I Twin paradox not including accelerations, it is wrong where?

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Mister T

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Han Solo´s line of world is not Mr. Spock´s line of world bended in a kick, is Han Solo´s straight line of world:
If you draw a spacetime diagram of the scenario you proposed in your original post, and then traced your finger along the paths used to determine the age of each photograph, you would find that the paths you traced are not both straight. For a path to match the path of an inertial motion, it must be a straight line.

Thus the reason that the photographs have different ages is because they traced out different paths through spacetime and that the lengths of those paths are unequal, or equivalently that the proper time associated with each path is different.
 
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Luis Babboni

If you draw a spacetime diagram of the scenario you proposed in your original post, and then traced your finger along the paths used to determine the age of each photograph, you would find that the paths you traced are not both straight. For a path to match the path of an inertial motion, it must be a straight line.

Thus the reason that the photographs have different ages is because they traced out different paths through spacetime and that the lengths of those paths are unequal, or equivalently that the proper time associated with each path is different.
Sorry, not understand.
This is, I think, an spacetime diagram of the scenario I proposed in my original post.
3-4 and 4-5 are straight in my point of view, I´m wrong?
tp08.jpg
 

jbriggs444

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3-4 and 4-5 are straight in my point of view, I´m wrong?
3-4 is straight. 4-5 is straight. 3-4-5 is not straight. That's the point.

3-4-5 is a valid world-line but it's not a straight world-line.
 
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Luis Babboni

3-4 is straight. 4-5 is straight. 3-4-5 is not straight. That's the point.

3-4-5 is a valid world-line but it's not a straight world-line.
There is no thing that moves along 3-4-5
 
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Luis Babboni

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Luis Babboni

At this moment what most concerned me is how to describe properly the "symmetrical" situation in respect to twin paradox.

May be it deservs another thread.
 
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At this moment what most concerned me is how to describe properly the "symmetrical" situation in respect to twin paradox.
The situation is not symmetrical. Not in the original twin paradox, and not in your version. In your version, the twin who stays on Earth just has one Polaroid and keeps it for the entire time. But the traveling "twins" (there is more than one person playing this role in your version) have to keep taking new Polaroids and switching them around. That is an observable difference--a lack of symmetry--between the traveling and the stay-at-home "twins".
 

Ibix

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At this moment what most concerned me is how to describe properly the "symmetrical" situation in respect to twin paradox.

May be it deservs another thread.
It isn't symmetrical. That's kind of the point.

All that really matters is the proper time, ##\sqrt {c^2\Delta t^2-\Delta x^2}##, along the line 3-5 and along the lines 3-4 and 4-5. Whether those lines are interesting to you because one person followed the path 3-4-5 or because one person followed 3-4 and handed a measure of elapsed time to someone else who followed 4-5 doesn't make any difference. The proper time for 3-4 plus the proper time for 4-5 is the same either way.
 

Mister T

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The length of the 3-4-5 path differs from the length of the 3-5 path.

Nothing travels along the 3-4-5 path but if something did it's age would equal the age of your less-aged photograph. Its motion would be noninertial.
 
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so there is no paradox without need to use aceleration to explain it! Right?
There is no actual paradox, period. What explains the different elapsed times, as has already been said in this thread, is the geometry of spacetime: different paths through spacetime can have different lengths. Sometimes the different paths are due to acceleration; sometimes they are due to something else, like people taking new Polaroids and exchanging them. But it's the different paths that matter.
 
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Nothing travels along the 3-4-5 path
That's not correct. Earlier it was said that "information" travels along the 3-4-5 path; but that "information" has a physical embodiment. There are physical interactions that have to take place for the chain of Polaroids along the 3-4-5 path to carry the information that gets compared with the stay-at-home Polaroid at the end. In other words, "information" really means a causal chain of physical processes along the 3-4-5 path, whose end result gets compared with the end result of a (much simpler) causal chain of physical processes along the 3-5 path. That causal chain of physical processes is not "nothing".
 

vanhees71

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It's all too complicated ;-)). Just keep in mind that a clock moving along a (necessarily timelike) worldline ##x^{\mu}(\lambda)## (where ##\lambda## is an arbitrary parameter) shows the time
$$\tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\dot{x}^{\mu}(\lambda) \dot{x}^{\nu}(\lambda) \eta_{\mu \nu}}.$$
That's all there is. So the proper time, shown by a clock (and measuring the aging of each of the twins) is simply a functional of the worldline each of the twin moves. It's like in ordinary geometry: The travel length between two points is a functional of the path taken to connect the points. There's nothing paradoxical or mysterious about this.
 
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Luis Babboni

OK, but what about this, is not the on Earth twin the younger at the reencounter?
tp09.jpg
 
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Yes, younger twin on Earth. But you need giant engine set on the Earth to accelerate the Earth in a moment.   ref. Japanese Si-Fic movie to build huge atomic rocket engines on the south pole continent to evacuate from an approaching gravity monster.  (^.^)
 
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Luis Babboni

Yes, younger twin on Earth. But you need giant engine set on the Earth to accelerate the Earth in a moment.   ref.  (^.^)
I do not think that is necesary, just need an engine able to change the direction of the spaceship and a frame* attached to it.

*: that frame could be many other numbered spacheships remaining at the same distance to the "traveler" twin´s spaceship and with coordinated clocks and instructions in a way all them change direction at the same time in proper spaceships time.
 
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jbriggs444

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I do not think that is necesary, just need an engine able to change the direction of the spaceship and a frame* attached to it.
You misunderstand what a frame is. It is not just a bunch of hypothetical rulers all laid out in a hypothetical right-angled grid. It also contains a bunch of hypothetical and synchronized clocks. Accelerate the "frame" and the synchronization does not work the way you imagine -- it is systematically skewed as a result of the acceleration.
 
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Luis Babboni

You misunderstand what a frame is. It is not just a bunch of hypothetical rulers all laid out in a hypothetical right-angled grid. It also contains a bunch of hypothetical and synchronized clocks. Accelerate the "frame" and the synchronization does not work the way you imagine -- it is systematically skewed as a result of the acceleration.
I bet you are right, but please let me know where is my mistake!
My proposed frame have content hypothetical and synchronized clocks.
If all spaceships, with synchronized clocks, planed before to accelerate at the same amount at the same time, the frame do not suffer any distorsion in proper time and proper distances. I´m wrong?
 

Ibix

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the same time
Almost every time when you write these words in relativity, that is where you are going wrong. "At the same time" assumes an absolute definition of simultaneity, which is wrong in relativity.

Your ships can't accelerate "at the same time" because before and after the acceleration they won't agree what "at the same time" means.
 
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Luis Babboni

Almost every time when you write these words in relativity, that is where you are going wrong. "At the same time" assumes an absolute definition of simultaneity, which is wrong in relativity.

Your ships can't accelerate "at the same time" because before and after the acceleration they won't agree what "at the same time" means.
Mmmm..... if all are at rest respect each other, I think they can cause all clocks run at the same speed.
For the on Earth twin instead, differents spaceships change theire direction at different time. I´m right?
But in this last case, for the traveller twin that goes to a near star and then return, the star change direction at the same time than Earth or not? :wideeyed:
 
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Luis Babboni

To start from the begining and in agree.

Is this way to see it correct?:
tp10.jpg
 

vanhees71

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OK, but what about this, is not the on Earth twin the younger at the reencounter?

View attachment 211994

Well, why don't you calculate it? Let's take the Earth's twins coordinate time as the parameter of the world line. Then we need to evaluate only one integral, namely the propertime of the travelling twin. I use natural units with ##c=1##. The world line of the traveling twin is
$$x(t)= \begin{cases} v t & \text{for} \quad 0 \leq t<t_0, \\
v (2 t_0-t) & \text{for} \quad t_0 \leq t \leq 2 t_0.
\end{cases}$$
Then the eigen time of the traveling twin is
$$\tau=\int_0^{2 t_0} \mathrm{d} t \sqrt{1-v^2}=2 t_0 \sqrt{1-v^2}.$$
The twin's proper time staying on Earth is obviously his coordinate time, i.e., ##2 t_0##. So the traveling twin is aged less by a factor of ##1/\gamma=\sqrt{1-v^2}##.
 

jbriggs444

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Mmmm..... if all are at rest respect each other, I think they can cause all clocks run at the same speed.
For the on Earth twin instead, differents spaceships change theire direction at different time. I´m right?
But in this last case, for the traveller twin that goes to a near star and then return, the star change direction at the same time than Earth or not? :wideeyed:
You used the words "at the same time" without specifying a reference frame. That's a non-starter.
 
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Luis Babboni

You used the words "at the same time" without specifying a reference frame. That's a non-starter.
I said "for the traveller twin... the star change direction at the same time than Earth or not?"
With it I tried to said for his reference frame.
 

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