# A New Simultaneity Method for Accelerated Observers in Special Relativity

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

1. ### phytiRegistered Senior Member

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Mike;

Apparently you still aren't convinced that there is no instantaneous knowledge of distant events.

In the 1905 paper, simultaneity par.1, the time of an event is indicated by a clock near the event. The awareness of the event E by a distant observer is always later, due to light transmission time. Thus the distant observer must know the distance to the event E to know when E occurred.

Another graphic for the long running saga, 'what did she know and when did she know it'.
On the left:
Ann stays home. Bill flies away at .6c, reverses direction (instantly) and returns at .8c.
The red lines indicate time dilation relative to the A-clock.
Bill sends light signals (blue) to Ann at very short intervals to get clock readings.
Because B changes direction at At=10 with a corresponding jump of his axis of simultaneity, it is easiest to view the upper part rotated 180° and run the film backward.
On the right:
While outbound, the last reading is At=4, which Bill assigns to Bt=5, thus calculating td of 4/5=.80, in agreement with SR.
While inbound the next reading is At=16, which Bill assigns to Bt=10, thus calculating td of 1.5/2.5=.60, in agreement with SR.
Bill observes (analyzes images, and calculates) that the A-clock rate increases from Bt=5 to 10, then decreases from Bt=10 to 12.5. He calculates the speed of A for the middle segment as .2c. The physical cause for the apparent behavior of the A-clock is, sending a signal while moving in one direction and receiving the return signal while moving in a different direction, the same scenario for curved or accelerated motion.
Since the initial conditions require constant clock rates for both, the strange behavior of the A-clock results from Bill's altered perception.
Motion of the observer cannot alter distant processes, but can alter his perception.
Notice:
Bill does not know that At=4 when Bt=5 until Bt=11.
Bill does not know that At=16 when Bt=10 until Bt=12.
No instant knowledge.

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3. ### Mike_FontenotRegistered Senior Member

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If the actions of the traveler (he) are unknown, the home twin (she) can't have any instant knowledge of his current age. But if they each know what the plan is, and assuming the plan is followed, they ARE able to correctly calculate each other's current ages during the trip, at least in the simplest case. The simplest case is when both people are perpetually inertial and moving at a constant relative velocity (say 0.866 ly/y, with gamma 2.0), whose mothers have always had that velocity difference (and their ancestors before them), with their babies bring born at the instant they were momentarily co-located. In that case, we know that each of those babies should conclude that the other is always ageing at a rate half as fast as they themselves are, and for a constant relative velocity, the other is always half as old as they themselves are. We (and they) know that because Einstein's time dilation equation tells us (and them) that, and Einstein was correct.

In more complicated scenarios, we can still determine what their two conclusions are about their current ages, assuming that the simultaneity method we choose to use is correct. If the home twin remains perpetually inertial, we DO know that her conclusion about the current age of the traveler is correct ... she can continue to use the more general time dilation equation, which says that the traveler's rate of ageing will be slower than hers, by the gamma factor. But he can't use the time dialation equation. He has to choose one of the four simultaneity methods for accelerated observers, and each method gives a different answer. He might decide to reject two of them (Dolbe and Gull's "Radar method", and Minguizzi's "fictitious twin" method) because they are both non-causal. That leaves the CMIF (Co-Moving Inertial Frames) simultaneity method, or else my simultaneity method. I prefer the CMIF method, but some people dislike it because it says that in some circumstances, the home twin can get younger, according to the traveling twin. Those people may prefer my method, which has no discontinuities and no negative ageing of the home twin. Which of them (CMIF or mine) is correct (if either) is unknown. I believe that there IS a correct simultaneity method (for philosophical reasons), but I don't know what it is.

Last edited: Apr 30, 2021
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5. ### phytiRegistered Senior Member

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Mike;

[If the speed profiles are specified in advance, they still need to verify IF experience matches calculations, as time advances. Spacex may plan a launch but sometimes it fails. The definition of 'know' includes facts, observations or recordings, that verify specific events occurred. If not, it's speculation. Assuming a distant clock has a specific reading at a specific observer time, can only be verified by measurement. No matter how you approach it, verification and conclusion are always after the fact.]

[The simplest case is the separation of two clocks resulting from relative motion. The history of the clocks is irrelevant. For simplicity, the clocks are synchronized and moving at a uniform speed and direction as in the graphic. This is a general case with both clocks A and B moving relative to the U ref. frame. Sending light signals IS the 'radar' method, just different frequencies. Both A and B send signals to the other at t=.68 to get a time reading of t=1.00, which returns at 1.47. Per SR clock synch convention, the time of the distant clock event is half of the round trip transit time.
Each computerized clock system calculates local time as (1.47+.68)/2=1.08. The distant clock is running slower than the observing clock. But that doesn't make logical sense, says the novice, and he is correct. The assignment rule meets the expectation of an observer at rest. The A and B system can only measure the difference in their speeds, and not their own speed in any absolute manner. SR postulate 1 states physics is the same in all inertial frames, and that is what is observed by A and B, meeting the requirement of reciprocity. Since they are diverging, the clock frequency will be perceived as slowing. The frequency is constant, the relative motion alters the perception of the frequency, which is true in Newtonia[ n physics. That's doppler effects, not aging. Aging considers the accumulated time on a clock, not its rate.]

[As to the Gull paper on 'radar method', it seems to be the case of replacing reality with the representations. The 2nd graphic is from fig.2 in the paper concerning the (green) axis of simultaneity in the direction opposite to acceleration. Those references are mathematical aides and are not magically there for the observer to interpret. They are measurements using light in the same manner as clock synchronization.
With object E at rest relative to A:
B sends a signal at t0 which returns at t2. B assigns e1 to before t1.
B sends a signal at t1 which returns at t3. B assigns e2 to to t2.
B sends a signal at t2 which returns at t4. B assigns e3 to after t3.
Each sample event occurs once with one assigned B-time. They are not detected multiple times as assumed in the paper, just because the aos rotates with the acceleration. While the signal is in transit, A, B, and E continue to move!]

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7. ### Mike_FontenotRegistered Senior Member

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I've worked out an example that shows an instantaneous negative ageing for her, according to him, when CMIF simultaneity is used ...i.e., she instantaneously gets YOUNGER according to him, when using the CMIF simultaneity method. And I'll show the results for this example for my simultaneity method, which gives no discontinuities and gives no negative ageing.

First, here's a description of the Minkowski diagram (with time tau on the horizontal axis and separation X on the vertical axis). That always has to be determined first.

The two "twins" in this case aren't really twins ... they are just babies who were born at the same time but 20 lightyears (ly) apart, and with zero relative velocity. This situation continues until they are both 40 years old. We represent this initial situation by drawing a horizontal line on the diagram at the point X = 20 on the vertical axis, extending from tau = 0 to tau = 40. That is the initial segment of his worldline. At the end of that first segment, write 40 immediately above the end of that segment of his worldline to show his age , and vertically below there write 40 immediately below the horizontal axis to show her age.

Then, he instantaneously changes their relative velocity to v = 0.57735 ly/y, and continues that velocity for the rest of their lives. This causes his worldline to slope upward toward the right at a slope of 0.57735 (and an angle wrt the horizontal axis of 30 degrees). Label that point where the second segment of his worldline starts as point T.

Next, draw a 45 degree line starting at the point 40 on the horizontal axis, and sloping upward to the right, representing the worldline of a light pulse that she transmits when she is 40, and that is moving toward him. We then write an equation giving X as a function of tau for that light pulse, and then we write another equation giving X as a function of tau for the upward sloping segment of his worldline. Then, we set those two equations equal (force their X values to be equal). The result gives the value of tau where those two lines intersect ... label that point Q. That point is vertically above the point tau = 87.32 on the horizontal axis. Write that value just below the horizontal axis, vertically below that point of intersection.

We also need to determine their separation according to her (the value of X) when she is 87.32. The answer is 47.32 ly.

Next, we need to plot two lines of simultaneity (LOS's) that show what "Now" is for him. (The LOS's for her are just vertical lines). His LOS's (anywhere for him when his velocity is 0.57735) have slope 1/v = 1/o.57735 = 1.73, and they make an angle of 60 degrees wrt the horizontal axis. The first LOS we need goes through point Q. That line intersects the horizontal axis at the point tau = 60. That is determined by writing the X(tau) equation for that LOS, and solving it when X is set to zero. So this tells us that when he is 78.63 years old, she is 60, according to him.

Next, we need to determine how old she says he is when she is 60. We know that according to her, he ages slower than she does by the factor gamma = 1.2247 (once he has changed his velocity to 0.57735). So according to her, while she ages from 40 to 60, he ages from 40 to 56.33. Mark that age on his worldline.

We also need to determine their separation according to her (the value of X) when she is 60. The answer is 31.55 ly.

Next, we do the same thing for the LOS that goes through the point T where his worldline starts sloping upward. The result is that her age when he changes velocity is 28.45, according to him. He was 40 then.

From the above information, we can draw the Age Correspondence Diagram (the ACD), which is a plot of her age (on the vertical axis), according to him, versus his age (on the horizontal axis).

During the first segment, their relative velocity is zero, so they each agree that they are ageing at the same rate. Therefore the first segment of the ACD is just a line of slope 1, sloping upward to the right, making a 45 degree angle wrt the horizontal axis. This first segment is the same, regardless of whether you are using the CMIF simultaneity method, or my method. Label the end of that segment point T.

In the CMIF method, at point T, when he changes velocity from zero to 0.57735, he says that she instantaneously gets younger by 11.55 years, from 40 to 28.45. So, for the CMIF case, we draw a vertical line downward from point T, of length 11.55 ly. Then, the next (last) segment slopes upward forever at a slope of 1/gamma = 0.8165.

What does the plot look like after the point T in the case of my simultaneity method? It is a straight line between the point T and the point Q. Point Q is where his age is 78.64 and her age is 60. It is a point on the third segment of the CMIF line we determined above. Point Q is where he received the pulse from her, and it is the end of the "Disagreement Interval" (DI) between him and the a perpetually-inertial observer who is co-located and co-moving with him. So, after point Q, the ACD for my method coincides with the CMIF method for the rest of their lives. I.e., after the end of the disagreement interval (DI), the CMIF method and my method agree thereafter in this example.

So, as was claimed, with my method, the ACD has no discontinuities, and no negative ageing (i.e., the ACD plot never slopes downward).

I personally prefer the CMIF method, because of its simplicity, and because I'm not bothered by discontinuities or by negative ageing. But for those people who ARE bothered by those characteristics of the CMIF method (and in my experience, that's a LOT of people), my method offers a safe refuge.

8. ### phytiRegistered Senior Member

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725
Mike;

On the left is your example in the standard Minkowski format. A minor change is B moving at .6c. Red line is time dilation, green line is Axis Of Simultaneity.
As in your previous examples, knowing the planned motions in advance is not certainty.
When NASA launches a mission, they monitor it from the beginning, based on past experience of failures.
In SR the aos has to be established, with light signals or a system of synchronized clocks, the latter not logistically possible. That means an observer cannot assign an 'a' or event time until he has a 'b' or detection time. I.e. there is a time lag, so it can't be instantaneous. B doesn't get an A-clock reading of 60 until Bt=120. The reading is not transmitted faster than light!
If you predict the corresponding times for A and B correctly, that only means the motions went as planned. After each reception of a signal, neither one knows if anything has changed for the other.
A's clock rate for B is simple as 1.00 for the first part and .80 for the second.

On the right, B's clock rate for A is 1.00 for the first part and .67 for the second.
Your reversal in the A-clock rate is erroneously based on the rotation of the aos as B instantaneously moves away. B would effectively miss At=28 to At=40, yet we know those events would be received later up to Bt=80. The discontinuous segments are not connected as they would be with a smooth curved transition from one to the other. The aos only applies to inertial motions.
It will work for short distances with small accelerations but with some distortion.

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10. ### phytiRegistered Senior Member

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Mike;

Seems my response was incomplete. I questioned why the td wasn't reciprocal for B.
By extending B's signals, after Bt=40, and using the aos for B, from Bt=80 to Bt=120, the At interval is 32, giving .80. There is a transition period when B is sending signals at v=0, and receiving at v=.6, which gives a misleading td of .67.

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11. ### phytiRegistered Senior Member

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Mike;
duplication!

Last edited: May 16, 2021
12. ### phytiRegistered Senior Member

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725
Mike;

duplication!

Last edited: May 16, 2021
13. ### phytiRegistered Senior Member

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I can only repeat, the aos is a simulated system of simultaneity. It is a mathematical tool, and only exists if you make measurements.
The synchronization of clocks must be done using light signals, which creates an aos. The aos pointing backward doesn't give you access to the past. There is more drama in 'moving in time' than 'my clock is running slow'.

14. ### river

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Yes .

Last edited: May 16, 2021

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16. ### Mike_FontenotRegistered Senior Member

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There is another argument that the home twin's (her) negative ageing must occur, according to the accelerating twin (he). Suppose in the standard twin paradox scenario that he instantaneously changes his velocity at the intended turnaround from v to -v. That causes an instantaneous increase (according to him) in her age, say by +T years. But suppose that he immediately changes his mind, and instantaneously changes his velocity from -v back to +v. The net effect on her age (according to him) of those two back-to-back instantaneous velocity changes is the same as if he never changed his velocity at all: his world line just continues at the constant slope +v, with no turnaround. But that requires that her first instantaneous age change of +T must be canceled by her second instantaneous age change of -T. If negative instantaneous age changes aren't allowed, then positive instantaneous age changes can't be allowed either.

17. ### phytiRegistered Senior Member

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mike;

[Clocks can run at slower rates but do not run in reverse, thus people do not get younger. Time accumulates.
In fig.1, imagine line 1 and the green Axis Of Simultaneity is missing. A sends a signal (line 2) at t1. B unexpectedly approaches A while wandering through space and receives signal 2. He does not know the distance to A.
Now let signal 1 be sent by B and signal 2 be his return signal. He has a round trip time as a basis for calculating a distance and assigns t1 to a B-time via the green aos.
That is the process of forming an aos.
Now consider fig.2, the oversimplified typical 'twin' scenario. The green aos points to t1 before reversal, then to t2 after reversal, which is instantaneous. Based on the aos, B is missing the history of A from t1 to t2 (the 'time jump'). He couldn't have made any measurements during the reversal. Any conclusion based on the aos is incorrect. Even if B had sent blue signals to A getting the image of t1, the graphic shows he would have assigned an earlier time (magneta).

The bizzare A-clock behavior of running in reverse is the result of B performing an impossible sequence of motions. A smooth transition from the outbound to inbound is the only way to remove the discontinuity.
B's motion cannot alter the functioning of the A-clock, but it can alter B's perception of that clock.]

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