The SR Question of the Century

[Nereid]

I'm surprised that this thread lasted so long. You say that Martin's gone! At which point did he give up? Or has he argued his case enough?

I'm involved with some local politics now and have had less time to argue a case against SR. But my view hasn't changed. I'm 100% convinced that a simple one-way light speed test using 2 clocks will falsify SR.

To what extent is VLBI not 'a simple one-way light speed test using 2 clocks'?

 

[outandbeyond2004]

One source of experimental error lies in the synchronization bw the two clocks. No two clocks are ever in exact synch. Every clock rate may vary unpredictably. Won't wisp explain the experiment in detail so that we can understand all the possible experimental errors and judge how well it must be done in order to falisfy SR? How much would it cost, etc? What overall experimental error can we achieve?

Wisp may possibly have a lack of understanding of the limits of science and technology. No one experiment can invalidate any theory ever. It's always possible that the experimenter overlooked something or did something wrong, for one thing. The results of anyone experiment should never be relied on 100%.

 

[outandbeyond2004]

It occurs to me that it may be rather more technologically more feasible to test for anisotropy than to measure the one-way speed directly. Would this be acceptable to Wisp, MM, and Eyesaw? Why not?

 

[wisp]

To what extent is VLBI not 'a simple one-way light speed test using 2 clocks'?

This test falls into the category of GPS. I had a discussion "SR and one-way speed of light tests" on Tom's Metaresearch forum some months back. After much discussion I concluded that GPS cannot be used to verify a one-way speed of light test. See
http://www.kevin.harkess.btinternet.co.uk/reasons_einstein_wrong/reasons_einstein_wrong.html
and go to the end of the page.
So on this basis VLBI equally cannot be use as a one-way test.
The reason a one-way test using two clocks is important is because it is simple to do and will prove one and for all if the speed of light is really constant, and hence if SR is true or false.

 

[Nereid]

This test falls into the category of GPS. I had a discussion "SR and one-way speed of light tests" on Tom's Metaresearch forum some months back. After much discussion I concluded that GPS cannot be used to verify a one-way speed of light test. See
http://www.kevin.harkess.btinternet.co.uk/reasons_einstein_wrong/reasons_einstein_wrong.html
and go to the end of the page.
So on this basis VLBI equally cannot be use as a one-way test.
The reason a one-way test using two clocks is important is because it is simple to do and will prove one and for all if the speed of light is really constant, and hence if SR is true or false.

Have you considered doing this test yourself? A laser can be purchased for <US$10, but two atomic clocks would be a bit more expensive. Still less than US$100k I should think, and prices are falling all the time. Then you may have missed a good opportunity; with the collapse of so many telephone/telecom start-up companies - to compete with BT, AT&T, FT, DT, Verizon, ... - there were surely bargains to be had as the receivers auctioned off their large numbers of very accurate clocks. There are probably (UK) high schools or (US) colleges which might be interested in organising such an experiment, if done as part of larger project into SR; they may not have the $/£ to buy the equipment, but they would have several good students who may be willing to do the leg work.

 

[outandbeyond2004]

Wisp, some questions:
1. (Just curious. This question may or may not be quite relevant to this topic.) The GPS satellites have to move through the Earth's atmosphere. It is extremely rarified at the GPS satellite altitude, but I believe space-and-time varying atmospheric drag is a significant and somewhat unpredictable factor as far as trying to determine the speed of light to high precision. Correct? You never mentioned it as a reason to reject GPS data for one-way light measurement.

2. Have you performed a detailed analysis of your proposed one-way light speed experiment using unsynch'd clocks that takes into account anisotropic gravitational effects, such as the rotation of the experiment relative to the Milky Way center?

3. What about ring lasers? For those who don't know about them, they use 3 mirrors in an equilateral triangle arrangement, with each mirror angled so that any correctly aimed light signal between them travels on the triangle formed by the mirrors. A laser is mounted between two mirrors. It shoots mono & coherent light out of either end, equally. One set of light rays travels the triangle one way. Let's label the mirrors A, B, and C. So the set goes A, B, C and finally back to where it started from. Another set goes the other way, C, B, A, and finally back to its start. This can be a table-top device. I don't know today's state of art, so I cannot say how sensitive advanced ring lasers are to anistropy.

 

[wisp]

Have you considered doing this test yourself?

I think a professional body should do this experiment. Amateurs like Roland DeWitte have carried out a similar experiment with positive results. But professional bodies have ignored this without good reason.

 

[Nereid]

Wisp, some questions:
1. (Just curious. This question may or may not be quite relevant to this topic.) The GPS satellites have to move through the Earth's atmosphere. It is extremely rarified at the GPS satellite altitude, but I believe space-and-time varying atmospheric drag is a significant and somewhat unpredictable factor as far as trying to determine the speed of light to high precision. Correct? You never mentioned it as a reason to reject GPS data for one-way light measurement.

2. Have you performed a detailed analysis of your proposed one-way light speed experiment using unsynch'd clocks that takes into account anisotropic gravitational effects, such as the rotation of the experiment relative to the Milky Way center?

3. What about ring lasers? For those who don't know about them, they use 3 mirrors in an equilateral triangle arrangement, with each mirror angled so that any correctly aimed light signal between them travels on the triangle formed by the mirrors. A laser is mounted between two mirrors. It shoots mono & coherent light out of either end, equally. One set of light rays travels the triangle one way. Let's label the mirrors A, B, and C. So the set goes A, B, C and finally back to where it started from. Another set goes the other way, C, B, A, and finally back to its start. This can be a table-top device. I don't know today's state of art, so I cannot say how sensitive advanced ring lasers are to anistropy.

For all readers, FYI:
1) One of the things that GRACE will determine, to great precision, is the distortion in radio signals through the ionosphere (inc those from GPS satellites)
2) ring lasers are the heart of many modern gyroscopes, including those in commercial airliners. To underline outandbeyond2004's point, if there were any anisotropies, or other deviations of the kind that wisp and Eyesaw have described, it would seem remarkable that they'd cancel out perfectly in devices like ring lasers. But maybe that's exactly what wisp et al expect?

 

[outandbeyond2004]

A ringlaser project that is able to detect the anistropy of the sort theorized by Wisp? http://www.wettzell.ifag.de/LKREISEL/CII/precise.htm

This group says it hopes to achieve resolution of the Earth's rotation to a few parts in a billion! That's much more than we need, if the wisp ether speed of the ring laser on the equator is totally due to Earth's rotation. speed of the r.l/c is about 1 per million or maybe much more. I forgot what the speed of the solar system (divide by c) is wrt the CRB, but it must be much more, I think.

 

[Nereid]

outandbeyond2004: I forgot what the speed of the solar system (divide by c) is wrt the CRB ...

Well, the speed is approx 385 km/sec, so the ratio you're looking for is >1:1,000.

 

[wisp]

2) ring lasers are the heart of many modern gyroscopes, including those in commercial airliners. To underline outandbeyond2004's point, if there were any anisotropies, or other deviations of the kind that wisp and Eyesaw have described, it would seem remarkable that they'd cancel out perfectly in devices like ring lasers. But maybe that's exactly what wisp et al expect?

The principle on which the ring laser gyroscope works is due to an effect that SR cannot explain clearly.

The Sagnac Effect
An article that explains this clearly, but from relativity's viewpoint is given at http://www.mathpages.com/rr/s2-07/2-07.htm.
The first few paragraphs explain the Sagnac effect (you can ignore the section showing loop and area calculations). The argument in support of relativity's explanation is summed up on the basis that the device centres around one particular system of inertial coordinates (centre of circle), and all other inertial coordinate systems are related to it by Lorentz transformations.

But the flaw in this argument is simply this: What happens to the measuring clock when the radius of the circle becomes very large and the clock's velocity small - a limit process?
The Sagnac effect still applies and the clock's motion becomes more linear. In this limit process it is not unreasonable to treat the moving clock as an inertial reference frame in its own right (the Sagnac effect has been tested to great accuracy and so it perfectly reasonable to use a limit process to make the moving clock's frame inertial). Now according to relativity, since this is an inertial frame, light must travel at speed c in both directions. But the Sagnac effect requires that the speed of light must be c+v and c-v respectively, and not c! This limit process shows that relativity contradicts itself, as the real measurements are made in the moving clock frame and not at the centre of the circle. An argument that focuses on one inertial frame that is the centre of the circle is the only way relativity can explain this effect, and so the case for relativity is very weak.

wisp

"particles of nothingness"

 

[wisp]

A ringlaser project that is able to detect the anistropy of the sort theorized by Wisp? http://www.wettzell.ifag.de/LKREISEL/CII/precise.htm

This group says it hopes to achieve resolution of the Earth's rotation to a few parts in a billion! That's much more than we need, if the wisp ether speed of the ring laser on the equator is totally due to Earth's rotation. speed of the r.l/c is about 1 per million or maybe much more. I forgot what the speed of the solar system (divide by c) is wrt the CRB, but it must be much more, I think.

I don't believe that this experiment will detect any galactic ether flow, because it will be affected by "jiggle effect" (wisp theory). It will suffer the same fate of the MM experiment carried out on the surface of the earth.

However, it will measure the rotation of the Earth with incredible accuracy.

wisp

"particles of nothingness"

 

[Hurkyl]

But the flaw in this argument is simply this: What happens to the measuring clock when the radius of the circle becomes very large and the clock's velocity small - a limit process?

You mean relativity isn't preserved by limits? *gasp* Guess what? Neither is mathematics.

Besides, if v is negligably small, then c - v and c + v are negligably different.

 

[outandbeyond2004]

Exactly how does this limit process that wisp writes about in post #151 work? It seems as though as the circle gets larger, the angular speed gets slower proportionally, so as to keep the linear velocity the same. If that is correct -- so what? The rotating frame becomes more and more like a inertial frame -- neglecting gravitational effects of course. If you keep the measurement time small enough, the rotating frame is indistinguishable from an inertial frame beyond the limits of instrumental technology, and that's far easier to do that with a huge and s-l-o-w-l-y rotating r.l.

One has to be careful in applying SR to situations where the curvature of spacetime cannot be neglected. The huge ringlaser is simply in reality not a subject for SR anyway.

There may be some worth in wisp theory, the explantion of how gravitational effects may arise from wisps is intriguing. However, I am less enthusiatic than before. I will continue to skim wisp's future posts to see if there is any significant change of some nature, but except for one thing, I do not think I will post any more on wisp theory.

 

[russ_watters]

Besides, if v is negligably small, then c - v and c + v are negligably different.

Cute, isn't it - wisp declares negligible the most important piece of the model in order to disprove the validity of the model.

 

[Hurkyl]

The problem is that as the ring's radius grows to infinity, so does the amount of time it takes for the light to travel around the ring. An infinite amount of time is plenty of time for those negligable errors to build up.

Basically, he's neglecting the fact that 0 * &infin; and &infin; - &infin; are indeterminate forms.

 

[outandbeyond2004]

I propose to derive the Sagnac effect from classical physics for the benefit of readers not familiar with the effect. I want then to incorporate reasonable wisp-theory changes. I hope to go on from there and then to show that we have reasonable grounds to expect that wisp theory does predict time-varying signals that ringlasers can detect, contrary to wisp's dismissal of them (post #151).

If my derivations pass muster (e.g. nobody can show any error or show that an assumption is unreasonable), I am going to unsubscribe this thread. Look for the next post . . . (How's that for salesmanship? <smile> )

 

[outandbeyond2004]

If wisp ever thought that my last post (#157) was an April Fool joke, no, he fooled himself.

The link that wisp gave in #151 on the Sagnac Effect would be helpful tho not necessary.

A laser can be viewed as a chamber that creates conditions for resonance. There must be a positive integer number, called the mode number, N, of whole cycles (no partial cycles) between the ends of the chamber:

N$$\lambda$$ = L, _________ EQUATION 1

where L is the "optical length" between the ends of the chamber, and $$\lambda$$ is the wavelength. That is the resonance condition for lasing.

A ringlaser is two chambers, one for counterclockwise traveling photons, and the other for cw photons. Let's consider in this thread only a circular ringlaser of radius R. Let it be at rest in an inertial frame of reference whose origin is at the center of the circle. On a point of the circle is a device that does several things:

1. shoot out photons of the same mode number in either end equally (often many modes appear in the ringlaser, but we can assume just one mode is excited);
2. combine the ccw signal with the cw signal to form a beat frequency;
3. and measure the beat frequency.

Experimentalists would see the above description as an overly simple and impractical model of the laser, but it should suffice for order-of-magnitude estimates.

Let's modify Equation 1 to get the frequency:

$$f$$ = N/T, _____________ EQUATION 2

where T is the time for a photon to go around the ring from the device and back to it.

Given $$T_+$$ for the trip time of a ccw photon, and $$T_-$$ for that of a cw photon, we have for the beat frequency $$f_b$$:

$$f_b$$ = N|$$\frac{1}{T_+} - \frac{1}{T_-}$$|. ________________ EQUATION 3

Of course if no anisotropy existed in the two chambers, and the ringlaser was still at rest ia inertial rf, the beat would be zero. Suppose the irf moved through wisp space at speed $$v$$, it will still not change anything. For every ccw photon going at c + w
we also have, on the other side of the circle, a cw photon going c + w; and likewise for photons going c - w, for any speed w <= $$v$$.

However, the situation changes when you rotate the ringlaser at a constant angular speed of $$\omega$$ wrttirf. Then the photon speeds cannot be matched up that way, canceling out all the w's.

Readers may find it helpful for later posts to derive the Sagnac Effect beat frequency.

 

[wisp]

Besides, if v is negligably small, then c - v and c + v are negligably different.

No, the radius of the ring increases, but v stays the same. So c - v and c + v are not negligibly different.
And if the radius were very large, why should errors come into this argument!

wisp

"particles of nothingness"

 

[wisp]

However, the situation changes when you rotate the ringlaser at a constant angular speed of $$\omega$$ wrttirf. Then the photon speeds cannot be matched up that way, canceling out all the w's.

Your reasoning is OK, but if no change occurs due to a galactic ether flow, SR supporters would argue that this was expected - just like in the MM experiment.

But wisp theory also predicts a no change. The argument is similar to that used in the MM experiment and is due to jiggle dilation effect.

So SR and Wisp predict no change due to an ether flow component.

However, if the device were operated away from the Earth's surface things would be different. SR would again predict no change, whereas wisp theory would predict change due to the reduced jiggle effect.

See section 7.7 / equation set 7.7 for the affect of jiggle on the MM experiment.
http://www.kevin.harkess.btinternet.co.uk/wisp_ch_7/wisp_ch_7.html

wisp

"particles of nothingness"

 

[outandbeyond2004]

Response to wisp's post #159:

I am sorry that other people misread your prior post. However, your infinitely large ringlaser hardly disproves SR. Did you indeed get the point that SR is only of limited validity (the spatio-temporal dimensions of the irf must be small enough that our instruments cannot detect gravitational/acceleration effects)? Your monstrous rl pushes poor SR out of its limits, shivering and shivering. O! Cruel Wisp!

 

[Michael F. Dmitriyev]

I think that the main problem of SR is the clock traveling together with light.
Any object cannot travel in c except of light. Including the clock.

 

[outandbeyond2004]

Just to make sure, Wisp: you think that your theory predicts that the motion of the rl through wisp space will not produce any beat whatsoever? No matter how fast it may be rotating and how it is oriented to the motion? If yes, I am surprised that jiggle would cancel things out or balance out so neatly. Oh well, we'll see. . .

You realize, of course, that if you just say that jiggle does this or that so that a null effect ALWAYS occurs, it is just too convenient an ad hoc hypothesis. Your only possible justification for this wondrous pheonomenon is experimental verification. I doubt you can get people to spend scarce time and money on ad hoc experiments to find jiggle without some substantial theoretical foundation.

One correction: I will ignore SR corrections to my calculations. Keep in mind that I specified "optical length." That is not necessarily the same as the circumference of the at-rest, nonrotating rl. That is about all the concession to SR that I have made and that I think I will make. Perhaps after I have presented my theory including the effects of jiggle as I understand it, we can discuss the effects of SR corrections.

 

[Hurkyl]

No, the radius of the ring increases, but v stays the same.

The argument to which I responded said that v was small.


Anyways, how did you manage to conclude that if R is large enough so that the observer is in an essentially inertial frame, then the speed of light around the ring was c+v or c-v?

 

[outandbeyond2004]

Here a ringlaser rotating at angular speed w is moving through the ether at velocity v. Jiggle and wisp space will be considered in later posts but not here.

Source independence is assumed, i.e. EM signals travel through empty ether at c. At the outset of calculating the modified SE beat frequency, let's try to get a clear picture of what is going on. In ether space, each ringlaser photon travels in ether space what we might call pseudo cycloids, but the photon generator/beater travels a true cycloid. It is worth noting that no ringlaser photon travels the same path in ether space as any other; in that sense we do have one-way travel (if this is not acceptable to MM et al. -- why not?).

The following point is not one I am sure ought to be made, but I will make it nevertheless. The photons that beat together in the beater are not photons that start out together, at the same time. Instead of my drawing a diagram, why don't you draw it following these instructions: Draw the following on a piece of paper:
A circle about the size of your fist on a piece of paper.
About the center of the circle, draw a curved arrow going ccw and label it w for angular speed.
A vertical arrow starting near the right side of the circle and label it v for the ring's velocity wrt the ether.
On the circle make a tick somewhere between zero and pi/2 from the direction of v (which we can take to be the x-axis also). Label the tick thetab, for the point on the circle where a ccw photon and a cw photon meet in the beater to create the beat frequency.

Now, the ccw photon had to have started earlier than the cw photon, so you have to draw their respective ticks on the circle as follows:
For the cw photon, simply go a little cw from the first tick and between that and zero angle, draw a tick. Label that theta-.
Again go a little cw and so on except label *that* theta+.

The ccw photon leaves theta+, goes through theta- and then thetab and on all around the circle, ending at thetab. The cw photon goes the other way and on *almost* all around the circle, ending at thetab.

I want to do the math in the ether frame of reference, and am finding it tough going. I never got really good at problems like that. It might be several days before I decide it is good enough to present. Or maybe I will decide on another approach.

 

[wisp]

outandbeyond2004

No, it is not acceptable. The ring laser device that controls the motion of the photons is moving through the ether at speed v.
To mimic the pseudo cycloid motion from the ether frame, a device would have to be made that could accelerate and decelerate the photons, which would make the photons feel forces that were variable.
In the moving frame the forces are of constant acceleration only.
Also in ether frames force devices operate differently, as force dilates as the speed v increases (but, Lorentz symmetry stays the same).
With SR you can swap frames about, but you cannot do this with ether frames.

To start of with a simple model, why not make a square sided laser device, which rotates by pi/2. The sides can be the x-y-axes, and later the model can be made circular.

wisp

"particles of nothingness"

 

[outandbeyond2004]

wisp, I do not understand your post, but I did imply that I will not consider wisp theory for now. Later I will consider it. Then I probably will need considerable amounts of discussion. Got the time for that?

I think I am getting the hang of the problem, anyway. In truth, I caught a blooper. The motion of the beater is pseudo cycloid, not true, because its rotation does not necessarily have the proper relationship to the velocity.

 

[outandbeyond2004]

My calculations made it obvious to me that the theory of linear addition of speeds plus the postulate of constant speed of light in the ether predicts a modification of the Sagnac Effect:

Multiply it by a factor like

sqrt ( 1 - (gamma*cos(theta))^2 ) - gamma sin(theta),

where theta is the angular position of the beater measured at the center of the ringlaser from the velocity direction and gamma = v/c, v being the speed of the ringlaser wrt the ether.

I am confident that if some ringlasers can resolve the modified Sagnac Effect to better than the order of gamma (v/c), they would disprove SR, provided gamma is known.

My calculations did not reach the point where the Sagnac Effect is modified, but surely time varying signals in it are obviously demonstratable. The calculations are now presented (skip to --------------- if you want) and some discussion follows.

The position P(vector) of a photon at time t starting from theta0 at t = 0 in the ether frame, given

R = radius of the ring
v = speed of the ring wrt the ether, in the x direction
theta = angular position, = theta0 at t = 0, measured at the center of the ring from the x direction:

P = (vt + R cos(theta) , R sin(theta))

Take the derivative wrt t:

dP/dt (vector) = (v - Rsin(theta)*dtheta/dt, Rcos(theta)*dtheta/dt)

Because the photon is supposed to move at the speed of light in the ether frame,

c^2 = |dP(vector)/dt| (square of the magnitude) =
v^2 + (R*dtheta/dt)^2 -2vRsin(theta)*dtheta/dt

Let w = angular speed = dtheta/dt, and solve for it:

w = (c/R)(+/-sqrt(1-(gamma*cos(theta))^2) -gamma*sin(theta))

The ccw photon requires the plus sign in the above equation; the cw photon the minus sign.
-------------------------

Note that the formula given above does not apply to a ringlaser attached to the surface of Earth, if gamma = Earth rotation speed/c, because the ringlaser does not go in a line, rather it revolves in a circle. I don't see, however, that the qualitative picture is any different.

The factor after c/R in the above equation does approximate 1 - gamma*sin(theta) for gamma << 1, but if you take gamma = 0.9 it is obviously asymmetric as a plot shows. Not that we would ever have ringlasers hurtling through ether at speeds like that, but it was interesting at least to me.

Remember the ringlaser that claimed to be able to resolve Earth's rotation to a few parts in a billion? I do not know the particulars of its design, so I cannot really say 100% positively that its operators ought to see ether anistropy effects generated by Earth's motion and rotation like predicted above, but I would be amazed if it were not so. If my calculations and conclusion are correct, I believe we would have seen reports of anisotropy in ringlaser experiments by now if gamma ~ 1/1000 as per Nereid. MM et al would have been happy. Happy now?

 

[wisp]

Let w = angular speed = dtheta/dt, and solve for it:

w = (c/R)(+/-sqrt(1-(gamma*cos(theta))) -gamma*sin(theta))

The ccw photon requires the plus sign in the above equation; the cw photon the minus sign.

I got a quadratic, which gives
w=(v*sin(theta)/R)+/-sqrt((v^2*sin^2(theta)-(v^2)+(c^2))/R)

Anyway, with the jiggle factor (slows light in the perpendicular direction to the Earth's motion through the ether) the v terms will cancel. It did exactly this with the MM formula, which had v terms.
I will use your formula and add in the wisp bits. It should give a formula for w without any v term.

 

[outandbeyond2004]

Wisp's quadratic result is correct. However, v^2 -(v^2)sin^2 = (v^2)cos^2. My result is the same.

edit - I did leave out the factor ^2, sorry, while copying from my paper. I edited it into that previous post.

 

[outandbeyond2004]

Anybody know how to solve this analytically?

$$\int \frac{d\theta}{\pm \sqrt{1-(\gamma cos\theta)^2} - \gamma sin\theta}$$

 

[outandbeyond2004]

The readers who want a neat formula for the modified Sagnac Effect:

There is not any, generally speaking. Numerical approximation looks necessary. This post assumes that you have followed the directions in post #165 https://www.physicsforums.com/showpost.php?p=177177&postcount=165

A general procedure goes as follows:

Obtain a formula relating initial angle and final angle for the ccw photon to the trip time:

$$T_+ = T_+(\theta_+, \theta_b)$$

Likewise for the cw photon:

$$T_- = T_-(\theta_-, \theta_b)$$

We also have

$$\omega_bT_+ = \theta_b - \theta_+$$

and

$$\omega_bT_- = \theta_b - \theta_-$$

where omega_b is the rotation speed of the beater.

Even if we had analytic formulas for the trip times as functions of the initial and final angles, these equations must still be solved simultaneously. This is in general impossible to do so analytically. I fear this is the case for the rotating ringlaser in translational motion. I am not sure whether I will do the numerical work necessary to show a time-varying signal. I can program my computer, but I must learn how to solve these equations simulataneously on it. I don't know how much time that would take.

It might be worth doing if it were original work and would be accepted in a journal like Physics Review.

 

[wisp]

outandbeyond2004

I can’t see a simple solution either, but I will use your formula to input into mathcad and calculate a set of series equations that will give an approximation answer.
I will do this with and without jiggle.
Jiggle slows light on the Earth's surface in the perpendicular direction to ether motion by
$$\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$$

Initial mathcad plots (without jiggle) of w against theta show that it is affected by the ether flow and varies in a sinusoidal pattern.
I haven’t added the jiggle effect yet, and initially said that the results would cancel any ether flow effect. However, there is a small asymmetry to the jiggle cancellation effect due to device rotation, which will result in an extremely small variation in w, which will be dependant on the ether flow, but its value may be too small to detected.

 

[outandbeyond2004]

Mathcad! O, you lucky fellow. What I was planning to do is simply to calculate the photon trajectories backwards until omega_b*deltatime = deltaangle for both photons. Probably have to do this several times, a few times integrating forwards rather than backwards like above, until you find a good step size that minimizes numerical error. Then run with the optimal step size for all theta_b. I guess the angle that would give the most trouble is when cos(theta) = 1, i.e. theta ~ pi/2. That's where I would look for optimal step size. No, come to think of it, I'd try pi/4, because the cos term is only of second order in gamma.

I cannot believe the jiggle only operates in the peripendicular direction. I think it has something like a sine dependence. Let v be the velocity (vector) of the rl and n be an unit vector in the peripendicular direction to c travel. then

$$\gamma_{jiggle} = \frac{1}{\sqrt{1-(v\cdot n)^2}}$$

Possibly wrong formula, but you get the idea.

 

[outandbeyond2004]

I looked at the case of a circular ringlaser attached to the spinning earth, ignoring Earth's revolution about the Sun.

Let

$$v_L =$$
velocity of photon measured in ether space;
$$v_E =$$
velocity of ringlaser center wrt inertial frame centered on Earth's center of mass or in ether space; both velocities are fractions of the inertial speed of light in vacuo.

For the case of a ringlaser on the equator, with its plane parallel to the equatorial plane, let

$$\theta_L =$$
angle of photon position measured wrt x-axis of the inertial frame of Earth used above at the center of the rl;
$$\theta_E =$$
angle of rl position measured wrt x-axis of the same irf at the Earth center of mass;

So,

$$v_L^2 + 2v_E cos(\theta_E + \theta_L)v_L - c^2 + v_E^2 = 0$$

Surprisingly simple. I am not sure what the cos term would lead to in the beat frequency.