A New Simultaneity Method for Accelerated Observers in Special Relativity

[Mike_Fontenot]

Until recently, I've been an avid proponent of the "co-moving inertial frames" (CMIF) simultaneity method (previously called the "CADO" method by me). I had claimed to have proven that the CMIF method is the ONLY method that agrees with the accelerated observer's own elementary observations and elementary calculations. But I recently concluded that there was a loophole in that proof, and therefore I had failed to prove what I thought I had proven. I decided to take a fresh look at the whole issue of simultaneity for an accelerated observer. In the course of doing that, I discovered a new simultaneity method that shows, with a very simple proof, that the CMIF method isn't correct. My new method says that when the accelerating observer instantaneously changes his velocity, the current age of the home twin DOESN'T instantaneously change. Instead, the slope of the age correspondence curve instantaneously changes its slope from a constant less than one to a constant greater than one. And then after a well-defined passage of time, the slope instantaneously switches back to the same constant less than one that occurs in the first segment. So the "curve" in the age correspondence diagram is always a continuous, piecewise-linear line of three straight line segments. Unlike the Dolby and Gull simultaneity method, and the Minguzzi simultaneity method, my method is causal, i.e., effects are always PRECEDED by causes. My new method is explained in detail on my webpage referenced below (in front of the old information on my webpage, which I now know to be incorrect).

Michael Leon Fontenot

--
https://sites.google.com/site/cadoequation/cado-reference-frame

All you ever need to know about the twin "paradox".

 

[Neddy Bate]

Until recently, I've been an avid proponent of the "co-moving inertial frames" (CMIF) simultaneity method (previously called the "CADO" method by me). I had claimed to have proven that the CMIF method is the ONLY method that agrees with the accelerated observer's own elementary observations and elementary calculations. But I recently concluded that there was a loophole in that proof, and therefore I had failed to prove what I thought I had proven. I decided to take a fresh look at the whole issue of simultaneity for an accelerated observer. In the course of doing that, I discovered a new simultaneity method that shows, with a very simple proof, that the CMIF method isn't correct. My new method says that when the accelerating observer instantaneously changes his velocity, the current age of the home twin DOESN'T instantaneously change. Instead, the slope of the age correspondence curve instantaneously changes its slope from a constant less than one to a constant greater than one. And then after a well-defined passage of time, the slope instantaneously switches back to the same constant less than one that occurs in the first segment. So the "curve" in the age correspondence diagram is always a continuous, piecewise-linear line of three straight line segments. Unlike the Dolby and Gull simultaneity method, and the Minguzzi simultaneity method, my method is causal, i.e., effects are always PRECEDED by causes. My new method is explained in detail on my webpage referenced below (in front of the old information on my webpage, which I now know to be incorrect).

Michael Leon Fontenot

--
https://sites.google.com/site/cadoequation/cado-reference-frame

All you ever need to know about the twin "paradox".

Mike,

I am glad to see that you are open minded enough to change your whole webpage based on this new insight. However, unfortunately I was not able to understand.

Here is the basic "twin paradox" with an outbound leg of uniform motion, a very brief turnaround, and an inbound leg of uniform motion, with gamma=2.

ACCORDING TO THE STAY-HOME TWIN:
t=0 t'=0
t=20 t'=10
t=40 t'=20
(Very brief turnaround)
t=40 t'=20
t=60 t'=30
t=80 t'=40
(Twins re-united, both agree on their ages)

ACCORDING TO THE TRAVELING-TWIN:
t'=0, t=0
t'=20 t=10
(Very brief turnaround)
t'=20 t=70
t'=40 t=80
(Twins re-united, both agree on their ages)
As you can see, this method has her age going from t=10 to t=70 (according to him) during his turnaround. What does your new method say?

Please use this example, and show me your new method, by filling in the blank "????"
ACCORDING TO THE TRAVELING-TWIN:
t'=0, t=0
t'=20 t=10
(Very brief turnaround)
t'=20 t=????
t'=40 t=80
(Twins re-united, both agree on their ages)

If ???? is not 70, then what is it?

 

[Mike_Fontenot]

Please use this example, and show me your new method, by filling in the blank "????"
ACCORDING TO THE TRAVELING-TWIN:
t'=0, t=0
t'=20 t=10
(Very brief turnaround)
t'=20 t=????
t'=40 t=80
(Twins re-united, both agree on their ages)

If ???? is not 70, then what is it?

???? = 10. In my new simultaneity method, there is NO change in the home twin's age immediately before and immediately after the traveler's instantaneous velocity change. THEN, he says that her age starts to increase at a constant linear rate greater than his rate of ageing, and at a later time, she again ages at half his rate until their reunion.

 

[Neddy Bate]

???? = 10. In my new simultaneity method, there is NO change in the home twin's age immediately before and immediately after the traveler's instantaneous velocity change. THEN, he says that her age starts to increase at a constant linear rate greater than his rate of ageing, and at a later time, she again ages at half his rate until their reunion.

Hmmm. That seems very problematic to me. The return trip is an inertial frame, so the traveling twin should not be treated any differently than the stay at home twin. To claim only one frame is entitled to use the time dilation formula to calculate the tick rate in the other frame is tantamount to claiming a preferred frame.

You claim your new method has to do with what each twin sees with their eyes, (or television broadcasts sent & received), so I would challenge you to start the scenario right after the turnaround, and to forget which twin is which. In fact you should be able to start the scenario there without even knowing that the twins started off at the same age at the beginning. Just put a random time on each one's clock right after the turnaround, and go from there.

In that case, how are you supposed to discern which frame is supposed to get your new treatment, and which frame is supposed to get the traditional treatment? If that information is not inherent in this perfectly symmetrical arrangement, then you simply cannot justify doing it. And I don't see how that information could possibly be inherent in this situation. Both inertial frames are equally valid, and the laws of physics should be the same in both frames, as per Einstein.

 

[Neddy Bate]

I would challenge you to start the scenario right after the turnaround, and to forget which twin is which. In fact you should be able to start the scenario there without even knowing that the twins started off at the same age at the beginning. Just put a random time on each one's clock right after the turnaround, and go from there.

I will attempt to take my own challenge using gamma=2. Let's say that neither twin knows much about the situation until after they are reunited. At that time, let's say that both of their clocks happen to display 100 by complete coincidence, (because the clocks were reset to random times that happen to match that result).

ACCORDING TO THE STAY-HOME TWIN:
(Very brief turnaround)
t=60 t'=80
t=80 t'=90
t=100 t'=100
(Twins re-united, both agree on their ages)

ACCORDING TO THE TRAVELING-TWIN:
(Very brief turnaround)
t'=80 t=90
t'=100 t=100
(Twins re-united, both agree on their ages)

Important things to notice:

1. Both twins say that the other twin's clock ticked at a rate of 1/2 their own, as they must because of the symmetry, and because all inertial frames are equally valid.

2. Both twins agree on the end result, as they must, because that is an event.

3. Both twins agree that the traveling twin's clock displayed t'=80 right after the very brief turnaround, as they must, because that is also an event.

4. The twins do not agree on what time was displayed by the home twin's clock right after the very brief turnaround, because simultaneity at a distance is frame-dependent. (Note that the home twin says her clock displayed t=60 at that time, but the traveling twin says her clock must have displayed t=90 at that time.

5. If you try to say that #4 can be false, (by saying that the twins can agree on what time was displayed by the home twin's clock right after the very brief turnaround), then #1 will not always be true.

 

[Mike_Fontenot]

Hmmm. That seems very problematic to me. The return trip is an inertial frame, so the traveling twin should not be treated any differently than the stay at home twin. To claim only one frame is entitled to use the time dilation formula to calculate the tick rate in the other frame is tantamount to claiming a preferred frame.

That's one of the startling results from mu new method: there can be LONG periods of time after the traveler instantaneously changes his velocity in which a perpetually-inertial observer sitting right next to the traveler will disagree with the traveler! And there is no inconsistency in that!

You claim your new method has to do with what each twin sees with their eyes, (or television broadcasts sent & received), [...]

My method is based on the traveler determining by how much the home twin ages from the time she transmits a special light pulse to him, until he receives that pulse.

My method is not trivial to understand. Unless and until you spend quite a bit of time working through my paper, it won't make sense to you. However, the proof that she doesn't age at all during his instantaneous velocity change is quite simple ... it occurs near the end of section 7. It is a very simple proof.

I suggest that if you are not willing to spend quite a bit of time studying my paper, you are better off ignoring it, instead of guessing what's in it.

 

[Neddy Bate]

That's one of the startling results from mu new method: there can be LONG periods of time after the traveler instantaneously changes his velocity in which a perpetually-inertial observer sitting right next to the traveler will disagree with the traveler! And there is no inconsistency in that!

There is plenty of inconsistency in that. You are saying that you and I can both be sitting side by side with no relative motion between us, and you can be correct in saying that a distant clock is ticking faster than our own clock, while I can also be correct in saying that same distant clock is ticking slower than our own clock.

My method is based on the traveler determining by how much the home twin ages from the time she transmits a special light pulse to him, until he receives that pulse.

My method is not trivial to understand. Unless and until you spend quite a bit of time working through my paper, it won't make sense to you. However, the proof that she doesn't age at all during his instantaneous velocity change is quite simple ... it occurs near the end of section 7. It is a very simple proof.

I suggest that if you are not willing to spend quite a bit of time studying my paper, you are better off ignoring it, instead of guessing what's in it.

Fair enough, but I should not have to study the paper just to get an answer to this important question, from post #4:

"In that case, how are you supposed to discern which frame is supposed to get your new treatment, and which frame is supposed to get the traditional treatment?"

If you have a satisfying answer to that, I might be more inclined to study your new method. But so far it seems to be just as bad as those other approaches that you and I both thought were way off track. SR does not need 'fixing' in this sense.

 

[Mike_Fontenot]

There is plenty of inconsistency in that. You are saying that you and I can both be sitting side by side with no relative motion between us, and you can be correct in saying that a distant clock is ticking faster than our own clock, while I can also be correct in saying that same distant clock is ticking slower than our own clock.

Yes! That is exactly what I'm saying. For some period of time after the traveling twin changes his velocity, he will NOT agree with a perpetually-inertial observer who is co-located and now stationary with him. During that period, the amount of their disagreement will slowly decrease. Eventually (at a well defined time), they will completely agree with each other thereafter (unless and until the traveler decides to instantaneously change his velocity again).

I have always been very impressed with your command of the Lorentz equations. But one deficiency in your understanding of the Lorentz equations is how they can be PLOTTED to form the Minkowski diagram ... i.e., the Minkowski diagram is just a graphical representation of the Lorentz equations, and it is often VERY useful. Since my new method is DEFINED via a graphical procedure using the Minkowski diagram, my method cannot be understood without an understanding of Minkowski diagrams. However, my simple proof that the CMIF simultaneity is incorrect requires only a minimal understanding of Minkowski diagrams. I encourage you to spend some time time developing at least that minimal understanding of Minkowski diagrams. One cavi0t: I draw them differently than most people do. I plot the home twin's age on the horizontal axis, and the distance of any specified object (or person) from the home twin on the vertical axis, whereas most people do the opposite. My paper assumes that the reader has only a minimal knowledge of Minkowski diagrams, and I walk the reader through the process of constructing the diagrams that are needed to understand my method. You should do that at least for the initial diagram that I describe, which is needed to understand my simple proof that the CMIF method can't be correct.

Fair enough, but I should not have to study the paper just to get an answer to this important question, from post #4:

"In that case, how are you supposed to discern which frame is supposed to get your new treatment, and which frame is supposed to get the traditional treatment?"

If you have a satisfying answer to that, I might be more inclined to study your new method.

My method specifies how the traveling twin (who instantaneously changes his velocity ONCE during his trip) can determine (at each instant of his life during his trip) the current age of the home twin (according to HIM, not according to HER). THAT is the objective. There is never any disagreement among physicists about the home twin's perspective, or about the perspective of any other perpetually-inertial observer. But there is PLENTY of disagreement among physicists about the accelerating twin's perspective. Some physicists believe that simultaneity at a distance is a meaningless concept for an observer who ever accelerates while separated from his home twin. Physicists who DON'T believe that simultaneity is meaningless most often adopt the CMIF simultaneity method (as long as they don't have to think about the fact that it implies that the home twin can get YOUNGER according to the accelerating twin). At least a couple of physicists (Dolby and Gull, and Minguzzi) have proposed other simultaneity methods. My new method shows that the CMIF method, and the Dolby and Gull and Minguzzi methods, are all incorrect.

 

[Neddy Bate]

Yes! That is exactly what I'm saying. For some period of time after the traveling twin changes his velocity, he will NOT agree with a perpetually-inertial observer who is co-located and now stationary with him. During that period, the amount of their disagreement will slowly decrease. Eventually (at a well defined time), they will completely agree with each other thereafter (unless and until the traveler decides to instantaneously change his velocity again).

If that is what you are saying, then you are saying that you think SR has some nonsense in it. Two people in the same location with no relative velocity between them cannot both be correct if one person says a distant clock ticks slower than their own clock, and the other person says a distant clock ticks faster than their own clock. That is not SR.

So that nonsense is not in SR at all. It is either something you made up, or something you think you derived without realizing that you have made an error someplace.

I have always been very impressed with your command of the Lorentz equations. But one deficiency in your understanding of the Lorentz equations is how they can be PLOTTED to form the Minkowski diagram ... i.e., the Minkowski diagram is just a graphical representation of the Lorentz equations, and it is often VERY useful.

Thank you. I am familiar with both the Lorentz equations and Minkowski diagrams. Since a Minkowski diagram is nothing more than a graphical representation of the Lorentz equations, it cannot contradict the equations. But there is nothing in the equations that says anything like what you are saying here.

Since my new method is DEFINED via a graphical procedure using the Minkowski diagram, my method cannot be understood without an understanding of Minkowski diagrams. However, my simple proof that the CMIF simultaneity is incorrect requires only a minimal understanding of Minkowski diagrams. I encourage you to spend some time time developing at least that minimal understanding of Minkowski diagrams. One cavi0t: I draw them differently than most people do. I plot the home twin's age on the horizontal axis, and the distance of any specified object (or person) from the home twin on the vertical axis, whereas most people do the opposite. My paper assumes that the reader has only a minimal knowledge of Minkowski diagrams, and I walk the reader through the process of constructing the diagrams that are needed to understand my method. You should do that at least for the initial diagram that I describe, which is needed to understand my simple proof that the CMIF method can't be correct.

I have more than a minimal understanding of Minkowski diagrams. Here is one I made and posted in a previous thread of yours:

Note that there are two gray lines of simultaneity, one horizontal for the stay-home frame (ground frame), and one diagonal for the traveling frame (or train frame). The slopes of these lines are determined in a specific way, and they have specific meanings which cannot be changed.

The meanings are as follows:
1. In the moment before Charlie jumps off the train, he uses the diagonal line of simultaneity and says Alice is 10 years old.
2. After he lands on the ground he uses the horizontal line of simultaneity and says she is 40 years old.
3. Then if he immediately jumps back on the train he uses the diagonal line of simultaneity again, and says Alice is 10 years old again.

There is no "new method" that says that that Charlie can think Alice remains 40 after he jumps back on the train, (or whatever else your new method would say about this). Charlie is not supposed to use the line of simultaneity of the ground when he himself is in the train frame.

My method specifies how the traveling twin (who instantaneously changes his velocity ONCE during his trip) can determine (at each instant of his life during his trip) the current age of the home twin (according to HIM, not according to HER). THAT is the objective. There is never any disagreement among physicists about the home twin's perspective, or about the perspective of any other perpetually-inertial observer. But there is PLENTY of disagreement among physicists about the accelerating twin's perspective. Some physicists believe that simultaneity at a distance is a meaningless concept for an observer who ever accelerates while separated from his home twin. Physicists who DON'T believe that simultaneity is meaningless most often adopt the CMIF simultaneity method (as long as they don't have to think about the fact that it implies that the home twin can get YOUNGER according to the accelerating twin). At least a couple of physicists (Dolby and Gull, and Minguzzi) have proposed other simultaneity methods. My new method shows that the CMIF method, and the Dolby and Gull and Minguzzi methods, are all incorrect.

I asked you to address a new thought experiment that begins immediately after the turnaround point. Forget which twin is which, and simply look that symmetrical situation. How do you determine which frame gets to use the time dilation equation and which frame doesn't get to use it? You can't, because the situation is symmetrical. So your method cannot even be applied to this situation. You have to allow both frames to use the time dilation equation.

There is nothing in SR that says anything like what you are saying here. Sorry but you must have made an error someplace.

 

[Mike_Fontenot]

If that is what you are saying, then you are saying that you think SR has some nonsense in it. Two people in the same location with no relative velocity between them cannot both be correct if one person says a distant clock ticks slower than their own clock, and the other person says a distant clock ticks faster than their own clock. That is not SR.

I disagree. There is no nonsense in SR. Two people co-located for a substantial amount of time (years) CAN both be correct in their conclusions about the current age of the home twin. One of those two people has never accelerated, and the other HAS accelerated in his recent past. That difference causes them to come to different conclusions about the home twin's current age.

I asked you to address a new thought experiment that begins immediately after the turnaround point. Forget which twin is which, and simply look that symmetrical situation.
How do you determine which frame gets to use the time dilation equation and which frame doesn't get to use it?

The two twins are certainly NOT symmetrical. One has NEVER accelerated, and the other one HAS. The one who has never accelerated gets to use the time dilation equation. The one who HAS accelerated recently CAN'T use the time-dilation equation for a while. The point in the traveler's life when he can again use the time-dilation equation is determined by a very special light pulse that is transmitted by the home twin and received by the traveler. When the traveler receives that pulse, he can start using the time dilation equation again. This is explained more fully in my paper. Read the paper.

 

[Neddy Bate]

I disagree. There is no nonsense in SR. Two people co-located for a substantial amount of time (years) CAN both be correct in their conclusions about the current age of the home twin. One of those two people has never accelerated, and the other HAS accelerated in his recent past. That difference causes them to come to different conclusions about the home twin's current age.

What you call "that difference" is not really a physical difference at all. If you did not have the knowledge of which person accelerated in the past, you would not be able to tell which one was which, by any experiment in the present. The physics and kinematics of both are identical. To claim that only one can apply the Lorentz equations, and the other cannot, is a departure from what SR says.

The two twins are certainly NOT symmetrical. One has NEVER accelerated, and the other one HAS. The one who has never accelerated gets to use the time dilation equation. The one who HAS accelerated recently CAN'T use the time-dilation equation for a while. The point in the traveler's life when he can again use the time-dilation equation is determined by a very special light pulse that is transmitted by the home twin and received by the traveler. When the traveler receives that pulse, he can start using the time dilation equation again. This is explained more fully in my paper. Read the paper.

Okay so you are refusing to acknowledge that if you start your thought experiment right after the turnaround, and you don't know which twin is which, then you cannot even determine which twin should use which equation. Fine, let's skip that, since it obviously makes applying your idea impossible.

Let's try something different. Here is my Minkowski diagram again:

1. I assume you agree that before Charlie jumps off the train he can say that Alice is 10 years old. Please correct me if you do not agree.

2. I also assume you agree that right after Charlie jumps off the train and lands on the ground, he can say that Alice is 40 years old. They are both in the same frame at that point, so there should be no problem there. But maybe you think Charlie should still say she is 10, and then later let her age faster? Even though they are in the same frame as each other? See the nonsense yet?

3. And I assume that right after Charlie jumps back on the train, you do not agree that he can say that Alice is 10 years old again. If you do, then your new idea accomplishes nothing, because I assume your goal was to avoid the idea that her age could go from 40 to 10 according to Charlie. So I assume you think when he jumps back on the train, he still thinks she is 40 right? And Charlie's assistant, who never jumped off the train, still thinks Alice is 10. So this is where you have Charlie and his assistant, both in the same frame, right next to each other, disagreeing on Alice's age. See the nonsense yet?

 

[Mike_Fontenot]

We're going to have to agree to disagree. Over and out.

 

[Neddy Bate]

Okay, then.

I think that means you believe that when Charlie jumps off the train, and lands on the ground, he still thinks Alice is 10, even though he is at rest in her own frame, where everyone else including herself agrees she is 40. After all, that would be consistent with what you wrote in post #3. And you don't see any problem with that, because you know his acceleration history is different from everyone else's acceleration histories, which you also know, because you kept track of everyone's acceleration history. And worse, you think that is what SR says.

Best of luck with your new theory.

 

[Mike_Fontenot]

Okay, then.

I think that means you believe that when Charlie jumps off the train, and lands on the ground, he still thinks Alice is 10, even though he is at rest in her own frame, where everyone else including herself agrees she is 40. [...] And worse, you think that is what SR says.

I haven't kept track of your Charlie scenario, but I think it is equivalent to me saying that if the traveler (he) instantaneously changes his velocity at the turnpoint from +0.866 to ZERO, he will say that that doesn't instantaneously change the home twin's (her) age AT ALL. But her age WILL immediately begin to increase linearly. You can get the slope of the age correspondence diagram analytically, using the equation near the end of Section 10. Or you can easily get that slope graphically, as described in the paper.

If you do nothing else, you should read the short proof I give that her age doesn't change at all during his instantaneous velocity change. It's very near the end of Section 7. It's a very simple proof.

 

[Neddy Bate]

I haven't kept track of your Charlie scenario, but I think it is equivalent to me saying that if the traveler (he) instantaneously changes his velocity at the turnpoint from +0.866 to ZERO, he will say that that doesn't instantaneously change the home twin's (her) age AT ALL. But her age WILL immediately begin to increase linearly. You can get the slope of the age correspondence diagram analytically, using the equation near the end of Section 10. Or you can easily get that slope graphically, as described in the paper.

You are not even using the most basic premises of SR, and as such, you are no longer doing SR at all.

It is one of the most basic premises of SR that two identical clocks at rest in the same reference frame tick at the same rate. This is easily derived from the premise that the speed of light is the same constant in all inertial frames. Simply consider a light-clock made from two mirrors a known distance apart, with a pulse of light bouncing between them. When two such identical clocks are at rest relative to each other, we conclude that they tick at the same rate. From this we can conclude that if the traveling twin changes his velocity at the turnpoint from +0.866 to zero, then after that moment, the home twin's clock and the traveling twin's clock will run at the same rate. There is nothing in SR about two identical clocks at rest in the same frame running at different rates, as it would violate the constancy of light speed if light-clocks are used.

Furthermore, it is one of the most basic premises of SR that two identical inertial clocks with uniform relative motion between them will each say that the other clock ticks at a rate of 1/gamma their own rate. This is easily derived from the premise that the speed of light is the same constant in all inertial frames. Simply consider a vertically oriented light-clock made from two mirrors a known distance apart, with a pulse of light bouncing up and down between them. When such a clock has a uniform relative motion, the light obviously follows a diagonal path instead of up and down, and we can use the Pythagorean theorem to derive the slowed rate that the "other" clock has. Note that this time dilation effect is 100% reciprocal, meaning that each clock will consider the "other" clock to be the one that is ticking slower, and we do not even need to know which one (or both) might have accelerated. Note that I am talking about the RATE of the clocks, not any particular time being displayed.

I will call this the premise of reciprocal time dilation rates, as I will refer to it again later.

If you do nothing else, you should read the short proof I give that her age doesn't change at all during his instantaneous velocity change. It's very near the end of Section 7. It's a very simple proof.

Thank you for pointing me to this, as now I have some idea where you went wrong. If the traveling twin (he) wants to calculate her current age from the special image pulse that she sends to him, he will need two things. First he will need to know the distance between the emission point and the reception point. Note that this is not based on her current location relative to him, but rather where he reckons she was located in the past when she sent the image. Remember, he considers her to be moving and himself to be still.

Once he knows that distance, he cannot simply add the transit time to the time he sees on her clock in her image pulse. He must consider what rate her clock has compared to his, and add in the amount of time that would have elapsed on her clock during the transit of the image. Luckily he knows her clock's rate from the premise of reciprocal time dilation rates. In fact, since the premise of reciprocal time dilation rates is a basic premise of SR, he knows what her clock's rate will be for the entire return leg of the journey. And since the journey ends with both twins located in the same place, he can take her age at the end, and work backwards to the moment just after the turnaround point, and figure out her age from the known rate he got from the premise of reciprocal time dilation rates. Note that this makes calculating her age from the image pulse unnecessary or redundant. At the very least, calculating her age from the image pulse must give the same answer as working backwards from the end of the journey, and both methods use the known clock rate from the premise of reciprocal time dilation rates.

You say, "The amount that she aged during the transit of that pulse (according to him) can’t have changed: it was (and forever is) an historical fact for him," but that is not entirely correct. Consider that the first thing he needs in order to calculate her current age from the special image pulse is the distance between the emission point and the reception point. Note that this is not based on her current location relative to him, but rather the location where he reckons she was located in the past when she sent the image. Remember, he considers her to be moving and himself to be still. Just before the turnaround point, he considers her to be moving away from him, so he would say that the location where she emitted the image (in the past) was closer to him than where she is currently located. And just after the turnaround point, he considers her to be moving toward him, so he would say that the location where she emitted the image (in the past) was when she was farther away than where she is currently located. This is how the same image pulse (just before/ just after) the turnaround point can result in two different calculations for her current age.

But again, calculating her age from the image pulse is not even necessary. At the very least, calculating her age from the image pulse must give the same answer as working backwards from the end of the journey using the known clock rate from the premise of reciprocal time dilation rates.

 

[Neddy Bate]

Earlier you urged me to use Minkowski diagrams, so I am posing one that I made for you:

I think these were the figures that you had on your CADO page before you updated it, or at least these were the figures we discussed in a different thread. Note that the lines of simultaneity of the traveling twin change at the turnaround point. Just before the turnaround the line of simultaneity of the traveling twin points to t=10, and just after the turnaround the line of simultaneity of the traveling twin points to t=70.

Note also that the RATE that he considers her clock to be ticking is exactly 1/gamma of the rate of his own clock, as must be the case do to the premise of reciprocal time dilation rates. This holds true for the entire outbound leg of the journey where she ages 10 years to his 20, and the entire inbound leg of the journey where she again ages 10 years to his 20.

And of course the rate that she considers his clock to be ticking is exactly 1/gamma of the rate of her own clock, as must be the case do to the premise of reciprocal time dilation rates. And of course this holds true for the entire outbound leg of the journey where he ages 20 years to her 40, and the entire inbound leg of the journey where he again ages 20 years to her 40. Her lines of simultaneity are not shown on the diagram, but they are simple horizontal lines.

 

[Mike_Fontenot]

It is one of the most basic premises of SR that two identical clocks at rest in the same reference frame tick at the same rate.

I will call this the premise of reciprocal time dilation rates, as I will refer to it again later.

Two identical clocks that have ALWAYS been inertial and which have always had zero relative velocity WILL always tick at the same rate. But it is NOT true for one perpetually inertial clock and one clock which has accelerated in the (recently-enough) past.

Point out the exact sentence in my proof which you think is incorrect.

 

[Neddy Bate]

Two identical clocks that have ALWAYS been inertial and which have always had zero relative velocity WILL always tick at the same rate. But it is NOT true for one perpetually inertial clock and one clock which has accelerated in the (recently-enough) past.

I just gave you a proof using light clocks. Two identical inertial light clocks at rest in the same reference frame must tick at the same rate, or else the speed of light inside one of them would have to be different than the speed of light inside the other one.

https://en.wikipedia.org/wiki/Time_dilation

Excerpt:
"Consider then, a simple clock consisting of two mirrors A and B, between which a light pulse is bouncing. The separation of the mirrors is L and the clock ticks once each time the light pulse hits either of the mirrors. In the frame in which the clock is at rest (diagram on the left), the light pulse traces out a path of length 2L and the period of the clock is 2L divided by the speed of light:"

My bold.

Point out the exact sentence in my proof which you think is incorrect.

Your sentence, "The amount that she aged during the transit of that pulse (according to him) can’t have changed: it was (and forever is) an historical fact for him," is not correct. As I explained, in order to calculate the amount that she aged during transit, you would need to already know the rate of her clock relative to his, which is the very thing you are trying to calculate. Plus, before the turnaround, the distance between emitter and receiver is reckoned to be different than it is after the turnaround.

Also, you appear to be using the same line of simultaneity before and after the turnaround, which you cannot do in a Minkowski diagram. If so then that is a mistake.

I have just completed a Minkowski diagram based on your instructions, except without the errors you made. I will post it shortly.

 

[Mike_Fontenot]

I have just completed a Minkowski diagram based on your instructions, except without the errors you made. I will post it shortly.

Also, post a Minkowski diagram based on my instructions, WITH the errors you think I have made.

 

[Neddy Bate]

Okay here it is:

The light message contains the time on her clock, which is t=5.36, and that is the message that will be received at the turnaround.

Before turnaround, he reckons the distance between emitter and receiver was 9.28 in his frame. But he cannot simply add 9.28 to 5.36 because that would assume that her clock runs at the same rate as his. He has to ALREADY KNOW that her clock ticks at half the rate of his (according to him), so that he can add (9.28/2) to 5.36 and that is how he calculates 10.

After turnaround, he reckons the distance between emitter and receiver was 129.28 in his frame (this huge distance is off the bottom right edge of the image, where those two lines would intersect). But he cannot simply add 129.28 to 5.36 because that would assume that her clock runs at the same rate as his. He has to ALREADY KNOW that her clock ticks at half the rate of his (according to him), so that he can add (129.28/2) to 5.36 and that is how he calculates 70.