Age-Changes Caused By Instantaneous Velocity-Changes

[Mike_Fontenot]

The CADO equation, that I have described previously, applies to ANY kind of accelerations by the traveling twin. Its only requirement is that the home twin remain inertial.

For the simple limiting cases where the only velocity-changes made by the traveler are INSTANTANEOUS velocity-changes occurring at various instants in the traveler's life, separated by periods of constant velocity by the traveler, the CADO equation becomes especially simple, and especially easy to use.

In that case, it's possible to quickly and easily calculate the instantaneous CHANGE in the age of the home twin, according to the traveler, caused by an instantaneous velocity CHANGE by the traveler.

Denote the instantaneous velocity change of the traveler, at some instant, as "delta(v)". This is just the new velocity v2, minus the old velocity v1. (Velocities are measured in lightyears per year, which are always numbers greater than -1 and less than +1. Negative velocities mean that the twins are moving TOWARD each other.)

The ONLY other thing you need to know is the distance between the twins (according to the home twin), at the instant when the velocity change occurs. Call that distance "L" (measured in lightyears).

The age of the home twin, according to the traveler, is denoted CADO_T. (The acronym "CADO" always refers to the HOME twin's current age, at some given instant of the traveler's life, and the subscripts "_T" and "_H" are added to indicate WHOSE conclusion about the home twin's current age (Traveler or Home twin) we're referring to).

Denote the instantaneous CHANGE in the age of the home twin (according to the traveler), at some given instant of the traveler's instantaneous velocity change, as "delta(CADO_T)". delta(CADO_T) is just the new value of CADO_T, minus the old value of CADO_T ... immediately before and immediately after the instantaneous velocity change.

Then, the simple equation is just

delta(CADO_T) = -L * delta(v).

That's all there is to it.

(In the above equation, I've omitted some factors of c (the speed of light). Since we are using units for which c = 1 lightyear/year, those factors can be ignored ... they are needed only for dimensional correctness).

Note that the magnitude of the instantaneous change in the age of the home twin (according to the traveler), caused by a given magnitude of instantaneous velocity change, is proportional to the separation L of the twins.

In particular, this means that if L = 0 (i.e., if the twins are co-located), the instantaneous velocity change has NO effect on the age of the home twin. That's why the initial and final accelerations by the traveler (in the case of the classic twin "paradox" scenario, where the twins are co-located at the beginning and end of the voyage) don't affect the outcome of the twin "paradox" problem at all ... the classic twin scenario can always be reformulated with no accelerations at the beginning and end of the voyage, without changing the overall outcome.

And because their separation shows up as a proportionality factor in the equation, that means that, for a given velocity change, the effect on the age change of the home twin becomes greater when their separation is larger. In fact, you can see from the equation that the home twin's age can suddenly change by (almost as much as) twice the separation (because delta(v) can have a magnitude as large as 2 minus an infinitesimal amount.).

Here's an example of the use of the equation:

Suppose the twins are 30 lightyears apart at some instant of the traveler's life, and that the traveler's velocity has been constant at v1 = -0.8 for some period of time up until that instant.

The negative sign means that the twins have been moving toward each other. (For simplicity, I'm not bothering to write the units of velocity (lightyears/year)).

Suppose that the traveler then instantaneously changes his velocity to v2 = 0.6. Then

delta(v) = v2 - v1 = (0.6) - (-0.8) = 1.4,

and so we get

delta(CADO_T) = -30 * 1.4 = -42 years.

So, with the given instantaneous velocity change, and with the given separation, the home twin gets YOUNGER by 42 years (according to the traveler), during the traveler's instantaneous change of velocity.

Once you've got the capability to quickly and easily calculate the instantaneous age-changes of the home twin (according to the traveler), caused by instantaneous velocity-changes by the traveler, you can easily get the complete solution for how the home twin's age changes (according to the traveler), during a complete (and fairly complicated) voyage by the traveler, consisting of any arbitrary sequence of instantaneous velocity-changes, separated by periods of constant velocity in between those velocity-changes. The ageing of the home twin, during each of the constant-velocity periods, is easily calculated from the well-known time dilation result, and the home twin's ageing during each of the instantaneous velocity-changes can be easily calculated using the above delta(CADO_T) equation ... the combination of all those amounts of ageing gives the complete solution.

Mike Fontenot

 

[JesseM]

The CADO equation, that I have described previously, applies to ANY kind of accelerations by the traveling twin. Its only requirement is that the home twin remain inertial (and even that requirement can be relaxed a bit under certain circumstances).

But it only applies if you choose to use a non-inertial frame for the traveling twin with the property that the frame's judgment about simultaneity at any point on the twin's worldline coincides with the definition of simultaneity in the inertial frame where he is instantaneously at rest at that point. There is no fundamental physical reason why this type of non-inertial frame should be viewed as the "perspective" of the traveling twin while other non-inertial frames should not be. When I have brought this up before you have claimed that this non-inertial frame is the only one that matches the twin's "elementary measurements" and "first-principle calculations", but as I said in post #11 here, you never define those phrases:

Well, you never responded to my questions in post #130 about how you can define terms like "elementary measurements" and "first-principle calculations" in a non-circular way which doesn't just presuppose that the non-inertial observer must at every moment have the same definitions of simultaneity as in their instantaneously comoving inertial frame. Since your claims are nonstandard and your paper isn't available online (and is from a journal that http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=PHESEM000019000001000006000001&idtype=cvips&gifs=yes&ref=no ), I think you should explain the detailed basis for your argument, either here or in the Independent Research forum, before promoting the conclusions on threads like this one.

 

[ghwellsjr]

...
Velocities are measured in lightyears per year
...
distance "L" (measured in lightyears).
...
Then, the simple equation is just

delta(CADO_T) = -L * delta(v).

That's all there is to it.

(In the above equation, I've omitted some factors of c (the speed of light). Since we are using units for which c = 1 lightyear/year, those factors can be ignored ... they are needed only for dimensional correctness).
...

Is the reason that your equation does not have "dimensional correctness" because you "omitted some factors of c"?

If so, could you put them back in, even though "c = 1 lightyear/year", so that we could see that the units are consistent because right now you have:

years = lightyears*lightyears/year

 

[Mike_Fontenot]

Is the reason that your equation does not have "dimensional correctness" because you "omitted some factors of c"?

Yes.

If so, could you put them back in, even though "c = 1 lightyear/year", so that we could see that the units are consistent because right now you have:

years = lightyears*lightyears/year

The actual equation divides the right-hand-side by the factor c*c.

Mike Fontenot

 

[Mike_Fontenot]

For the simple limiting cases where the only velocity-changes made by the traveler are INSTANTANEOUS velocity-changes occurring at various instants in the traveler's life, separated by periods of constant velocity by the traveler, the CADO equation becomes especially simple, and especially easy to use.

In that case, it's possible to quickly and easily calculate the instantaneous CHANGE in the age of the home twin, according to the traveler, caused by an instantaneous velocity CHANGE by the traveler.

In the above posting, I should have elaborated a bit, on why the CADO equation is especially easy to use for the special case of instantaneous velocity changes (and why the equation can be re-formulated into an especially simple "delta-CADO" equation).

The complete (and general) CADO equation is

CADO_T = CADO_H - L*v.

The left-hand-side (CADO_T) is the current age (in years) of the home twin, at any given age t of the traveler, ACCORDING TO THE TRAVELER. (The subscript "_T" on the root "CADO" stands for the Traveler's conclusion, as opposed to the Home twin's conclusion).

So the CADO equation says that we can easily compute CADO_T, provided that we know the values of three quantities at the given time t.

The first of those three quantities that we need to know, CADO_H, is the age of the home twin, when the traveler's age is the given value t, ACCORDING TO THE HOME TWIN. So why does the CADO equation help us determine CADO_T, if we have to know CADO_H in advance? The reason is, the home twin never accelerates, and because of that, it is relatively easy to calculate all sorts of things when we use her inertial reference frame. In particular, when the traveler's velocity changes are all instantaneous, CADO_H can be easily determined purely from the simple and well-known time dilation result.

The second quantity that we need to know, L, is the separation (in lightyears) of the twins, when the traveler's age is the given time t, ACCORDING TO THE HOME TWIN. Again, this quantity is relatively easy to compute, because the home twin never accelerates. And, again, for instantaneous velocity changes by the traveler, L is almost trivial to compute.

The third quantity that we need to know, v, is the velocity (in lightyears per year) of the traveler, relative to the home twin, when the traveler's age is the given time t, ACCORDING TO THE HOME TWIN. v is positive when the twins are moving apart, and negative when they are moving closer together. Because of the units we are using, v is always between -1 and +1 (because v is just the relative velocity of the twins, expressed as a fraction of the velocity of light). Again, this quantity is relatively easy to calculate, because the home twin never accelerates. And for instantaneous velocity changes by the traveler, v (as a function of t) is part of the given statement of the problem ... it's known from the outset.

Now, where did the especially simple "delta(CADO_T)" equation come from, in the special case where all velocity changes are instantaneous? The "delta(CADO_T)" equation just gives the CHANGE in the value of CADO_T (as given by the basic CADO equation), when the traveler instantaneously changes his velocity. What makes things simple, is that, according to the home twin, the quantities CADO_H and L don't change AT ALL during an instantaneous velocity change by the traveler. So when we compute CADO_T immediately after the velocity change (call that instant t+), it is EXACTLY the same as it was immediately before the velocity change (call that instant t-). So, when we compute CADO_T(t+) - CADO_T(t-), the first term on the right-hand-side of the CADO equation just cancels out, so we don't even need to know its value at all (when we're using the "delta(CADO_T)" equation.

Similarly, when we compute the quantity L*v at both t- and t+, the quantity L is the exactly same in each case. So the difference in the quantity L*v during the instantaneous velocity change is just

L * delta(v) = L * ( v(t+) - v(t-) ).

That's where the "delta(CADO_T)" equation comes from, and that's why it's so simple.

For some other examples of the use of the CADO equation (both for situations where the accelerations are all +-1g, and for situations where the velocity changes are instantaneous), look at the posting

https://www.physicsforums.com/showpost.php?p=2934906&postcount=7

and my webpage

http://home.comcast.net/~mlfasf Mike Fontenot

 

[Passionflower]

Here's an example of the use of the equation:

Suppose the twins are 30 lightyears apart at some instant of the traveler's life, and that the traveler's velocity has been constant at v1 = -0.8 for some period of time up until that instant.

The negative sign means that the twins have been moving toward each other. (For simplicity, I'm not bothering to write the units of velocity (lightyears/year)).

Suppose that the traveler then instantaneously changes his velocity to v2 = 0.6. Then

delta(v) = v2 - v1 = (0.6) - (-0.8) = 1.4,

and so we get

delta(CADO_T) = -30 * 1.4 = -42 years.

So, with the given instantaneous velocity change, and with the given separation, the home twin gets YOUNGER by 42 years (according to the traveler), during the traveler's instantaneous change of velocity.

Then I personally do not find this method very useful.

Personally I find the idea to have a clock aboard the spaceship that shows the spacetime distance between the departure event and the current event more useful. Aging would always be in the same direction (e.g the clock would never go back) regardless of movement or acceleration.

 

[Mentz114]

Instantaneous changes in velocity are unphysical so it seems rather pointless.

 

[Dale]

Personally I find the idea to have a clock aboard the spaceship that shows the spacetime distance between the departure event and the current event more useful. Aging would always be in the same direction (e.g the clock would never go back) regardless of movement or acceleration.

That's a pretty cool idea. So, in an inertial reference frame your lines of constant time coordinate would form a family of hyperbolas. What would you use for your spatial coordinates?

 

[Passionflower]

That's a pretty cool idea. So, in an inertial reference frame your lines of constant time coordinate would form a family of hyperbolas. What would you use for your spatial coordinates?

At each point on the traveler's curved worldline we can span a cord from the event of origin to the current point, this cord represents the geodesic between these events and could, if you like, represent the path of a hypothetical inertial twin.

This idea is worked out by Minguzzi, he calls it differential aging. Using the Cauchy-Schwarz inequality he shows that this aging is always in the positive direction (e.g. clocks do not go backwards in this model).

See for instance: http://arxiv.org/abs/physics/0412010 for a two dimensional and http://arxiv.org/abs/gr-qc/0611076 for a full dimensional treatment.

 

[ghwellsjr]

Mike, it would be one thing if you were promoting your scheme as an easy way to calculate the final result of the ages of the twins after they reunite but that is not what you are doing. You are claiming in your paper at the very end that:

"Under a specific definition of 'meaningfulness' and 'reality', the bizarre behavior of the CADO, for an accelerating observer, must be regarded as being fully meaningful and real".

And by bizarre behavior, you mean clocks that go backwards. There is no frame of reference that will have any clocks going backwards. This does not comport with any legitimate definition of reality.

Clocks accumulate time and things age at different rates as they spend time at different velocities. They do not accumulate all their time differences simply by accelerating. In any frame that you want to analyze a situation, a clock moving at a constant velocity accumulates time at some constant rate. When it experiences acceleration, the clock changes it tick rate (for lack of a better term) and it accumulates time at a different rate and so on. I repeat, this is true no matter what frame of reference you use to analyze the situation and this corresponds to physical reality. Clocks physically run at a constant rate if they are not accelerating and when they are accelerating, the clock rate physically continuously changes in a smooth manner until the acceleration stops and the clock physically is left running at a new rate. This is reality.

When doing thought problems, to simplify matters, it is OK to claim instantaneous changes in velocity but a clock will not change its time during this process. It only changes its ticking rate--it is different after than it was before. Until the clock spends time at the new rate, it will not show a difference in time than it would have without the experience of acceleration.

So I don't think you are going to get much acceptance of your scheme unless you refocus it as simply a quick way to calculate the final result of the twins.

 

[Mike_Fontenot]

Instantaneous changes in velocity are unphysical so it seems rather pointless.

It is true that instantaneous velocity changes are idealizations. They are limiting cases that can't actually happen in real life. But when the separation between the twins is large enough, finite accelerations (for example, +-1g accelerations) produce results that are qualitatively very similar to the results produced by instantaneous velocity changes. For a specific numerical example with +-1g accelerations, see my webpage (that I referenced in my previous posting).

Mike Fontenot

 

[Mike_Fontenot]

Mike, it would be one thing if you were promoting your scheme as an easy way to calculate the final result of the ages of the twins after they reunite but that is not what you are doing.

My CADO equation gives the home twin's age for each instant of the traveler's life (according to the traveler). The result it gives, when the twins are reunited, agrees exactly with the home twin's conclusions about their ages when they are reunited. The CADO equation is not an easier way to get that particular piece of information, but it IS a very easy way to determine the home twin's current age at ANY instant during the traveler's voyage (according to the traveler) ... THAT is why the CADO equation is useful.

Mike Fontenot

 

[Mike_Fontenot]

You are claiming in your paper at the very end [in the abstract of my paper] that:

"Under a specific definition of 'meaningfulness' and 'reality', the bizarre behavior of the CADO, for an accelerating observer, must be regarded as being fully meaningful and real".

And by bizarre behavior, you mean clocks that go backwards. There is no frame of reference that will have any clocks going backwards.

There is no INERTIAL frame of reference that will conclude that any clock (whether in relative motion or not) is running backwards. But there IS a NON-INERTIAL frame of reference that DOES come to that conclusion ... it is the frame of the traveler, as calculated by the CADO equation.

This does not comport with any legitimate definition of reality.

Actually, it does. The results calculated from the CADO equation always agree with the traveler's own elementary measurements, and with his first-principle calculations. You can't get any more "real" than that. Any other proposed frame for the traveler will contradict the traveler's own measurements, and must be rejected.

Mike Fontenot

 

[ghwellsjr]

So you are claiming that the stay-at-home twin, who is not experiencing any acceleration, is experiencing a changing rate of aging, just because his twin, off in never-never land, is firing his rockets, correct?

 

[Dale]

Any other proposed frame for the traveler will contradict the traveler's own measurements, and must be rejected.

Mike, we have been over this many many many times before, it is getting rather tiresome. Any convention is equally acceptable, and there are some very good mathematical reasons for rejecting the CADO convention. You are certainly justified in using your convention if you like as long as you are aware of its mathematical problems and as long as you realize that it is merely a convention and not some "gospel truth".

Btw, you still have not addressed JesseM's point.

 

[Mike_Fontenot]

Personally I find the idea to have a clock aboard the spaceship that shows the spacetime distance between the departure event and the current event more useful. Aging would always be in the same direction (e.g the clock would never go back) regardless of movement or acceleration.

Ok ... Here's a challenge (for you, or for anyone else who proposes any frame for the traveler other than the one I have given):

Start with no acceleration at all: there are two inertial frames IF1 and IF2, moving at a relative velocity of 0.866, giving a gamma value of 2. Two pregnant mothers (one stationary in IF1 and the other stationary in IF2) each give birth at the instant they happen to be momentarily co-located. Call the IF1 baby Sue, and for convenience we will refer to her as "the home baby". Call the other baby, Tom, "the traveler", even though the situation is obviously initially completely symmetrical.

When the traveler is 20 years old, he instantaneously changes his velocity to -0.866 (so that he is moving back toward Sue). He continues at that velocity until he is 22 years old, and then instantaneously changes his velocity back to +0.866, and thereafter continues at that velocity.

According to your proposed frame for Tom, describe (in detail) the plot of Sue's age, versus Tom's age, according to Tom. Also, describe the plot of Sue's age, versus Tom's age, according to Sue.

Mike Fontenot

 

[Mike_Fontenot]

So you are claiming that the stay-at-home twin, who is not experiencing any acceleration, is experiencing a changing rate of aging, just because his twin, off in never-never land, is firing his rockets, correct?

No. The home twin generally comes to different conclusions about their corresponding ages (except when their separation is zero, or at instants when their relative velocity is zero). The home twin's perception, of the progression of her own life, is in no way influenced by any other person's motion.

Mike Fontenot

 

[Mike_Fontenot]

Btw, you still have not addressed JesseM's point.

Anyone who really wants to understand the elementary measurements and first-principle calculations that the traveler can perform, to confirm the results given by the CADO equation, will have to read my paper ... there just isn't any shortcut. The reference to the paper is:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.


Mike Fontenot

 

[Passionflower]

Anyone who really wants to understand the elementary measurements and first-principle calculations that the traveler can perform, to confirm the results given by the CADO equation, will have to read my paper ... there just isn't any shortcut.

It is not a matter of understanding. I do not believe anyone claims here that there is anything wrong with this method. It appears the issue is your claim that this is the only valid method.

 

[ghwellsjr]

No. The home twin generally comes to different conclusions about their corresponding ages (except when their separation is zero, or at instants when their relative velocity is zero). The home twin's perception, of the progression of her own life, is in no way influenced by any other person's motion.

Mike Fontenot

So you are not claiming that the stay-at-home twin is experiencing a changing rate of aging, correct? You are merely saying that when one person looks at another person's clock that is running at a different tick rate, it will appear that the slower clock is running backward. Is that what this is all about?

 

[Mike_Fontenot]

It is not a matter of understanding. I do not believe anyone claims here that there is anything wrong with this method. It appears the issue is your claim that this is the only valid method.

The CADO reference frame is the ONLY frame of the traveler that does NOT contradict the traveler's own elementary measurements and first-principle calculations. I contend that any frame of the traveler which contradicts those measurements and calculations is invalid.

If the traveler is a physicist, he simply couldn't do any physics if he had to ignore his own such measurements and calculations ... a traveling used-car salesman could probably tolerate that situation, but no true physicist could.

In order to understand my contention that the traveler's measurements and calculations (which confirm the CADO results) are "elementary" and based only on "first-principles", you've got to spend some quality time with my paper.

Mike Fontenot

 

[Mike_Fontenot]

You are merely saying that when one person looks at another person's clock that is running at a different tick rate, it will appear that the slower clock is running backward. Is that what this is all about?

It is misleading, and ambiguous, to use the terms "looks at" and "appears" as you did above. If the traveler looks at an image of the home twin (holding a sign which gives her current age at the instant the image was transmitted), he should understand that the age reported on the sign is NOT the current age of the home twin when the traveler receives that image ... because of the finite velocity of light, the home twin will have aged during the transit of the image. If the traveler CORRECTLY allows for the home twin's ageing during the transit of the image, he will obtain the CORRECT current age of the home twin, at the instant when he receives the image. That result is the same result given by the CADO equation. That result is NOT any kind of "apparition" ... it is as real as real can be.

Mike Fontenot

 

[Passionflower]

In order to understand my contention that the traveler's measurements and calculations (which confirm the CADO results) are "elementary" and based only on "first-principles", you've got to spend some quality time with my paper.

I am getting the impression that you are advertising this 'CADO thing' as if it is some trademarked toothpaste. Claiming that one only can understand it if one reads your paper, which btw is not freely available, I find nonsense.

If your motive to post here is to help people understand SR then please do so, however if you are only here to promote this paper I think you are misusing this forum.

 

[ghwellsjr]

It is misleading, and ambiguous, to use the terms "looks at" and "appears" as you did above. If the traveler looks at an image of the home twin (holding a sign which gives her current age at the instant the image was transmitted), he should understand that the age reported on the sign is NOT the current age of the home twin when the traveler receives that image ... because of the finite velocity of light, the home twin will have aged during the transit of the image. If the traveler CORRECTLY allows for the home twin's ageing during the transit of the image, he will obtain the CORRECT current age of the home twin, at the instant when he receives the image. That result is the same result given by the CADO equation. That result is NOT any kind of "apparition" ... it is as real as real can be.

Mike Fontenot

OK, but you didn't address my question. Is the reason why you can consider a clock to be going backward because you are comparing the time on two clocks that are running at different rates? The first clock will claim the second one is going backward in time simply because it is running at a slower rate and the second clock will claim the first one is going forward in time simply because it is running at a faster rate. Is that what this is all about?

 

[grav-universe]

The sudden difference in age that the traveller conceives of the stay at home twin, the instantaneous aging of the stay at home twin upon turning around and heading back to the stay at home twin, as well as the instantaneous reversing of age when the traveller turns away from the stay at home twin, is due to the simultaneity convention used within each frame of observers. In other words, it is purely a coordinate effect determined by how the clocks are set. Let's say Alice is the stay at home twin and Bob and Carl pass Alice in a ship at a speed of v, Bob at the front. When Bob passes Alice, they each synchronize their clocks to T=0, as well as all other clocks within their respective frames. That means that Carl synchronizes his own clock to Bob's as well, but if they are using the Einstein simultaneity convention, whereas Bob and Carl will measure the same speed of light traveling from Bob to Carl as from Carl to Bob, then Alice will say that Carl's clock is set to some time of tl greater than Bob's, even though Bob and Carl claim their clocks to be perfectly synchronized.

Now let's say that at the point where Carl reaches Alice, Bob and Carl instantly reverse the direction of their ship and Bob is now traveling at -v back toward Alice. However, upon doing this, Bob and Carl realize that their clocks are no longer synchronized. In order to re-synchronize their clocks to measure the same speed of light between them in either direction, once again providing the Einstein simultaneity convention, Carl must now either set his clock to -tl less than the reading of Bob's or Bob must set his own clock to greater than Carl's by tl. Since we are comparing clock readings between Alice and Bob, for when Bob returns to Alice, we would want to leave Bob's clock alone, so we set Carl's clock back by 2 tl, setting it from previously tl greater than Bob's clock to -tl less than Bob's clock, so that the clocks are now re-synchronized within their own frame. However, in the process, since Carl coincides with Alice at the point of turn-around, Carl will say that Alice's clock has now suddenly aged an additional 2 tl, although Carl really knows it was he that set his own clock back by 2 tl. Since Bob's clock is now re-synchronized to Carl's within the same frame, Bob will also say Alice has instantly aged by 2 tl, although he also knows better, that it is really just a coordinate effect, the result of setting clocks according to a simultaneity convention.

 

[Dale]

Anyone who really wants to understand the elementary measurements and first-principle calculations that the traveler can perform, to confirm the results given by the CADO equation, will have to read my paper ... there just isn't any shortcut

I am going to call BS on this. The level of complexity discussed on a routine basis in these forums is high enough that even a very technical definition should be readily understood. This is just an excuse to avoid defining your terms. JesseM is correct, the only way for you to make your claims is by defining "elementary measurements" and "first principle calculations" such that your claim is a tautology.

 

[Dale]

I do not believe anyone claims here that there is anything wrong with this method. It appears the issue is your claim that this is the only valid method.

Exactly. He preaches his CADO equation like the "one true gospel". Although, I can't tell if he is trying to patent it and sell license rights or if he just wants to drum up business for the journal.

 

[Dale]

I contend that any frame of the traveler which contradicts those measurements and calculations is invalid.

There is no coordinate system that will contradict any measurement that the traveler (or anyone else) will make. All frames using any synchronization convention will agree on the results of any measurement as long as you use the correct metric for the given coordinate system. Can you come up with even one counter-example?

That is such a basic and easy concept that there is no excuse for you to have not understood it after it has been so clearly pointed out to you on so many occasions by so many different people.

 

[PAllen]

It seems to me that the only objective statements you can make about the twin scenario (not paradox) are what is literally observed. Other statements are interpretations of observations with respect to various assumptions and models. So far as I understand, in the instant turnaround scenario, what the moving twin literally observes is as follows:

1) Before turnaround, they see a reddened image of the other twin's clock moving slowly, with rotation of the image as well.

2) After turnaround, they see a blue image of the clock moving faster (and rotated), until they meet the other twin. They see no jump whatsoever, in the time on the image of the clock. For instant turnaround, they do see a discontinuity in rate: image moving slower to image moving faster.

Anything you want to deduce about distance and what time (for the turning twin) an image corresponds to is based on model and is not a direct measurement.

I personally would attach no significance whatever to the Lorentz coordinate distance. To me, you should adopt some model of distance/time determination. For example:

1) Parallax: calculate, for two light rays emitted at infinitesimal angle to each other by the stationary twin, what angle they appear to have between each other by the turnaround twin, in relation to their separation. In the limit as angle goes to zero, you have a measure distance as perceived by a pair of 'ideal eyes'. I have never done such a calculation, but would definitely not assume it has anything to do with Lorentz coordiane distance. Then, time might (by assumption, not objective reality) be defined as this distance over c from signal receipt time.

2) Similarly for a normalized luminosity distance (factor out red/blue shift etc to estimate the apparent distance of a source instantaneously in the same position as the stationary twin but moving so that their is no red/blue shift for the turnaround twin).

3) Radar ranging distance.

I believe these could be different from each other, all defensible, and possibly none match Lorentz coordinate position, which would be the most meaningless.

 

[grav-universe]

My last post gives a basic idea about why the difference in the age of the stay at home twin would occur according to the traveller upon turn-around, but it does not give an accurate measure for the difference in age that Bob actually measures for Alice, complicated by the fact that while Carl reaches Alice simultaneously with Bob turning around from Alice's point of view, Carl has not reached Alice yet when Bob turns around according to the frame of Bob and Carl, so I will go ahead and work it through more thoroughly.

In this scenario, a space station is at a distance of L from Alice as she measures it and is stationary to her. Bob passes Alice at a speed of v1, travels to the space station, then instantly turns around and travels back to Alice at v2. According to Alice, then, the time it takes for Bob to reach the station is t1 = L / v1 and t2 = - L / v2 to return (with negative v2), so Alice's age when Bob travels back to Alice is t1 + t2 = L (1 / v1 - 1 / v2).

According to Bob's perspective, when Alice passes Bob, Alice and the station are both moving at v1, so the distance between them is sqrt[1 - (v1 / c)^2] L, and the station reaches Bob in a time of t1' = sqrt[1 - (v1 / c)^2] L / v1, whereas with time dilation, Bob says a time of (1 - (v1 / c)^2) L / v1 has passed for Alice. Then Bob turns around so that Alice and the station are now traveling at v2 in the opposite direction, Bob still coinciding with the station and now measuring sqrt[1 - (v2 / c)^2] L for the distance to Alice. Alice meets back up with Bob in a time of t2' = - sqrt[1 - (v2 / c)^2] L / v2, so with time dilation, Bob says a time of - (1 - (v2 / c)^2) L / v2 has passed for Alice. The total time that Bob says has passed for Alice away and back, then, is L [(1 - (v1 / c)^2) / v1 - (1 - (v2 / c)^2) / v2].

Bob and Alice, of course, must both agree upon Alice's age when they meet back up, so the discrepency between what Alice and Bob measure for Alice during the journey away and back is the simultaneity shift of Alice's age that Bob conceptualizes has passed during his turn-around. The simultaneity shift is

ΔtA

= L (1 / v1 - 1 / v2) - L [(1 - (v1 / c)^2) / v1 - (1 - (v2 / c)^2) / v2]

= L [(v1 / c)^2 / v1 - (v2 / c)^2 / v2]

= L (v1 - v2) / c^2

= L Δv / c^2

 

[ghwellsjr]

We do know that whenever any observer, clock, object, or any other thing experiences acceleration and there is a changing speed, that thing experiences a real, physical change in the rate its clocks tick and the rate it ages and it experiences a real, physical change in dimension along the direction of acceleration.

 

[Mentz114]

We do know that whenever any observer, clock, object, or any other thing experiences acceleration and there is a changing speed, that thing experiences a real, physical change in the rate its clocks tick and the rate it ages and it experiences a real, physical change in dimension along the direction of acceleration.

I understood that the 'ideal clock' of special relativity does not change its tick rate when accelerated.

 

[ghwellsjr]

Of course every clock changes its tick rate, ideal or not, when it is accelerated, you just can't tell that it is changing because you are constantly reassigning a new frame of reference for it as you go along. But if you analyze an accelerating clock from any single frame of reference, you must conclude that it is changing its tick rate.

 

[Mentz114]

Of course every clock changes its tick rate, ideal or not, when it is accelerated, you just can't tell that it is changing because you are constantly reassigning a new frame of reference for it as you go along. But if you analyze an accelerating clock from any single frame of reference, you must conclude that it is changing its tick rate.

So it's not

... a real, physical change in the rate its clocks tick ...

but frame dependent ?

 

[PAllen]

. But if you analyze an accelerating clock from any single frame of reference, you must conclude that it is changing its tick rate.

You mean inertial frame. There is no requirement in SR to use inertial frames. Why not use an accelerated frame for an accelerated observer? Then this statement is false.