# New Arguments In A Possible Proof That Negative Ageing Doesn't Occur In Special Relativity

Discussion in 'Physics & Math' started by Mike_Fontenot, Aug 7, 2021.

1. ### SsssssssRegistered Senior Member

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302
It's the region covered by Rindler coordinates which if you have an inertial frame with coordinates $t,x,y,z$ and an observer with constant proper acceleration $a$ who passes through $x=c^2/a$ at $t=0$ is the region $x>|ct|$.

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5. ### Neddy BateValued Senior Member

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2,329
So change the "light years" units into "light nanoseconds", and you have a much smaller version of the same thing. SR is not about waiting for information to reach a particular observer -- the time coordinates of events can be transformed from one reference frame to another using the Lorentz transformations. Since you tend to prefer Minkowski diagrams, it just means that you can draw the simultaneity lines as per the Minkowski diagram rules. You do not have to make round-trip light signals if you don't want to.

That is correct, and that is an important part of SR. Events which are simultaneous using the train frame's time coordinates will not be simultaneous using the track frame's time coordinates. That is why the event of him jumping off the train is simultaneous with the event of her turning 10 years old, using the train frame's time coordinates. And that is why the event of him jumping off the train is simultaneous with the event of her turning 40 years old, using the track frame's time coordinates.

Last edited: Sep 7, 2021

7. ### Mike_FontenotRegistered Senior Member

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359
Thanks, I'll take a look. Are there any other writings of his on this subject that I can get?

8. ### phytiRegistered Senior Member

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Mike & Neddy;

Your imaginary system of synchronized clocks doesn't work on a universal scale.
left:
B is moving toward A at .866 and passes A1 when all three indicate t=0.
B would have to send a signal to A at t'=-130 to get a return at t'=10. That assigns a simultaneous t'=-60 to At=0. A and A1 are definitely not synched according to B.
right:
B's perception of events.
A and B both perceive the other clock rate as 1/2 their own.
From A(35, 0) the coordinate transforms result in B(-70, -60).

9. ### phytiRegistered Senior Member

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625
Mike & Neddy;

Your imaginary system of synchronized clocks doesn't work on a universal scale.
left:
B is moving toward A at .866 and passes A1 when all three indicate t=0.
B would have to send a signal to A at t'=-130 to get a return at t'=10. That assigns a simultaneous t'=-60 to At=0. A and A1 are definitely not synched according to B.
right:
B's perception of events.
A and B both perceive the other clock rate as 1/2 their own.
From A(35, 0) the coordinate transforms result in B(-70, -60).

View attachment 4406

10. ### Neddy BateValued Senior Member

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phyti,

System A can have its own array of synchronised clocks which are stationary with respect to system A. In system A, those system-A-clocks all display the same time t simultaneously, but system B does not agree with that. System B can also have its own array of synchronised clocks which are stationary with respect to system B. In system B, those system-B-clocks all display the same time t' simultaneously, but system A does not agree with that.

For any event, it can be assumed that there are both a system-A-clock and a system-B-clock in essentially the same location as the event, so that signal propagation times are not a factor. The time coordinates of that event t and t' are defined by the times displayed on those clocks when the event occurs. It does not matter how much time it takes for any observer to receive the information of what times were displayed on those clocks at the time of the event. Those times for any event are not going to change once they are recorded.

Furthermore, it is the standard arrangement in SR that the system-A-clock located at x=0 displays the time t=0 at the same time and location that the the system-B-clock located at x'=0 displays the time t'=0.

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11. ### phytiRegistered Senior Member

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625
Neddy;
Yes, in the A frame.
Mike says B passing A1t=0 informs him that At=0 also.
As the graphic shows, it does not, since 2 simultaneous events in A, are not in B.
B cannot know instantly what the A clock reads.
There is no instant knowledge, since the discovery of finite light speed.

12. ### phytiRegistered Senior Member

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625
Neddy;
B doesn't get an image of the A-clock until Bt=10. How would he know before then?

13. ### Neddy BateValued Senior Member

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2,329
phyti;

We are given a scenario where something is moving with a constant velocity (v). We can predict where it will be located (x) at any time (t) by using the formula x=vt. That means that anyone can predict where it will be located at any time, even if they are light years away.

But you seem to be saying that they have to wait until they get a photograph of it? SR factors that part out, to save you time in your calculations.

14. ### phytiRegistered Senior Member

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625
Neddy;

This is a section from the previous graphic stopped at A's now, t=0.
The brown lines are outside A's light cone, so A is not aware of those events.
How does A know what the B-clock reads?

15. ### Neddy BateValued Senior Member

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phyti;

That is the very purpose of the equation at the top of this wiki page about the Lorentz transformation:
https://en.wikipedia.org/wiki/Lorentz_transformation

t' = γ(t - (vx / c²))
For the case where c=1.000 and v=0.866c we have γ=2.000 and that equation becomes:
t' = 2.000(t - 0.866x)
t' = 2.000t - 1.732x
Your question seems to be specifically for t=0.000 so:
t' = -1.732x

If your chosen value of x is -60, (I can't really tell from your diagram sorry), then you get t'=103.92. If your observer is 60 light years away, he will have to wait 60 years to actually see that clock reading through his telescope, but that does not change the fact that it should be t'=103.92.

Of course you might say that the clock might have stopped working before then, so his prediction would be wrong, but the question never should have been whether or not the clock malfunctioned. The question should be what time it was in that location, in that inertial frame.

16. ### phytiRegistered Senior Member

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625
Neddy;

Referring to drawing in post 185:
A concludes t'=0 when t=0 and x=35.
True if A measures distance to A1 with a roundtrip signal from t=-35 to t=35,
an assumption otherwise based on nothing changes.

B concludes At=0 when t'=-60, based on blue light signal from t'=-130 to t'=10.
A1 time is different from A time, thus B will not know the A time from the A1 clock.
A calculates using the LT:
x'=2(35-.866*0)=70
t'=2(0-.866*35)=-60
The LT allows checking accuracy of spacetime graphics, which agrees with the right side for B's perception.
After reflectors were left on the moon, scientists sent signals there and back. They discovered the moon is currently receding at 1" (2.5 cm) per year.
An awareness only due to measurement, the verification tool of science.

17. ### Neddy BateValued Senior Member

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2,329
You appear to be transforming (x, t) = (35.000, 0.000) to (x', t') = (70.000, -60.621) using the LT, and that is fine. If an observer located at x=10 wants to know what time is displayed on the moving clock at x'=70.000 at t=0.000, he can make that same calculation, even though he will not be able to see that event actually happen in his telescope until 10 years later.

It is as if the observer located at x=10 has "instant knowledge" because he was given v=0.866c so he can safely assume that never changes. He can also assume the clocks are all fully functioning, and that never changes. And he knows the speed of light never changes. So he does not have to wait 10 years to verify the event with his telescope and his own eyes. The event is as predictable as the sun rising tomorrow morning.

Last edited: Sep 13, 2021
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18. ### Neddy BateValued Senior Member

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Sorry, I misspoke. An observer located at x=10 would not see that event until 25 years later, not 10.

19. ### phytiRegistered Senior Member

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625
Neddy;

The drawing in 185 may have been confusing since the common origin was in the future, so I made a new one diverging from the origin.
Red lines indicate time dilation from B path to the ct axis.
A concludes Bt=20 at At=40, and x=34.6.
For A position coordinates are A(0, 40).
What are A position coordinates for B ?
Using LT and rounding,
x'=2(0-.87*40)=-70
t'=2(40-.87*0)=80
B cannot use A1 time to determine A time, which is the purpose of this graphic.
The basic reason, simultaneity is relative to the ref. frame.
In the general case, without predefined motions, measurements are necessary.

20. ### Neddy BateValued Senior Member

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In the standard configuration of SR, the two origins are in the same place when time is 0.000 on both of those clocks, so that event has coordinates (x, t) = (0.000, 0.000) and (x', t') = (0.000, 0.000). That condition is necessary in order to use the LT in the first place.

I have no idea what you mean by "common origin in the future".

Again, I have no idea what you mean. Time in either system, at any location, can be calculated using the LT.

That is already accounted for in the LT.

Okay, but when you are given a constant velocity of v between the two systems, and when you follow the standard configuration of SR regarding the locations of the origins, then the LT lets you calculate everything without any actual measurements.

21. ### phytiRegistered Senior Member

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625
Neddy;

You are considering a programmed motion that is completed and successful.
If either A or B didn't follow the programmed motion, the outcome would not agree with the prediction. If B didn't reverse direction for whatever reason, how would A know that? It's not covered by the LT.

NASA continuously monitors launches because there is the possibility of failure.

In air traffic control, an aircraft does not know how distant another aircraft is unless radar or GPS measurements are used. They cannot assume everything is in its correct place.

The other issue is the distributed system of synched clocks.
In #196, why doesn't B claim At=40 when he passes At1 which reads 40?
Why not At=10 according to the simultaneity axis (x') for B?

22. ### Neddy BateValued Senior Member

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Yes. I am considering a constant velocity of v. That is usually a given in SR thought experiments. When that is a given, there is no need for your round-trip-light-signals.

Right. But if the velocity is constant and unchanging, then the LT covers everything, which is my point.

The clock synch weirdness is already factored in the LT. I can't read your diagrams, so if you just give me the coordinates that you are trying to figure out in clear simple English, then I will show you how to solve it using the LT. Be sure to use a constant velocity v, and make sure that (x,t)=(0,0) is in the same place as (x',t')=(0,0) as per standard configuration.

23. ### Neddy BateValued Senior Member

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I didn't go back to look at #196, but the familiarity of these numbers makes me think I might understand what you are asking. When the traveler's age is t'=20 he is passing one of her clocks which displays the time t=40. The inverse LT tells us that as follows:
t = γ(t' + (vx' / c²))
Where:
c = 1.000
v = 0.866c
γ = 2
t' = 20
x' = 0
Thus:
t = 2(20 + (0.866*0)) = 40

But x'=0 represents the location of the traveler, and the stay-home twin is not in that location at that time. Instead she is located at x'=-vt'=-17.32 at that time, so the LT calculation required to find her clock reading in her location is this:
t = γ(t' + (vx' / c²))
Where:
c = 1.000
v = 0.866c
γ = 2
t' = 20
x' = -17.32
Thus:
t = 2(20 + (0.866*-17.32)) = 10
That is the time on her clock which is simultaneous (according to the traveler) with his clock displaying t'=20, but this is only the case while he is traveling because then he considers all the train-clocks to be displaying the same times t' simultaneously. Note that there can be a photograph of her celebrating her 10th birthday with a train clock passing close by her in the background and displaying the time t'=20.

As you can see, the LT takes the clock synch problem into consideration, and gives you different times based on different locations.

Last edited: Sep 17, 2021