The discussion centers on the implications of the twin paradox within the context of a toroidal universe. Participants explore whether both twins can be considered equivalent when one travels through a compact space, examining the effects of spacetime curvature and topology on their relative aging. The conversation touches on theoretical aspects, potential inconsistencies with general relativity, and the nature of preferred frames in cosmological models.
Participants express a range of views, with no consensus on whether the twins can be considered equivalent in a toroidal universe. Disagreements persist regarding the implications of general relativity, the nature of preferred frames, and the validity of Mach's Principle in this context.
The discussion highlights limitations in understanding the implications of compact spaces and the potential inconsistencies that arise when applying general relativity to cosmological models. The assumptions regarding the nature of spacetime curvature and the role of mass in determining reference frames remain unresolved.
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.touqra said: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?
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,.
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.Garth said: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 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 said:We find that the paradox is resolved only at the expense of the consistency of GR!
Garth
I don't know Garth I think you may want to go back and reread some of Einstin's stuff.Garth said: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,
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:Garth said: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 said:Compact space = closed universe, finite yet unbounded.
Garth
JesseM said: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;
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.selfAdjoint said:"Inertial" means non-accelerated. Your example has a highly nonlinear curved worldline, hence an observer experiencing it is accelerated, therefore not inertial.
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".selfAdjoint said:Inertial observers in flat Minkowski space have straight worldlines, in GR their worldlines are geodesics.
Thank youselfAdjoint said: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.
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?Garth said: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
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.jcsd said: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 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 said: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.
JesseM said: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.
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?
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.jcsd said: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 separated) 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.
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 said:The equations of GR look the same in all coordinate systems in GR as they are tensorial.
JesseM said: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.
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?
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}.The Bob said:\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt
Out of interest, what would this give?
The Bob (2004 ©)
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?jcsd said:You just differentiate a timelike coordiante wrt a spacelike coordinate, usually the choice of which coordinates are due to phsycial concerns
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 said: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.
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.JesseM said: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?
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?
Garth said: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 said:You do have to worry about the unievsre as a whoole tho' as the spatial slices are different (psuedo-Riemannian) manifolds to each observer.