Twin paradox not including accelerations, it is wrong where?

[Mister T]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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]

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.

 

[Luis Babboni]

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

 

[sweet springs]

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. 　(^.^)

 

[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.

 

[jbriggs444]

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.

 

[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]

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.

 

[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?

 

[Luis Babboni]

To start from the beginning and in agree.

Is this way to see it correct?:

 

[vanhees71]

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 traveling 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]

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?

You used the words "at the same time" without specifying a reference frame. That's a non-starter.

 

[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.

 

[Janus]

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?
View attachment 211955

Your space-time diagram is only drawn for the rest frame of the Earth. In order to get a true picture of what is going on, you need to draw the space-time diagrams for the other two frames as well.
Here's your diagram redrawn with markers showing how time passes for each of the three participants according to the Earth frame. I gave the two spaceships a velocity of 0.6c relative to the Earth. (I chose this value because it makes the later diagrams easier to read.)

The blue line is the Earth observer, the green line is our traveling twin and the red line the inbound spacecraft . 0 marks where the Earth and traveling twin cross paths (I did not extend the lines prior to this point, even though you can imagine them as being so.)
If each number represents 1 year.
For every year the Earth photo ages, the traveling twin's photo ages 0.8 years. In four years Earth time, the two space ships meet, and the traveler's photo has aged 3.2 years. The incoming ship takes a photo of this photo. The new photo (of a 3.2 year old photo) also ages 0.8 years for every Earth year, so that upon its reaching Earth it ages 3.2 years while the Earth photo ages 8 yrs. The numbers on the red line represent the 3.2 years that the traveler's photo aged, plus the additional aging of its appearance during the return leg. (I need to note here that the 5 on the red line is misplaced, it should be level with the 5 on the green line. I didn't catch it right away, and with the software this was drawn with, the only way to fix it was to to go back and start from scratch. It is only a small mistake and doesn't really effect the general gist of the example.)

Now we look at events as they occur for the traveling twin. For that we use the following space-time diagram. (keep in mind that these are the exact same events as in the last diagram, just drawn according to a different frame.)

Note that the Earth and traveling twin still pass each other when their respective time marks are zero. Now however, it is the Earth photo that ages 0.8 years for every year that the traveling twin's clock ages. When the traveling twin meets the incoming ship it the photo has aged 3.2 years and the Earth photo has aged only a little more than 2 1/2 years. The relative speed between the traveling twin and the incoming ship is ~0.882c. in the same direction as the Earth is traveling with respect to the traveling twin. Thus according to the traveling twin, the other ship has to chase after and catch up to the Earth. This is take a lot longer by the traveling twin's clock than it took between his passing Earth and meeting up with the other ship. During that time, even though it is aging slower than his photo, the Earth photo will age an additional 5.5 years for a total of 8 yrs. Since the photo on the inbound ship is moving at 0.882c relative to the traveling twin, his will measure this photo as aging at a rate of 0.47 years for every one of his years and will measure it as having aged 3.2 yrs upon reaching Earth. We end up with exactly the same answer as we got according to the Earth frame.

Finally, we look at the incoming ship frame.

The thing to note here is that by this ship's time reckoning, the Earth and the traveling twin pass each other almost 7 years before he meets up with the traveling twin's ship. In that time, the Earth photo ages over 5 years at 0.8 years per year and the traveling twin's photo ages 3.2 yrs at 0.47 years per year. It takes an additional 3.2 years for him to meet up with the Earth, while the Earth photo ages to 8 years.

No matter which of the three frames you are working with, you end up with the same results, even though they will disagree as whose photo was aging slower throughout the exercise.
It also illustrates what people having been saying about the difficulty with defining "at the same time" in Relativity.
In the Earth frame "at the same time" that the traveler's photo has aged 3.2 years, the Earth photo has aged 4 years.
However, according to the traveling twin, "at the same time" that the traveler's photo has aged 3.2 years, the Earth photo has aged 2.56 years.
And, according to the inbound spaceship, "at the same time" that the traveler's photo has aged 3.2 years, the Earth photo has aged 5.44 years

 

[Luis Babboni]

WOW WOW Janus!
Very nice and clear!

Thanks for your very detailed and complete work!

What software you use?

 

[Janus]

WOW WOW Janus!
Very nice and clear!

Thanks for your very detailed and complete work!

What software you use?

It's called Minkowski2. I downloaded several years ago. I don't know if it is still available.

 

[Luis Babboni]

Thanks. I googled for half hour without luck.

 

[vanhees71]

Let me just mention that a space-time diagram is not referring to one and only reference frame. To the contrary, it's frame-independent, and you can draw world lines of observers easily, defining frames. That's the great advantage of space-time diagrams.

The disadvantage is that you tend to think in terms of Euclidean geometry, because we are used to it from elementary school on. That's, however, a wrong point of view. The affine plane is not used as a model of Euclidean but of Minkowskian/Lorentz symmetry here, and this difference is of utmost importance to understand relativity.

My attempt to explain this, can be found here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

For me, if in doubt, formulae in terms of manifestly covariant equations with vector and tensor components are much more safe!

 

[Mister T]

Earlier it was said that "information" travels along the 3-4-5 path; but that "information" has a physical embodiment.

Ah... Yes, I see that now. So what the OP has done is construct a version of the twin paradox where no single object follows the 3-4-5 path, but nevertheless something follows that path. Otherwise, there would be no way to construct the paradox.

 

[sweet springs]

I think not actual force on the body of a traveler twin but change of time correspondence or synchronicity time line far away is essential. As for information transfer by clock adjustment or so, the former does not apply but the latter does by transferring frames of inertia.

 

[PeterDonis]

I think not actual force on the body of a traveler twin but change of time correspondence or synchronicity time line far away is essential. As for information transfer by clock adjustment or so, the former does not apply but the latter does by transferring frames of inertia.

I'm not sure what you're trying to say here. What does "transferring frames of inertia" mean? As for "change of time correspondence or synchronicity", that looks to me like "choice of simultaneity convention", which can't affect any physics.

 

[sweet springs]

I apologize my poor English and wrong use of physics words might have caused a confusion.
Ref: https://upload.wikimedia.org/wikipe.../333px-Twin_Paradox_Minkowski_Diagram.svg.png
Re:my post #7 when rocket A and rocket B pass by, the clock adjustment of pilot B means change of simultaneity plane from blue one (trip out) to red one (ret. trip). Adjustment of clock B is just an information transfer and no force appear here. Force take place if pilot A would toss his clock to rocket B or jump by himself from rocket A to rocket B (what a tough guy he is!). Time dilations with the Earth time are same in the two ways.

 

[PeterDonis]

the clock adjustment of pilot B means change of simultaneity plane from blue one (trip out) to red one (ret. trip)

Yes, but it's important to understand that this is a convention, not required by the laws of physics. And it's not what causes the two clocks to have different readings at the end--the cause of that is the different lengths of the two paths through spacetime.

Adjustment of clock B is just an information transfer and no force appear here. Force take place if pilot A would toss his clock to rocket B or jump by himself from rocket A to rocket B (what a tough guy he is!). Time dilations with the Earth time are same in the two ways.

Yes, this is true: the important thing is the transfer of information about the elapsed proper time along the path, not the acceleration of anything.

 

[sweet springs]

Thanks for your clarification.

 

[sweet springs]

I think that gravity is more appropriate term in explanation of twin-paradox than acceleration. An egocentric traveler twin who believes he never move and keep still, would interpret that time dilation is due to a sudden appearance of gravitation field in all over the space including the Earth though he cannot explain why it appears.

 

[PeterDonis]

I think that gravity is more appropriate term in explanation of twin-paradox than acceleration.

The best explanation, as has already been said in this thread, is the different lengths of the two paths through spacetime.

though he cannot explain why it appears

Which is a good reason not to say that gravity is a "more appropriate" explanation.

 

[sweet springs]

Yea, the best explanation, right.
I would like to just add transfer of inertial frames is more appropriate (not the best) word than acceleration in twin-paradox explanation. When a massive body keep still during its changing the frame of reference, acceleration or force always work on it and vice versa. These are two equivalent sayings. But for information transfer as in post #7 the former applies, the latter does not.

 

[PeterDonis]

transfer of inertial frames is more appropriate (not the best) word than acceleration

"Transfer of inertial frames" is vague. Also it's not physical--"inertial frames" is a term for an abstract model made by humans, not a physical happening. Also it's unnecessary: you can just say "path through spacetime" and not have to worry about frames at all.

When a massive body keep still during its changing the frame of reference

I think I understand what you're trying to say, but this is not a good way to say it. A better way would be to say that if an object has nonzero proper acceleration, any frame in which it is at rest must be a non-inertial frame; it can't be at rest in an inertial frame for more than an instant. That puts the physical observable (proper acceleration) before the abstract model (a frame).

for information transfer as in post #7 the former applies, the latter does not.

See above.

 

[f todd baker]

The twin paradox absolutely does not need any mention of acceleration. And it is very easy to understand the asymmetry to boot. The traveling twin sees the distance to the destination to be length contracted, the earth-bound twin does not.

 

[Nugatory]

The twin paradox absolutely does not need any mention of acceleration. And it is very easy to understand the asymmetry to boot. The traveling twin sees the distance to the destination to be length contracted, the earth-bound twin does not.

You are right that the acceleration is not required to explain the twin paradox, but mistaken about the role of length contraction. If we use coordinates in which the traveling twin is at rest on the outbound leg, the earthbound twin will find that leg is shorter than the traveling twin; and likewise for the return leg when we use coordinates in which the traveller is at rest on the return leg.

It is not possible to link too often to the twin paradox FAQ: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[f todd baker]

You are right that the acceleration is not required to explain the twin paradox, but mistaken about the role of length contraction. If we use coordinates in which the traveling twin is at rest on the outbound leg, the earthbound twin will find that leg is shorter than the traveling twin; and likewise for the return leg when we use coordinates in which the traveller is at rest on the return leg.

It is not possible to link too often to the twin paradox FAQ: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

If the traveler is taken to be at rest, there is no length contraction because the earth-bound twin and the destination are in the same frame.

 

[jbriggs444]

If the traveler is taken to be at rest, there is no length contraction because the earth-bound twin and the destination are in the same frame.

There is length contraction because the frame in which the Earth bound twin and the destination are at rest is one you have chosen to consider as being in motion.

 

[f todd baker]

There is length contraction because the frame in which the Earth bound twin and the destination are at rest is one you have chosen to consider as being in motion.

Sorry, I am not getting your point. Regardless of which frame you choose to be at rest, the traveling twin sees the distance from Earth to destination to be contracted and the Earth bound twin does not see that distance contracted. What we have here is a failure to communicate!