or at least only weakly periodic (ie almost linear).

(15) The incline, ie the accumulation of fringes, is due to the fringes.
Its like this. One end mirror in one arm or the other has to be turned slightly off 90 deg so that the two beams cross at the detector, to form fringes. The best way to think of it is that the beams are made of long trains of waves, & that these waves are stationary, whilst the MMMX rotates. Or at least that the waves are stationary in the beam itself, but that the beams of course rotate with the MMMX (keep a constant angle in each arm). There is no need here to complicate things by trying to invoke the reality that an aetherwind blows throo the lab. My simple model is ok.

During rotation each end mirror will either move into the waves or will move away. When moving into the waves it can be said to be eating waves, & when moving away it is vomiting waves. It is of course possible to be neither moving into nor away (in this thort-X).

With perfect symmetry the eating & vomiting will be equal in both arms. However, remember, we have turned one mirror a little. So, now the eating-vomiting in each arm must be different, because the radius from the axis of rotation to the beam (the perpendicular distance to the beam) will not be the same in each arm.

If the eating-vomiting is different in each arm, then the fringes will move as the MMMX rotates, for ever. If rotation stops the fringes don't go backwards, they just stop moving, & they wait.

deHaan's optical fibre MMMX was stationary during all readings, yet he suffered a severe incline, so much so that he slowed the overall rotation to only one revolution per 40 minutes. He needn't have bothered. Incline is not affected by the rate of rotation (ie rpm), it is affected by the revs only.

The rate of rotation will of course yield lots of other effects, but these will all be periodical, & a slower rotation might help here, but it wont help u re incline.

Miller's MMMX had 4 mirrors on each end of the two arms (the two arms forming a cross), ie 16 mirrors, but only one mirror (No8A) was turned off 90 deg (No8B, on arm B, remained 90 deg).

In addition, Miller's MMMX had a horrid design, placing the splitter mirror approx 350 mm away from the axis. But i don't think that we are forced to consider Demjanov's (3) etc. I haven't given it a lot of thought, but i think that Demjanov's (3) etc are in effect a second way of analysing my (15), ie they are a faux(15) or a quasi(15) or a pseudo(15), & my (15) is the primary real cause.

I have corrected & improved the above wording coloured red with the following wording coloured blue......

In Miller's MMMX end mirrors No1 2 3 4 5 6 & 7 each reflect the beam twice, & mirror No8 reflects just the once. If the outgoing beam & the returning beam share the exact same point of reflexion on mirrors No1 2 3 4 5 6 & 7 then there will be no incline. But mirror No8 has been turned-in (or out), in the telescope arm, hencely in that arm the returning beam reflects at a different point on them seven mirrors, & we get incline, because the seven radii from the axis of rotation to the seven returning beams (the perpendicular horizontal distances to the beams) must be different to the radii to the seven outgoing beams, & there will be a non-zero nett eating or vomiting (of waves). In the other arm (the lamp arm) the nett eating & vomiting is always zero if the mirrors have been properly adjusted so the returning beam follows its outgoing path.

Due to the turn-in in mirror No8, & the consequent non-zero eating or vomiting (of waves), the fringes will move as the MMMX rotates, for ever. If rotation stops the fringes don't go backwards, they just stop moving, & they wait.

[crawler]

Today we abandon Miller's MMMX (two arms in the shape of a cross) & we go back to the traditional simple design, ie single 90 deg mirrors on the ends of two equal arms at 90 deg, with a single 45 deg splitter mirror at the junction, & a laser on one arm & a fringe-target on the other arm. I will call this an MX.

A simple MX has a potential problem, it has a built-in incline, even when the end mirrors are at 90 deg.

Look at a static MX. A say single line of photons (beam 2AB) emerges from the laser (at point A) & travels along arm No2 to the 45 deg splitter-mirror, hitting the silver of the mirror at point 2B, & then a half of the photons pass through the glass of the 45 deg mirror (beam 2BC), emerging at point 2C, & then travel along arm No2 (beam 2CD), & hit the 90 deg mirror No2 at point 2D, & reflect back to the 45 deg mirror (beam 2DE), hitting the glass at point 2E, & pass through the glass (beam 2EF), & hit the silver of the 45 deg mirror at point 2F, & reflect, & pass through the glass (beam 2FG), emerging at point 2G, & travel along arm No1 to the fringe-target (beam 2GH), hitting the target at point 2H.

The 90 deg end mirrors have their silvered surfaces facing the beam, as recommended by Demjanov, because Demjanov reckoned that the glass creates a number of problems.

Points 2B & 2F coincide (on the silver of the 45 deg mirror), as do points 2C & 2E (on the back of the 45 deg mirror). But, the MX has a problem, because beam 2CD has a different angle to beam 2EF. Beam 2CD intersects with beam 2EF at point 2C (ie point 2E). Hencely if we locate the axis of rotation of the MX at point 2C then that solves the problem for beam 2 in arm No2. The problem being that we want the 45 deg mirror & the 90 deg mirror to eat & vomit the same numbers of waves when the MX is rotating, so that the nett eating & vomiting is zero (so that the incline is zero). This requires that the two radii from the axis to the two beams (beam 2CD & 2EF)(hitting the two mirrors) need to have the same length. By having the axis at point 2C (on the back of the 45 deg mirror) not only are the two radii the same length, but both radii are zero mm (which might possibly be even better in some yet to be discovered way)(but not necessarily).

If we wanted we could locate the axis anywhere along the bisector of angle BCD (which is identical to angle DEF), or along the extension of that bisector on the far side of BCD. The axis would be the same distance from the two beams, thusly giving zero incline (term coined by Miller).

However the trouble is that having the axis at point 2C gives us an incline problem for arm No1.

Look at arm No1. A line of photons (beam 2AB) emerges from the laser (at point A) & travels along arm No2 to the 45 deg splitter-mirror, hitting the silver of the mirror at point 2B, & a half of the photons reflect along arm No1 (beam 1BC), & hit the 90 deg mirror No1 at point 1C, & reflect back to the 45 deg mirror (beam 1CD), hitting the silver at point 1D, & pass through the glass (beam 1DE), & emerge at point 1E, & travel along arm No1 to the fringe-target (beam 1EF), hitting the target at point 1F. (Just for interest, beam 1DEF shares the same path as beam 2FGH).

The axis for arm No1 needs to be an equal distance from beams 2AB & 1BC. Hencely the axis can be located at the junction of the two beams, ie at point 2B. But the axis cant be placed anywhere along the bisector of angle 2A-2B-1C, or the extension of that bisector. The bisector gives an equal radius to each beam, but the 90 deg mirror & the 45 deg mirror will both be vomiting waves, giving a double dose of incline. And if the axis is anywhere along the extension of the bisector then both mirrors will be eating waves, once again giving a double dose of incline. Luckily locating the axis anywhere along a line at 90 deg to the above bisector will give an equal radius whilst causing one mirror to eat waves whilst the other mirror is vomiting waves, & that line happens to be the silver surface of the 45 deg mirror.

If we draw a drawing we can see that the aforementioned extension of the bisector for arm No2 crosses the silver surface approx 21.626 mm from point 2B (measured along the silver surface). We can call this point O. If the axis of rotation for the MX is located at point O then the MX shouldn't suffer any unwanted incline signal during operation of the MX, because the rogue unwanted incline signal will be zero in each arm. The 21.626 mm is based on the 45 deg mirror being 10 mm thick, with a refraction index of 1.5, the 45 deg incidence giving a 28.1 deg beam in the glass.

An overall nett zero incline signal can also be obtained if the incline in each arm is equal, eg if arm No1 suffers say one unit of eating, & if arm No2 suffers one unit of eating, then they will cancel. Likewise if each has one unit of vomiting. There must be a number of lines or curves where the axis could be located to achieve such a cancelling, but i haven't given it much thought.

In all of the above both end mirrors are at 90 deg, ie we wont have a visible fringe, & we would need to use a light intensity meter to measure interference, ie to get our desired half period signal (which is what Cahill did, & Galaev, & Marinov, albeit using a different kind of apparatus).

[crawler]

Today i revisited a youtube video by Martin Grusenick (2009) re his horizontal MMX, & another by Frank Pearce (2010) re his horizontal MMX.
https://www.youtube.com/user/Gruma09/vi ... _polymer=1
https://www.youtube.com/watch?v=iJodqg5XPdk
https://www.youtube.com/watch?v=aNEryiOKkrc

Both say that their MMXs had zero fringe-shift, but i could see that Grusenick had a 0.3 fringe shift, & Pearce had a 0.6 fringe shift. However their fringe-shifts looked to be periodic in a full turn, whereas a proper MMX signal is periodic in a half turn. Hencely they must have had something strange happening, some kind of bending, due to bad design or workmanship or something.

I did see one other such youtube video last week, showing a supposedly zero fringe-shift (it was difficult to see to check if zero or not), but now i cant find that video. Likewise i saw another one a few years ago. I think that most universities etc have such MMXs for students, but no videos. If anyone knows of any such videos i would like to view them.

Michelson's MMX in 1881 had two 1200 mm arms & he predicted a fringe-shift of only 0.04 fringes. Modern-day MMXs i think have small arms & yet they consider the result null if the fringe-shift is less than i think only say 0.05 fringes (which is a bit unrealistic). Grusenick & Pearce had 0.3 & 0.6 & they called that null!

Anyhow what i realised today is that after one rotation the fringes didn't go back to where they started, the fringes finished on 0.3 left (Grusenick) & 0.25 right (Pearce). Years ago i thought that their MMXs must have been horribly built, but today i realised that Miller's incline (ie Demjanov's linear drift of zero)(ie my own linear ever-growing non-periodic fringe-shift) was responsible for their non-zero endings.

I would like to see some university MMX videos, to see if they too suffered any incline.