A New Simultaneity Method for Accelerated Observers in Special Relativity

[Neddy Bate]

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.

This is getting ridiculous. The light from her clock is constantly broadcasting her current age. That light could reach his eyes (with aid of a powerful telescope), and just before he stopped, it would show her clock displaying 16.91 YEARS. Once he stops, he will still be able to use his telescope to look at the flow of light coming from her clock. If his stopping takes a fraction of a second, he will see her clock displaying 16.91 years plus a fraction of a second. It makes no difference, and the fact that you are going down this path of argument makes me wonder if you even understand anything I am saying.

So he stops moving, and sees her clock saying 16.91 years (plus a fraction of a second!), knowing that the distance between them is 23.09, and calculates her age to be 16.91+23.09=40.000. You are arguing that he can't do that, because he calculated a different value a fraction of a second earlier?

Just the fact that your end result has non-constant speed of light should be an immediate sign that you are wrong. But if you think this is just a new way of doing SR, I will leave you to your delusions.

 

[QuarkHead]

we're physicists,

Oh really? The evidence is not compelling.

 

[phyti]

U observes identical clocks (1 and 2) moving at .4c and .6c in the x direction.
Using the gamma factor,

t1 = t*sqrt(1-(v1)^2)

t2 = t*sqrt(1-(v2)^2)

Eliminating t,

t2 = t1*sqrt[(1-(v2)^2)/ (1-(v1)^2)]

t2 = t1*sqrt[.64/.84]

t2<t1

The graphic shows all clocks lose time while moving through space. The difference in loss is what is measured by observers with the clocks.
C2 moves a greater distance than C1, in the same time t.
If C2 rejoins C1 (green), this is a 'twin' scenario.

 

[origin]

The graphic shows all clocks lose time while moving through space.

So the clock on your phone is "losing time"? Compared to what?

 

[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.

But we already know and agree on the age he calculates for her just before the turnaround. What we do not agree on is the age he calculates for her just after the turnaround.

So let there be another pulse which he will receive just after the turnaround. Let it arrive to him a few seconds after the turnaround, to clearly establish that this is a completely different pulse.

Now let him use the same technique you let him use before, in order to calculate how much time elapsed for her during the transit of that second pulse. Just as before, he should obtain her current age from a line of simultaneity the slope of which depends on his velocity, and then he should subtract the time he sees in the pulse from that value. After all, that is how he did it before the turnaround.

So, which line of simultaneity are you going to tell him to use? Remember, this is after the turnaround, and you are quoted here as saying that anything he does after the turnaround has nothing to do with the pulse that he received before.

For reference, here is a Minkowski diagram showing the two different lines of simultaneity that you can choose from:

So, please tell me which simultaneity line he should use after the turnaround, and why.

 

[Mike_Fontenot]

So let there be another pulse which he will receive just after the turnaround. Let it arrive to him a few seconds after the turnaround, to clearly establish that this is a completely different pulse.
[...]
So, please tell me which simultaneity line he should use after the turnaround, and why.

First of all, the simple proof I give (near the end of Section 7) doesn't require that ANY calculations be made at either the instant of velocity change, or at any time after that. The proof only requires calculations made up to and including the instant that he receives her pulse, which occurs an infinitesimal time before the velocity change. So your most important job (finding an incorrect statement in my proof) doesn't require analyzing any subsequent pulses. All that's necessary in the proof depends on the facts that happen up to, and including, the instant that he receives that first pulse, PLUS the fact that the causality principle requires that nothing that happens after the pulse is received can have ANY effect on the facts about her ageing while the pulse was in transit.

Secondly, your question indicates that you haven't yet read Section 8 of my paper, which shows how pulses that are partly in both halves of the Minkowski diagram must be handled. That section answers your question.

 

[phyti]

So the clock on your phone is "losing time"? Compared to what?

compared to someone with an identical watch with a velocity relative to me.

 

[phyti]

First of all, the simple proof I give (near the end of Section 7) doesn't require that ANY calculations be made at either the instant of velocity change, or at any time after that. The proof only requires calculations made up to and including the instant that he receives her pulse, which occurs an infinitesimal time before the velocity change. So your most important job (finding an incorrect statement in my proof) doesn't require analyzing any subsequent pulses. All that's necessary in the proof depends on the facts that happen up to, and including, the instant that he receives that first pulse, PLUS the fact that the causality principle requires that nothing that happens after the pulse is received can have ANY effect on the facts about her ageing while the pulse was in transit.

Secondly, your question indicates that you haven't yet read Section 8 of my paper, which shows how pulses that are partly in both halves of the Minkowski diagram must be handled. That section answers your question.

The event At=16 would correspond to B2t=48, using the aos for B2.
The aos for a frame of ref. requires a constant speed and a common point of reference (origin). For B2 that is the reunion at At=80.

Interval for A is 80-16=64.

Interval for B2 is 128-48=80.

B2 concludes A clock td rate is .8.

 

[Neddy Bate]

First of all, the simple proof I give (near the end of Section 7) doesn't require that ANY calculations be made at either the instant of velocity change, or at any time after that. The proof only requires calculations made up to and including the instant that he receives her pulse, which occurs an infinitesimal time before the velocity change. So your most important job (finding an incorrect statement in my proof) doesn't require analyzing any subsequent pulses. All that's necessary in the proof depends on the facts that happen up to, and including, the instant that he receives that first pulse,

Yes, I know. Your "proof" regarding his assessment of things just after the turnaround does not involve using any calculation from after the turnaround. You just make a calculation from before the turnaround, and then say it is "obvious" that that calculation still holds true after the turnaround. That is the problem.

You could similarly "prove" that relativity of simultaneity does not exist by showing that two events are simultaneous in one reference frame, and then just say it is "obvious" that they are also simultaneous in another reference frame, while refusing to do the actual calculation to find out for sure. But if you actually did the calculation in the other reference frame, and found out that the events were not simultaneous, then that would show that your proof was not a proof at all, it was just you being wrong about what you thought was "obvious."

PLUS the fact that the causality principle requires that nothing that happens after the pulse is received can have ANY effect on the facts about her ageing while the pulse was in transit.

Let's assume for the moment that Minkowski wasn't as stupid as you apparently think he was, and let's assume that the correct line of simultaneity for the traveling twin to use the moment after the turnaround is the one that points down-and-right to her current age being 53.33, rather than the one that points down-and-left to 26.67. That would not mean that the 26.67 was not correct before the turnaround. So that is not changing anything that happened in the past, contrary to your also incorrect claim that it would somehow violate causality.

That is why I was trying to get you to consider another pulse a few seconds after the first one, so that you could see that the 26.67 was correct to be used to do the calculations for the first pulse, and the 53.33 was correct to be used to do the calculations for the second one. But you are too stubborn to even consider trying it, so never mind.

Secondly, your question indicates that you haven't yet read Section 8 of my paper, which shows how pulses that are partly in both halves of the Minkowski diagram must be handled. That section answers your question.

Funny you should mention that. It is true that I hadn't gone any further than section 7, because that is where your mistake is located, so I stopped there to try to get you to see your mistake. Now that I know that you will never see your mistake, and have no intentions of reconsidering, I went ahead and read more, for the entertainment value.

I see now that you have divided the types of pulses into categories, and that you even allow for the correct lines of simultaneiity which point down-and-right, but only on the far right end of the diagram.

Now that I understand that, maybe it would help if I mentioned to you that you do not really have two identically constructed clocks in the same frame ticking at different rates. So you can rest assured that is not what your idea says, which would be extremely ridiculous.

But what you do have is also ridiculous. What you have is all perpetually inertial observers in the same frame agreeing on the tick rate of a particular clock, with one person who is also at rest with respect to those observers saying that the tick rate of that particular clock is different than what they say, because he alone had accelerated recently. All of the others could try to tell the one guy that he is wrong, but he is too stubborn, so he goes on believing what he thinks is true, even though it does not match reality. They could even tell him that if his idea were true, and if that particular clock were a light clock, then what he is claiming would violate the constancy of the speed of light. But that one guy is just too stubborn to listen to reason. I see now why you thought it up, and put your name to it.

I look forward to you getting some responses from the physics community. Don't say I didn't try to warn you.

 

[Neddy Bate]

Yes, I know. Your "proof" regarding his assessment of things just after the turnaround does not involve using any calculation from after the turnaround. You just make a calculation from before the turnaround, and then say it is "obvious" that that calculation still holds true after the turnaround. That is the problem.

You could similarly "prove" that relativity of simultaneity does not exist by showing that two events are simultaneous in one reference frame, and then just say it is "obvious" that they are also simultaneous in another reference frame, while refusing to do the actual calculation to find out for sure. But if you actually did the calculation in the other reference frame, and found out that the events were not simultaneous, then that would show that your proof was not a proof at all, it was just you being wrong about what you thought was "obvious."

Sorry to reply to my own post, but I just realised that what I wrote there is in fact exactly what Mike Fontenot is doing. He thinks he is proving that two events which are simultaneous in the out-bound frame (before the turnaround) are also simultaneous in the inbound frame (after the turnaround) by simply stating that it is "obvious."

In the out-bound frame (before the turnaround), the event of his clock displaying t'=32.66 and her clock displaying t=26.67 are simultaneous events that are separated in space. He calculates that correctly, but he never explicitly calculates that in the in-bound frame (after the turnaround), the event of his clock displaying t'=32.66 and her clock displaying t=53.33 are simultaneous events that are separated in space. And in this thread he has made up every excuse to ensure that he will not be doing that calculation any time soon. In fact he avoids it like the plague!

Instead, he simply says it is "obvious" that the events of his clock displaying t'=32.66 and her clock displaying t=26.67 are SIMULTANEOUS IN BOTH FRAMES!

Could it be possible that this guy has come this far in learning SR, without ever learning about the of relativity of simultaneity? It sure seems to be the case.

From this wikipedia article:
https://en.wikipedia.org/wiki/Relativity_of_simultaneity

"According to Einstein's special theory of relativity, ... If one reference frame assigns precisely the same time to two events that are at different points in space, a reference frame that is moving relative to the first will generally assign different times to the two events (the only exception being when motion is exactly perpendicular to the line connecting the locations of both events)."

 

[phyti]

Mike;

This graphic relates to the 'radar method' (Dolby & Gull) you cite in your cado paper. The aos (green) depends on direction, so the B aos differs from the C aos.
If B, the aos is B3-E.
If C, the aos is C7.5-E.
The frame can only follow one of 3 paths.
If BC, the observer will experience acceleration, his clocks will be out of synch, thus he has no usable aos. The issue is using the aos in a case where it doesn't apply.

 

[Mike_Fontenot]

On my webpage, I just added some comments to my proof that the home twin's age (according to the traveling twin) doesn't change instantaneously during the instantaneous velocity change. These comments are essentially the same additional information that I have given Neddy in my recent responses to him on this forum.

 

[Mike_Fontenot]

This graphic relates to the 'radar method' (Dolby & Gull) you cite in your cado paper.

Can you qualitatively describe what the Age Correspondence Diagram (ACD) looks like for your Dolby&Gull solution? I.e., how does her age, according to him, change as a function of his age?

 

[Neddy Bate]

On my webpage, I just added some comments to my proof that the home twin's age (according to the traveling twin) doesn't change instantaneously during the instantaneous velocity change. These comments are essentially the same additional information that I have given Neddy in my recent responses to him on this forum.

It is not a proof to simply say that after the velocity change he can "obviously" use the same calculation from before the velocity change. That is an SR beginner's error called mixing frames. The incoming frame is a completely different frame than the outgoing frame, and it has a different line of simultaneity. The events of his clock displaying t'=32.66 and her clock displaying t=26.67 cannot be simultaneous in both of those frames.

Once again I refer you to this wikipedia article:
https://en.wikipedia.org/wiki/Relativity_of_simultaneity

"According to Einstein's special theory of relativity, ... If one reference frame assigns precisely the same time to two events that are at different points in space, a reference frame that is moving relative to the first will generally assign different times to the two events (the only exception being when motion is exactly perpendicular to the line connecting the locations of both events)."

 

[Mike_Fontenot]

It is not a proof to simply say that after the velocity change he can "obviously" use the same calculation from before the velocity change.

You still haven't explained how his change in velocity at time "T" can possibly affect the amount of her ageing during the transit of the message, when that message transit was completed at "T-", which occurred before time "T". That would be a gross violation of the principle of causality, which holds that effects cannot precede causes.

 

[Neddy Bate]

Can you qualitatively describe what the Age Correspondence Diagram (ACD) looks like

I can show you the age correspondence that SR would have, based on the Lorentz transformations, and/or a properly drawn Minkowski diagram:

According to her
HER, HIM
0.00, 0.00
40.00, 32.66
80.00, 65.32
(80.00 - 40.00) / (65.32 - 32.66) = 1.225 = gamma

According to him (outbound half)
HER, HIM
0.00, 0.00
13.33, 16.33
26.67, 32.66
(32.66 - 16.33) / (26.67 - 13.33) = 1.225 = gamma

According to him (inbound half):
HER, HIM
53.33, 32.66
80.00, 65.32
(65.32 - 32.66) / (80.00 - 53.33) = 1.225 = gamma

 

[Neddy Bate]

You still haven't explained how his change in velocity at time "T" can possibly affect the amount of her ageing during the transit of the message, when that message transit was completed at "T-", which occurred before time "T".

His change in velocity does not affect the amount of her aging during the transit of the message that was already received a fraction of a second before the turnaround. There can be a different message that is received a fraction of a second after the turnaround, and that is the message which will have a different amount of her aging during its transit.

I asked you to consider a second pulse just a fraction of a second after the turnaround, but you refused. Considering your units are in years, it would be easy to have two pulses just a fraction of a second apart, and it would not affect the results.

So, before the turnaround, you get t'=32.66 simultaneous with t=26.67. To find the amount she aged while the pulse was in transit, subtract the age he sees in the image pulse (16.91) from 26.67, as you did on your website.

And after the turnaround, you get t'=32.66 simultaneous with t=53.33. To find the amount she aged while the pulse was in transit, subtract the age he sees in the image pulse (16.91) from 53.33, as you did not do on your website.

 

[Mike_Fontenot]

I can show you the age correspondence that SR would have [my color emphasis (mlf)], [...]

It's NOT the ACD "for SR". There are four simultaneity methods that I am aware of in special relativity, and they each have a different ACD for the given scenario.

The CMIF simultaneity method has an ACD that starts out with a slope of 1/gamma, then rises vertically from 26.67 to 53.33, and then finishes with a slope of 1/gamma again.

The Dolby& Gull simultaneity method starts out with a slope of 1/gamma, but partway to the velocity change, starts increasing linearly with a slope greater than 1, which continues until partway home, and finally resumes with a slope of 1/gamma.

The Minguizzi simultaneity method starts with a slope of gamma (not 1/gamma), and then has a complicated non-linear curve fro the rest of the diagram.

And my simultaneity method starts and ends with a slope of 1/gamma, and in the middle section has a slope of approximately 3.0.

There could be other simultaneity methods that have been published, but I'm not aware any.

My proof has shown that the CMIF simultaneity is incorrect. And both the Dolby&Gull and the Minguizzi methods are non-causal, which makes them non-sense, in my opinion. Of the four simultaneity methods, mine is the only one left standing. The only alternative is to just assume that simultaneity at a distance is a meaningless concept, and shouldn't even be contemplated ... a fair number of physicists hold that belief.

 

[Mike_Fontenot]

His change in velocity does not affect the amount of her aging during the transit of the message that was already received a fraction of a second before the turnaround.

Glad that you understand that.

There can be a different message that is received a fraction of a second after the turnaround, and that is the message which will have a different amount of her aging during its transit.

That subsequent message has NOTHING to say about how old he says she is at the instant "T+". I've already clearly shown that he says her age is the same at "T+" as it was at "T-". Any other answer would violate the principle of causality, as I've conclusively shown. Are you a denier of causality in special relativity?

 

[Neddy Bate]

It's NOT the ACD "for SR". There are four simultaneity methods that I am aware of in special relativity, and they each have a different ACD for the given scenario.

No, the only one that is correct under SR is the one that maintains a constant speed of light in all non-accelerating reference frames.

If you start with a vertically oriented light clock (light bouncing up and down between two parallel mirrors and ticking at that rate), you can derive that the only way there can be a constant speed of light in all non-accelerating reference frames is if all non-accelerating reference frames consider their own light clocks (which are stationary with respect to their frame) to tick at a rate of one second per second. AND if all non-accelerating reference frames consider light clocks which are moving at constant speed with respect to their frame to tick at a rate of one second per second divided by gamma.

That leaves one, and only one, simultaneity method. The one which you think you have proven wrong.

My proof has shown that the CMIF simultaneity is incorrect.

No it doesn't, and it is not even a proof. You are simply making an error when you use the same calculation for both before and after the turnaround point. The calculation you show for the amount of her aging during pulse transit, 26.67 - 16.91 = 9.76, is only valid for the outbound leg of the trip, not the inbound leg. The reason is because the line of simultaneity you used to get 26.67 is not valid in the reference frame of the inbound leg. The line of simultaneity for the inbound leg would point to 53.33 and so your calculation for the amount of her aging during pulse transit for AFTER THE TURNAROUND should be 53.33 - 16.91 = 36.42.