# Does the twin paradox work in an almost empty universe?

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Background

The twin paradox is the popular thought experiment involving twins, one of whom makes a journey to a nearby star in a high-speed rocket (travelling at a velocity close to the speed of light) and when he reaches the star, changes direction and returns home at the same speed. When he gets home, he finds that time has slowed down during his journey and that his twin who has remained on Earth has aged more. (The paradox being that because both twins have travelled at the same speed with respect to each other, why has one aged faster than the other?)

My understanding is that the solution is as follows. When he arrives at the star and changes direction, the travelling twin switches from a reference frame travelling away from Earth at a velocity near the speed of light to another reference frame travelling towards Earth with a velocity of the same magnitude but in the opposite direction. In contrast the non-travelling twin stays in the same reference frame. It is often pointed out that the travelling twin experiences an acceleration when he switches from one reference frame to another, whereas the non-travelling twin does not experience an acceleration. (This acceleration could be a Dirac delta function i.e. an instantaneous switch of direction)

Questions

My first question is what is the travelling twin's acceleration with respect to? Since in special relativity there are no absolute reference frames. I assume it is the Earth??

Secondly, and this is the question which is intriguing me the most, suppose hypothetically there were only two objects in the Universe namely the two twins. If, in this hypothetical universe, one twin travelled on a rocket for a distance of say ten light years and then returned back to meet the original twin. Would the travelling twin's clock still run slow with respect to the other twin? since there would be no distant background stars/galaxies in this hypothetical universe to measure the acceleration against? Is it just as valid to say that the “stationary twin” is the one in motion? Additional comment

In a various replies to this question, it has been mentioned that the moving twin feels an acceleration and the stationary one does not, but in this hypothetical Universe containing only the two twins I cannot see how this can be the case

As Richard Feynman said about a similar situation, we can't do the experiment of removing all of the matter from the universe to see what would happen, so we don't really know.

But from what we do know, it seems like there would be spacetime even if there was hardly any matter in it. The geometry of spacetime defines the meaning of acceleration and elapsed time. The twins move "relative to spacetime" and that determines who's really accelerating and what their ages are at the end. There's nothing suggesting it's possible to have some sort of pre-spacetime that the twins could move in, but that would leave it ambiguous which one was accelerating.

Velocity is relative, but acceleration is not: you can feel it (and it feels just like gravity). But as I said in the question comments, acceleration is a red herring.

The real issue is that the travelling twin occupies two (or more) distinct reference frames. That means their worldline (their trajectory through spacetime) is not a straight line, and any inertial observer measuring that worldline will agree on that, although different observers will disagree on the lengths of the segments of that trajectory, depending on their speed relative to the traveller.

So the important factor is that the worldline has a bend in it, and so it cannot be transformed into an unbent line. It doesn't make much difference whether it's a perfectly sharp bend (your Dirac delta function), or a smooth gradual curve.

There's a variant of the twin paradox that requires no acceleration, but it has 3 participants, A, B, and C. A is based on Earth, C is based on Alpha Centauri (which we can assume is at rest relative to Earth).

B's rocket takes off. It's moving at constant speed $$v$$ (relative to Earth) as B passes A, and at that moment they use radio to synchronise their clocks to zero. B then travels towards Alpha Centauri.

One year later (according to B's shipboard clock), C sees B approaching. C starts their rocket and heads towards Earth. C's rocket has a speed of $$v$$ relative to Alpha Centauri. (According to the relativistic law of composition of velocities, B & C have velocity relative to each other of $$2v/(1+v^2)$$, using natural units, where the speed of light $$c=1$$). Just as C passes B, C synchronises their clock to B's clock. That is, C "clones" the value of "1 year" from B's clock the instant they pass one another.

One year later, according to that clock, C arrives at Earth, and as C passes A, they compare clocks. C's clock says "2 years". A's clock says "9 years".

## How does the twin paradox work with wormholes?

Wormholes are connections of different points in space-time which occur due to bending of the space time geometry. Like a piece of paper can be folded and a hole can be created to connect two points on the paper, which causes the distance between those two points to reduce to zero by creating a pathway directly from one point to another point. There are no experimental evidences to prove the existence of wormholes, however there are some theoretical solutions for wormholes.

I will describe a thought experiment which questions the existence of wormholes. I haven’t yet found out the flaw in the thought experiment or proper explanation other than that wormholes can’t exist. It would be great if I can get suggestions regarding the matter.

The idea goes thus:

Suppose two twins Howard and Raj are at the same point A in space-time. One day Howard gets the chance to travel in space (Basically the same setup as in the twin paradox). Howard boards a spaceship and starts moving away from point A in a straight line at a speed comparable to the speed of light. When he reaches point B in his journey (still travelling) the statements recorded by both of them are:

Raj: Howard has reached point B and time on my clock is 5 hours and time on Howard’s clock must be 3 hours (assume that Howard is moving with velocity corresponding to the above time intervals). So he is younger than me. Howard: Raj is moving away from me and hence he is younger than me. Howard is still moving at constant speed and hence both the frames are inertial. This is a paradoxical situation that who is correct and who is wrong. But on closer observation we find that there is no paradox. The two statements are not simultaneous according to relativity of simultaneity. Suppose event P is the event when Raj’s clock ticks 5 hours and event Q be the event in which Howard reaches point B. These two events are simultaneous in reference frame of Raj, but they are not simultaneous in Howard’s reference frame i.e. when Raj’s clock ticks 5 hours according to Howard, he is not at point B.

Now comes the main part of this thought experiment. Suppose if there is a wormhole at point B which connects directly to point A. Now Howard reaches the same point in space-time where Raj is, and still moving with constant velocity so that his frame is still inertial. Now what happens? Both of them again give the same statement that the other one is younger than me. As they have reached the same point in space-time the two statements must be simultaneous!

If we consider that wormhole connects two points only through space and not through time, than who is younger? If wormholes connects two points through both space & time, does that mean they have time travelled with respect to one another?

## Back in 1905, in Einstein special theory of relativity, there are three important conclusions-

### Time dilation:

Time becomes slow for an object when it has a velocity comparable to the speed of light.

### Length Contraction:

Length contracts for the observer when the object gains a speed comparable to light.

### Relative mass gain:

we all know that mass is a constant physical property of an object, but it’s true until it has no speed almost the same as the light. When any object runs with a velocity comparable t light, it gains its mass, and then its mass does not remain constant and it depends on its speed.

However, we have to remember that these three phenomena, i.e. special theory of relativity take place for an inertial frame of reference. Now, what is an inertial frame of reference? An inertial frame of reference is a frame of reference which is either slow or has a constant velocity. Just when it accelerates or retards, it becomes the non-inertial frame of reference and for this frame of reference special theory of relativity is not acceptable.

Now, come to the point to explain Twin Paradox-

Let’s imagine two twin brothers-A & B. Again assume A for astronaut and B for businessman. At the age of 20, A becomes an astronaut and he gets a chance to go to space, where on the other hand at this age B is a successful businessman.

After spending 50 years on the earth at the age of 70 years B sees A to return back. but, looking at him, A seems to be younger than B. Asking A, B comes to know that A is 50 years old. B comes to know that A is 50 years old. B becomes astonished. He says, brother, I know, you have spent 50 years I space but did you forget you went at the age of 20 when we were of the same age. A smiled and says, no brother, I know, I have spent only 30 years in the space. B did not believe it. A starts to explain it to him. We also should listen to him.

A says that his space-ship has a velocity of 0.8c i.e. 80% of the speed of light. The speed was almost comparable to the speed of light and at this speed, he and his space-ship felt time-dilation, i.e. normal time for B became slow for A. That’s why A spent 30 years when B spent 50 years on the earth. He also showed his calculations-

t=t0/√(1-(v/c)²) equation, where t was 50 years, the time with respect to the earth frame, v was his space-ship speed=0.8c, and c is the speed of light.

However, after calculation we get t0, the time A spent with his frame of reference= 30 years.

Now, you can ask, what was if we consider space-ship as rest frame. i.e. if we observe the whole event from the space-ship. The earth looks going away when A started his journey and the earth looks returning back when A was returning back.

No, you can not. You can not observe the scene from the space-ship frame, because it is not an inertial frame of reference. When it returns back it retards, then it Is a non-inertial frame.

I hope, you enjoy reading. Please like, share, comment if you really enjoy the reading. Stay with us along the journey.

This is a hard question to answer.

There is no paradox of course, but I'm sure that's not what you meant by it cannot be real.

A "preferred frame of reference" means a frame of reference where the physics works, but it doesn't work in others. For intance, in the rotating earth frame, in Newton's mechanics, we must take into account so-called fictitious forces to account for coriolis effects, for instance. Newton's laws do not work in that rotating frame, without adding additional terms.

Not so, with relativity. It just gets more complicated

I think the answer you're looking for is that the twin in the rocket turns around to come back home. This is the acceleration that makes the difference. It's usually depicted in Minkowski diagrams (and is usually simplified with 'instant' acceleration etc.) where you can see how the worldline of the twin in the rocket 'sweeps across' the world line of the other.

EDIT: the other way to look at it is that there are already three frames of reference in the twin paradox:
1.earthbound twin
2.outbound twin
3.inbound twin

Yes, sorry, my first answer was really just referring to twin one and twin two.

I hope someone more knowledgeable will confirm this (and I hope I'm picturing you question correctly), but if you look at the diagram in the wiki article and imagine the third twins outbound and inbound legs are twice as long, there will be a simple geometric relationship with the portion of the first twins world line that is 'swept' when twin three turns around (the part of the first twins worldline between the red and blue simultaneity lines. The third twin will have 'aged' more than the second, and 'much more' than the first.

Don't take that as gospel, though, unless someone better qualified gives it the thumbs up!

Grant Hutchison is a wizz at these diagrams. This thread descibes them quite well.

I hope someone more knowledgeable will confirm this (and I hope I'm picturing you question correctly), but if you look at the diagram in the wiki article and imagine the third twins outbound and inbound legs are twice as long, there will be a simple geometric relationship with the portion of the first twins world line that is 'swept' when twin three turns around (the part of the first twins worldline between the red and blue simultaneity lines. The third twin will have 'aged' more than the second, and 'much more' than the first.

Don't take that as gospel, though, unless someone better qualified gives it the thumbs up!

Yes, that's how it works. The simultaneity shift the accelerated twin experiences is greater the farther she is from "home" when she makes here turnaround. The effect is visible in the geometry of the spacetime diagram of the twins' relative motion.

I've seen very few discussions of "the twin paradox in GR" virtually all of the discussions I've seen are set in flat spacetime. The twins themselves are assumed to have negligible gravity.

I have seen some discussion of "twin paradox" scenarios in curved spacetime for example, one twin stays at the starting point, floating in empty space, and the other travels inertially (i.e., zero proper acceleration) to a star, "slingshots" around the star (still in free fall, zero proper acceleration), and comes back to the starting point. The presence of the star allows the traveling twin to reverse his direction of motion without any proper acceleration. But the final result is similar to the flat spacetime case: the traveling twin ends up younger when the two meet up again. The twins are also assumed to have negligible gravity themselves in this scenario.

As others have mentioned, it's not necessary to consider the presence of some large mass, and I don't recall offhand many treatments that do consider a large mass to be present.

The distance from A to B never changes. The distance from A to P=B to P and we assume the pulse reaches A and B simultaneously. A's velocity through space at .6c causes A to slow down in time, literally everything in this non-inertial frame evolves slower than B and everything around B is normal except for A. These cameras record the video of the pulsar providing a moment of simultaneity for both to sync at :00 and for the next pulse which video B stamps the time :05 and video A stamps the time :04.

This seems like a logical setup and we all can agree there is no paradox, so the paradox occurs when A records B's time and B records A's time?

This is impossible by simple geometry. (I'm assuming that you intend A to be moving in a circle around B.)

Simultaneously according to whom? Simultaneity is relative.

No, it doesn't. Only A "evolves slower" than B, because only A is moving. A's motion can't make anything else evolve slower, nor can A make anything else evolve slower by adopting a particular set of coordinates.

No, they don't, because the cameras are never co-located in space they are always spatially separated. That means relativity of simultaneity is always present the two cameras will never agree on a notion of simultaneity, at least not if they both adopt the "natural" notion of simultaneity for their state of motion, which is what you appear to be assuming.

(Note: in the other thread where we are discussing a similar scenario, I had assumed you intended A to be moving in a straight line back and forth, passing B during each straight-line segment of motion. In that case, when A and B pass each other and are co-located, they can use the pulsar signals as a common reference to compare their elapsed times. But this only works when A and B are at the same point in space.)

So now we have an omni-directional equidistant pulse where A and B sync at t=0. A's trip around P is 1000 pulses in which t=4000 for A and t=5000 for B. This obviously complicates it slightly because A's spacetime velocity is not identical to B's relative motion away from A's unique frame of rest. Would it be acceptable to break this trip into 3 phases depicted by red, green and blue?

My only goal here is to isolate and comprehend where the paradox occurs and how relativity explains it.

I'm gripping very tightly to this as a given.

Ok, that clarifies what you meant. But my remarks still apply (at least, I think they do--see below), because A and B are never spatially co-located in this scenario.

However, I'm still confused because this does not look like what you described in post #8. Was that supposed to be a different scenario? If so, it would really help to stick to one scenario at a time.

I'll assume this is a different scenario from the previous one (see above), because here it looks like A and B are spatially co-located, every time A returns around to B's position on the circle. If so, you are correct that both A and B will count the same number of pulses (1000) between two successive times that they meet, and that B's clock will record more elapsed time than A's clock (5000 vs. 4000) for that number of pulses. This is the invariant sense in which A's clock "runs slower" than B's.

The general explanation in relativity is that different paths through spacetime between the same two events can have different lengths. In the second example above, A and B take different paths through spacetime between the same two events (two successive instants at which they are spatially co-located), and A's path is shorter than B's, so A records less elapsed time. It's just geometry.

There is no single place "where the paradox occurs", because the two different path lengths between the two events are global properties of the paths, not local properties of some particular point on the paths. Asking "where the paradox occurs" is like asking "where" the difference in distance is between two different routes from New York to Los Angeles. It's not at any particular point on the routes it's a global property of the two routes as a whole.

I strongly advise you to take my repeated hints and think about this in terms of spacetime geometry. Imagine a "stack" of circles like the one you drew each one represents the space you are describing at an instant of Bob's time. (Why Bob's time? See below.) Stacking all of them together creates a diagram representing the geometry of spacetime, and the worldlines of Alice and Bob as curves in that geometry. It should be easy to see that Bob's worldline is just a vertical line, while Alice's worldline is a helix, winding around the cylinder, and crossing Bob's every time Alice passes Bob. The pulsar is at the center of the circle, and its worldline is just a vertical line as well (or a vertical tube, if you want to take into account the pulsar's size, but that's not needed for this discussion).

Now, consider two events, E1 and E2, representing two successive meetings of Alice and Bob. Each of these events is a point in spacetime, where the two worldlines cross. In between the two, Bob and Alice each follow a curve--again, Bob's is just a vertical line segment between E1 and E2, while Alice's is a helical segment that winds once around the cylinder, starting at E1 and coming back around to E2. Then the length of Bob's segment is 5, and the length of Alice's segment is 4. These are invariant geometric facts they don't depend on whose frame we pick to describe them.

Now, what about Alice thinking that Bob's clock should be running slower, because in her (non-inertial) rest frame, he is moving and she is not? The key point is this: in order to draw this conclusion, Alice has to look at the length of Bob's worldline between a different pair of events, not between E1 and E2. It's harder to visualize which pair of events this would be, because Alice's motion is circular instead of in a straight line, but the basic reason is, again, geometry: if Alice thinks that Bob's clock advanced by only 3.2 "at the same time" as her clock advanced by 4, she must be comparing the length of her worldline with the length of his worldline between some different pair of events, because we know that the length of Bob's worldline between E1 and E2 is 5, not 3.2. (That "at the same time" should also be a cue that relativity of simultaneity is involved.)

So the resolution of the "paradox" is, again, geometry: if Alice thinks Bob's clock is "running slower", it's because she is comparing lengths of worldlines between different pairs of events than the pair (E1, E2), which clearly shows Alice's clock running slower. But the events E1 and E2 are not picked out arbitrarily: they are picked out by the geometric fact that the two worldlines cross at those events. (This is another way of saying that Alice and Bob meet--are co-located--at those events it should be clearer now why I was focusing on that as being important.) So there is a good reason why we use these events to judge whose clock is running slower.

Picking out the worldline crossings also helps to explain why we used instants of Bob's time to define the "slices" of space at an instant that we then "stacked up" to form our diagram of spacetime: this is what makes the scenario look the simplest, because only one object (Alice) moves. Any other "slicing" would make things look more complicated. Other slicings are still valid, and you could try using them to draw a diagram, for example, in which Alice's worldline was a vertical line. But the geometric facts would remain the same, just as the geometric facts about distances on the Earth's surface remain the same whether we look at a Mercator projection or a stereographic projection.

This is an interesting question. If the universe is finite due to spacetime curvature, I would guess that it works out OK as the situation is non-symmetrical. The curvature is due to mass in the universe, and one of the twins is moving relative to that mass, the other one isn't.

However, you can assume finiteness due to the topology of the universe, rather than the geometry (this is like the universe of the Asteroids game). This point of view is put forward by Janna Levin, in the book How the universe got its spots. Then there's more of a problem. It's difficult to see how to get consistent Newtonian physics in such a universe, and maybe the inconsistency isn't solved by General Relativity.

This is the cosmological twin paradox and has been discussed on these Forums here in depth.

Again I quote from a paper by Barrow and Levin "The twin paradox in compact spaces" http://arxiv.org/abs/gr-qc/0101014

You do not have to have an exotic topology like a torus for this paradox, an ordinary spherical closed universe will do just as well. Two inertial observers, travelling at speed relative to each other, pass closely and set their clocks. After a very long time they pass each other closely again and compare clocks a second time.

One observer has circumnavigated the universe at speed and so her clock has recorded less proper time lapse than the other one who has remained ‘stationary’. But which one of the two is this, and how do you tell?

The resolution of this paradox is that the presence of matter in the universe has closed the universe making it a 'compact topological space'. Furthermore this topology has “introduced a preferred frame” so that one definite observer ends up younger than the other. This is because the preferred frame is determined by the mass in the universe so the stationary observer is stationary relative to the centre of momentum of the rest of the matter-energy in the universe.

However is the existence of this ‘preferred frame’ not inconsistent with the principles of GR?

We find that the paradox is resolved only at the expense of the consistency of GR!

One observer has circumnavigated the universe at speed and so her clock has recorded less proper time lapse than the other one who has remained ‘stationary’. But which one of the two is this, and how do you tell?

paradox is resolved only at the expense of the consistency of GR!

I've seen a lot of circumnavigated the universe posts & most wonder about how to tell which one traveled - I don't see the paradox.

This shouldn’t be such a mystery - send me on the trip in my sleep and I’ll figure out if I'm the traveler or not.

Assuming a uniform universe - when I start noticing Galaxies coming at me and zipping by me, I’ll at least guess I’m not in Kansas any more! And Ask to use the Wizard’s equipment - making a quick check of the background radiation should prove that what seems to be coming at me and going by so fast must be closer to “stationary” (but not preferred now) than I.

Seems to me it would be obvious I’m moving, in any shape universe, as long as it had a CBR like ours. Observers should be able use the same references and SR to figure out the same. In a similar manner that a GPS satellite and Earth station can tell each other apart by looking at the background star positions.
RB

We find that the paradox is resolved only at the expense of the consistency of GR!

RandallB I quite agree that it will be possible to work out which is travelling and which is not, the problem is the equivalence principle would have us believe you should not be able to.

The issue I am getting at is GR ought to include Mach's Principle whereas in fact MP is inconsistent with the equivalence principle, that is with the principle of 'no preferred frame'.

The equivalence principle was set up in SR in flat, empty space-time, and works in, and is appropriate for, that scenario however once you extend the theory to curved space-time and introduce matter, you introduce a means of determining a 'preferred' frame, that which co-moves with the Centre of Momentum.

In so doing you break the conditions required for the equivalence principle. This frame is preferred in the sense that it is the one in which one of our twins in stationary and records the greatest elapsed proper time between such encounters.

JesseM In a similar way the notion of "no preferred reference frame" applies to the equivalence principle which applies wherever GR is applied, i.e. cosmologically and not just at a local level.

What I am actually arguing is that this cosmological twin paradox exposes an inconsistency in GR cosmology. The topology of a compact cosmological model is in conflict with the principles of 'no preferred frames' and therefore equivalence. It is this conflict that I have tried to address in A New Self Creation Cosmology.

RandallB I quite agree that it will be possible to work out which is travelling and which is not, the problem is the equivalence principle would have us believe you should not be able to.

The issue I am getting at is GR ought to include Mach's Principle whereas in fact MP is inconsistent with the equivalence principle, that is with the principle of 'no preferred frame'.

The equivalence principle was set up in SR in flat, empty space-time,

I don't know Garth I think you may want to go back and reread some of Einstin's stuff.

He did not use "the equivalence principle" for SR at all. He only agreed with and followed Newtons lead - that the Laws shouold apply the same everywhere. Einstin was first to figure what that meant to apply them to near light speeds. SR even gave him E=mc^2

It was not till the equivalence principle that he started on GR.

And without opening the window on the ship the non move twin, the earth station or the GPS satilite I think the equivalence principle would stand up OK.

Let me reiterate Einstein founded the theory of SR on the 'no preferred frames' concept.

In the presence of gravitational fields the Einstein Equivalence Principle (EEP) is a necessary and sufficient condition for the Principle of Relativity, (PR). Here I summarise PR as the doctrine of no preferred frames of reference. In the absence of such fields the EEP becomes meaningless, although then the PR does come into its own and is appropriate in Special Relativity (SR), which was formulated for such an idealised case. However the presence of matter and the gravitational fields that it generates allows a particular frame to be identified, the co-moving centroid or centre of momentum.

The question is, "Does this constitute a 'preferred frame'?" The cosmological twin paradox reveals that it does. Of all inertial observers passing and re-passing each other after circumnavigations of the compact space of a closed universe one will have an absolutely longest proper time passage between encounters. This will be that observer in the co-moving frame, stationary wrt 'the rest of the universe,' in apparent contradiction to the Principle of Relativity.

Maybe what I'm missing is the definition of "Compact Space" What is that.

In the case of a GPS satellite and it's base station set on the North pole. Does compact space mean that neither would be able to look at the rest of the earth nor the stars behind each other?
Under those conditions the Earth station would seem to be preferred by both as long as they never become aware of the other parts of earth or stars.
Is that the meaning of "compact space".

Compact space = closed universe, finite yet unbounded.

Garth, I'm not sure the notion of a "reference frame" is even meaningful in GR, because I don't think there's a unique way of defining the "relative velocity" of two objects which aren't in the same local neighborhood. In SR, your reference frame is defined by the readings on a network of rulers and clocks which are at rest relative to you in GR you can define various abstract global coordinate systems, but I don't think they'd necessarily have this sort of physical meaning. Note that even in SR, it's certainly not true that the laws of physics work in any coordinate system where the origin is moving inertially, they only work the same in coordinate systems defined by a network of rulers and synchronized clocks. For example, say you have defined coordinates x,y,z,t based on readings on rulers and synchronized clocks at rest relative to yourself, and then I define a new coordinate system x',y',z',t' using the following abstract mathematical transformation:

Since there's a one-to-one mapping between the two coordinate systems, then any object with a fixed coordinate x',y',z' must also have a fixed x,y,z coordinate, so it must be moving inertially but this does not qualify as a valid "inertial reference frame" because it isn't based on physical measurements, and the laws of physics certainly would not look the same if expressed in this coordinate system. In GR I don't think there's this sort of natural distinction between "physical" coordinate systems and "abstract mathematical" ones (although locally there is of course, since GR reduces to SR in local neighborhoods).

Compact space = closed universe, finite yet unbounded.

It could be bounded, within the definition of compact, provided it included the boundary. What it can't be is open. Compact requires every infinite sequence of points to have a cluster point (the advanntage of "compact" is that you don't have to say "bounded sequence" the topology bounds it for you). So in a flat or hyperbolic universe you could define a sequence going "out to infinity" wth each point a fixed step away from all the preceding ones, so it would never cluster.

Since we tend to assume the universe has no boundary, we also tend to skip over the fine points of the definition.

For example, say you have defined coordinates x,y,z,t based on readings on rulers and synchronized clocks at rest relative to yourself, and then I define a new coordinate system x',y',z',t' using the following abstract mathematical transformation:

Since there's a one-to-one mapping between the two coordinate systems, then any object with a fixed coordinate x',y',z' must also have a fixed x,y,z coordinate, so it must be moving inertially

Why don't we ask a similar question:

If, say on my basketball at home, we had a two dimensional finite yet unbounded universe: how does the twin paradox play out there? Well, obviously, we can equate a twin travelling around a 2d universe (sphere) to me walking in a circle. The twin that goes around ends up younger because he is 'accelerating' around the ball

It could be bounded, within the definition of compact, provided it included the boundary. What it can't be is open. Compact requires every infinite sequence of points to have a cluster point (the advanntage of "compact" is that you don't have to say "bounded sequence" the topology bounds it for you). So in a flat or hyperbolic universe you could define a sequence going "out to infinity" wth each point a fixed step away from all the preceding ones, so it would never cluster.

Since we tend to assume the universe has no boundary, we also tend to skip over the fine points of the definition.

The qualifier "unbounded" is not necessary.

Alkatran Alternatively Use a cylinder instead.
The long axis represents the time axis and the circumference represents space.

Have two pins on the outer surface at either end and connect with two elastic strings, one of which is straight between the pins and the other twists round the cylinder between the two.

Allow one end of the cylinder to be rotated relative to the other end.

The straight string represents the world-line of a 'stationary' observer and the twisted string a moving observer. Obviously the model is set up in the frame of reference of the first observer. If we rotate the cylinder the twisted string can be made straight and the other now twists around the cylinder. We are now in the frame of the second observer.

You cannot straighten out both strings, there is always a difference of one complete twist between them, this is the 'winding number’, which is a topological invariant.

However how do you tell the difference between the two? As we have set it up you cannot, each scenario is equivalent to the other and there is no preferred frame or observer. However in a real closed or 'compact space' universe one observer will definitely have run up a longer elapsed time between encounters than the other, her frame can therefore be said to be 'preferred'.

My point is that this preferred frame is introduced by the presence of mass in the universe. It is the distribution of matter in motion that determines this special frame of reference and that is in accordance with Mach's Principle rather than those of Einstein's relativity.

JesseM A 'physical' coordinate system, as opposed to a merely 'mathematical' one, is defined by a system based on physical measurements, i.e. scales, clocks and rulers. Nothing you have said has convinced me that this paradox does not reveal an inconsistency in GR. The observer is in a physical preferred frame in the sense that it is determined by the measurement of her clock.

JesseM A 'physical' coordinate system, as opposed to a merely 'mathematical' one, is defined by a system based on physical measurements, i.e. scales, clocks and rulers. Nothing you have said has convinced me that this paradox does not reveal an inconsistency in GR. The observer is in a physical preferred frame in the sense that it is determined by the measurement of her clock.

Well, how do you propose an observer should set up a network of clocks and rulers throughout space to define a "physical" coordinate system? There are all kinds of problems that will crop up in GR--for example, in curved spacetime you can't assume that clocks in different locations will all tick at the same rate, because gravitation causes time dilation (not to mention the stretching of rulers). What's more, in SR we assume that all the clocks and rulers are at rest relative to each other, but as I said before there is no unique way to define the "relative velocity" of two distant objects in GR the only way to compare distant velocities is by doing a parallel transport of the velocity vector at one point along a geodesic to another point, but there can be multiple geodesics between two points (think of gravitational lensing), so the result of the parallel transport is path-dependent. Imagine if you tried to define a coordinate system in SR using clocks which were not at rest relative to one another--the laws of physics certainly wouldn't look the same in such a coordinate system as they do in normal inertial coordinate systems, even if each clock in the system is moving inertially. So despite the fact that this coordinate system is still based on "physical measurements" in some sense, it doesn't qualify as a valid inertial frame. In GR there doesn't seem to be any way to define the notion of a network of clocks which are all at rest wrt one another, so how do you distinguish between coordinate systems that qualify as "reference frames" and those that don't?

One more point: gravitational lensing shows that in GR, unlike SR, light from an event can take multiple paths to reach you. In SR, different clocks are synchronized using light signals--If I look through my telescope and see a clock one light-year away that reads "12:00, Jan. 1, 2105" then at the moment I receive that light my clock should read "12:00, Jan. 1, 2106" if the two are "synchronized" according to the SR definition. But if light from a single event can reach me at two different times, how are clocks to be synchronized in GR? I suppose if you have some unique way of defining the "distance" that light travelled along a geodesic, you could divide the distance along a particular geodesic by c to define "how long ago" the event happened, but I don't know if such a unique definition of "distance" is possible, and I'm pretty sure that the answer you'd get for "how long ago" the event happened using this procedure would be not necessarily be the same along two different paths (As an extreme case, imagine a wormhole that allows an event to send light into its own past light cone, something that is permitted by GR although quantum gravity may rule it out in this case, an observer might first receive light that travelled 'backwards in time' through the wormhole, then later receive light from the same event that travelled along a more normal path.)

JesseM -The coordinate system doesn't have any real physical significance (even thoguh it's time axis corresponds to the worldline of an inertial observer) as the basis vector fields don't correspond to anything of real physical signifcance. In SR arbiartry coordiante transformations are not particularly interesting as the laws of SR refer to specifc sets of coordinate sytems only and it only really makes sense to call these coordinate systems inertial.

In GR it certainly does make sense to tlak of reference frames, though two obsrevres are usually considerd to be in different refernce frmaes not only if they are travelling at different velocities, but if they are spatially separted. The laws of GR apply to all coordiante systems, so in this sense there is not distinction between coordinate systems with physical and those without signifacnce, though it doesn't mean that coordinate systems with physicla signifcance don't exist.

Garth - the term compact manifold is probably what you're looking for as it implies usually that the manifold is boundaryless (I think the term boundaryless is better than unbounded as in actually fact these universe are bounded metric spaces!).

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This is actually an excellent question and it bothered me for an incredibly long time because I thought exactly what you did: acceleration is relative, is it not? One twin will see the other's speed is changing, and vice-vera. So how does the universe know which once is actually accelerating? The answer is that acceleration is detected using a scale. If you stand on a scale on the Earth, you see a number appear: your weight is non-zero. This means that you are accelerating, and certainly you must be, because the Earth is rotating, and because you are on the Earth, you must be moving with it. Thus gravity acts as a centripetal force that pulls you down, and you measure the existence of this force on a scale. Even if the Earth were not rotating, you are still in a non-inertial frame because there is still a force felt and a measurement made on a scale. Consider throwing a ball horizontally on the Earth: it will fall to the ground. If there Earth were an inertial frame, there would be no way for said ball to gain a vertical component of velocity relative to the Earth. If you step on a scale and measure a weight, your motion is said to be non-inertial.

Consider yourself in free fall above the Earth, and ignore wind resistance (to be specific, consider just the tug of gravity). If you stood on a scale moving with you, it would read zero. Thus, as far as we are concerned in relativity, you are NOT accelerating, and there is no force acting on you. This might be a difficult concept to come across because you might think otherwise. Indeed you are gaining speed as you move toward the Earth, but in your own reference frame you feel no force, and thus we say you are not accelerating, and you are in an inertial rest frame. Actually, inertial reference frame are an idealized concept, only a single point can actually be inertial: all extended bodies can only be approximated as inertial. But this is off-topic, so for our purposes we make the rather good approximation that a person in free fall is in an inertial reference frame.

The final answer is then that during the acceleration, the twin in the rocket could stand on a scale and the reading on the scale would reflect the acceleration process, whereas the other twin's scale should theoretically read zero because they should be in an inertial reference frame. It is the twin in the non-inertial reference frame that actually experiences the acceleration, and this is the twin that has a reading on the scale.

What you should take home from this is that forces and acceleration are detected via a scale. Certainly in many physics courses you might take that the acceleration is the time derivative of velocity, and that works but, we see here that thinking of acceleration in this way leads to some ambiguity and makes the twin paradox seem more daunting than it should be.

Indeed, you hit the nail on the head: how are they able to have lunch together?

The answer is because they have matched velocities and are in the same rest frame. Thus, they measure the universe identically.

Sure, while on their trips they made all sorts of measurements. (Hey look, that restaurant passing by my windows at .999c looks massively longitudinally contracted - as does my local universe! And everything is now moving verrrry slowly.)

But those measurements are relative. When they slow their ship and turn around, they will get different answers. They know they are changing their velocities, so they will certainly know their measurements will vary accordingly.

(Ah, now that I am in the same rest frame as the restaurant, it looks quite normal, as does my local universe.)

13.8 billion years old and is such and such a diameter across. Bob says, no no no, the universe is

13.8 byo + 5 years and has an accordingly larger diameter.

Bob is correct, because (by hypothesis) he has remained at rest in "comoving" coordinates--physically, he has continued to see the universe as homogeneous and isotropic. To relate how old you think the universe is to its diameter in the way you are assuming, you have to be such an observer--one who has always seen the universe as homogeneous and isotropic. Since Alice is not such an observer, she has to apply a correction when deriving the diameter of the universe from how old she thinks it is with the correction applied, she gets the same answer Bob does.

(Note that the above applies to Alice after she has come back and met up with Bob again, so she is now at rest relative to Bob. During Alice's trip, the "diameter" she assigns to the universe in a frame in which she is at rest will be different because of her motion relative to Bob, and the relationship between this "diameter" and the age Alice assigns to the universe will also be different. I put "diameter" in quotes here because Alice's surfaces of simultaneity while she is in motion are different from Bob's, so the "diameter" she is assigning belongs to a different slicing of spacetime into space and time. Discussions of cosmology almost always assume "comoving" observers, like Bob, and the corresponding slicing so if you deviate from that, you have to be careful not to make assumptions that are only valid for that slicing.)

(Also, I assume that the "diameter" you mean here is the diameter of the observable universe. According to our best current model, the universe itself is spatially infinite.)

A short answer - there is no absolute time in special or general relativity, so there is no true "absolute age" of the universe. We do have certain conventions, those conventions are used when "the age of the universe" is given. These conventions are that we measure the age in a co-moving frame, one in which the CMB is isotropic.

Seems simple to me, but it appears people have difficulties with giving up the idea of "absolute time".

Yeah, that's exactly my point. My point is that, as they are having lunch together and look through the telescope the waiter brought them, they each look through the lens and see a different universe? With a different age and diameter? (visible universe, that is).

PeterDonis says Bob is correct. Does that mean that when Alice looks through the telescope, she sees the universe as Bob sees it? Is that how this is reconciled?

Of course not. They are both at the same point in spacetime, with the same velocity, so they see everything the same. How they got there does not affect what they see at that moment it only affects the elapsed times on their respective clocks.

That doesn't really make complete sense to me. If the age (of the visible universe I'm assuming) would be different for Bob and Alice, then they would not see the same thing looking through the telescope, would they?

Ok, sounds like there may be differences, but.

Ok, that sounds like there are no differences.

Ok, let's take the standard twin paradox, Alice leaves on a trip in her rocketship near the speed the light, and comes back to Earth some time later to find herself 5 years younger than her twin, Bob.

Now they go out to lunch and strike up a conversation as to how old the universe is. Alice says it's

13.8 billion years old and is such and such a diameter across. Bob says, no no no, the universe is

13.8 byo + 5 years and has an accordingly larger diameter. Which one is correct? How do we reconcile this? I mean, if each are living in a different-sized universe, how are they able to have lunch together? For instance, how are the physical properties of the restaurant not affected, etc.?

What is special about universe, so that universe is not just another aging object?

Alice: "During my 1 year trip universe and Bob aged so quickly that they aged 6 years."

Bob: "During Alice's 6 year trip universe and me aged 6 years. But Alice aged so slowly that she aged 1 year."

Nothing. The universe IS just another aging object, but like Bob and Alice, and everything else, how you measure the age of the universe depends on your path through spacetime.

What you seem to not be getting is that the age of an object ALWAYS happens at one second per second for that object, BUT . when compared to the age of a different object, then what matters is the path through spacetime that each have taken.

The age of the universe is not something you directly observe it's something you calculate, and the calculation depends on your own past history, not just on what you're currently observing. Bob and Alice each have the same current observation, but they have a different past history, so they calculate a different age for the universe.

What this really means is that the term "age of the universe" is really something of a misnomer for what we are all talking about. See below.

You're imagining it way too simply. Alice's watch registers the proper time elapsed along her worldline, and Bob's registers the proper time elapsed along his worldline. If we suppose (which of course can't be the case in a real scenario) that both watches were set to time zero at the Big Bang, then the reason the two watches read differently is simply that Alice and Bob followed different worldlines to get to the same current event in spacetime. So the readings on their watches aren't really telling you "how old" the universe is, because the readings don't depend on the universe itself--both worldlines are in the same universe. They're just different worldlines, and the watch reading depends on the worldline. So Alice should not expect a simple relationship between her watch reading and what she sees through the telescope, and she certainly should not expect to see the same relationship as Bob does.

When cosmologists use the term "age of the universe", what they really mean is the proper time elapsed since the Big Bang along a "comoving" observer's worldline, like Bob's. The special property that picks out "comoving" worldlines from all other worldlines (like Alice's) is that "comoving" observers always see the universe as homogeneous and isotropic observers following any other worldline will, for at least some portion of their history, see the universe as non-isotropic or non-homogeneous (like Alice does during her trip). So the "age of the universe" in cosmology is really "the proper time elapsed since the Big Bang along a comoving worldline". A non-comoving observer, like Alice, can still calculate this number from their watch reading and what they see through a telescope, but the calculation won't be as simple as it is for a comoving observer, who can just read it directly off his watch (again, assuming an idealized case where the watch is set to time zero at the Big Bang).

Originally i just wanted to look at how much analogy can be made between light and sound waves using all that math has to offer to depict them in most similar framework possible - just so as to have a different perspective to understand some things better. Anyhow, no matter how well one tries to hide the (sound) medium (and that can be done pretty well), it will always be there and not allow for a perfect equivalence of inertial frames.

In order to make comparisons with light/SRT of some specific aspects i was curious about i went back to the twin paradox and looked as some special modifications to make them stand out. In particular the twin paradox in a closed world (cylinder manifold which is geometrically perfectly flat) and luckily the internet already had something on it: https://physics.stackexchange.com/questions/353216/twin-paradox-in-closed-universe
I'm not sure if the explanation is fully correct, but it surprised me as it breaks the equivalence of frames (i picked the example specifically to understand how SRT maintains the equivalence under non trivial circumstance).

The problem can be traced back to the issue that in a closed world the one way speed of light seems to be partially measurable: send two signals around the world in opposed directions and if they don't come back at the same time there is a difference in the one way speed of light along that axis. In the example in the link this causes Betties plane of const time to twist such that in her frame her other instances are at a different age on her plane of const time (due to clock synch convention).

And I don't understand how to fix this, because all waves travel at a fixed speed i.e. independently from the source they were emitted from. So if Betty and Albert emitted signals to measure the one way speed of light from Alberts place they will travel together and an asymmetry becomes visible making Alberts frame to stand out (unless a many worlds approach is taken). So Albert's frame seems to be more at rest then Betty's.

Anyhow, so i thought maybe this is a issue unique to a closed world setup. But in all versions of the twin paradox there is a age asymmetry between the "traveling" and "home" twins. I looked at the the "out and back" twin resolution at Wikipedia: https://en.wikipedia.org/wiki/Twin_paradox#A_non_space-time_approach. It needs the 3 twins A (Albert), O (Outgoing) and B (Back) i.e. trins. So to make the example easier let's copy/clone each of the trins indefinitely and arrange them in evenly spaced grids A, O, B. Now whenever a trin passes by another the times of all clocks can be recorded. In particular of interest are the time intervals ##Delta t_## measured by a clock at grid ##I## for the time between two meetings with a trin from grid ##J##. Now let all trins be at that at ##x,t=0## point our initial condition. Furthermore the start of the return trip can be set to when O meets B the first time since ##t=0##. Then I gather that the resolution implies $frac 1 2 Delta t_ + frac 1 2 Delta t_ < Delta t_ = Delta t_$
So that would seem to imply that frames O and B are still not fully equivalent to A since ##Delta t_ eq Delta t_## even without the closed world assumption. So it would seem clocks tick indeed faster in one grid then in the other but in an absolute sense (after all the ##Delta t_## just compare two times of the same inertial frame clock between events at its exact location - i.e. independent of any clock synch and whatever). I also considered viewing grid O as the "stay home trin" and A as the outgoing one. That needs introducing another grid C that serves the travel back trin role for O. But doing so is a bad idea because that seems to shatter logic consistency a bit (so far i could not resolve the contradictions between the ##Delta t_## relations of all 4 grids).

Anyhow, the closed world case provides a mean to associated the ##Delta t_## asymmetry with one way speed of light difference and having that, one can add more grids moving at different velocities just to probe the asymmetry. But does that not give a means of deriving the one speed of light in all directions and deduct a frame specific offset velocity vector?

So as you can see I am at a loss now since the one way speed of light is not allowed to have any measurable effect therefore all the asymmetries cannot exist. Or to put it more provocatively: a detectible offset velocity in the one way speed of light is basically an aether wind - so i expected these asymmetries only to pop up for my sound waves analogon but not for light. I don't get where the mistake in all this is and the more I try to understand SRT the less I actually do.