Twin paradox in a closed universe

[phyti]

I`ve thought about a special sort of twin paradox.
I know the usual explanation of the twin paradox but give me please the answer to this special case:

Imagine:
A static universe (non-expanding) with a closed geometry and a circumference of one lightyear. The twins start their journey in different direction from their planet (EARTH2) with nearly light speed.


<-------- [TWIN1] [EARTH2] [TWIN2] -------->


Here is my question:
When they will met again after one year on EARTH2 --
which twin is the younger one?

You're making it overly complicated.
Given your initial conditions, each twins clock would record the same amount of time, experiencing the same physics (whatever it is), on equivalent journies.

 

[Daniel42]

Are these clocks at rest in the Earth frame? If so, are they also evenly spaced, so the first is 0.01 light years from Earth in the Earth frame, 100th is 1 light years from Earth, the 500th is 5 light years from Earth, etc.?

No, they are in the system of twin 1 just in the same way of my first question (#1):

Twin 1 had placed 1000 clocks along the circumference of the universe.
The clocks have all the same distance between them - and they also move exactly in the same direction and with the same velocity like twin 1 does.
Imagine now twin 1 synchronises all these moving clocks with his own clock. They all show the same time in the system of twin 1.

The circumference was one lightyear but you are free to use 10 lightyears if you want.

But the numbers would be different in other frames.

Of course from the perspective of the closed universe all these B twins are "really" the same person, but from the perspective of the infinite universe with repeating patterns of matter/energy, they are different copies at different locations in space.

In my opinion there is a contradiction between those two sentences.

Imagine the twin write the number down and say:
"Oh, this number look likes the birthday of my wife."
So he writes a letter to her and they meet after the journey and later a son namely Peter is born.

In a different frame is the number different and no Peter is born.

I can`t believe that history is different in different frames.
We live in a relativistic universe but not in a schizophrenic.

 

[JesseM]

No, they are in the system of twin 1 just in the same way of my first question (#1):

So you're asking what time each clock would read at the moment twin 2 passes them? If so, can you take a stab at trying to figure this out yourself as DaleSpam suggested? As a hint, it's probably easiest just to use twin 1's rest frame since the clocks are all at rest and synchronized in this frame...

The circumference was one lightyear but you are free to use 10 lightyears if you want.

I was talking about the example I had given (and the one you were talking about too when you said Everything work well in your calculation if you assume and calulate in agree with the SR that: >>Instead, in twin B1's frame each copy leaves his own Earth 5.7735 years before the copy to the left...<<).

But the numbers would be different in other frames.

Of course from the perspective of the closed universe all these B twins are "really" the same person, but from the perspective of the infinite universe with repeating patterns of matter/energy, they are different copies at different locations in space.

In my opinion there is a contradiction between those two sentences.

Imagine the twin write the number down and say:
"Oh, this number look likes the birthday of my wife."
So he writes a letter to her and they meet after the journey and later a son namely Peter is born.

In a different frame is the number different and no Peter is born.

I can`t believe that history is different in different frames.
We live in a relativistic universe but not in a schizophrenic.

The second quote of mine above wasn't talking about different frames at all, it was talking about different (equivalent) ways of representing a closed universe. In the infinite-but-with-matter repeating version, there are multiple copies who have different coordinates in a standard infinite inertial frame, which is equivalent to the fact that in a closed universe the coordinate axes of any given frame will keep wrapping around the universe and thus assigning multiple combinations of coordinates to the same event.

As for the first quote, when I talked about the numbers being different in other frames, I was referring to the time between events at different locations in spacetime--in the Earth's frame the events are simultaneous, in the twin's frame the events occurred 5.7735 years apart. That's just how relativity of simultaneity always works in SR, different frames disagree about the time between events at different locations. All frames agree about local events, like the time the twin will see on a given clock at the moment he passes it--so if that time is predicted to be the date of the birthday of his wife in one frame, then all frames will agree that that's the time he sees when he passes that clock. But they may disagree about whether the event of the clock showing that time is actually simultaneous with the distant event of his wife having a birthday back on Earth, some frames will say that this clock is simply out-of-sync with Earth clocks. But since that issue of simultaneity doesn't affect the local question of what reading he actually sees on the clock when he passes it, it's not going to affect other events which are causally influenced by what he saw, like the fact that he was inspired to write a letter.

 

[Daniel42]

You're making it overly complicated.
Given your initial conditions, each twins clock would record the same amount of time, experiencing the same physics (whatever it is), on equivalent journies.

You are right, if relativity is right, then relativity is right.

One have to proof that the principle of relativity is in no contradiction with the constance of light velocity.
There is relativity of simultaneity, time dilation and so on.
There only is no contradiction if the following statement is true:

The moving observer view you just in the same way as you view him (per example time dilation).
And one point more:
You have to calculate if he view you in the same way only with your view/knowledge of him.
This face in the mirror looks at you - just as you looks at him.

I understood the SR first after I calculated this.
Later found I the calculation of this proof in an Einstein paper.

There is only one reason, why my question and experiment is so complicated:

I want to examine if the face in the mirror looks at me just as I look at him.
I want to examine if relativity is also true in a closed universe.
I want to know, if twin 1 view twin 2 just in the same way...

 

[Dale]

In my opinion there is a contradiction between those two sentences.

It seems like you may not understand the relativity of simultaneity, it is one of the most difficult concepts for new students to grasp.

Imagine the twin write the number down and say:
"Oh, this number look likes the birthday of my wife."
So he writes a letter to her and they meet after the journey and later a son namely Peter is born.

In a different frame is the number different and no Peter is born.

Time coordinates are frame variant, so in different frames a given event (e.g. wife's or Peter's birthday) will be assigned different coordinates. But if a given event occurs in one frame it will occur in all frames.

 

[heldervelez]

The stronger contradiction is to pretend that the happenings are observer dependent. 'Happenings' are always independent of 'observers'. Of course observers have a particular 'view' of events, indexing then in their own local proper time, but they are not determinants. We must get the correct meaning of 'observer' as one that is not interacting.

It occours to me several hypotesis to test, and the results shoud be the same:
Forget the 'mirror' strategy:
make 2 independent trips, each in different way, then compare the elapsed time in each (you have only 1 twin at a time and Earth).
Center the observer strategy:
put the Earth in the geom center as observer strategy:
the central observer is constantly looking up, and rotating his head as long the trip progresses, as a geosynchronos sattelite, and answer:
what is the relative motion between them? maybe 0? (ignoring gravitational time delays).
Instant observer strategy:
make the observer not depend on light to acknowledge the position (godlike). Shouldn't use this approach?
Lorentz-Fitzgerald strategy:
use physical clocks instead of idealized clocks (as Einstein did).

The whole problem share the problematic of the one of circular rotating disk, and Sagnac effect is useful to understand what is said by Mr Phyti, and myself:
if they leave Earth in identical conditions and travel in identical conditions then they will age equally, time as Lorentz dilation, provided that Earth does not move in relation to CMB frame (rotational stabilyzed and more).

 

[Daniel42]

All frames agree about local events, like the time the twin will see on a given clock at the moment he passes it--so if that time is predicted to be the date of the birthday of his wife in one frame, then all frames will agree that that's the time he sees when he passes that clock. But they may disagree about whether the event of the clock showing that time is actually simultaneous with the distant event of his wife having a birthday back on Earth, some frames will say that this clock is simply out-of-sync with Earth clocks. But since that issue of simultaneity doesn't affect the local question of what reading he actually sees on the clock when he passes it, it's not going to affect other events which are causally influenced by what he saw, like the fact that he was inspired to write a letter.

Glad you agree with this.

As for the first quote, when I talked about the numbers being different in other frames, I was referring to the time between events at different locations in spacetime--in the Earth's frame the events are simultaneous, in the twin's frame the events occurred 5.7735 years apart.

OK. That`s my fault. I misunderstood this. Please excuse me.



So I try to calculate the different times (what times the clock shows) of the clock-clock-meeting.

Twin 1 had placed 1000 clocks along the circumference of the universe.
The clocks have all the same distance between them - and they also move exactly in the same direction and with the same velocity like twin 1 does.
Imagine now twin 1 synchronises all these moving clocks with his own clock. They all show the same time in the system of twin 1.

To make things more simple I assume that clock1000 is in the rocket of twin 1 and is therefore identical with twin 1 own clock.
The speed of twin 2 in the frame of twin 1 is 0.8c according to:

Now you can look at the same series of events in the frame of one of the twins, say twin B1 who departs from Earth 1. In his frame, all the A-twins are moving at (0.5c + 0.5c)/(1 + 0.5*0.5) = 1c/1.25 = 0.8c according to the relativistic velocity addition formula, which means in his frame the aging of the A-twins is slowed down by a factor of 0.6.

OK. So the if the time is slow down with the factor 0.6 and the speed of twin 2 is 0.8c in the frame of twin 1 and the circumference of the universe is 1 lightyear.



--> twin 2 starts his journey at clock1000 (= twin 1-own-clock in his rocket)

twin 2-clock shows:
0 seconds

clock1000 shows:
0 seconds


--> twin 2 meets clock1

twin 2-clock shows:
o.6 x 0.001 x 0.8 x 1 year

clock1 shows:
0.01 x x 0.8 x 1 year


--> twin 2 meets clock10

twin 2-clock shows:
0.6 x 0.01 x 0.8 x 1 year

clock10 shows:
0.01 x 0.8 x 1 year


--> twin 2 meets clock100

twin 2-clock shows:
0.6 x 0.1 x 0.8 x 1 year

clock1000 shows:
0.1 x 0.8 x 1 year


--> twin 2 meets clock500

twin 2-clock shows:
0.6 x 0.5 x x 0.8 x 1 year

clock1000 shows:
0.5 x 0.8 x 1 year


--> twin 2 meets clock1000

twin 2-clock shows:
0.6 x 0.8 x 1 year

clock1000 shows:
0.8 x 1 year


I believe that both twins must have the same age at their second meeting.

It seems that everybody agree with this:

I agree with Ich and Dalespam that the twins will be the same age on the second meeting (providing Earth 2 is not moving).

I think that both twins must have the same age at their first meeting too.
The situation is totally symmetric, so i don`t see any cause of none-symmetric results if the Earth is at rest in this closed universe.

But the result of the simple time dilation calculation is not symmetric. This is a contradiction.

twin 2-clock shows:
0.6 x 0.8 x 1 year

clock1000 shows:
0.8 x 1 year



This is actually not the same result. It may be my fault, but I don`t see were the fault is.
May be my calculation is wrong. you sure can make it better then me :((

I even don`t understand how the relativity of simultaneity can solve this problem.

Instead, in twin B1's frame each copy leaves his own Earth 5.7735 years before the copy to the left, so twins A2 and B2 leave Earth 2 5.7735 years before A1 and B1 leave Earth 1, and twins A3 and B3 leave Earth 3 5.7735 + 5.7735 = 11.547 years before A1 and B1 leave Earth 1 (and all twins are age 40 at the moment they leave their own Earth).

Actually the twin 2-clock and clock1000 shows the following when the journey starts:

twin 2-clock shows:
0 seconds

clock1000 shows:
0 seconds


They don`t show this:

twin 2-clock shows:
-11.547 years

clock1000 shows:
0 seconds

All frames agree about local events, like the time the twin will see on a given clock at the moment he passes it--so if that time is predicted to be the date of the birthday of his wife in one frame, then all frames will agree that that's the time he sees when he passes that clock.

If the clock-clock-meeting in the frame of twin 1 shows this result at the start of the jouney:

twin 2-clock shows:
0 seconds

clock1000 shows:
0 seconds


then this result is the same in all frames.

It seems like you may not understand the relativity of simultaneity, it is one of the most difficult concepts for new students to grasp.

I think I understood relativity of simultaneity.
But how do you explain the different time the clocks show (caused by the simple time dilation in the frame of twin 1)?

 

[Dale]

The situation is totally symmetric, so i don`t see any cause of none-symmetric results if the Earth is at rest in this closed universe.

But the result of the simple time dilation calculation is not symmetric. This is a contradiction.
...
May be my calculation is wrong. you sure can make it better then me :((

I even don`t understand how the relativity of simultaneity can solve this problem.
...
I think I understood relativity of simultaneity.
But how do you explain the different time the clocks show (cause of simple time dilation in the frame of twin 1)?

I showed an example of this explicitly back in https://www.physicsforums.com/showpost.php?p=2568118&postcount=47" demonstrating how the simple time dilation calculation is wrong and showed how the relativity of simultaneity is essential to obtaining the correct results. Please re-read my post number 47 and if something is unclear please ask for clarification. Once you have understood the approach I used there then you should be able to follow the same steps to obtain the correct calculations for any other clock.

 

[Mentz114]

...if they leave Earth in identical conditions and travel in identical conditions then they will age equally, time as Lorentz dilation, provided that Earth does not move in relation to CMB frame (rotational stabilyzed and more).

This is true until you mention the CMB. To work out this problem one needs to consider relative motion between the travellers frames and the Earth's frame. Neither the CMB nor any other frame is necessary. I wish you would stop talking absolutist nonsense.

But you believe that that the apparaent contraction of length predicted by the LT between IFRs is 'physical', so its no surprise you're obsessed with absolute motion.

 

[JesseM]

--> twin 2 starts his journey at clock1000 (= twin 1-own-clock in his rocket)

twin 2-clock shows:
0 seconds

clock1000 shows:
0 seconds--> twin 2 meets clock1

twin 2-clock shows:
o.6 x 0.001 x 0.8 x 1 year

clock1 shows:
0.01 x x 0.8 x 1 year--> twin 2 meets clock10

twin 2-clock shows:
0.6 x 0.01 x 0.8 x 1 year

clock10 shows:
0.01 x 0.8 x 1 year--> twin 2 meets clock100

twin 2-clock shows:
0.6 x 0.1 x 0.8 x 1 year

clock1000 shows:
0.1 x 0.8 x 1 year--> twin 2 meets clock500

twin 2-clock shows:
0.6 x 0.5 x x 0.8 x 1 year

clock1000 shows:
0.5 x 0.8 x 1 year--> twin 2 meets clock1000

twin 2-clock shows:
0.6 x 0.8 x 1 year

clock1000 shows:
0.8 x 1 year

The problem with this calculation is that you're not taking into account what I and DaleSpam mentioned earlier, that there are multiple copies of everything: multiple copies of twin 1, multiple copies of twin 2, and multiple copies of each of those 1000 clocks, with the different copies of a given clock not synchronized in twin 1's frame because twin 1 is not in the preferred frame. So let's say the numbers on clocks increase from left to right, so that clock999 is to the left of clock1000 and both read 0 years simultaneously, clock998 is to the left of clock 999 and both read 0 years simultaneously, etc. So what happens when we get all the way down to clock1, and then look to the left of that? Well, to the left of clock1 will be a different copy of twin 1 with a different copy of clock1000, which we can call twin 1B and clock 1000B. And the key point about synchronization is that clock1000B will not by synchronized with clock1 immediately to its right--you can figure out how far it's out-of-sync by going back to the preferred frame where twin 1 is going at 0.5c, since all identical events that happen to different copies, like clock1000 reading 0 seconds and clock1000B reading 0 seconds, do happen simultaneously in the preferred frame. If the distance between copies of twin 1 is 1 light year in his own frame as your calculations seem to suggest, the distance between them (and thus the 'size of the universe') must be less in this preferred frame due to Lorentz contraction. So, first figure out the distance D between clock1000 and clock1000B in the preferred frame, then you can find the coordinates of two identical events on the worldline of these clocks which are simultaneous in the preferred frame (for example, you can set the event on clock1000B's worldline to have coordinates x=0,t=0 in the preferred frame, in which case the identical event on clock1000's worldline has coordinates x=D,t=0 in the preferred frame), then you can use the Lorentz transformation to find out how far apart these identical events must have occurred in the rest frame of those two clocks, similar to the calculation I did at the beginning of post #60.

If you imagine the two identical events are clock1000 and clock1000B reading 0 seconds, then if you find that in their own frame the event of clock1000B reading 0 seconds happened T seconds before the event of clock 1000 reading 0 seconds, that implies that according to this frame's definition of simultaneity, at the moment that clocks 1-1000 read 0 seconds, clock1000B already reads T seconds. And if twin 2 departs from clock1000 traveling to the right while twin 2B departs from departs from clock1000B traveling to the right, then the reading on clock1 when twin 2B passes it will actually be much less than the reading on clock1000B when twin 2B departed from it (with twin 2B's clock originally showing the same reading as clock1000B at the moment of departure, since twin 1B and twin 2B were supposed to start at the same age), because clock1000B and clock1 are out-of-sync by so much. And it'll be twin 2B that eventually travels to the right far enough to meet up with twin 1--twin 1 will never meet with the copy that departed from him (twin 2) again. Try redoing your calculations taking all this into account and see what happens.

 

[JesseM]

This is true until you mention the CMB. To work out this problem one needs to consider relative motion between the travellers frames and the Earth's frame. Neither the CMB nor any other frame is necessary. I wish you would stop talking absolutist nonsense.

It seems unclear whether heldervelez was talking about circumnavigating a closed universe or just making an ordinary round trip of the type that could be taken in an infinite SR spacetime. In the latter case of course you're right that there'd be no need to consider the CMB frame, but in the case of circumnavigating a closed universe you do have to keep track of which frame is the preferred one where copies of identical events happen simultaneously.

 

[heldervelez]

We can detect the absolute rotation using a laser-ring interferometer.
IMO in a closed universe, and in a universe with a finite set of components, a preferred frame can be defined and choosen at will provided it encloses all the universe. The symmetry of this problem apeeals centering a preferred observer at the center. In any rotational motion the center point/line is a preferred reference.
we can think of the travelers at the rim of a rotating circle (or two) rotating clockwise and anticlockwise.
------------------
we can think on a synchronization based on the center, emitting rays to the rim and use mirrors to reflect the light back to the center.
whith this configuration the center will have a 'master-clock' and the whole rim segments (travelers) became synchronous with the central allowing the central observer to know what he needs to monitor the evolution of a traveller allong the trip.
this kind of synchronization is no good to comunicate events between distinct travelers, but it is not the question in this problem.
Ahah, it occurs to me now that each traveller can know is position in space and time using the time,common to all, and use a ring-laser interferometer or a Foucault pendullum to know the 'angle' and add it to the common time, like a gps. (does it works ?)
------------------
I do not know, at this time, how to continue the exploration of this reasoning.
I think I did no errors until now and I will try to continue later.
------------------

 

[JesseM]

We can detect the absolute rotation using a laser-ring interferometer.

True, but in an infinite non-closed universe that doesn't imply a preferred inertial frame--the rotation is absolute because all inertial frames agree whether something is accelerating, and rotation is a form of acceleration. Hopefully you agree that the ordinary Sagnac experiment doesn't pick out a preferred inertial frame?

 

[Mentz114]

It seems unclear whether heldervelez was talking about circumnavigating a closed universe or just making an ordinary round trip of the type that could be taken in an infinite SR spacetime. In the latter case of course you're right that there'd be no need to consider the CMB frame, but in the case of circumnavigating a closed universe you do have to keep track of which frame is the preferred one where copies of identical events happen simultaneously.

Thanks for the clarification, I guess I wasn't paying attention.

The necessity for a preferred frame other than, say the earth, isn't clear to me. If we assume the Earth isn't going to accelerate, that is.

I withdraw my previous remarks, in any case.

 

[heldervelez]

...Hopefully you agree that the ordinary Sagnac experiment doesn't pick out a preferred inertial frame?

Indeed the Sagnac effect is not sufficient.
------------------------

 

[Daniel42]

The problem with this calculation is that you're not taking into account what I and DaleSpam mentioned earlier, that there are multiple copies of everything: multiple copies of twin 1, multiple copies of twin 2, and multiple copies of each of those 1000 clocks, with the different copies of a given clock not synchronized in twin 1's frame because twin 1 is not in the preferred frame. [...]Try redoing your calculations taking all this into account and see what happens.

I can`t do this. So please let me participate on your knowledge.
It`s your argumentation and so it`s your turn.

 

[JesseM]

I can`t do this. So please let me participate on your knowledge.
It`s your argumentation and so it`s your turn.

OK, let me get you started and see if you can carry on from there. In the twin1 frame you wanted the size of the universe--which is the same as the distance between copies of twin 1--to be 1 light year. If two copies of twin 1 are 1 light year apart in their rest frame, then in the Earth frame where they are moving at 0.5c, the distance between them must be sqrt(1 - 0.5^2)*1 light year = 0.8660254 light years due to length contraction. Since we want to measure time on the clocks in seconds it's more convenient to have distances in light-seconds (that way c=1 light second/second), so since 1 year = 31536000 seconds, 0.8660254 light years = 27310977 light-seconds.

Since the Earth frame is the preferred frame, this is the only frame where identical events happening to different copies are simultaneous. So, take the event of twin 1 departing from twin 2 when the clock twin1 carries (clock1000) reads 0 years. In the Earth frame this happens simultaneously with the event of twin 1B departing from twin 2B when clock1000B reads 0 seconds. If the second event happening to twin 1B is assumed to happen at coordinates x=0 light-seconds, t=0 seconds in the Earth frame, then the first event happening to twin 1 must happen at x=27310977 light-seconds, t=0 seconds in the Earth frame (x=27310977 light-seconds because that's the distance between copies in the Earth frame as mentioned in the first paragraph, t=0 seconds because these identical events must be simultaneous in the Earth frame).

Now we know the coordinates of these identical events (event #1 is twin 1 and twin 2 departing from one another with clock 1000 reading 0 seconds, event #2 is twin 1B and twin 2B departing from one another with clock 1000B reading 0 seconds) in the Earth frame. Can you use the Lorentz transformation to find the coordinates of these same events in twin 1's own rest frame, given that twin 1 is traveling at v=0.5c relative to the Earth? Remember, the equations of the Lorentz transformation are:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c^2)
with gamma = 1/sqrt(1 - v^2/c^2)

It's not actually necessary to find the x' coordinates of these events in the twin 1 frame (if you do you'll find that they take place 1 light year = 31536000 light-seconds apart), but what you want is the t' coordinates of both these events in twin 1's frame. That way you'll know how much earlier event #2 happened than event #1 in twin 1's frame. And since you know clock1000 read 0 seconds at event #1 and clock1000B read 0 seconds at event #2, that'll tell you how out-of-sync the two clocks are. Can you try this calculation, and see if you get an answer or if you run into problems? Then if you get an answer but don't know where to go from there I can give you some more tips (hint: in twin 1's frame, clock1 - clock999 should all be synchronized with clock1000, so if clock1000B is out-of-sync with clock1000 by a certain amount, then clock1000B should be equally out-of-sync with clock1 which is immediately to its right).

 

[Daniel42]

OK, let me get you started and see if you can carry on from there.

I now give it up. It seems that you really want to help me.
But there is one thing you have to know:
It is of no real help to get an question answered with an question (although there may be the best motives) but only frustrating.

Thanks.

 

[JesseM]

I now give it up. It seems that you really want to help me.
But there is one thing you have to know:
It is of no real help to get an question answered with an question (although there may be the best motives) but only frustrating.

Thanks.

Did you at least try plugging in the coordinates of the events I mentioned into the Lorentz transformation to see what the time difference would be in twin 1's frame? This is nothing more than algebra, and if you have trouble with it, you can show your work so I can see where you got stuck and help you out. The advantage of answering questions with hints as to how to figure it out yourself is that you'll probably learn better and gain confidence with using and understanding the equations if you follow the hints, whereas if I just give you the answer you probably won't have much understanding of the logic behind that answer, you'll basically just be taking whatever numbers I give you on faith and having your eyes glaze over at the explanation I give for where the numbers came from.

Also, like I said, if you can figure out the time difference in twin 1's frame but don't know where to go from there, I'm happy to give some additional hints.

 

[LukeD]

So I haven't done the calculations on paper, but what I've done in my head suggests to me that in a closed dimension, velocities are no longer anti-symmetric.

This means that if I see you moving at 0.9c, you might not see me moving at 0.9c in the opposite direction.

I wonder if my intuition is correct...

This should occur because the size of the dimension changes with the velocity.

 

[JesseM]

So I haven't done the calculations on paper, but what I've done in my head suggests to me that in a closed dimension, velocities are no longer anti-symmetric.

This means that if I see you moving at 0.9c, you might not see me moving at 0.9c in the opposite direction.

I wonder if my intuition is correct...

This should occur because the size of the dimension changes with the velocity.

In a closed but flat spacetime velocities would still work the same way as they do in SR. As I've said, a closed but flat spacetime should be exactly equivalent to an infinite flat spacetime where patterns of matter and energy just happen to repeat in a periodic way, like a hall of mirrors but with each image equally real.