A possible proof that negative ageing doesn't occur in special relativity?

[Mike_Fontenot]

Even in the CMIF simultaneity method, no one ever visually sees anyone getting younger, ever.

In my proof, I'm not assuming ANY simultaneity method. The only thing I assume is that Einstein's gravitational time dilation equation is valid. Here is a Wiki link that gives that equation:

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

Scan down to the first section ("Definition"), and find the first equation. In that equation, assume that "g" is constant, so the integration is trivial, and the time dilation equation simplifies to

T_d(h) = exp(gh),

(where I have also eliminated the "c" factor by using units of years and light-years). That equation says that the clock that is closest to the source of the gravitational field will run slower than the clock "h" lightyears farther from the source, by the factor exp(gh).

Then, I use the equivalence principle to replace the constant gravitational field g by a constant acceleration "A" of the two observers (the "HP" and "he") who are separated by the distance "d", giving me

T_d(d) = exp(Ad),

which says that the "HP", who is located "d" light-years behind "him", will age more slowly, by the factor exp(Ad), than "he" will.

That is the equation I use at the beginning of my proof.

In answer to your statement "Even in the CMIF simultaneity method, no one ever visually sees anyone getting younger, ever.", I say "No one has ever made use of the HP before, to find out!" But as I said in the proof, it would be absurd for anyone co-located with another person to ever contend that he sees the other person getting younger.

What I've really done, by doing a simultaneous and equal constant acceleration of "him" and the "HP", is to construct a coordinate system for the accelerating observer, analogous to the coordinate system that Einstein constructed for a perpetually-inertial observer. Of course, the clocks for "him" and the "HP" don't run at the same rate as the perpetually-inertial clocks do, but if "he" knows how to correct for the "HP" clock rate, it can serve as a coordinate system.

 

[Neddy Bate]

In answer to your statement "Even in the CMIF simultaneity method, no one ever visually sees anyone getting younger, ever.", I say "No one has ever made use of the HP before, to find out!"

In our favorite scenario, the traveling twin stops when he is 20 years old, and using the CMIF method, he calculates the home twin is 20*2=40 years old. At that same moment, he looks through a powerful telescope, and sees her and her clock displaying 5.359 years. Knowing that he is 34.641 light years away from her, he can calculate she must have aged 34.641 years during the time this light was in transit to him. So he double checks his CMIF calculation, and sure enough, 5.359+34.641=40.000.

If instead he had instantaneously jumped back on his spaceship and resumed traveling away from her at 0.866c, he would still see the same light entering his telescope, so he would see her and her clock displaying 5.359 years. But using the CMIF method this time, he would calculate she is 20/2=10 years old. Knowing that her clock ticks at half the rate of his own, he knows that this visual image came from when his own time was 5.359*2=10.718 years ago his time. And knowing that she has been moving away from their original co-location at constant speed 0.866c, he can calculate that this visual image came from when the distance between them was 0.866*10.718=9.282 light years. So he knows the light has been traveling for 9.282 years his time, but he also knows that her clock has been ticking half the rate of his, so she must have aged 9.282/2=4.641 years during that time. So he double checks his CMIF calculation, and sure enough, 4.641+5.359=10.000.

He calculated her age to be 40, then a moment later, he calculated her age to be 10. In both cases, he was looking at the exact same light image in his telescope. This is a proof that THE CMIF METHOD DOES NOT REQUIRE AN OBSERVER TO SEE A CLOCK RUNNING IN REVERSE, because it works with the very same light in the telescope, and of course it works with later arriving light as well. Give it up, man.

 

[Bells]

Mod Note

Moving this mess to a more appropriate sub-forum..

 

[phyti]

Neddy Bate;

Knowing that he is 34.641 light years away from her, he can calculate she must have aged 34.641 years during the time this light was in transit to him.

Keeping it simple, all clocks will emit an encoded time signal when triggered by a light signal from a distant observer. B has the primed variables.

Length contraction comes with time dilation.

left:
B thinks he is at the small circle, x=17.3, t=20, resulting from time dilation (red line). He receives a signal t=5.4 via his signal sent at t'=1.6.

right:
At t'=20, he assigns t=5.4 to t'=10.8. This agrees with the predicted gamma value of 2. On that assumption and inertial motion, he predicts t=40 will correspond t'20.
He won't be able to verify it until he is approx. 150.

 

[Neddy Bate]

phyti;

As usual, I have no idea what you are trying to say.

left:
B thinks he is at the small circle, x=17.3, t=20, resulting from time dilation (red line). He receives a signal t=5.4 via his signal sent at t'=1.6.

I don't know what you mean when you say B "thinks" he is at x=17.3, t=20.0, because B should be able to calculate his own location (in A's coordinates) with certainty.

right:
At t'=20, he assigns t=5.4 to t'=10.8. This agrees with the predicted gamma value of 2. On that assumption and inertial motion, he predicts t=40 will correspond t'20.

From what you say first, (t=5.4 and t'=10.8) it looks like you are saying B can use t'=2*t

So shouldn't B predict t=40 will correspond to t'=2*40=80?

What a mess.

 

[phyti]

phyti;

As usual, I have no idea what you are trying to say.



I don't know what you mean when you say B "thinks" he is at x=17.3, t=20.0, because B should be able to calculate his own location (in A's coordinates) with certainty.



From what you say first, (t=5.4 and t'=10.8) it looks like you are saying B can use t'=2*t

So shouldn't B predict t=40 will correspond to t'=2*40=80?

What a mess.

B thinks he is at the small circle, since 20*.866=17.3.
Yes, aging is reciprocal, so t=t'/2, A is 40 when B is 80.
Never said I don't makes mistakes!

 

[Write4U]

Question : Both A and B do age, but A only ages slower relative to B's aging, is that what all this means?

The term negative aging is confusing me? That would imply "getting younger", but that is a false impression, correct?

Moreover, can a case be made that both age at the same rate but cover a different distances?

 

[Neddy Bate]

Question : Both A and B do age, but A only ages slower relative to B's aging, is that what all this means?

The term negative aging is confusing me? That would imply "getting younger", but that is a false impression, correct?

Moreover, can a case be made that both age at the same rate but cover a different distances?

Hi Write4U,

You are correct that it is a false impression to think that anyone "gets younger" in Special Relativity (SR).

What really happens in SR is that different reference frames can have a different idea of what "now" means. For example, you and I both agree that it is 2021 right now on earth, because you and I are both stationary in the same reference frame, (that of earth).

But some distant person moving at high speed relative to us might calculate that it is currently 1999 on earth, and they would be correct as far as their idea of "now" is concerned. That does not change anything for you and I, all it really means is that that point in the other reference frame's "now" corresponds to a time when you and I were younger. (By the way, this is not because distant objects visually appear as they had been at an earlier time, due to the speed of light being finite. That is well known and calculated for in SR, but this is what remains after those calculations.)

It would not matter to you and I whether some other reference fame calculates that the current "now" on earth is a time before you and I were born, or if a different reference frame might calculate that the current "now" on earth is a time after you and I both died of old age. They cannot calculate what is actually happening on earth, or even whether the planet still exists, but they can calculate the time that it must be in this location of space, at least as far as their reference frame is concerned.

So if we look at a specific scenario, someone might accelerate so that they change from the earth reference frame to some other reference frame. And then it might seems as though they would have to say that you and I went from older to younger, or vice versa. But of course that is to be expected if different reference frames have a different idea of what "now" means. Otherwise there would have to be a "universal now", which is what everyone thought there was, until SR came along and said otherwise.

Anyway, you and I, in our own earth reference frame, do not ever get younger in SR. Regardless of what other reference frames think is the current "now" for us, you and I will never get younger, even in SR.

 

[Neddy Bate]

B thinks he is at the small circle, since 20*.866=17.3.

So, when B "thinks" he is located at the unprimed coordinates x=17.3 and t=20.0, he really is located at the unprimed coordinates x=17.3 and t=20.0, right? So why do you say he "thinks" that is where he is located?

Oh, I guess you means he calculates that, right?

 

[phyti]

So, when B "thinks" he is located at the unprimed coordinates x=17.3 and t=20.0, he really is located at the unprimed coordinates x=17.3 and t=20.0, right? So why do you say he "thinks" that is where he is located?

Oh, I guess you means he calculates that, right?

Yes. His conclusion is based on his analysis of images, and calculations.
Notice in the graphic, B's perception is of a smaller world than that of A.

 

[phyti]

Question : Both A and B do age, but A only ages slower relative to B's aging, is that what all this means?

The term negative aging is confusing me? That would imply "getting younger", but that is a false impression, correct?

Moreover, can a case be made that both age at the same rate but cover a different distances?

The aging rate is 'less', would be a better word choice.
All clocks lose time due to motion.
We can only measure differences.
When SR claims each observer measures the other cock rate to be less than their own, it is only their perception of the other clock due to the clock synch method.

 

[Write4U]

 

[DaveC426913]

Phyti. You might want to clear your predictive text cache on your phone after conversations with your dates.

 

[Neddy Bate]

Yes. His conclusion is based on his analysis of images, and calculations.
Notice in the graphic, B's perception is of a smaller world than that of A.

Sorry, but I still don't understand. Let's use the stay-home twin's unprimed coordinates, for simplicity. Let's look at the time t=40.00, when the traveling twin is located at x = vt = 40.00*0.866 = 34.64.

Are you claiming that the traveling twin "thinks" he is located at x=17.32 instead? Is that why you have a red diagonal line going from t=40.00 to t=20.00? If so, that is very wrong.

Of course the traveling twin was located at x=17.32 at time t=20.00 but that is a totally separate event from him being located at x=34.64 at time t=40.00.

 

[exchemist]

Phyti. You might want to clear your predictive text cache on your phone after conversations with your dates.

Very good. Though I'm not sure what physical meaning to attach to cock rate, anyway.

 

[phyti]

forum;

Reciprocal time dilation.

In the graphic, A and B move relative to U.
A and B synchronized their clocks at t=0.
The red calibration curves are constant time values wherever they intersect the time lines of the moving observers.
The A-clock ticks slower than the ref. frame U-clock due to its relative velocity.
The B-clock ticks slower than the ref. frame U-clock and the A-clock due to its relative velocity to both.
Both A and B send blue light signals to retrieve time values from the others clock.
Each receive a return time of t=1.47. SR states a ref. frame with inertial motion may be considered as a rest frame. The expectation of an observer at rest is, signal time out equals signal time back. Both A and B assign the distant clock event of 1.47 to local time (1+2.15)/2 =1.58 (later), indicated by green axis of simultaneity. Each interprets/perceives the distant clock rate slower than their own.
------------------------------
The skeptic argues it's illogical that each claims the other clock runs slower than their own, and that is true. Logic however was not the purpose of SR, but a principle of relativity that provided a common/universal interpretation of physical phenomena.
SR correctly predicts each of the observers records the same physics.
Both A and B calculate the speed v of the other as
v=(2.15-1.00)/(2.15+1.00)=.365,
1/gamma=.931,
the other clock should read .931*1.58=1.47.
------------------------------
What SR fails to explain is their description is perception (in the mind of the observer).
The distant clock is not actually running slower. The user of the distant clock will disagree with the others assessment of a slower rate.
The approaching siren has not changed its frequency as the observers ear indicates, according to the observer moving with the siren.
Reality rule 1. An observers motion cannot alter distant physical processes, but it can alter the observers perception of those processes.
U sees/measures/perceives/concludes; A-clock running slower than his, B-clock running even slower than his, and both clocks running at constant rates.
The source of the logic problem is time. On the Ut scale, time is linear (equal increments). On the time scale of the moving clock, it is exponential.
The red curves do not divide the roundtrip time exactly in half for moving clocks.
We must remember, the clock synch convention provides a simulated system of synchronization, that only applies to the ref. frame where it is established.
A's clock synchronization does not include the B-clock.
When Einstein defined the outbound and inbound light path as equal, it was for the purpose of theoretical consistency, not light propagation. We might consider the clock behavior an artifact or anomaly.

 

[Neddy Bate]

So phyti, you seem to be avoiding answering the questions in post #34.

 

[Write4U]

I don't know why you can't view that link. It works fine for me. If you still have your copy of Greene's "The Fabric of the Cosmos", he does essentially the same scenario on page 134, starting at the last paragraph.

I believe that Nova productions viewed on a PBS channel are not always accessible to all viewers, but should still be available on YouTube. One of those little extras YouTube offers, when used with a little discretion.

When I try a NOVA production on PBS, it always asks me if "T.H.I.S." PBS channel is in my area.

 

[phyti]

Neddy;

The spacetime graphics are geometric representations of activities within the SR environment. A picture is still worth many words, and a page of math. Also the brain is an image oriented organism. In the endless process of learning, the realization that because the st graphic is essentially an historical account of events, it might be difficult for some viewers to correctly track time of events, actual or predicted.

Here is a revision of the previous graphics, a comparison of 'now'.
Left:
At At=40, she knows at At=20, Bt=10 at x= 17.3.
Right:
At Bt=20 he knows at B=10.8, At=5.4 at x= .9.
That is all the real time info they have, and it's all historical.
These are simultaneous conclusions.
All this is the point repeatedly made to Mike, the calculations/predictions are all conditional depending on a successful experiment. Just calculating an event doesn't make it happen. If true NASA would never have had failures! There is no instant knowledge.
The observers rely on their local time for calculations and conclusions.
_______________
The spacetime graphic results can be checked with the coordinate transforms.
Using the A and B 'perception' graphics:
For the event B is 20,
A(x, t)=A(34.6, 40)
x'=2(34.6-40*.866)=0
t'=2(40+34.6*.866)=20
giving B(0, 20).

 

[Neddy Bate]

phyti,

It is given that v=0.866c, so...

1. At time t=20 all she has to do is calculate 20*.866=17.32 to know where he is located at that time.
2. At time t=40 all she has to do is calculate 40*.866=34.64 to know where he is located at that time.

So, x=17.3 is definitely NOT where he is located at time t=40.