A New Simultaneity Method for Accelerated Observers in Special Relativity

[Neddy Bate]

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

I think you are not including the two lines which run off the bottom and right edges of my image. One is a worldline and the other is a simultaneity line. After the turnaround, the traveling twin has to determine where those two lines intersect, in order to reckon the distance between emission and reception.

I told you earlier that after the turnaround, he reckons that distance to be even greater than the current distance to her, because he reckons she's been moving toward him instead of away from him. Since the image pulse comes from the past, he would reckon she was even farther away than currently, (but only after the turnaround).

 

[Neddy Bate]

For reference, and because we are now on page 2, here are both of my Minkowski diagrams:

 

[Mike_Fontenot]

I think you are not including the two lines which run off the bottom and right edges of my image. One is a worldline and the other is a simultaneity line. After the turnaround, the traveling twin has to determine where those two lines intersect, in order to reckon the distance between emission and reception.

I told you earlier that after the turnaround, he reckons that distance to be even greater than the current distance to her, because he reckons she's been moving toward him instead of away from him. Since the image pulse comes from the past, he would reckon she was even farther away than currently, (but only after the turnaround).

All of the calculations, and data, and diagrams needed for my proof concern ONLY the outbound leg, BEFORE the change in velocity at the turnaround. During that leg, both the traveler and the home twin can use the time-dilation equation (the home twin because she is perpetually inertial, the traveler because the distance between them was zero when he instantaneously changed his velocity at the beginning of the scenario). And quit changing the scenario. The initial velocity is 0.57735 ly/y, NOT 0.866. And draw the Minkowski diagrams like I describe them, with the home twin's age on the horizontal axis, not the vertical axis. Unless and until you do that, we're done.

 

[Neddy Bate]

All of the calculations, and data, and diagrams needed for my proof concern ONLY the outbound leg, BEFORE the change in velocity at the turnaround. During that leg, both the traveler and the home twin can use the time-dilation equation (the home twin because she is perpetually inertial, the traveler because the distance between them was zero when he instantaneously changed his velocity at the beginning of the scenario).

You wrote, "Then, he instantaneously changes his velocity to -0.57735 ly/y, now heading back toward her. How much does he say she aged during the transit of that pulse now? The answer is obviously 9.76 years. 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."

That is after the turnaround, and you are still using the same calculations you made from before the turnaround, which is what I pointed out was wrong.

And quit changing the scenario. The initial velocity is 0.57735 ly/y, NOT 0.866. And draw the Minkowski diagrams like I describe them, with the home twin's age on the horizontal axis, not the vertical axis. Unless and until you do that, we're done.

No need to get upset. I am just trying to help you see where you went wrong. It doesn't matter what the velocity is, or how you draw the Minkowski diagrams. The lines of simultaneity are supposed to change at the turnaround, contrary to everything you have been saying here lately.

And don't forget that you still haven't accounted for the light clock proof I gave you, which tells you that all identical clocks at rest in the same frame tick at the same rate. I suggest you change your webpage back to the old CADO equation, at least that was consistent with SR.

 

[LaurieAG]

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.

Hi Mike, are you referring to a PLOT in a way similar to how Øyvind Grøn used SR to generate his solution for Figure 9 part C in his paper "Space geometry in rotating reference frames: A historical appraisal"?

I have bolded the things that distinguish Grøn's PLOTTED PLANE from a traditional frame in the passage below and have provided a link to one worked version of the solution posted on a now defunct, but still viewable, forum below. The x and y dimensions are measured in c and a straight line drawn from any emission point to the camera reflects the actual time/distance of travel of the photon between those 2 points.

If your PLOT/PLANE is similar then the problem may lie in the terminology you use to describe your solution.

The positions of points on a rolling ring at retarded points of time were calculated with reference to 0 K by Ø. Grøn [111]. The result is shown in Fig. 9. Part C of the figure shows the “optical appearance” of a rolling ring, i.e. the positions of emission events where the emitted light from all the points arrives at a fixed point of time at the point of contact of the ring with the ground. In other words it is the position of the points when they emitted light that arrives at a camera on the ground just as the ring passes the camera.

https://www.thephysicsforum.com/spe...elativistic-rolling-wheel-ii-3.html#post12704

 

[Mike_Fontenot]

Hi Mike, are you referring to a PLOT in a way similar to how Øyvind Grøn used SR to generate his solution for Figure 9 part C in his paper "Space geometry in rotating reference frames: A historical appraisal"?

I don't see any similarity in his plot and mine. And it sounds like he is considering circular motion, which is much more complicated than linear (single-dimensional) motion. The case of linear motion and instantaneous velocity changes must be fully understood before considering two 0r three dimensional motion.

BTW, I make use of two completely different plots. One is the Minkowski diagram, and the other is the age correspondence diagram (ACD). The latter seems to be used only by me. The ACD is a plot of the perpetually- inertial home twin's age, according to the traveler (who sometimes accelerates), versus the traveler's age, for all his ages during his trip. It is the whole objective of any simultaneity-at-a-distance method ... the Minkowski diagram is just a means of getting the ACD.

 

[Mike_Fontenot]

I should add that the age correspondence diagram can also be drawn for any two perpetually-inertial observers, say observer_1 and observer_2. The ACD for observer_1 is just an upward sloping straight line (that goes on forever) with a slope of 1/gamma. Observer_2's age (according to observer_1) is plotted on the vertical axis, and observer_1's age is plotted on the horizontal axis. It is just a plot of the time-dilation equation from observer_1's point of view. And because of the symmetry of time-dilation for inertial observers (i.e., because they each say the other observer is ageing gamma times slower than they themselves are), the ACD for observer_2 looks exactly like the ACD for observer_1 (except that the two axes are labeled oppositely).

 

[Mike_Fontenot]

Now that I have described the ACD's for perpetually-inertial observers, I'll describe how those diagrams compare with the ACD for the traveler in the classic twin paradox scenario (for the specific choice of velocity, etc., chosen for the scenario given in the paper), using my new simultaneity method.

For the outbound leg, the ACD "curve" is an upward-sloping straight line of slope 1/gamma ~= 0.82, just like the ACD for the two perpetually-inertial observers. That first segment of the ACD goes from the origin (0.0, 0.0) to the turnaround point at (32.66, 26.67).

The middle segment starts at the end of the first segment, and is a straight line that slopes upward with a slope of a bit over 3.0. The end of the middle segment is at the point (44.61, 63.09).

The final segment starts at the end of the middle segment, and is a straight line of slope 1/gamma (just like the first segment). The final segment ends at the point (65.32, 80).

By comparison, the ACD for the CMIF simultaneity method (which I now know to be incorrect) has the same first segment as in my method, but the middle segment is a vertical line going from the end of the first segment to the beginning of the final segment (which is a straight line of slope 1/gamma, and whose final point is the same as for mine: (65.32, 80)).

 

[Mike_Fontenot]

I got a new result from my simultaneity method, and have added it near the end of the new material on my webpage. I was able to prove that in my new method, the home twin's current age, according to the traveling twin, can never DECREASE. I.e., she can never get YOUNGER, according to him. That is a nice property for a simultaneity method to have, because the prospect of the home twin getting younger is repugnant to many physicists. In contrast, the well-known co-moving inertial frames (CMIF) simultaneity method implies that she DOES get younger (according to him) when he accelerates in the direction away from her (when their separation is sufficiently great).

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

 

[Neddy Bate]

Mike,

Have you given any thought to the idea that your simultaneity method does not maintain a constant speed of light in all inertial reference frames?

After all, SR was built entirely from two postulates, that the laws of physics are the same in all inertial reference frames, and that the speed of light is the same constant in all inertial reference frames. I would think many physicists would find it repugnant that your method claims to be compatible with SR, yet does not even adhere to its first and second postulates. You might want to think about that, if you haven't yet. Unless you don't care that your method is incompatible with SR.

 

[Mike_Fontenot]

Mike,

Have you given any thought to the idea that your simultaneity method does not maintain a constant speed of light in all inertial reference frames?

After all, SR was built entirely from two postulates, that the laws of physics are the same in all inertial reference frames, and that the speed of light is the same constant in all inertial reference frames. I would think many physicists would find it repugnant that your method claims to be compatible with SR, yet does not even adhere to its first and second postulates. You might want to think about that, if you haven't yet. Unless you don't care that your method is incompatible with SR.

The speed of light is not constant for the accelerating observer in my simultaneity method (and it is not constant for the accelerating observer in the CMIF method either). The accelerating observer's frame is not an inertial frame. My method has nothing unusual to say about inertial frames.

 

[Neddy Bate]

The speed of light is not constant for the accelerating observer in my simultaneity method (and it is not constant for the accelerating observer in the CMIF method either). But my method has nothing unusual to say about inertial frames.

I am talking about inertial frames. You said your method allows for two identically constructed clocks at rest in the same inertial frame to tick at two different rates. That is highly unusual.

Imagine those two clocks are "light-clocks" (light bouncing between two parallel mirrors, and ticking at that rate). Now calculate the speed of light in each of the two identically constructed light-clocks, which you claim tick at different rates. You will calculate two different speeds of light, surely.

 

[Mike_Fontenot]

I am talking about inertial frames. You said your method allows for two identically constructed clocks at rest in the same inertial frame to tick at two different rates. That is highly unusual.

One of the two clocks (or people) that are rest in the same inertial frame (specifically, the traveling twin) has not always been at rest wrt that inertial frame. Previously, he has accelerated. He is NOT like the clocks or observers that Einstein described when he defined his inertial frames. His clocks and observers were perpetually inertial. The traveling twin in the twin "paradox" is NOT perpetually inertial. Just because he stops accelerating for a while (even for years) DOESN'T make him "inertial" while he is not accelerating. "Inertial" isn't equivalent to "not currently accelerating". "Inertial" means "hasn't ever accelerated".

If you want to prove my simultaneity method wrong, you need to identify a specific error in the proof I give that the home twin (she), according to the traveling twin (he), doesn't age at all during his instantaneous velocity change. It's a very simple proof. It is near the end of section 7. Let me know when you've done that.

 

[Neddy Bate]

One of the two clocks (or people) that are rest in the same inertial frame (specifically, the traveling twin) has not always been at rest wrt that inertial frame. Previously, he has accelerated. He is NOT like the clocks or observers that Einstein described when he defined his inertial frames. His clocks and observers were perpetually inertial. The traveling twin in the twin "paradox" is NOT perpetually inertial. Just because he stops accelerating for a while (even for years) DOESN'T make him "inertial" while he is not accelerating. "Inertial" isn't equivalent to "not currently accelerating". "Inertial" means "hasn't ever accelerated".

All experimental evidence points to a constant speed of light in non-accelerating reference frames, regardless of whether they had accelerated in the past or not. That is why SR says the speed of light is constant in all non-accelerating reference frames, without any distinction between those which may or may not have accelerated in the past.

If you want to prove my simultaneity method wrong, you need to identify a specific error in the proof I give that the home twin (she), according to the traveling twin (he), doesn't age at all during his instantaneous velocity change. It's a very simple proof. It is near the end of section 7. Let me know when you've done that.

No, it's the other way around. If you want your simultaneity method to be taken seriously at all, you need to show experimental evidence that points to a non-constant speed of light in a non-accelerating reference frame, simply because it had accelerated at some time in the past. Let me know when you've done that.

 

[Neddy Bate]

Furthermore...

If (as you claim) two non-accelerating light-clocks which are stationary with respect to one another can tick at two different rates, then ANYONE who happens to be stationary with respect to those two clocks could measure two different speeds of light. So even someone who had never accelerated in the past would measure two different speeds of light. Not just the one person who had accelerated in the past, but EVERYONE would be able to measure two different speeds of light. Give it up.

 

[Neddy Bate]

Mike,

As you requested, I have drawn the Minkowski diagram in the way you described. However, I drew two lines of simultaneity, whereas you only described drawing one. You need a different line of simultaneity for after the turnaround, and that results in a different calculation for how much she aged while the image was in transit according to him.

Your description and calculations were fine when you wrote, "So, when he receives her image immediately before his turnaround, we’ve determined that he knows that she aged 9.76 years during the pulse’s transit."

That is correct because the line of simultaneity that you used to calculate it is from just before the turnaround. That line tells you her current age according to him just before the turnaround (26.67), and from there it is only a matter of subtracting her known age at the time when she sent the image (16.91) to determine that the she aged 26.67-16.91=9.76 years during the pulse’s transit, according to him BEFORE the turnaround.

Your description and calculations went wrong as soon as you wrote, "Then, he instantaneously changes his velocity to -0.57735 ly/y, now heading back toward her. How much does he say she aged during the transit of that pulse now? The answer is obviously 9.76 years."

That is incorrect because the line of simultaneity that you used to calculate it is from BEFORE the turnaround, and yet you are still using it in your calculations for AFTER the turnaround. Surely you must agree that in that statement you are describing him as having already turned around, with his velocity now negative instead of positive. The rules of drawing a Minkowski diagram say that different velocities have different lines of simultaneity.

Therefore you must use the line of simultaneity from after the turnaround. That line would have told you her current age according to him just AFTER the turnaround (53.33), and from there it would have only been a matter of subtracting her known age at the time when she sent the image (16.91) to determine that the she aged 53.33-16.91=36.42 years during the pulse’s transit, according to him AFTER the turnaround.

 

[phyti]

Mike and Neddy;

This is the scenario from Neddy, post#2.
v= .866, with gamma = 2.
On the left side of the graphic, A remains static, while B departs outward, reverses direction and returns. The reversal lasts for zero time, thus is not an acceleration, but the path is discontinuous. This can be resolved by using two observers, one outbound and one inbound (B1 and B2) to make it more realistic. A clock event is a 'tick'.
A observes the 20 B1 ticks in 40 A ticks and 20 B2 ticks in 40 A ticks (red line).
With blue light paths, B1 observes 5.4 A ticks in 20 B1 ticks, and B2 observes the remaining 74.6 A ticks in 20 B2 ticks.
The doppler shift for B1=5.4/20=.27, and for B2=74.6/20= 3.7. This is only a perceived clock rate due to relative motion of the observer, since all clocks are running at constant rates. It is NOT an indication of aging. Collectively, the B path sends 40 ticks vs 80 ticks for the A path. The B clock accumulated less time than the A clock.

The correct interpretation requires knowing that the aos is a fictional mathematical tool resulting from the SR clock synch convention. The physical x axis of both frames are still parallel. Its function depends on a round trip signal from the observer, and a constant velocity.
Referring to the graphic in post#16, event At=70 hasn't occurred yet, and won't be detected until approx. Bt=38. The aos as shown would require a signal from B2 to be sent to A before the reversal at At=5.4 This means involving B2 for the entire duration of the experiment. The same requirement would apply to B1.
Using the first half of B1 and the second half of B2, requires a shorter aos, as shown on the left of the included graphic. Each B can only poll the short 5.4 interval at both ends of the A path, and assign corresponding B times.
B1 calculates 5.4/10.7=.50, reciprocal td.
B2 calculates, after observation, (80-74.6)/(40-29.3)=.50, reciprocal td.
At reversal, B1 calculates forward At=20(5.4/10.7)=10.
At reversal, B2 calculates backward At=80-10=70.
The B perception is shown on the right.

You were both correct if you match your answer with the right observer.
Both are inertial observers, but with a relative velocity. We can't expect both descriptions to be identical.

Will next visit the new and improved cado site.

 

[Mike_Fontenot]

Mike,

As you requested, I have drawn the Minkowski diagram in the way you described. However, I drew two lines of simultaneity, whereas you only described drawing one. You need a different line of simultaneity for after the turnaround, and that results in a different calculation for how much she aged while the image was in transit according to him.

Your description and calculations were fine when you wrote, "So, when he receives her image immediately before his turnaround, we’ve determined that he knows that she aged 9.76 years during the pulse’s transit."

That is correct because the line of simultaneity that you used to calculate it is from just before the turnaround. That line tells you her current age according to him just before the turnaround (26.67), and from there it is only a matter of subtracting her known age at the time when she sent the image (16.91) to determine that the she aged 26.67-16.91=9.76 years during the pulse’s transit, according to him BEFORE the turnaround.

Your description and calculations went wrong as soon as you wrote, "Then, he instantaneously changes his velocity to -0.57735 ly/y, now heading back toward her. How much does he say she aged during the transit of that pulse now? The answer is obviously 9.76 years."

That is incorrect because the line of simultaneity that you used to calculate it is from BEFORE the turnaround, and yet you are still using it in your calculations for AFTER the turnaround. Surely you must agree that in that statement you are describing him as having already turned around, with his velocity now negative instead of positive. The rules of drawing a Minkowski diagram say that different velocities have different lines of simultaneity.

Therefore you must use the line of simultaneity from after the turnaround. That line would have told you her current age according to him just AFTER the turnaround (53.33), and from there it would have only been a matter of subtracting her known age at the time when she sent the image (16.91) to determine that the she aged 53.33-16.91=36.42 years during the pulse’s transit, according to him AFTER the turnaround.

Look at the yellow line you drew on the diagram. That is the pulse that she sends, and which reaches him immediately before he changes his velocity. So that pulse is in his past when he has changed his velocity. So anything he does later in his life has nothing to do with that pulse, or with any conclusions he came to when he received the pulse. As I say in my paper,

"So, when he receives her image immediately before his turnaround, we’ve determined that he
knows that she aged 9.76 years during the pulse’s transit. Then, he instantaneously changes his
velocity to -0.57735 ly/y, now heading back toward her. How much does he say she aged during the
transit of that pulse now? The answer is obviously 9.76 years. 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. And since her age when she transmitted the image (16.91 years old) is also an
unchanging fact, it is an unchanging fact for him that she is

16.91 + 9.76 = 26.67 years old

before and after he instantaneously reverses course. Therefore there is no change in her age,
according to him, when he instantaneously changes his velocity. The CMIF simultaneity method is
incorrect! That is a very startling fact, given that the CMIF simultaneity results are found in many
(maybe most) text books on special relativity."

 

[Neddy Bate]

Look at the yellow line you drew on the diagram. That is the pulse that she sends, and which reaches him immediately before he changes his velocity. So that pulse is in his past when he has changed his velocity. So anything he does later in his life has nothing to do with that pulse, or with any conclusions he came to when he received the pulse.

Your paper says to write his time as 32.66 both before and after the turnaround. That is correct for the hypothetical case where the turnaround is instantaneous. At his time 32.66, he receives the image pulse that she sends, so he is receiving that same image both before and after the turnaround, because the turnaround take no time whatsoever.

Let's consider the simpler case where at his time 32.66 he instantly decelerates to v=0.000c. Keep everything else the same, so that he is receiving her image pulse showing her to be 16.91 years old both before and after he stops moving. After he has stopped, he can easily calculate her current age by simply adding the distance between the twins to the time he sees in her image pulse. The calculation looks like this:

16.91 + 32.09 = 40.00

In this case, the distance between the twins after he stops moving is not length contracted. Before he stopped, he considered that distance to be length contracted. In order for him to still calculate her age to be 16.91 after he has stopped, he would have to continue to consider the distance between them as length contracted, even though it isn't. By what stretch of the imagination can you consider it required in physics for him to ignore that the distance between them has changed now that he has stopped? You are also requiring him to consider himself to be moving away from her, when he no longer is doing so, just because of the fact that he was moving away from her in the past. That is the only way he could calculate her age to still be 16.91 after he has stopped.

Look, I don't mind that you are making up your own ideas. I just want you to understand that they are not SR. The reason the lines of simultaneity are supposed to change when the velocity changes, is to keep everything consistent with SR. Please note that if he uses a vertical line of simultaneity after he stops, it correctly points to her age being 40, which is consistent with the above simple calculation based on the known static distance, and the known constant speed of light. That is why Minkowski diagrams require the line of simultaneity to change when the velocity changes. What you have done is you tried to build a "proof" using a Minkowski diagram, while not following its rules.

 

[Mike_Fontenot]

Your paper says to write his time as 32.66 both before and after the turnaround.

There is no finite time lapse between the instant immediately before the instantaneous turnaround, and the instant immediately after the instantaneous turnaround. But there IS an infinitesimal amount of time between them. They are NOT the same instant. Mathematicians sometimes go to the trouble of denoting that precisely, but physicists don't bother. I COULD have denoted the turnaround as "T", the instant immediately before the turnaround as "T-" and the instant immediately after the turnaround as "T+", or perhaps as "T-epsilon" and "T+epsilon", for arbitrarily small epsilon, but I don't believe that is necessary ... we're physicists, not mathematicians.

When he receives the pulse, and before he has done his instantaneous velocity change, there's no way to know if he will actually change his velocity. So anything that happens after the instant the pulse is received can't possibly change how much she aged during the transit of the pulse. That is required by the principle of causality.