Are Both Twins Equivalent in a Torus Universe with the Twin Paradox?

[touqra]

Let say, the universe is a torus, a doughnut shape. Regarding the twin paradox, it is possible, that the moving twin is not in an accelerating frame, and can still come back to Earth to compare her time with the twin stationary on Earth.
Then, both the twins are equivalent in a torus universe.
Right?

 

[chronon]

Let say, the universe is a torus, a doughnut shape. Regarding the twin paradox, it is possible, that the moving twin is not in an accelerating frame, and can still come back to Earth to compare her time with the twin stationary on Earth.
Then, both the twins are equivalent in a torus universe.
Right?

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.

 

[Garth]

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

In a compact space, the paradox is more complicated. If the traveling twin is on a periodic orbit, she can remain in an inertial frame for all time as she travels around the compact space, never stopping or turning. Since both twins are inertial, both should see the other suffer a time dilation. The paradox again arises that both will believe the other to be younger when the twin in the rocket flies by. The twin paradox can be resolved in compact space and we will show that the twin in the rocket is in fact younger than her sibling after a complete transit around the compact space. The resolution hinges on the existence of a preferred frame introduced by the topology,.

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, traveling 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!

Garth

 

[RandallB]

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!

Garth

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

 

[JesseM]

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

Garth

I think in GR, the notion of "no preferred reference frame" only applies at a local level, there is no guarantee that if you create different coordinate systems which cover large regions of spacetime, the laws of physics will work the same in each coordinate system.

 

[Garth]

RandallB I quite agree that it will be possible to work out which is traveling 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 http://www.kluweronline.com/oasis.htm/5092775.

Garth

 

[RandallB]

RandallB I quite agree that it will be possible to work out which is traveling 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.

RB

 

[Garth]

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.

Garth

 

[RandallB]

Compact Space?

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".

 

[Garth]

Compact space = closed universe, finite yet unbounded.

Garth

 

[JesseM]

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.

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:

$$x^{'} = 12x$$
$$y^{'} = \pi y$$
$$z^{'} = z^9$$
$$t^{'} = e^t$$

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).

 

[selfAdjoint]

Compact space = closed universe, finite yet unbounded.

Garth

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.

 

[selfAdjoint]

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;

"Inertial" means non-accelerated. Your example has a highly nonlinear curved worldline, hence an observer experiencing it is accelerated, therefore not inertial. Inertial observers in flat Minkowski space have straight worldlines, in GR their worldlines are geodesics. In SR Lorentz transforms obtain between all inertial frames.

 

[JesseM]

"Inertial" means non-accelerated. Your example has a highly nonlinear curved worldline, hence an observer experiencing it is accelerated, therefore not inertial.

What do you mean "experiencing it"? An observer whose x',y',z', coordinates don't change as t' varies (one who is at rest in this coordinate system) will also have x,y,z coordinates that don't change as t varies, no? That means the observer will have a straight wordline as seen in any inertial frame, so he is moving inertially. But while you can call the x',y',z',t' system an "inertial coordinate system" in the sense that an observer at rest in these coordinates must be moving inertially, it's not a valid inertial reference frame, as I understand the definitions.

Inertial observers in flat Minkowski space have straight worldlines, in GR their worldlines are geodesics.

I know that, but I wasn't talking about worldlines, I was talking about coordinate systems. My point is that in GR I don't think it would make sense to ask whether a global coordinate system (say, Hubble coordinates) is "inertial" or "non-inertial".

 

[Alkatran]

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 traveling 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

 

[Garth]

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.

Thank you

compact = not open; finite.

The qualifier "unbounded" is not necessary.

Garth

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.

Garth

 

[JesseM]

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.

Garth

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 traveled 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 traveled 'backwards in time' through the wormhole, then later receive light from the same event that traveled along a more normal path.)

 

[jcsd]

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 systems 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 traveling 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!).

Cleraly this cosmological twin paradox doesn't actual breach GR in any tangible way, the worse it could be claimed is that such a result brecahes the philosphical aims of GR, howver even then I would disagree as the phislophy of GR does not extend to measuring everything the same in all coordinate sytems. Look at it this way: just like in the original twin paradox the two apparently symmertical obsrevers are not symmertical, in the cosmological version this is due to the fact that they both see different universes i.e. their spatial slices are different, so it should be no supridse when they measure different times for the round trip.

 

[JesseM]

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 traveling at different velocities, but if they are spatially separted.

But what I was saying is that there is no global notion of an observer's "reference frame" in GR, in the sense of a global coordinate system which tells you stuff like how fast a distant object is moving "in your reference frame". You seem to be saying that it only makes sense to say two objects are in "the same reference frame" if they are in the same local neighborhood, which agrees with that.

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.

But when you say the same laws of GR apply to all coordinate systems, does that actually mean the equations you'd use to make physical predictions would look the same in different coordinate systems? In the case of the twins circumnavigating a flat finite universe, if you define each twin's coordinate system the same way you would in SR, it doesn't seem like both can use the same equations to predict the other twin's time dilation as a function of speed in their own coordinate system, since each one sees the other moving at the same constant speed v in their coordinate system. Maybe the answer is that in GR you can't assume the equations describing things like time dilation are just functions of the coordinate values for position, time and velocity, maybe the equations would also have to be functions of the metric at each point in space...but in the case of a flat space whose topology allows it to be finite, doesn't the metric look the same in both coordinate systems?

 

[jcsd]

But what I was saying is that there is no global notion of an observer's "reference frame" in GR, in the sense of a global coordinate system which tells you stuff like how fast a distant object is moving "in your reference frame". You seem to be saying that it only makes sense to say two objects are in "the same reference frame" if they are in the same local neighborhood, which agrees with that.

Well an obsrever can observe distant objects and assign them a velocity based on these obsrevations (Though this velcoity is most defintelty diffefernt from the concept of the relative velocity of two objects which are not spatially seperated) and you cna construct a coordinate system to reflect the obsrevations of an obsrever (though it's entirely possible that the coordianate system assigns multiple sets of coordinate sto the same points or that it contains singularities). generally speaking you can't regard spatially seprated observers as having the same reference frame.

But when you say the same laws of GR apply to all coordinate systems, does that actually mean the equations you'd use to make physical predictions would look the same in different coordinate systems? In the case of the twins circumnavigating a flat finite universe, if you define each twin's coordinate system the same way you would in SR, it doesn't seem like both can use the same equations to predict the other twin's time dilation as a function of speed in their own coordinate system, since each one sees the other moving at the same constant speed v in their coordinate system. Maybe the answer is that in GR you can't assume the equations describing things like time dilation are just functions of the coordinate values for position, time and velocity, maybe the equations would also have to be functions of the metric at each point in space...but in the case of a flat space whose topology allows it to be finite, doesn't the metric look the same in both coordinate systems?

The equations of GR look the same in all coordinate systems in GR as they are tensorial.

For a universe that is geometrically flat, but finite the universe does not look the same to both twins, though it would appear flat to both of them (e.g. let's say this universe is a cylinder, they both still see the universe as a cylinder, but one of the twins would measure it's circumference to be less than the other one would)

 

[JesseM]

Well an obsrever can observe distant objects and assign them a velocity based on these obsrevations (Though this velcoity is most defintelty diffefernt from the concept of the relative velocity of two objects which are not spatially seperated) and you cna construct a coordinate system to reflect the obsrevations of an obsrever (though it's entirely possible that the coordianate system assigns multiple sets of coordinate sto the same points or that it contains singularities). generally speaking you can't regard spatially seprated observers as having the same reference frame.

But is there a unique physical procedure for an observer to assign a distant object a "relative velocity", or are there many equally valid procedures? Given the fact that the parallel transport of velocities is path-dependent, I'd think there could be no uniquely good way to assign distant objects a velocity, as there is in SR.

The equations of GR look the same in all coordinate systems in GR as they are tensorial.

I guess I'd have to know more about tensorial math to really understand this. But here's another question--in SR, if we know the position and velocity of a clock as a function of time in our own coordinate system, then we can calculate the time that elapses on the clock between two moments $$t_0$$ and $$t_1$$ (in our own time-coordinates) by integrating a function F that tells us the rate the clock is ticking at a given moment, as measured by our own time coordinate. In SR this function would just be $$\sqrt{1 - v(t)^2 / c^2}$$, so to find the elapsed time you'd evaluate the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt$$. Is it also possible to calculate the elapsed time in a given coordinate system using a similar integral in GR? If so, in this case the integral $$\int_{t_0}^{t_1} F \, dt$$ would not have an F that was a function only of the velocity v(t), presumably it would also have to be a function of the metric at the clock's current position. But if you can calculate the elapsed time by integrating some function F of the velocity, the metric, etc., then would this function have the same equation in different coordinate systems?

 

[jcsd]

But is there a unique physical procedure for an observer to assign a distant object a "relative velocity", or are there many equally valid procedures? Given the fact that the parallel transport of velocities is path-dependent, I'd think there could be no uniquely good way to assign distant objects a velocity, as there is in SR.

You just differentiate a timelike coordiante wrt a spacelike coordinate, usually the choice of which coordinates are due to phsycial concerns

I guess I'd have to know more about tensorial math to really understand this. But here's another question--in SR, if we know the position and velocity of a clock as a function of time in our own coordinate system, then we can calculate the time that elapses on the clock between two moments $$t_0$$ and $$t_1$$ (in our own time-coordinates) by integrating a function F that tells us the rate the clock is ticking at a given moment, as measured by our own time coordinate. In SR this function would just be $$\sqrt{1 - v(t)^2 / c^2}$$, so to find the elapsed time you'd evaluate the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt$$. Is it also possible to calculate the elapsed time in a given coordinate system using a similar integral in GR? If so, in this case the integral $$\int_{t_0}^{t_1} F \, dt$$ would not have an F that was a function only of the velocity v(t), presumably it would also have to be a function of the metric at the clock's current position. But if you can calculate the elapsed time by integrating some function F of the velocity, the metric, etc., then would this function have the same equation in different coordinate systems?

What you would do is take the path integral (just like SR) of the worldline of the obsrever between the two events. The time between two events is different for different obsrevers as they are calculating different things. Any equation written in terms of general tensors is independent of the coordinate system choosen.

 

[The Bob]

$$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt$$

Out of interest, what would this give?

The Bob (2004 ©)

 

[JesseM]

$$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt$$

Out of interest, what would this give?

The Bob (2004 ©)

That depends on what the function v(t) is along the particular path the clock is moving. For a clock whose velocity is constant, so $$v(t) = v_0$$, $$\sqrt{1 - {v_0}^2 / c^2}$$ is just a constant that's not a function of time, so the integral would just work out to $$(t_1 - t_0)\sqrt{1 - {v_0}^2 / c^2}$$.

 

[JesseM]

You just differentiate a timelike coordiante wrt a spacelike coordinate, usually the choice of which coordinates are due to phsycial concerns

But when you say "due to physical concerns", don't you just mean something like the coordinate system that makes the math easiest in the particular physical situation you're looking at? Unlike in SR, there's no notion that you must use a particular type of coordinate system if you want to describe things in a particular observer's "reference frame", is there?

What you would do is take the path integral (just like SR) of the worldline of the obsrever between the two events. The time between two events is different for different obsrevers as they are calculating different things. Any equation written in terms of general tensors is independent of the coordinate system choosen.

OK, so could you indeed find the time elapsed on the moving clock by integrating some function F(v(t), M(t)) where v(t) is the velocity of the clock at any time-coordinate t, and M(t) is the spacetime metric at the position the clock is at any time t? (Maybe you'd also need F to be a function of the stress-energy tensor at the clock's position, I don't know.) And if so, then would the general form of F (before you plug in a specific v(t) or M(t)) be the same in every coordinate system?

 

[jcsd]

But when you say "due to physical concerns", don't you just mean something like the coordinate system that makes the math easiest in the particular physical situation you're looking at? Unlike in SR, there's no notion that you must use a particular type of coordinate system if you want to describe things in a particular observer's "reference frame", is there?

But you're still going to prefer coordinate systems which [rproduce numbers which correspond to the results of physical measuremnts. You use whichever coordinate system is most convinete.

OK, so could you indeed find the time elapsed on the moving clock by integrating some function F(v(t), M(t)) where v(t) is the velocity of the clock at any time-coordinate t, and M(t) is the spacetime metric at the position the clock is at any time t? (Maybe you'd also need F to be a function of the stress-energy tensor at the clock's position, I don't know.) And if so, then would the general form of F (before you plug in a specific v(t) or M(t)) be the same in every coordinate system?

The metric in the purest sense is a function whose domain is the Cartesian product of the set of points on the mainfold and itself whose value is analogus to the distance between the two points. The metric tensor field defines the metric on the mainfold and is not necessarily (infact only in 3 special cases IIRC) a homogenous field. The time elapsed on a clock inbetween two events on the clock's worldline is independent of coordinate system full stop.

 

[Garth]

It's me again!

The only measurements made in this paradox are the local neighbourhood measurement of time made by both observers as the pass close by each other twice, with one of them circumnavigating the universe in the meantime. One of them will record a longer elapsed time, but which one?

You don't have to worry about constructing a cosmological coordinate system, just a local one.

How is the 'older' 'twin' selected?


Garth

 

[jcsd]

It's me again!

The only measurements made in this paradox are the local neighbourhood measurement of time made by both observers as the pass close by each other twice, with one of them circumnavigating the universe in the meantime. One of them will record a longer elapsed time, but which one?

You don't have to worry about constructing a cosmological coordinate system, just a local one.

How is the 'older' 'twin' selected?


Garth

You do have to worry about the unievsre as a whoole tho' as the spatial slices are different (psuedo-Riemannian) manifolds to each observer.

 

[yogi]

Garth - if you the author of the Self Creation Cosmology paper - I have a question.

 

[Garth]

You do have to worry about the unievsre as a whoole tho' as the spatial slices are different (psuedo-Riemannian) manifolds to each observer.

Certainly the two observers have different planes of simultaneity, that is why they record different times, however they can pass each time sufficiently close that the problem of synchronizing clocks is negligible compared to the time of passage around the universe.

One of them will record the longer elapsed time, but which one? Each has remained in an inertial frame of reference. The principle of relativity would have it that each observer is equivalent. Obviously we have to consider the rest of the universe in deciding which one is the stationary observer, the question is does this then violate the principles of GR? The selected frame of reference being a 'preferred one' in the sense that its clock is unique, it measures the maximum interval between all such twin encounters.

yogi Yes I am the author of the Self Creation Cosmology paper, ask away!

Garth

 

[jcsd]

Certainly the two observers have different planes of simultaneity, that is why they record different times, however they can pass each time sufficiently close that the problem of synchronizing clocks is negligible compared to the time of passage around the universe.

Yes I don't dispute that.

One of them will record the longer elapsed time, but which one?

I don't dispute that one observer will meausre time a longer time, infact I don't dispute (that their exist situations in which) their is one single observer out of the class of all possible observers who are locally inertial who experinces a maximal time.

Each has remained in an inertial frame of reference. The principle of relativity would have it that each observer is equivalent.

In GR though all obsrevers are equivalent, so when considering the equivaelnce principle why shouldn't we consider all the other obsrevers whose worldlines are coincidental to the two events? The only inetersting thing about our observers is that they travel along geodesics; clearly in GR are there are locally-length minimizing paths hich can be regarded as soem sense special, but I do not see the fact that under some topolgies more than one of these paths can pass through the same two points as a violation of GR! Surely you recognize that GR does assert absolute equiavalence in all ways for all observers (if that were the case we wouldn't get different measuremnts in different frames), only the equvalance of the laws of physics in each frame.

Obviously we have to consider the rest of the universe in deciding which one is the stationary observer, the question is does this then violate the principles of GR?

The universe can be regarded as geometrically different for each observer and the source of that geomertical difference can be traced to the different arrangement (caused by relative motion) of the matter in that universe (though of course both observers still inhabit the same spacetime).

 

[JesseM]

It's me again!

The only measurements made in this paradox are the local neighbourhood measurement of time made by both observers as the pass close by each other twice, with one of them circumnavigating the universe in the meantime. One of them will record a longer elapsed time, but which one?

You don't have to worry about constructing a cosmological coordinate system, just a local one.

How is the 'older' 'twin' selected?


Garth

If neither one's frame extends beyond their local neighborhood, and they don't have any type of global coordinate system, then neither one has any way to predict how fast the other one's clock is ticking while they're apart, so they can't predict how old the other will be when they meet again. You might as well imagine two twins that separate, one flies close to a black hole, then they reunite (both having traveled along geodesics, so locally they always observe the laws of physics working the same way as in SR), and the one that flew close to the black hole is younger thanks to gravitational time dilation...do you think this violates the SR principle that the laws of physics should work the same in all reference frames as well?

 

[yogi]

Garth - actually have several questions that relate to varying G - will send private message. Thanks Yogi