# A New Simultaneity Method for Accelerated Observers in Special Relativity

Discussion in 'Alternative Theories' started by Mike_Fontenot, Dec 26, 2019.

1. ### HalcRegistered Senior Member

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That's what I would have said. The traveller knows he's traveling and where he's going to return. When I drive to the corner market, I assume a frame in which I'm the one moving, which makes more sense than a frame that has the whole Earth accelerating each time I take a corner.
If I'm not going back somewhere, then there is no reason to assume a frame like that, and there's no particular reason to concern ones self with the 'current' age of some distant relative. Nobody has a current age. It's spacetime, which doesn't posit a preferred moment.

So what advantages do all these methods claim? What is the advantage of yours? I certainly don't see one.

They do no such thing, and you've failed to support this assertion after I pointed it out. All they're doing is a computation, not actually causing any effect at all.

Predicts? I understand that it computes this, but it is hardly a prediction. A prediction anticipates some empirical difference that the other methods do not.
Anyway, the Andromeda 'paradox' has long demonstrated the age of distant things changing relative to a mildly accelerating object, such as pacing back and forth. There's nothing magical about continuously changing the frame in which you are stationary.

If a physicist is bothered by the fact that multiple non-parallel lines converge on the same point and even (gasp) cross each other, maybe they're in the wrong business.

3. ### Neddy BateValued Senior Member

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2,548
The advantage of the "CMIF method," which is really just the Einstein/Minkowski/SR method, is the following:

It satisfies the requirement that, for all inertial clocks, a relatively moving clock ticks at a rate of 1/γ = 1/gamma the rate of a relatively stationary clock. Where γ = gamma = 1/√(1-(v²/c²)) and v is the speed of the relative motion. That is one of the first and easiest things that one can derive from the postulates of SR.

I don't know whether some of the other methods satisfy that requirement, but I'm sure Mike_Fontenot's current method does not do so.

5. ### HalcRegistered Senior Member

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Understood, but of course the traveler is not inertial, so multiple frames are involved if the traveler insists on always considering himself stationary despite all the acceleration. Yes, it makes the age of distant things get younger as you accelerate away from them, which some people apparently find offensive.
The other methods (D&G, Minguzzi) definitely do not satisfy this 'requirement' you state above. I don't have a full description of D&G, but it has her age suddenly going faster without any inertial change of the traveler. The Minguzzi thing defines simultaneity relative to an event rather than to an inertial reference frame, so the foliations resulting are not planar, but curved, and are completely dependent on the reference event.

My conplain about Fontenot's wordings is his metaphysical stance that seems to imply that there is a correct answer, even if it cannot be known:
There only reason that a method might be a 'correct' one is that her age is in fact this one specific age, and no other. That's an absolutist interpretation (which I think is Fontenot's unstated 'philosophical consideration'), and under absolutism, relative speeds have nothing to do with dilation. Dilation is a property under absolutism, and not a relation, and thus the rate of aging of either twin has zip to do with what the other is doing, in which case the D&G, MInguzzi, and the 'Fontenot' methods are all wrong.

7. ### Neddy BateValued Senior Member

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I wouldn't say the traveler is "not inertial". He is not inertial at the turnaround point, but in the specific scenario we have been discussing, he is inertial for the whole outbound leg of the trip, and he is also inertial for the whole inbound leg of the trip. That means he has to consider her clock to be ticking at a slower rate than his own clock by the factor 1/gamma for the whole outbound leg, and also the whole inbound leg, in order to satisfy the 'requirement' I stated earlier. This is the key to understanding why the CMIF method is the correct one, and when I say correct, I mean only that it maintains all of the postulates of SR perfectly intact and unchanged.

Again, the only reason I say the CMIF method is the correct one is because it keeps all of the postulates of SR perfectly intact and unchanged. To Mike Fontenot's credit, he does not seem to want to believe that his method violates the postulates. Unfortunately that means he is willing to believe and state outright that the postulates say things which they do not say at all, or that the postulates do not say things which they clearly do say.

8. ### HalcRegistered Senior Member

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Perhaps your requirement stated earlier is insufficiently precise then.
I assume this is the thing in question, and it would be better worded as: "for all stationary inertial clocks", since the statement simply isn't true without that.

Hence my comment above that said traveler is considering himself stationary both before and after the turnaround acceleration. In this idealistic view, it is the universe which has accelerated, not himself, contradiction the empirical fact that acceleration is absolute, not relative. All these non-standard methods seem to violate this, while the standard method (pick a frame, any frame, and everybody uses it for the duration) delivers consistent and continuous (no discontinuities in ages of distant things) results across the board.

Actually, none of them seem to violate SR at all. SR has no explicit problem with looking at the same situation from as many frames as you like, or from non-inertial frames like Minguzzi uses. None of that violates the very simple postulates of SR, but it also doesn't claim to provide an answer to how hold Alice back home actually is when Bob turns around. SR just says that her age simultaneous with that event is frame dependent.

They don't say much. Light speed is measured at a constant speed in any frame. That and Galilean relativity is it. That's the beauty of it. Only two postulates, and both empirical in nature. I can think of only one theory even more simple, and its sole postulate isn't really empirical, so it's just interpretational.

9. ### Neddy BateValued Senior Member

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I don't understand. My stated requirement involves relative motion between inertial clocks. The clocks are not stationary with respect to one another.

In other words, if I add the words you suggest, I would not even understand the meaning of the statement anymore. Could you please connect your words and mine together in one sentence or phrase, and then tell me what you think it means? Also, please tell me stationary with respect to what?

Yes, I agree SR lets you look at a situation from any perspective. You had mentioned earlier that it would be easier if the traveler used the stay-home frame. The correct way to do this is by transforming the coordinates from one frame to another. So, in the traveler's frame, he is stationary & she is moving, his clock ticks at the normal rate & hers tick at a slowed rate. But he can transform coordinates from his frame to her frame. In that case she is stationary & he is moving, her clock ticks at the normal rate & his tick at a slowed rate.

You are mistaken if you think the reason that the traveler ends up younger is because acceleration is absolute. There is no variable for acceleration in the SR equations. It doesn't matter that acceleration is absolute. What matters is that there is relative motion between the two frames. And relative motion is relative, and reciprocal.

You could make something similar to the "twin paradox" in which both twins accelerate equally. Let Bob and Alice be twins born at the same time, and let Bob accelerate away from Alice and then coast away at constant speed. Due to relative motion alone, Alice says Bob's clock ticks slower than her own, and Bob says Alice's clock ticks slower than his own. After awhile, Alice accelerates in the exact same way that Bob did, and then she starts to coast in the same exact way. Now both twins are stationary with respect to one another, although they are some distance apart. Now Alice is younger than Bob, even though they both accelerated identically, albeit at different times.

It is not the "standard" way in SR for everyone to agree to use one frame. It is standard for everyone to use their own rest frame, and then they can transform to other frames. Using your example again where the traveler uses the stay-home frame, that is not standard. Because it is not standard for the traveler to think his own clock is ticking slow.

Last edited: Feb 12, 2020
10. ### Neddy BateValued Senior Member

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The only reason there appears to be discontinuities in ages of distant things is because we let the turnaround acceleration be instantaneous, which creates an undefined slope at the turnaround point, like this:

But that is easily rectified by making the function continuous, so there is no undefined slope. Put a radius at the turnaround, and make it as small as you like, like this:

There. No more discontinuities in ages of distant things.

11. ### Mike_FontenotRegistered Senior Member

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But the home twin can still get YOUNGER with CMIF simultaneity, even with finite accelerations. That bothers a lot of people (although it didn't bother me, back when I was a CMIF proponent). The home twin never gets younger with my simultaneity (and I think that is also true for the D&G method, and for Minguzzi's method).

12. ### Mike_FontenotRegistered Senior Member

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Also, for 1g accelerations lasting a couple of years, the CMIF's ACD is qualitatively very similar to the instantaneous velocity change case. The home twin's age doesn't increase by 60 years or so instantly, but it increases by about 60 years while the traveler's age increases by only a couple of years. The instantaneous velocity change case is actually a good and useful approximation to the 1g acceleration case (and lots easier to compute, of course).

In contrast, the increased ageing rate for the home twin with my method, with an instantaneous velocity change, occurs over about 9 years of the traveler's life, during which the home twin's age increases linearly by about 64 years. (You can see that on the ACD diagram I posted to this forum, which is also now on my webpage). And with a 1g finite acceleration, over a couple of years, in my method, the ACD changes very little compared to the instantaneous velocity change case ... the sharp slope changes of the ACD just get rounded off a bit. That is another nice characteristic of my method ... little need to compute the finite acceleration case.

13. ### Neddy BateValued Senior Member

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But she doesn't actually get younger in the way that one might think, "Oh that must be wrong, she can never get younger!" It just means that during a "double-turnaround," the traveler would consider his current clock time to be simultaneous with when she was an older age, and then his simultaneity changes so that he would consider his current clock time to be simultaneous with when she was a younger age. But his concept of simultaneity cannot dictate anything regarding her physical metabolism, and cannot force her to age backwards in her own frame. She can be whatever age she actually is in her own frame, and he can still think that his current time is simultaneous with an earlier time in her life. The two things are not connected in the way that we are used to, as we live our non-relativistic lives.

The different concepts of simultaneity only belong to the different inertial reference frames that he becomes co-moving with. As Halc said, she is still older in the other inertial frame, but the traveler is no longer co-moving with that frame. I can understand why some people have trouble with it, but I don't see it as a problem myself.

What is more important to me is that, whatever simultaneity method you choose, it should have clocks ticking at the rates that SR says they should. Namely, a clock should tick at the rate of one second per second in its own frame, but at a slower rate in some other inertial frame which is in uniform translation relative to that clock. That means the traveler, using his own inertial frame as his frame, should say that the home twin's clock ticks at a slower rate than his own, during both the outbound and inbound inertial legs of the journey. I know you do not currently agree that is what SR says, (nor with the idea that he is inertial on the way back in!) but I am quite sure that is what SR says, so the CMIF method is the only one that satisfies me.

14. ### Neddy BateValued Senior Member

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I understand that the instantaneous acceleration case is a very good approximation. If you look at my two diagrams, the lines of simultaneity pointing toward 26.67 and 53.33 are still almost identical in both diagrams. The difference is that there really should not be any age discontinuity there, and there isn't in any realistic case.

The discontinuity is an artifact produced by the assumed instantaneous acceleration. In reality you cannot go from v=+0.577c to v=-0.577c without also passing through v=0.000 in between. And it is interesting to see that, in the realistic case, the line of simultaneity at the midpoint does not point toward either 26.67 or 53.33. It points to 40.00.

15. ### HalcRegistered Senior Member

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350
No, of course not, and I didn't say so. For any stationary clock, relatively moving clocks are moving, and 'tick at the rate of 1/γ'. But suppose we have two clocks moving north and south at equal speeds. They'd tick at the same rate despite their relative motion. So the validity of your requirement depends on a coordinate system in which the one clock is stationary.

I added one word. That so bad? Other clocks don't tick slower at the rate you specify relative to a non-stationary clock. I've given a simple example illustrating the point.

Well, stationary wrt/ the coordinate system chosen. Without that coordinate system, such statements are meaningless, and relative to a different choice of coordinate system such statements are wrong. So to add a little more than just one word:

You want a method that satisfies the requirement that, "for all inertial clocks, relative to an inertial coordinate system in which that clock is stationary, a moving clock ticks at a rate of 1/γ = 1/gamma of the rate of the stationary clock."

That's a thin requirement, and I'm not sure if D&G or Minguzzi methods violate that since they're not necessarily using an inertial coordinate system at all when making their calculations. I'm unclear about how D&G works, but Minguzzi is definitely using a totally different sort of coordinate system than the inertial one typically used for Minkowski spacetime. In fact, in a way it resembles (and can be reduced to) the non-inertial comoving coordinate system used by cosmologists for non-local descriptions of things. I digress I think...

Ach no! Just use the one system the whole time and don't translate anything at all. Of course Bob's going to need a clock that displays time and speed and such, but it's a digital clock hooked to an accelerometer so that shouldn't be unworkable.

My suggestion was to completely ignore the traveler frame.

He has almost no coordinates to translate since in his frame he's always 'here' and stopped. That's useless. He needs something like a speedometer and odometer, just like a car has, neither of which he would have if he used his own frame. No car attempts to use its own coordinates (I'm always here and stationary!) and then translate to a standard one. It uses the standard coordinates the whole time. Our hero is in a car and he knows it.

One does not need to select Alice's frame as the standard one. Any arbitrary frame will do, but if Alice is not stationary, she'll need a speedometer, odometer, and funny clock just like Bob.

I didn't claim that acceleration was the reason, which is easily verified right here on Earth by comparing a couple atomic clocks that differ only in acceleration, but not speed. If you asked, I'd say the reason is moment-of-acceleration, but I've not mentioned that until now. It's a term not often mentioned in the articles I find, but it's explicitly computed in the CMIF method. That's not 'the' correct answer since there are several different explanations that are all correct. One simply involves computing intervals between events.

I brought it up because the CMIF method assumes Bob is stationary both before and after a significant acceleration, which seems to violate a different but reasonable requirement, one that the pick-a-frame method doesn't violate.

Not in Earth's frame she isn't. That's why specification of frame is important, and your statement that Alice is younger than Bob omits any explicit frame. Seatmates on an airplane are stationary relative to each other but not necessarily stationary because of that. Almost all the confusion, often deliberate confusion, results from lack of frame specification when making such statements.
Oh, and thanks for picking up on the perpetual victims Alice and Bob, (and occasionally their mutual nemesis Victor).

Buck lack of the specification makes for ambiguous statements. I'm find with any choice of frame as long as it's explicit.

For doing local things in and about the ship, sure, but for interactions with home, neither frame is the best choice. He personally has no need of it. He cares about what age she appears now, which is whatever age she was when each photo-message was sent. She'll appear to age quite slow on the way out and stupid fast on the return leg. Both of them get that, and at the same two rates as well. Nice and symmetrical.
For example, they're talking about Betelgeuse in the news. Just google the one word. First hit today says: "Huge red star might explode soon and next few weeks are critical"
That's the standard that is used, the only one with practical value. Not our frame, and not its frame (not that it's a whole lot different). They use what they're seeing now. The star is 642+ light years away but no headlines reads "Huge red star might have exploded 642.4 years ago and the weeks just before that are critical".
I know. With any finite acceleration, you get a very steep slope with something like the CMIF method.

16. ### Neddy BateValued Senior Member

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Okay, that works also, but it seems a little redundant to me.

My original statement was, "For all inertial clocks, a relatively moving clock ticks at a rate of 1/γ = 1/gamma the rate of a relatively stationary clock. Where γ = gamma = 1/√(1-(v²/c²)) and v is the speed of the relative motion."

Note that I already restricted the topic to "inertial clocks", and it seems pretty clear that that I am comparing the rate of a "relatively moving clock" to the rate of a "relatively stationary clock" so I am not sure why you feel it is necessary to specify "relative to an inertial coordinate system in which (the relatively stationary clock) is stationary."

Let's take your counter example, two clocks moving north and south at equal speeds, which I assume is relative to the earth. You say they would tick at the same rate as each other, which is true in the frame of the earth. But remember, my statement involves one "relatively stationary clock". So it is clear that the clock you are choosing to be the "relatively stationary clock" is one that is stationary with respect to the earth, because you say that both of the other clocks are moving, and tick at the same rate as each other. In that case, my statement still holds true, because both of those relatively moving clocks will tick at the rate of 1/gamma times the rate of your chosen stationary clock.

Yes, you can do that, but it is not standard SR. In standard SR thought experiments we can give the stay-home twin a coordinate system filled with an infinite number of hypothetical synchronised clocks everywhere. They can record the relative location of the traveling twin at all times. They can even record the times displayed on his clock, for comparison purposes.

And in standard SR thought experiments, we can also give the traveling twin (on the outbound half of the journey) a coordinate system filled with an infinite number of hypothetical synchronised clocks everywhere. They can record the relative location of the stay-home twin at all times. They can even record the times displayed on her clock, for comparison purposes. And it is in this frame that the age of the stay-home twin is recorded as 26.67 at the moment just before the turnaround point.

The traveling twin is not necessarily isolated in a space ship. For the outbound half of the journey, he can be on a very long train which extends far enough so that the stay-home twin is always co-located with part of it. Likewise for the inbound half of the journey. In fact, all he has to do to at the turnaround point is jump off one train and onto the other. (It hurts, but that's what happens with assumed instantaneous accelerations.)

And in standard SR thought experiments, we can also give the traveling twin (on the inbound half of the journey) a coordinate system filled with an infinite number of hypothetical synchronised clocks everywhere. They can also record the relative location of the stay-home twin at all times. They can even record the times displayed on her clock, for comparison purposes. And it is in this frame that the age of the stay-home twin is recorded as 53.33 at the moment just after the turnaround point.

No, in standard SR our hero at any given time (other than during accelerations) is only in his inertial rest frame, which is just another inertial frame, with no special status over any other inertial frame. The only reason he chooses to use the frames he does is because they are his rest frames. Doing so lets him have his own clock, (stationary with respect to him), tick at the rate of one second per second. Not the artificial rate that you would have his clock tick at in your non-standard scenario.

I meant in the frame in which they end up relatively stationary, and you are correct that I should have specified it clearly. I did say, "Now both twins are stationary with respect to one another, although they are some distance apart" right before my claim about their ages, so I was hoping you would understand that was the frame I was using. But I should have said so explicitly.

17. ### HalcRegistered Senior Member

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I tried to be more careful and formal, and cross all my i's and dot the t's and so forth. Normally such wording isn't fully necessary, but when discussing such subjects, people take very incorrect meanings from sloppily worded statements.

That doesn't make the statement true, as in my equal speed north/south example. A clock stationary relative to the northbound clock would be northbound itself, and thus tick at the same rate of the southbound clock. It's the coordinate system that matters, and its specification needs to be explicit for the statement to be correct.
That's my point to Fontenot as well. His wording is always 'according to Bob' and not 'according to the frame in which Bob is stationary', and Bob has no particular reason to consider himself stationary when he's in a vehicle that can and does significant acceleration at times. If he's going to make a statement that assumes he's stationary, then that assumption needs to be stated, because it just isn't the standard for dealing with non-local events (anything not inside the vehicle).

Relative to the coordinate system chosen. I suppose Earth is a good example of that since there's not particularly a north and south to a random nowhere, but it is a convenient way to label the six arbitrary directions, N S E W U D.

Relative to the first clock, the 'for all inertial clocks' one. That makes it moving north as well.

No, your statement wording was in relation to the first clock, not in relation to the coordinate system, which was never mentioned. Read it. You are envisioning a coordinate system in which the first clock is stationary, but not actually stating that assumption.

You didn't say stationary clock. You said relatively stationary, and all speeds were phrased as relative to the first clock.

Agree. SR doesn't try to assert a method for determining Alice's age at any event along Bob's worldline. At best it says it is frame dependent, and no more.

Essentially you're envisioning physical ways to lay out a grid for various coordinate systems.
Cannot envision that without this coming up:

SR does acceleration just fine, even if calculus must suddenly get invoked. It just doesn't do gravity.

Of course. As I said above, one uses the local frame when doing local things, like anything in and about the ship. My coffee is stationary in the cup holder next to me, but the car in the lane next to me is going 100 km/hr despite being stationary relative to the coffee.

OK, you get the point.

18. ### phytiRegistered Senior Member

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Neddy Bate #146;
Mike F.#148;
The scenario proposed by Neddy seemed interesting. The red lines are for time dilation, the green lines are axis of simultaneity, relative to U. When A acquires the same velocity as B, at At=4, she re-synchs her clocks resulting in the same aos as B. She actually gets less younger than B, remaining one t unit younger until the next velocity change. B's description in on the right.

My response is still the same as the 1st encounter with CADO. There is no instant communication over distance.

19. ### Mike_FontenotRegistered Senior Member

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When Bob instantaneously changed their relative velocity from zero, their separation was zero. An instantaneous velocity change by Bob in that case has no effect on Alice's current age ... the EFFECT on current age by a given instantaneous velocity change depends linearly on the twins' separation. When Alice instantaneously changes her velocity, they have a large separation, and it produces a large instantaneous effect on Bob's age.

The above is for the CMIF simultaneity method. In my simultaneity method, the effect of an instantaneous velocity change when their separation is non-zero is to increase Bob's RATE of ageing for a while (compared to Alice's rate of ageing, and according to Alice), but it doesn't instantaneously change Bob's age.

Last edited: Feb 16, 2020
20. ### Mike_FontenotRegistered Senior Member

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My new simultaneity method, which I call Fontenot's simultaneity method, produces an age correspondence diagram (ACD), which is a plot of the home twin's current age (according to the traveling twin), as a function of the age of the traveling twin. My ACD is piecewise-linear, with no discontinuities. It consists of three connected straight lines. The first portion has a slope of 1/gamma_1. The last portion has a slope of 1/gamma_2. gamma_1 and gamma_2 are the gamma factors corresponding to v_1 and v_2. The middle section of that diagram (starting immediately after the traveler's instantaneous velocity change from v_1 to v_2) has a constant upward slope that, in the standard twin paradox scenario with v_2 = -v_1, is greater than one. That slope can be determined graphically. It can also be computed analytically. Denote the slope of the middle section of the ACD curve as S. Then the S equation is

S = (1 / gamma_2) + gamma_2 * (1 - v_2) * (v_1 - v_2),

where v_1 is the relative velocity before the velocity change, v_2 is the relative velocity after the velocity change, and gamma_2 is the gamma factor corresponding to v_2. Velocities are positive when the twins are diverging, and negative when the twins are converging. It is possible to show, by using the S equation, that my simultaneity method never produces a negative value for S. I.e., in my method, the traveling twin never says that the home twin is getting younger. That is in contrast to the CMIF method, which predicts negative aging of the home twin for some scenarios.

I call the above S equation Fontenot’s equation.

More information, including a plot of the ACD for the twin paradox with v = +-0.866 ly/y and gamma = 2.0, in available on my webpage:

21. ### HalcRegistered Senior Member

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The above equation seems to only work for a narrow and unrealistic set of cases. Where's the general solution? Suppose Bob is continuously accelerating this way and that as he runs his errands, and thus there is no v_1 to reference, only the current v. What if v does not lie on the x axis as it does in all the simplified examples? Suppose Bob turns around by retaining constant speed in a half circle. At what point does v_2 become negative if the gamma_2 remains a constant?
Minguzzi had a general solution that readily yielded an answer for any random path that Bob might take. I don't see it in yours.

22. ### Write4UValued Senior Member

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I don't know if this is relevant to the conversation so I'll just link the video and let you decide in case you are not aware of this experiment.
It records light; Filming the Speed of Light at 10 Trillion FPS

Last edited: Feb 18, 2020
23. ### Mike_FontenotRegistered Senior Member

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The Fontenot equation works for all possible scenarios with instantaneous velocity changes and one-dimensional motion. And one nice thing about the Fontenot simultaneity method is that analyzing scenarios with finite accelerations is almost always unnecessary, because 1g accelerations for a few years make very little difference in the age correspondence diagram (ACD): the finite acceleration just rounds off the sharp corner at the turnaround of the ACD. In contrast, in the CMIF simultaneity method, 1g finite accelerations produce an ACD that is quite different from the instantaneous velocity change case.

But on my webpage ( https://sites.google.com/site/cadoequation/cado-reference-frame ), I DO show how to do a numerical integration to handle finite accelerations. That CAN have some value in the case of a space voyage using a low-powered rocket continuously. Using that numerical integration procedure, I produced an ACD for the case of a 0.2g scenario where the rocket is continuously running, and is just reversed in direction twice during the trip: once on the outbound leg, half-way to the turnaround point, and later on the return leg, half-way to the reunion. Even in that case, the resulting ACD is qualitatively fairly similar to what you get with instantaneous velocity changes.

I HAVEN'T thought about how to handle two and three spatial dimensions with my method. Years ago, I DID generalize my CADO equation (which simplifies computing CMIF simultaneity) to two and three spatial dimensions. But I suspect it would be much more difficult for my method. It's not a high priority for me.

The ONLY part of his paper that addressed simultaneity at a distance was in his brief description of the use of the imaginary twins, near the end of his paper. He didn't describe how that would work with multi-spatial-dimensional motion, or with finite accelerations.

Even for the classic twin paradox scenario, Miguzzi's method is VERY tedious to apply, because the ACD after the turnaround is a curved line with constantly varying curvature. I carried out that process for just a few points on the curve, to get a rough qualitative idea of what his curve looks like. I wouldn't want to do that often. In contrast, my ACD just has three connected straight line segments, with easily determined slopes for all three of the straight lines. And, perhaps even more important, my method is causal (like the CMIF method), unlike Minguzzi's method and Dolby&Gull's method, which are clearly non-causal.

The most bizarre situation with Minguzzi's method, is that he now claims that his method has NOTHING to do with simultaneity at a distance, even though it's clear from his paper that it did. The imaginary twins serve no purpose otherwise.