Twin paradox in a closed universe

[madhatter106]

did I miss something, if they are traveling around a closed universe then on the return trip (1/2 way) they would be no longer heading away from each other but towards each other. the net effect is that they are the same age, each would measure a difference between them that would equal the same. the Earth in it's own frame would see them travel away and thus age slower but the return trip look as though they were aging faster and it should cancel out to match Earth's time frame too.

It made me think of the how far can you travel into the woods? 1/2 way answer. in a closed system such as that it's going to be 1/2 way then your advancing upon the origin again.

 

[JesseM]

did I miss something, if they are traveling around a closed universe then on the return trip (1/2 way) they would be no longer heading away from each other but towards each other. the net effect is that they are the same age, each would measure a difference between them that would equal the same.

Well, time dilation itself only depends on speed, not direction--maybe you're talking about the different Doppler shift for objects moving away or towards you? The Doppler effect influences how fast their clocks appear to be ticking visually (as opposed to how fast they are 'really' ticking in your frame). In this case, there's no change at the midpoint, instead each twin sees different copies of the other twin on either side of him--the "hall of mirrors" effect I mentioned before--with one visual copy continually moving away from him after he leaves Earth, visually aging slower the whole time (this is the copy he actually saw leave Earth next to him), another visual copy continually moving towards him until he meets that one at Earth upon his return, visually aging faster the whole time (that copy appeared to depart from a different Earth, and appeared to do so at a later time than the twin watching him left his own Earth).

 

[madhatter106]

edit that.. 1/4 to the half point of crossing paths. so it would be providing they could observe each other twin1 would observe twin2 not age nearly static until the 1/4 point in the curve then increase in age at an alarming rate then cross and slow to a near static then speed up until they arrive back at the origin. due to symmetry they both observe the same events.

 

[JesseM]

edit that.. 1/4 to the half point of crossing paths. so it would be providing they could observe each other twin1 would observe twin2 not age nearly static until the 1/4 point in the curve then increase in age at an alarming rate then cross and slow to a near static then speed up until they arrive back at the origin. due to symmetry they both observe the same events.

Huh? There's no "curve" here, the premise is that both twins are traveling at constant speed in a closed universe, so each one sees the other age at a constant rate throughout the trip (though like I said, in a closed universe each one sees multiple copies of the other who visually appear to age at different rates from one another)

 

[madhatter106]

Well, time dilation itself only depends on speed, not direction--maybe you're talking about the different Doppler shift for objects moving away or towards you? The Doppler effect influences how fast their clocks appear to be ticking visually (as opposed to how fast they are 'really' ticking in your frame). In this case, there's no change at the midpoint, instead each twin sees different copies of the other twin on either side of him--the "hall of mirrors" effect I mentioned before--with one visual copy continually moving away from him after he leaves Earth, visually aging slower the whole time (this is the copy he actually saw leave Earth next to him), another visual copy continually moving towards him until he meets that one at Earth upon his return, visually aging faster the whole time (that copy appeared to depart from a different Earth, and appeared to do so at a later time than the twin watching him left his own Earth).

Yes I was using a visual cue to quickly figure the time dilation only due to the description of a closed system with a small curvature. in sitting back and thinking about it in terms of actual visual perspective for each twin it gets interesting. if they both move at absolute c=1 then the twin leaving would be static and the twin approaching would be static as well. back off of c and then you can resume slow/fast.

 

[madhatter106]

so a hypothetical instantaneous c would equate to no age for either twin.

 

[JesseM]

Yes I was using a visual cue to quickly figure the time dilation only due to the description of a closed system with a small curvature. in sitting back and thinking about it in terms of actual visual perspective for each twin it gets interesting. if they both move at absolute c=1 then the twin leaving would be static and the twin approaching would be static as well. back off of c and then you can resume slow/fast.

It's not possible for a massive object to move at exactly c. But the idea is that both twins are moving at constant velocity in this closed universe (which we can assume to have zero curvature), so the rate that each one sees another one's clock ticking should be constant, it won't vary over the course of the trip.

 

[madhatter106]

agreed.
this example though is descriptive for a particle accelerator. and in the reference of the one particle time is nearly at a stop where it looks as though it hasn't left until it explodes with speed at the collision and deceleration.

the fabric of space is bunched up to close the gap until you slow down, giving you the visual of you being static while traveling at nearly c. so by that the fastest way to travel and not age is not to move.

 

[madhatter106]

I think for me it's an easy grasp of the time dilation in reference to speed because I've done high speed distance runs in race cars and even though I'm only dealing with velocities of about 200mph your brain processing speed increases in order to operate at that velocity and a strange effect of things slowing down occurs esp. when it goes sideways and the pucker factor goes thru the roof! you get this slow motion sensation that suddenly accelerates bringing you back.

 

[heldervelez]

This is the Sagnac effect with an added issue, the travellers.
it can dectect the absolute motion of Earth thru the background.
the Earth beeing 'stoped' is the least probable hypothesis against beeing in motion.
Again this length conctraction, and time dilation is much more simple, free of paradoxes, if one's consider a 'Real Lorentz contraction'.

 

[Dale]

Have you come up with a good definition of "real" yet? If so please post it in your thread on real lorentz contraction and stop contaminating other threads with irrelevancies. That is called "hijacking" and is considered rude.

 

[Dale]

Let`s look what this means in detail in the system of twin1:
...
Can anybody explain me this?

Sorry about the delay, I thought this deserved a little math rather than just some hand-waving. Before I begin, I want to state emphatically that my previous derivation is all that is needed. The spacetime metric, by definition, is preserved under any coordinate transform. However, I understand that sometimes it helps to do it a different way and see that you get the same result.

So, if we do a Lorentz transform of the equivalency relationship I posted earlier into one of the twin's frames we get:
$$t' = 1.25 t - 0.75 x = 1.25 t - 0.75 (x+n)$$
$$x' = -0.75 t + 1.25 x = -0.75 t + 1.25 (x+n)$$
which simplifies in terms of the primed coordinates into the new equivalency relationship
$$(t',x',y',z') = (t'-0.75n,x'+1.25n,y',z')$$

What this essentially says is that, due to the way the Lorentz transform mixes up space and time, in the twin's frame the universe is not only periodic in space but also in time.

Now, by velocity addition the other twin is traveling at -0.88 c and therefore has a time dilation factor of 2.125. The first twin wraps around to meet the Earth at t' = 1.333. At this time, a naive application of time dilation would say that the other twin's clock should read 1.33/2.125 = 0.63. However, remember the periodicity of the universe in both space and time. There are a infinite number of mirror copies of each twin in this universe, and the copy that the first twin encounters on his wrap around is not the one that left at the same time as he did.

In fact, by looking at the equivalency relationship above we can determine that the copy of the other twin that the first twin meets when he returns to Earth actually started his clock early in the first twin's frame. The one that he meets on his return to Earth is the one for n=2 (he crossed paths with n=1 halfway through the journey, on the opposite side of the universe from earth). So, this twin started his clocks at t' = 0 - 0.75*2 = -1.5. So then the correct application of time dilation would say that this copy of the other twin's clock should read (1.33+1.5)/2.125 = 1.33 as expected.

So basically, everything works out correctly as long as you remember the periodicity in space and time in the traveling twin's frames.

 

[heldervelez]

It is the same Sagnac geometry problem.
no gravity, no cilindrical space.
check 'laser-ring', etc.
IMO.

 

[JesseM]

It is the same Sagnac geometry problem.
no gravity, no cilindrical space.
check 'laser-ring', etc.
IMO.

But here you're talking about circular motion, which is non-inertial, so the "paradox" is not the same since you can't assume that in a non-inertial frame a moving clock will run slower than a clock at rest. The interesting feature of the closed but flat spacetime is that both twins can move inertially between successive meetings, so in each twin's own frame the other twin's clock should be running slower (but if you take into account the multiple-copies explanation that I and DaleSpam gave, you see that they indeed do, but they can still both agree that they're the same age when they reunite because the copy they departed from is different from the copy they meet on the second crossing of paths, and that second copy left its own copy of Earth at an earlier time in this frame which compensates for the slowed-down aging)

 

[heldervelez]

Sagnac effect is not about circular motion but closed loop path.
Actually the first experiment was done on a rectangular field. One Ly long does not change the geometric constraint of the problem.
can someone explain me, pls, the notion behind Lorentz invariance 'broken'?

 

[JesseM]

Sagnac effect is not about circular motion but closed loop path.
Actually the first experiment was done on a rectangular field. One Ly long does not change the geometric constraint of the problem.

But a rectangle still requires turning at the corners, so the point is the same--the path is not an inertial one (if you try to consider the 'point of view' of an object traversing the path by constructing a coordinate system where the object is at rest throughout the entire loop, then this coordinate system would be a non-inertial one). Perhaps you could consider a variant of the Sagnac effect taking place in a closed universe, but that's obviously not an experiment that has ever been performed in reality.

can someone explain me, pls, the notion behind Lorentz invariance 'broken'?

I think it basically just refers to the idea that the laws of physics could be such that different observers could perform identical experiments in the different inertial frames given by the Lorentz transformation ('identical' that each frame sets up their own experiment such that the coordinate positions and velocities of the initial setup of the experiment in that frame are the same as the coordinate positions and velocities of the initial setup of other observer's experiments in those observer's own frames) and not all get identical results.

 

[heldervelez]

Mr. JesseM I tank you for the explanation on 'broken Lorentz invariance'.
--------------------------------
Experimentation in a 'closed universe':
Closed loop wave-guides,ring-laser interferometers,
cold neutrons/electrons/atoms/coherent beams of atoms interferometers,
that have been used to test the Sagnac effect.
http://www.hep.princeton.edu/~mcdonald/examples/optics/anderson_ajp_62_975_94.pdf"
R.Anderson : section II pag 997 of Am.J.Phys., vol 62, n. 11 November 1994

IMO A 'closed universe' is a system from where the light/matter can not evade.
The light only goes out of the Sagnac apparatus because we made it so, 'on purpose'.

In a closed universe we can observe a 'closed path'.

In a rotating disk we can observe a 'closed path'.
Einstein synchronization of clocks around Earth equator can not be done without problems.
In closed loop wave-guides,ring-laser interferometers, cold neutrons/electrons/atoms/coherent beams of atoms interferometers we have a 'closed universe'
In Sagnac experiment we observe a 'closed path'.
The recent news on 'holographic knots' I suspect also a yes ( I didn't read the inners )
also electromagnetic cavity, cavity resonators and, a long shot: electrons, nucleons...

If one's consider the internals of the optics fiber it will be represented by a cylindrical geo where a varying core 'epsilon material' (c speed) makes light procceed macroscopically in a 'straight line' wathever curved it appears at our eyes.
(note : 'cylindrical geo is only relevant to maintain the light focus, it is a construction detail. It does not represent a 'cylindrical universe' as in previous posts).

---------------------------------------
---------------------------------------
I will take back the mention to the paper because when I checked again, now, the publication I realized that I've messed up with other paper with similar title in that Journal (the title also started by NONINVARIANT...and from Bari Univ." .
My apologies to everyone.
-----------------------------------------------------
-----------------------------------------------------
After reviewing the literature I mantain that the geometry of Sagnac effect is adequate to represent and analyse the closed universe OP problem.
-----------------------------------------------------

 

[JesseM]

IMO A 'closed universe' is a system from where the light/matter can not evade.

But that's not what physicists mean by "closed universe", that's more like a "closed system" in thermodynamics. "Closed universe" means that 3D space is actually finite, in the same way that the 2D surface of a sphere is a finite 2D manifold, so if you go far enough in anyone direction you'll return to your starting point.

In a closed universe we can observe a 'closed path'.

Even if we adopt your definition of "closed universe", the fact remains that you are talking about non-inertial paths, so the twin paradox, where each twin should expect the other one to be aging slower than themselves between the two meetings, doesn't apply. Only in inertial frames can we say that a moving object should be aging slower than an object at rest. In a finite space each twin does have an inertial rest frame in which they are at rest between both meetings while the other twin is traveling at constant velocity, which is the source of the "paradox", while this is not true when the two objects are taking closed non-inertial paths.

If one's consider the internals of the optics fiber it will be represented by a cylindrical geo where a varying core 'epsilon material' (c speed) makes light procceed macroscopically in a 'straight line' wathever curved it appears at our eyes.

But in any inertial frame as defined by SR, the light is not moving in a straight line--its path is curved in all inertial frames.

a nice paper about Sagnac effect/circular motion/accelerated frames (with relativistic corrections):
"[URL VELOCITY OF LIGHT AND CLOCK
SYNCHRONISATION IN ACCELERATED SYSTEMS
[/URL]publ. Fev/1997 Foundations of Physics Letters
Abstract.

This looks like a crackpot paper. They claim that the clock hypothesis (which 'states that the rate of an accelerated ideal clock is identical to that of the instantaneously comoving inertial frame') somehow conflicts with SR: "The opinion of the author is that the Clock Hypothesis, added to special relativity in order to extend it to accelerated systems leads to logical contradictions when the question of synchronisation is brought up." (p. 2) Then they claim on p. 3 that "The fact that only an ether theory is consistent with accelerated motion give strong evidences that an ether exist". And the claim that "theories based on the Einstein’s clock synchronisation procedure are unable to explain, for example, the Sagnac effect on the platform" is also a crackpot one, since the behavior of all clocks in this situation can be correctly predicted from the perspective of an inertial frame in SR.

 

[heldervelez]

First of all my apologies because when I checked again the publication I realized that I've made a mistake ( The title on that Journal was similar and from the same Bari Univ.)
I've edited my previous post and made a proper correction.
Sorry.

----
Quoting Mr. JesseM, from last post:
"Closed universe" means that 3D space is actually finite... so if you go far enough in anyone direction you'll return to your starting point."
Ok with the definition.
It is the same in a ring-laser or http://en.wikipedia.org/wiki/Fibre_optic_gyroscope"
In a usual 'closed' universe, we associate to gravitational bound of light/matter, because we are tinking in cosmology,
There are more ways to constrain light/matter in a closed loop that correspond to the definition of closed universe, as I showed, doping a material to change 'c' inside.
The net effect is a changing epsilon (and 'c') thru space, as with gravitation, but we do not say that we bend the space,there is no need. Can GR take the same approach ?

Beeing unsual or never heard (I simply do not know) does not invalidate the correctness of the analogy.

----------------
" ... light procceed ... in a 'straight line' wathever curved it appears at our eyes. "

Light is unaware of the exterior curvy path of the optical fibre. In the perspective of light there is only a 'go ahead', it can not choose a different path (like in bobsleigh sport).

 

[JesseM]

Quoting Mr. JesseM, from last post:
"Closed universe" means that 3D space is actually finite... so if you go far enough in anyone direction you'll return to your starting point."
Ok with the definition.
It is the same in a ring-laser or http://en.wikipedia.org/wiki/Fibre_optic_gyroscope"

Huh? The 3D space surrounding the ring laser is not finite unless the universe itself is finite. If the universe is infinite, then the space which any particular system like a ring laser sits in must be infinite as well.

There are more ways to constrain light/matter in a closed loop that correspond to the definition of closed universe, as I showed, doping a material to change 'c' inside.

But you aren't actually "constraining light/matter" in that loop in the sense that it's literally impossible for any particle inside the loop to escape it. All you're doing is making it so a typical photon will remain in the loop, but an arbitrary particle with arbitrarily high energy wouldn't automatically stay in it (the future light cone of an event inside the loop does not remain entirely in the loop)

The net effect is a changing epsilon (and 'c') thru space, as with gravitation, but we do not say that we bend the space,there is no need. Can GR take the same approach ?

c is defined as the speed of light in a vacuum, so you aren't actually changing c. Besides, if you take a quantum view then there are plenty of vacuum regions between the particles that make up the material of the loop, and rather then traveling continuously the photons are being repeatedly absorbed and re-emitted by the lattice of particles that make up the material (see ZapperZ's post #4 on this thread)

Light is unaware of the exterior curvy path of the optical fibre. In the perspective of light there is only a 'go ahead', it can not choose a different path (like in bobsleigh sport).

I don't know what you mean by "light is unaware". Light doesn't have a mind, it can't be aware of anything. Physically the path the light takes on the loop is not a geodesic path, though. And if we're talking about flat spacetime, that means it's not an inertial path, so as I said before:

the fact remains that you are talking about non-inertial paths, so the twin paradox, where each twin should expect the other one to be aging slower than themselves between the two meetings, doesn't apply. Only in inertial frames can we say that a moving object should be aging slower than an object at rest. In a finite space each twin does have an inertial rest frame in which they are at rest between both meetings while the other twin is traveling at constant velocity, which is the source of the "paradox", while this is not true when the two objects are taking closed non-inertial paths.

 

[Daniel42]

Thanks a lot to JesseM (#35) and DaleSpam [#47) for their inspiring answer.
I was at first very happy to see a solution and a calculation that seems completely answer all my questions. Everything goes well.

And because of the relativity of simultaneity, in frames other than the preferred Earth frame, the event of different copies of twin A leaving their respective Earths do not happen at the same moment in twin B1's frame. 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). So although A3 is aging slower than B1 in B1's frame, when A3 and B1 meet at Earth 2, in B1's frame B1 has been traveling 17.32 years while A3 has been traveling 17.32 + 11.547 = 28.867 years. So, if A3 has been aging at 0.6 the normal rate than A3 will have aged 28.867*0.6 = 17.32 years since departing Earth 3 at age 40, so A3 will be 57.32 years upon reaching Earth 2 just like B1, in spite of the fact that in B1's frame A3 has been aging more slowly than B1 throughout the journey.

But later I try to take this solution to the example with the 1000 clocks and found no way how it could work. (Sorry).
Now I rather make a fool out of me and say:
>>I don`t understand<<
then remain a fool who remains not understanding.

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...<<

If this is the solution it must work in every inertial frame.
But I don`t get it work.

Actually in the inertial system of twin 1 both twins start at the very same moment from Earth 2.

At the start their own clock (in the rocket) shows 0 seconds and then are counting.

Now twin 2 is moving in the frame of twin 1.
When he reach clock1 his own clock ("twin 2-clock") shows less seconds then clock 1 cause of time dilation.

You can really compare clocks if they are at the same location and get absolute results. This is nothing relative. The relativity of simultaneity only applies if you want to compare clocks of different locations. (Please excuse - you sure know that).

As Einstein wrote in his famous 1905 Paper:

"It might appear possible to overcome all the difficulties attending the definition
of “time” by substituting “the position of the small hand of my watch” for
“time.” And in fact such a definition is satisfactory when we are concerned with
defining a time exclusively for the place where the watch is located; but it is no
longer satisfactory when we have to connect in time series of events occurring
at different places, or—what comes to the same thing—to evaluate the times of
events occurring at places remote from the watch."

So I cannot see any technically difficult to compare the twin 2-clock (in his rocket) with the time of the 1000 clocks of the inertial system of twin 1 just in the very moment they pass by on his journey.


Cause of time dilation the twin 2-clock slows down on the journey and finally shows a different time then the clock 1000 or the twin 1-clock (which is almost the same).

I agree that cause of the relativity of simultaneity (when you look in direction of the past) you have to say that he starts earlier.

But when you look in the direction of the future twin 1 knew that his own twin 1-clock starts at the same moment as the rocket of twin 2 passes by.


I think I would understand it if you are so friendly to calculate some solutions.

I please want to know the following results of clocks the clocks-meetings:

--> twin 2 meets clock1
--> twin 2 meets clock10
--> twin 2 meets clock100
--> twin 2 meets clock500
--> twin 2 meets clock1000

please be patient with me. I don`t want to bother someone.
I just didn`t get it.

 

[madhatter106]

here's how I look at it without the math equations...

if they are both at the same velocity how would they measure a difference between them? what is the other reference point? if it's the origin then they could always point to common frame. but each twin could only do that provided they both know they started at the same origin. otherwise the other reference point to each one would have another velocity.

I believe the coordinates are the variable giving the value that one started before the other. when they both start they both have the same coordinates and thus the same frame, if the rest of the frames relate to that frame then the order of events will be that same. however once off the coordinate shared frame there is no longer a shared reference and thus the inability to determine based upon each others own location which occurred first to whom.

twin 1 at constant velocity of say .5c would not know he's traveling at that velocity because light still travels in his frame at the same speed as it would if he was at velocity 0. looking over to twin 2 he observes a difference in the clocks because he can see that twin 2 has a velocity different to his own if and only if twin 2 is not traveling at his velocity. in this case since it's in the opposite direction twin 2 would seem to be traveling away at c=1 with no other frame of reference, add in Earth and then Earth would be traveling away at .5c and then twin 1 could remove the velocity of Earth from twin 2 to determine his velocity. but without another reference his frame would remain static and twin 2 would be the one doing the traveling.

this is the simultaneity point, without a shared origin point you can not determine the order of events between the twins.

this seems very odd, but because light always travels at the same velocity independent of your velocity the velocities between all observers can be different and thus time.

 

[Dale]

http://www.hep.princeton.edu/~mcdonald/examples/optics/anderson_ajp_62_975_94.pdf"
R.Anderson : section II pag 997 of Am.J.Phys., vol 62, n. 11 November 1994

It is the same in a ring-laser or http://en.wikipedia.org/wiki/Fibre_optic_gyroscope"

Neither of these references are relevant to the topic of a (closed) universe spanning interferometer, however I believe that you are correct anyway. In the moving twin's frame the different copies are not simultaneous with one another as I showed above, but they are equidistant. This will cause an identical signal leaving the left copy and the right copy to arrive at different times therefore resulting in an interference pattern. I am not terribly confident about that conclusion since I haven't worked out the math completely, but I think it is correct.

 

[Dale]

I think I would understand it if you are so friendly to calculate some solutions.

I please want to know the following results of clocks the clocks-meetings:

--> twin 2 meets clock1
--> twin 2 meets clock10
--> twin 2 meets clock100
--> twin 2 meets clock500
--> twin 2 meets clock1000

I believe this will be of more value to you if you work it out yourself. Write down the expression for each clock's worldline and determine when it intersects twin 2's worldline. Then just integrate the metric from when they read zero to the intersection in order to determine what each clock reads. I know it sounds complicated, but you can follow my examples above.

 

[JesseM]

But later I try to take this solution to the example with the 1000 clocks and found no way how it could work. (Sorry).
Now I rather make a fool out of me and say:
>>I don`t understand<<
then remain a fool who remains not understanding.

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...<<

If this is the solution it must work in every inertial frame.
But I don`t get it work.

But the numbers would be different in other frames. Remember, in the Earth frame, each copy of Earth is 10 light years from its neighbors, and all copies of twins leave their own Earth simultaneously. So if you want to figure out how far apart twins on neighboring Earths (or Earths farther apart) leave in another frame, you have to use the Lorentz transformation. The 5.7735 number was based on assuming that in the unprimed Earth frame, one twin departure took place at x=0, t=0 while the nearby one to the right took place at x=10, t=0. So to find the time between these events in the frame of the twin moving at 0.8c, plug v=0.5c into the Lorentz transformation:

x' = 1/sqrt(1 - 0.5^2)*(x - (0.5c)*t)
t' = 1/sqrt(1 - 0.5^2)*(t - (0.5c)*x/c^2)

Plugging x=0, t=0 into this gives x'=0, t'=0. Plugging in x=10, t=0 gives:

x' = 1.1547*(10) = 16.667
t' = 1.1547*(-5) = -5.7735

So, you can see this right twin departure took place 5.7735 years before the left twin departure. If you want to change the velocity or the distance you have to change the variables in the Lorentz transformation equations.

Actually in the inertial system of twin 1 both twins start at the very same moment from Earth 2.

But my whole point was that there are multiple copies of each twin A and B, and multiple copies of each Earth. Twin A2 on Earth 2 does of course leave at the same moment as twin B2 on Earth 2 (in all frames, since these events happen at the same point in spacetime), but in twin A2's frame twin A2 does not leave at the same time as twin B1, or B3, or B4, or B100, etc. 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.

So I cannot see any technically difficult to compare the twin 2-clock (in his rocket) with the time of the 1000 clocks of the inertial system of twin 1 just in the very moment they pass by on his journey.

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.?

 

[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.