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

SR time dilation is there in any Lorentz-invariant model e.g. sine-Gordon ( https://en.wikipedia.org/wiki/Sine-Gordon_equation ), which can be realized with lattice of pendulms:


Below is its oscillating solution ( https://en.wikipedia.org/wiki/Breather ): on the left resting, on the right traveling - we can see it has reduced number of "ticks":

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Neddy;

With A1(0, 40) relative to A1,
A1t=40, not At=40.
A1 and B have a common origin.

With A(0, 40) relative to A,
LT calculates B's description as
x'=2(0-.866*40)=-70
t'=2(40-.866*0)=80
His clock reads 80 when hers reads 40.
The graphic in #196 shows the green x' (simultaneity) axis has a slope close to blue light speed, i.e. the gap closing speed is (1-.866) = .134, requiring a very long return.

With B(34.5, 40) relative to A,
LT calculates B's description as
x'=2(34.5-.866*40)=0
t'=2(40-.866*34.5)=20

Those values satisfy reciprocity, each observes the other clock rate as 1/2 of their own rate.

 

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Okay, here you are transforming the stay-home coordinates x=0 and t=40, which is the event where & when the stay-home twin celebrates her 40th birthday. So you have calculated that the train-clock passing her at that moment displays t'=80, and it is located on the train at x'=-70. It is actually x'=-69.28 but for some reason you don't care about accuracy.

That is fine. It is just one example of the fact that every train-clock which passes the stay-home twin displays a value that is twice her own age. But that does not mean that she thinks her traveling brother is 80 when she is 40. She thinks he is 20, not 80.

Okay, there you are transforming the stay-home coordinates x=34.5 and t=40 from the stay-home frame to the train frame, and you are finding x'=0 and t'=20. That is a completely different event than above, of course. Where did you get the coordinates x=34.5 and t=40? I think I know.

This exercise gives us x'=0 as the location of the traveling twin in the train frame, and it also gives us t'=20 as his age when we want to know what birthday he thinks his sister is celebrating at that time. But of course her birthday event must be located at x=0 to do it correctly, so you are clearly not trying to do that.

Nevertheless, the inverse Lorentz transform would have given you the coordinates you chose to use, if you had transformed x'=0 and t'=20 from his frame to her frame:

t = γ(t' + (vx' / c²))
t = 2(20 + (0.866*0)) = 40

x = γ(x' + vt')
x = 2(0 + (0.866*20)) = 34.64

That tells us that when he is t'=20 years old, he would be passing by one of her synchronous clocks located at x=34.64 which is displaying t=40.00. But true reciprocity would tell us that when he is t'=20 years old, he would say his stay-home twin sister is t=10 years old, not 40. Where is your calculation for him thinking she is 10 years old? Hint: I already posted it for you in post #200:

t = γ(t' + (vx' / c²))
t = 2(20 + (0.866*-17.32)) = 10

Where did I get x'=-17.32? That was explained in post #200.

 

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Neddy;

It's what B thinks.
We have all the A events on the ct axis.
We use LT to calculate B's coordinates of A events!
The notation is A(x, t) transformed to B(x', t').
To calculate A's coordinates of B events, v is reversed.

With A(0, t),
LT calculates B's description as
x'=2(0-.866*t)=-17.3t
t'=2(t-.866*0)=2t
His clock reads 2x what hers reads.

Those are the A(x, t) of B, when she is 40, which yields B(0, 20).
She would know the distance to B as 34.6 through measurement, when she is 40.

A1 and B have a common origin when A1t=40 and Bt=20.
With A1(x, t), (with t relative to A1, a new ref. frame),
x'=2(0-.866*t)+34.6=34.6-1.73t
t'=2(t-.866*0)=2t

At=40 and A1t=40 are not simultaneous in the B frame.

 

Yes, that is right. Except you don't really need the A or B because one set of variables is already primed, and the other set isn't. We know the unprimed belong to her coordinate system, and the primed belong to his.

So (x, t) = (0, 40) is the event where and when the female twin is celebrating her 40th birthday. She is located at x=0 and her clock displays t=40 years.

When you transform the coordinates of that event from the unprimed frame to the primed frame, you get the primed coordinates of that same event, (in that same location, at that same time).

So (x', t') = (-69.28, 80) represents the time and location of the moving clock which is passing close by her while she celebrates her 40th birthday.

Of course some day the traveling twin will be 80 years old and he will say his sister was 40 years old then. But we were not talking about him being 80 years old, we were talking about him being 20 years old! You claimed the LT could not handle him calculating her age to be 10 years old when he is 20 years old. But by your own method, all you would have to do is enter the pertinent coordinates (x, t) = (0, 10) and you would get (x', t') = (-17.32, 20) and now you have t'=20 and t=10. So the LT does work, even using your own methods.

Yes, just as he would know the distance is 17.32 when he is 20. So you could transform (x', t') = (-17.32, 20) and solve for (x, t) = (0, 10). You just tend to do it the other way around.

 

Neddy;

Never said that, but questioned what if B didn't reverse.
There is a difference in knowing and speculating.
You are using ideal scenarios where nothing goes wrong. After the experiment is verified successful, you claim measurements were not necessary, even though you had to have coordinates to calculate!

In real world scenarios, after experience departs from prediction, analysis is necessary for corrections/revisions.
In Einstein's original paper, he defines light transit time to be equal out and return, just as for a rest frame. The local time the observer assigns to the distant event is 1/2 the round trip time. How do you know that without an emission and detection?

Taking another look at the original 'twin' triangle, the LT cannot apply to a discontinuous motion. The outbound and inbound segments aren't long enough to record the history of the at rest twin. The only answer in that case is each counting the time signals from the other.

If you and a second anaut were each in a spaceship with a laser and a clock, moving independently, how would you know where the other is and what their clock reads?

 

Good. So you have no problem with the inverse LT being applied to (x', t') = (-17.32, 20) in order for him to conclude (x, t) = (0, 10) so that he concludes that when he is 20 years old, she must be 10 years old.

You yourself applied the LT to (x, t) = (34.64, 40) in order for her to conclude (x', t') = (0, 20). You assumed a constant velocity v, just like I did. Everyone does. Constant v is in all basic SR thought experiments.

It is the same thing with applying the inverse LT to (x', t') = (-17.32, 20) in order for him to conclude (x, t) = (0, 10). Constant velocity v is assumed, just like all basic SR thought experiments.

Wrong. Let me explain, and you will see:

Let there be two very long trains passing close by the location where the twins will be born. The velocity of one train is v=0.866c and the velocity of the other train is v=-0.866c. So their speeds are equal, just in opposite directions. Let both trains be filled with clocks which are Einstein-synchronised long before the twins are born.

Let the coordinates of the twins' birth be (x, t) = (0, 0) which also happens to be (x', t') = (0, 0) in both of the trains' coordinates. As soon as the twins are born, the male twin is thrown onto the train with velocity v=0.866c and the female twin is kept in place where she was born.

Whenever she is celebrating a birthday, they take a picture of her at the party. In the background of the picture, there is always a train clock displaying a time that is twice her own age. So when she is 10, there is a train clock in the picture displaying 20. When she is 70, there is a train clock in the picture displaying 140, etc.

Now, the traveling twin is living on the train, and he learns about SR from his tutors. They teach him that he and his twin sister are in an ideal SR scenario, that she is the "stay-home twin", he is the "traveling twin" and there are two trains going in opposite directions so he can do a round-trip to get back to her. He learns and understands that his twin sister's current age can always be obtained by looking at the clock in his train car, and dividing by 2.

When he is 20 years old, he looks at the clock in his train car, and it displays t'=20. He knows SR says his sister is 10 at that time. He knows they phtograph her every birthday, so later when he gets back to her, there should be a photograph showing her celebrating her 10th birthday. Knowing that the trains are very long in length, and that their velocities are constant, he predicts that in the background of that photo there will be a train clock displaying t'=20. If he sees that train clock displaying t'=20 in her 10th birthday photo, he will use it as proof that this prediction of t'=20 and t=10 was correct.

So, at 20 years old, he decides he wants to go back to her. He jumps from one train to the other. When he lands in the other train, he looks at the clock on the wall, and it says t'=140. He calculates that his twin sister is now 70 years old instead of 10. He rides the train back to her, during which time he ages from 20 to 40 years old. He arrives to find that she is 80 years old, just as he learned in his SR classes.

She shows him a picture of her 10th birthday, with a train clock in the background displaying t'=20. He tells her he predicted she was 10 at that time, and that is proof that he was correct. She also shows him a picture of her 70th birthday, with a train clock in the background displaying t'=140. He tells her he predicted she was 70 at that time, and that is proof that he was correct. They take a current picture of them together, which is her 80th birthday. In the background of that photo is a train clock displaying 160.

And we all lived happily ever after.

 

This is an important question::

That is Einstein's clock synch procedure. Once it is completed, those two clocks are considered to be synchronised, in their own rest frame. "Synchronised" means that each clock tender always knows what the other clock displays, at any time, without having to wait for more return-trip light signals!

 

Neddy;

Not true.
When B receives the image of At=10, the passing train time is the same as his time, Bt=5.
When B receives the image of At=70, the passing train time is the same as his time, Bt=35.
All that is conditional, requiring A and B to successfully complete the experiment.

Not true.
The train clocks have the same rate. If B transfers to the other train, its clock would also read 20. He would age another 20 to be 40 at reunion.

Why would the train clock run 2x her clock rate?

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How does the observer know distances of other objects?

 

Neddy;

Here is your planned 'twin' scenario to reunite at At=80.
A expects 20 time units from B on his return in 5 time units on her clock.
B's retro rockets malfunction, and he continues outbound.
She receives those same 20 time units in 75 time units on her clock.
Obviously the actual times don't agree with the predicted times.
They didn't know, they assumed/speculated.

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When flights are planned, the air traffic control system monitors those flights using radar and gps. They do not know or assume the flights will be successful. Medical emergencies, mechanical issues, (today, unruly passengers), can be handled when needed. I.e., they work in real (actual) time.

The 'twin' scenario you are using is 'fantasy physics', especially the stunt double who leaps from one train to the other!
Where is your fan Mike?

 

It is the standard twin scenario, with the traveling twin going out-and-back:

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All of the coordinates match what I said they would be.

 

Hi James, I am very glad for you to contribute here. Finally someone with some brains, lol.

No, I am not saying the diagram has any problems, I think it is fine. It is meant to correct all of phyti's irrelevant diagrams.

During the turn-around: The outgoing frame says x=0 is located at x'=-17.32. The incoming frame says x=0 is located at x''=-121.24. I just wanted to make a different symbol.

 

For my incoming frame, I probably have the direction of my x'' axis backwards compared to how most people would have it. Instead of thinking of it as a train moving right-to-left, I put myself on the opposite side, and think of it as a train moving left-to-right.

At any rate, I thought that the traveler jumped from (x',t')=(0,20) in outgoing-frame-coordinates, to (x'',t'')=(-138.56,140) in incoming-frame-coordinates. That jump is what makes him change from thinking his twin sister's age was t=10 to thinking she is t=70. Of course she would not be located at x''=-138.56 she would be located a distance of 17.32 from there, at x''=138.56+17.32=-121.24.

James, would it make more sense to you if those x'' coordinates were positive instead of negative? I think that is probably the correct way to do it, with the velocity then being negative. Is that the convention that you would use?

 

I updated my spacetime diagram to have the correct orientation of the x'' axis:

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The traveler jumps from (x',t')=(0,20) in outgoing-frame-coordinates, to (x'',t'')=(138.56,140) in incoming-frame-coordinates. To get the age of his sister at that time, he can just divide the time he sees on the train clock wall by 2. So before he jumps off the outgoing-frame train, he divides t'=20 by 2 and says she is t=20/2=10. And after he lands in the incoming-frame train, he divides t''=140 by 2 and says she is t=140/2=70.

The longer calculation in the outgoing frame would be:
t = γ(t' + vx'/c²)
t = 2(20 + (0.866*-17.32))
t = 10

The longer calculation in the incoming frame would be:
t = γ(t'' + vx''/c²)
t = 2(140 + (-0.866*121.24))
t = 70
I had to use v = -0.866c for the incoming frame, appropriately enough.

 

I updated my spacetime diagram to show the incoming-frame-coordinates (x'',t'')=(138.56,140):

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Mission complete.