A variation on the twin paradox

[JesseM]

I'm not using two laboratories.

Well, you're at least using one laboratory at two significantly different points in time, as opposed to one arbitrarily small laboratory looked at during a single arbitrarily small time interval.

Barrow and Levin say in The twin paradox in compact spaces

The resolution hinges on the existence of a preferred frame introduced by the topology

Yes, but they aren't talking about a local preferred frame, they're just saying that if you create global coordinate systems for each twin using the same procedure that's used in SR, then one of these coordinate systems is in some sense "preferred" (for example, they'll be only one such coordinate system where each apparent copy of the same clock is synchronized with every other copy). Also, my understanding is that because of diffeomorphism invariance, if you express the laws of physics in different global coordinate systems using tensor equations, the laws will be the same in each coordinate system--it's only when you try to express how things look without using tensor equations that they may look different in different coordinate systems. And this would probably be true in any curved spacetime, it wouldn't be specific to the compact universe scenario.

When the two observers pass close by the first time in the single local laboratory already one of them is maked off as being in the 'preferred frame'.

Not by any experiment that can be observed within that laboratory. If you look only at the results of experiments in the lab, you see complete symmetry.

It is true that you have to wait until the second encounter to do the experiment and discover which observer it is, or you could simply look out and see what the matter in the rest of the scenario is doing and discover which observer is at 'rest'.

Yes, and because you have to do these things, I don't think you are testing the "principle of relativity" as physicists define it. As I understand it, the principle of relativity in GR only says that within a single small region of spacetime, ignoring everything outside that region, the laws of physics look just like they do in SR, including the symmetry between different inertial observers.

 

[pervect]

Therefore, I would like to redo the calculation in the Post Newtonian approximation. Remember in the calculation of the deflection of light the spatial curvature term, g₁₁, makes an equal contribution as the time dilation term, g₀₀, to the total result.

Garth

I believe that g11 is 1+2U in the PPN approximation? Any g11 > 1 will tip the balance towards favoring staying still, but the effect is extremely small - because the velocities involved are extremely small (nowhere near lightspeed, where the contribution does become equal).

Changing the density profile of the planet will probably have a bigger effect than including g11.

The most important point, IMO, is that if one changes the problem so that the clock is allowed to reach an altitude above the surface of the planet, rather than being confined to the interior, I would predict that the lowest time would definitely be seen by the "moving" clock.

 

[Garth]

I believe that g11 is 1+2U in the PPN approximation? Any g11 > 1 will tip the balance towards favoring staying still, but the effect is extremely small - because the velocities involved are extremely small (nowhere near lightspeed, where the contribution does become equal).

Changing the density profile of the planet will probably have a bigger effect than including g11.

The most important point, IMO, is that if one changes the problem so that the clock is allowed to reach an altitude above the surface of the planet, rather than being confined to the interior, I would predict that the lowest time would definitely be seen by the "moving" clock.

I concur.

Garth

 

[Garth]

Well, you're at least using one laboratory at two significantly different points in time, as opposed to one arbitrarily small laboratory looked at during a single arbitrarily small time interval. Yes, but they aren't talking about a local preferred frame, they're just saying that if you create global coordinate systems for each twin using the same procedure that's used in SR, then one of these coordinate systems is in some sense "preferred" (for example, they'll be only one such coordinate system where each apparent copy of the same clock is synchronized with every other copy). Also, my understanding is that because of diffeomorphism invariance, if you express the laws of physics in different global coordinate systems using tensor equations, the laws will be the same in each coordinate system--it's only when you try to express how things look without using tensor equations that they may look different in different coordinate systems. And this would probably be true in any curved spacetime, it wouldn't be specific to the compact universe scenario. Not by any experiment that can be observed within that laboratory. If you look only at the results of experiments in the lab, you see complete symmetry. Yes, and because you have to do these things, I don't think you are testing the "principle of relativity" as physicists define it. As I understand it, the principle of relativity in GR only says that within a single small region of spacetime, ignoring everything outside that region, the laws of physics look just like they do in SR, including the symmetry between different inertial observers.

It depends on what you are prepared to call a "law of physics". I understand about the Principle of Equivalence! IMHO the 'compact space twin paradox' is interesting because clock rate is fundamental to physical experiment and it seems that at the second encounter, also determined by observation of the rest of the universe, one clock measured duration will definitely be greater than the other.

Garth

 

[ralfsmith]

Oh its very interesting. Could you provide me more information ?

johntvery@operamail.com

johntvery@hotmail.com

 

[yogi]

IMHO the 'compact space twin paradox' is interesting because clock rate is fundamental to physical experiment and it seems that at the second encounter, also determined by observation of the rest of the universe, one clock measured duration will definitely be greater than the other.

Garth

But can't the same reasoning be applied to any pair of clocks that pass with relative velocity v - there is no guarantee that the proper rate of the two clocks is the same - and in general it will not be. For example if at some point in the distant past two clocks had been synchronized and put in relative uniform motion toward one another by a short duration acceleration, which could be applied to one or the other (or both but with a different magnitude), then upon passing each could mark the time on the other clock and after they had separated some distance L either could be decelerated to bring their relative velocity to zero - at this point both clocks are in the same frame and they can be stopped and compared - depending upon which clock accumulated the most time between the point at which they passed, the difference between the proper rates would be revealed.

 

[dicerandom]

yogi: I believe what you have described is essentially a more complicated version of the flat space twin paradox. Which clock reads more is determined by the nature of the accelerations and can be determined using k-calculus on spacetime diagrams.

 

[Garth]

Oh its very interesting. Could you provide me more information ?

johntvery@operamail.com
johntvery@hotmail.com

Hi ralfsmith! Have you read all the material already posted and linked? on
http://www.findarticles.com/p/articles/mi_qa3742/is_200108/ai_n8954244
cosmological twin paradox,
The twin paradox in compact spaces
Twin paradox and space topology?

Read these and if you have any further questions come back and ask again!

Garth

 

[yogi]

dicerandom - correct - what i was attempting to illustrate was a common misconception at the outset that leads to an apparent paradox - the reasoning goes like this - two clocks pass each other and each observes the other clock to be running slow - but it is impossible that each can be actually running slower than the other - yet when this initial assumption is followed by a round trip analysed which includes a turn-around accceleration, there is a real time difference when the clocks are ultimately brought to rest and compared in the same frame. I maintain you cannot synchronize clocks in relative motion - you can read a passing clock - but you cannot be assured that the proper rate of two clocks in relative motion will be equal. In the cosmological twin paradox, there is a similar ambiguity at the outset (first passing) and the clocks will continue to run at different proper rates until they re-encounter, ergo there is no difference in the cosmological case than the local round trip case - each involved an initial difference in velocity due to some past acceleration; the root of both the local and global paradox lies in the failure to properly consider this initial condition.

 

[Garth]

Hi yogi! I also agree with dicerandom, the point about the cosmological twin paradox is that the two observers are always in inertial frames of reference, there is no acceleration, they are both on geodesics all the time.

I don't understand what you mean by your 'initial condition'. You don't synchronise the clocks, just read them twice, which you can do so while in relative motion by passing a light signal, as long as they pass arbitrarily close to each other.

Consider two clocks, one freely floating, 'sitting', in a cave at the COM of the Earth and the other traveling at high speed through that cave along a tunnel in a vacuum tube on a rectilinear orbit, they compare times at the encounter.

They both think the other is moving and they are stationary, so each thinks the other clock is running 'slow'.

Then they encounter each other again and compare times. Each thinks they have recorded the greatest time elapse between encounters and yet only one can actually do so, so which one is it?

The truly 'stationary' clock must be defined by the mass of the Earth.


Garth

 

[yogi]

If the truly stationary clock is defined by the mass of the Earth - then you are implying the Earth's mass creates a preferred frame - this as you already know smacks of LR ...the notion that MMX and other experiments are better explained by considering the Earth mass as creating a preferred frame as opposed to SR which treats all inertial frames as equivalent. LR is a competing theory of SR usually associated with relativity dissidents ...that is not a problem from my perspective, but it is a yet to be verified theory
But let's take the case you have proposed - what were the initial conditions? - was the oscillating clock originally at rest at the center of the Earth and synchronized with the clock which remained at the center - then raised to the surface potential and dropped to be forever in oscillatory motion? And is this any different than a GPS satellite clock that is in a free float inertial frame. In the latter case its clear that, if we ignor the height factor, the Earth centered clock will run faster than an uncompensated GPS clock - in other words the proper rate of the satellite clock is slower than the proper rate of the Earth centered clock - if we take the Earth centered clock to a tower at the North pole, each time the GPS clock passes by, it will believe the Earth clock to be running slow by making overservational experiments as it passes by - and likewise the clock at the top of the tower will believe the satellite clock to be running slow - but the conflict as I have maintained in other threads lies in the fact that these measurements determine what is apparent - neither is actually measuring the proper rate of the other clock - they are only measuring the apparent rate of the other clock in relation to their own.

 

[yogi]

Garth - to further embellish on what I think I am trying to say, consider a non rotating Earth centered clock A and a GPS satellite clock B that is compensated only for its height. If A sends out radar pulses every 10 microseconds according to A's measurement of time 10 and B sends out radar pulses every 10 microsceonds according to B's measurment of time, and both A and B measure the times of arrival of the return pulses - then since the distance between A and B is always constant for a circular orbit, A will be able to determine that the proper rate of B clock is always less than the proper rate of A clock. In other words, this geometry provides a convenient method of determining the difference in the proper rate of both clocks. Real time dilation involves a difference between proper rates - there is no paradox here. I do not see why the same reasoning cannot be applied to the cosmological case i.e., assuming the universe to be a Hubble sphere - put a clock at some point C and establish two other clocks J and K at R = c/H that pass each other as they circumscribe the universe following a geodesic - although we would have to wait a long time for the signals to be returned - in our imagination we could surmise that the C clock would reveal the difference between the proper rates of J and K

 

[Garth]

they are only measuring the apparent rate of the other clock in relation to their own.

That is all that they can ever measure. However they can each make the two measurements of time at consequtine encounters. And then radio each other the results. One result will definitely be a longer duration than the other, but which one and how is that to be determined.

I do not see how the inital conditions resolve the paradox in GR, for the gedanken experiment both clocks are simply dropped, one at the COM and the other down the vacuum tube from some height, well above the Earth's surface if necessary. They do not have to be synchronised, just working accurately.

They each record t₁ & t'₁ at the first encounter and t₂ & t'₂ at the second. Once the results have been exchanged, by radio, then they can each compare t₂ - t₁ against t'₂ - t'₁ and see which is greater.

I agree that if the Earth determines which inertial frame of reference records the longer duration then that would not be GR. But how else would the result be determined? My POV is that GR needs to fully include Mach's Principle which is what I have done in http://en.wikipedia.org/wiki/Self_creation_cosmology at this moment.

Garth

 

[Garth]

Garth - to further embellish on what I think I am trying to say, consider a non rotating Earth centered clock A and a GPS satellite clock B that is compensated only for its height. If A sends out radar pulses every 10 microseconds according to A's measurement of time 10 and B sends out radar pulses every 10 microsceonds according to B's measurment of time, and both A and B measure the times of arrival of the return pulses - then since the distance between A and B is always constant for a circular orbit, A will be able to determine that the proper rate of B clock is always less than the proper rate of A clock. In other words, this geometry provides a convenient method of determining the difference in the proper rate of both clocks. Real time dilation involves a difference between proper rates - there is no paradox here. I do not see why the same reasoning cannot be applied to the cosmological case i.e., assuming the universe to be a Hubble sphere - put a clock at some point C and establish two other clocks J and K at R = c/H that pass each other as they circumscribe the universe following a geodesic - although we would have to wait a long time for the signals to be returned - in our imagination we could surmise that the C clock would reveal the difference between the proper rates of J and K

Our posts crossed.

IMHO the difference in this example is the issue of a local laboratory. It is because the two inertial observers pass arbitrarily close to each other that the paradox arises, both should be equivalent, yet they are not.

This seems to raise questions about the Equivalence Principle.

If as you correctly suggest the difference in clock rate is due to the geometry of space-time, or topology in the cosmological case, what is it that determines that geometry if is not the mass of the Earth?

Garth

 

[Hurkyl]

It is because the two inertial observers pass arbitrarily close to each other that the paradox arises, both should be equivalent, yet they are not.

They are equivalent.

Equivalent things are allowed to give different answers when they're asked different questions.

It almost seems like you're suggesting "they travel inertially" is a complete specification of the two problems -- which would explain why you think there's a paradox.

 

[Garth]

They are equivalent.

Equivalent things are allowed to give different answers when they're asked different questions.

It almost seems like you're suggesting "they travel inertially" is a complete specification of the two problems -- which would explain why you think there's a paradox.

Hi Hurkyl!
Do you think they are equivalent in the cosmological case of a closed 'compact' space?

And what are the "different questions" you refer to in the Earth centred case? The only question I have asked each observer is: "How much time has elapsed between encounters?"

Garth

 

[Hurkyl]

Hi Hurkyl!
Do you think they are equivalent in the cosmological case of a closed 'compact' space?

Yes. But again, you're asking two different questions, so it is not a paradox that you get two different answers.

The whole thing seems trivial (to me) if you ask it in geometric terms:

(1) We draw two non-parallel straight lines L and M on a cylinder.
(2) The lines intersect multiple times. Pick two consecutive intersections and call them P and Q.
(3) Compute the length of the line segment PQ along L.
(4) Compute the length of the line segment PQ along M.
(5) Gasp in confusion when the two lengths aren't equal!

Of course, to really be an exact description of the cosmological twin paradox, we need to integrate the Minowski metric along the line segments instead of the Euclidean metric.

The only question I have asked each observer is: "How much time has elapsed between encounters?"

Yes: you've asked two different questions.

"Observer 1: how long was it?"
"Observer 2: how long was it?"

These questions are not identical -- there's no reason to think they should have the same answer.

 

[Garth]

Yes. But again, you're asking two different questions, so it is not a paradox that you get two different answers.

The whole thing seems trivial (to me) if you ask it in geometric terms:

(1) We draw two non-parallel straight lines L and M on a cylinder.
(2) The lines intersect multiple times. Pick two consecutive intersections and call them P and Q.
(3) Compute the length of the line segment PQ along L.
(4) Compute the length of the line segment PQ along M.
(5) Gasp in confusion when the two lengths aren't equal!

Of course, to really be an exact description of the cosmological twin paradox, we need to integrate the Minowski metric along the line segments instead of the Euclidean metric.

In agreement with your example let us take the compact space to be Einstein's static cylindrical model for the sake of the argument.

There is no surprise in the fact that the two lengths/space-time intervals are different, the problem is: "In whose frame of reference is the diagram drawn?"

You can draw it in either observer's frame, they both think they are the one that is stationary and the other is the one that is moving.

So which one actually does have the longer duration, and how is that observer to be selected?

Yes: you've asked two different questions.

"Observer 1: how long was it?"
"Observer 2: how long was it?"

These questions are not identical -- there's no reason to think they should have the same answer.

Well I call that asking the same question to two different observers.

I do not think they do have the same answer!

That is the problem, again if, as you say they are equivalent, whose answer gives the longer duration and how is that observer selected?

GArth

 

[pervect]

In agreement with your example let us take the compact space to be Einstein's static cylindrical model for the sake of the argument.

There is no surprise in the fact that the two lengths/space-time intervals are different, the problem is: "In whose frame of reference is the diagram drawn?"

My understanding is this:

The winding number is a topological quantity that is not dependent on any particular choice of reference frame (i.e. choice of coordinates).

I'm not quite sure of the mathematical details.

Given that we assume that the above statement is true, the answer to the question becomes clear - the winding number distinguishes the obsevers.

 

[Hurkyl]

So which one actually does have the longer duration, and how is that observer to be selected?

That is the problem, again if, as you say they are equivalent, whose answer gives the longer duration and how is that observer selected?

By computation! You integrate the metric along the worldline.

In the SR analysis of the classic twin paradox, we have a very clever way of avoiding direct computation: we can invoke the Minowski version of the triangle inequality.

There isn't a general purpose shortcut, though. Unless you have a specialized theorem to invoke for the situation at hand, you have to compute.


Unfortunately, I don't know of a good way of actually computing things without picking a coordinate chart. But the process -- and the answer -- is the same no matter what chart we use.

My understanding is this:

The winding number is a topological quantity that is not dependent on any particular choice of reference frame (i.e. choice of coordinates).

I'm not quite sure of the mathematical details.

Given that we assume that the above statement is true, the answer to the question becomes clear - the winding number distinguishes the obsevers.

The winding number will allow you to make statements such as:

"He's gone around the universe one more time than me in that direction!"

which is, of course, equivalent to "I've gone around the universe one more time than him in the opposite direction!"

But this is still not a complete description of their paths on the cylindrical space-time: it still contains insufficient information to figure out which one measures more time between meetings.

 

[Garth]

By computation! You integrate the metric along the worldline.

In the SR analysis of the classic twin paradox, we have a very clever way of avoiding direct computation: we can invoke the Minowski version of the triangle inequality.

There isn't a general purpose shortcut, though. Unless you have a specialized theorem to invoke for the situation at hand, you have to compute.


Unfortunately, I don't know of a good way of actually computing things without picking a coordinate chart. But the process -- and the answer -- is the same no matter what chart we use.

The problem is that the oscillating observer thinks that as she suffers no inertial forces she can take herself to be stationary. In a totally equivalent calculation you have to integrate the metric along her worldline using the Schwarzschild metric transformed into her system of coordinates in which she remains at the centre and it is the other observer, and the Earth that is moving.

In that case her duration is easy: as dx' = dy' = dz' = 0 then

$$\int d\tau' = \int dt'$$

the problem is working it out for the other observer in these coordinates.

The winding number will allow you to make statements such as:

"He's gone around the universe one more time than me in that direction!"

which is, of course, equivalent to "I've gone around the universe one more time than him in the opposite direction!"

But this is still not a complete description of their paths on the cylindrical space-time: it still contains insufficient information to figure out which one measures more time between meetings.

I concur, I think this problem exists in both problems, the answer to the conundrum in my understanding must be that the extra required information is given by the distribution of the other mass in the universe/Earth. i.e. It is a paradox that is only resolved by application of Machian principles.

Garth

 

[yogi]

That is all that they can ever measure. However they can each make the two measurements of time at consequtine encounters. And then radio each other the results. One result will definitely be a longer duration than the other, but which one and how is that to be determined.
Garth

When measurements of the clocks are made during close fly-by by reading the other clock - you are actually obtaining the actual time dilation (the difference between the proper time accumulated by one clock in comparison to the proper time accumulated by the other clock - the accumulated proper times in general will not be equal - even though both clocks are following cosmic geodesics (or a local orbit). But that is not paradoxical - it only becomes paradoxical if one attempts to incorporate w/i the same argument, the apparent slowing of the other clock that would be observed by each observer when he considers himself stationary. This latter measurement is a distortion of reality - in actuality we never make this experiment - we simply take it as a true. Moreover it is tacitly assumed that since both clocks are in inertial frames, each clock is equally capable of making a measurement and that the measurment would be the same if the roles are reversed - but where is it written that both clocks will be running at the same proper rate - and if they are not then they would not make identical measurements of the slowing of the other clock

While the twin problem can be analysed using apparent observations - and this methodology leads to the same time loss as that obtained directly by differencing the accumulated proper time logged by each clock, the mathematical statements cannot logically be both true at the same time.

 

[yogi]

Garth - i was wondering if Mach's principle were applied to a universe where all matter exists in a concentrated lump ..e.g., at the center of a Hubble sphere - the metric for the surrounding space is spherically symmetrical as determined by the central mass - Would a clock orbiting the central mass and a clock following a geodesic determined by the central mass be governed by the same relativistic relationships?

 

[Garth]

When measurements of the clocks are made during close fly-by by reading the other clock - you are actually obtaining the actual time dilation (the difference between the proper time accumulated by one clock in comparison to the proper time accumulated by the other clock - the accumulated proper times in general will not be equal - even though both clocks are following cosmic geodesics (or a local orbit). But that is not paradoxical - it only becomes paradoxical if one attempts to incorporate w/i the same argument, the apparent slowing of the other clock that would be observed by each observer when he considers himself stationary. This latter measurement is a distortion of reality - in actuality we never make this experiment - we simply take it as a true. Moreover it is tacitly assumed that since both clocks are in inertial frames, each clock is equally capable of making a measurement and that the measurment would be the same if the roles are reversed - but where is it written that both clocks will be running at the same proper rate - and if they are not then they would not make identical measurements of the slowing of the other clock

While the twin problem can be analysed using apparent observations - and this methodology leads to the same time loss as that obtained directly by differencing the accumulated proper time logged by each clock, the mathematical statements cannot logically be both true at the same time.

I concur yogi, although we cannot actually make this measurement, though I suppose in future you might use an asteroid and its field as the base for an experiment, it is a useful and instructive scenario for a 'gedanken'.

If the mathematical statements cannot be true at the same time, which of course I agree with, it is where I come in, then what needs to be changed, the fact that equivalent inertial clocks are not equally capable of making a measurement? If that is the case, what principle do you use to decide between them?

As far as a 'lump' universe is concerned both inertial clocks would be telling their own time, and it would be possible to transform from one time scale to the other. A third clock that might be considered to be in a privileged position would be the one at the Centre of Mass and the one 'at infinity' from the mass, and comoving with it. That clock in my understanding would be recording the greatest proper time between any contrived inertial encounters.

Garth

 

[Hurkyl]

The problem is that the oscillating observer thinks that as she suffers no inertial forces she can take herself to be stationary.

Why is that a problem?

I concur, I think this problem exists in both problems, the answer to the conundrum in my understanding must be that the extra required information is given by the distribution of the other mass in the universe/Earth.

Well, it's wrong. The mass distribution doesn't contribute anything to the problem. The metric is all that matters.


Now, it might be possible, by watching all matter get pushed around for all time, to solve for the metric and then have enough information to work out the proper time along a path. I don't know enough about GR to know if the Einstein field equations are that strong.

However, I do know it is possible for two different metrics to push all matter around in the exact same way. So knowing the mass simply might not be enough to work out the time everybody experiences.

However this does lead to a measurement problem (but not a paradox): the difference between idealized and physical clocks. If two different metrics (meaning different readouts for idealized clocks, because the proper time is different!) have identical action on matter, that should extend to physical clocks.

 

[George Jones]

Up until now I have agreed with what Hurkyl has said in this thread, but now he's lost me somewhat.

The mass distribution doesn't contribute anything to the problem. The metric is all that matters. ... However, I do know it is possible for two different metrics to push all matter around in the exact same way.

I don't understand this. A solution to Einstein's equation includes a pair (g, t), where g is the metric and T is the energy-momentum tensor.

Are you saying that it's possible for (g, T) and (h, T), where g =/= h (but T is the same), to both be solutions to Einstein's equation, including the same boundary conditions?

Could you give an example?

Regards,
George

 

[Garth]

Why is that a problem?

Because, in their own inertial coordinate systems, they both can take themselves to be stationary. Therefore, if as you say the two observers are equivalent, why should one have the 'right' answer and the other the 'wrong' one?

To reiterate, they both think that their clock, and not the other clock, should have recorded the greater time elapse. Yet obviously if the two time elapses are not equal only one obsever will be correct, but which one?

On what basis do you select between the two?

Well, it's wrong. The mass distribution doesn't contribute anything to the problem. The metric is all that matters.

I disagree.

In agreement with what Bernard said later, what is it that determines the metric if it is not the distribution of mass and energy?

However this does lead to a measurement problem (but not a paradox): the difference between idealized and physical clocks. If two different metrics (meaning different readouts for idealized clocks, because the proper time is different!) have identical action on matter, that should extend to physical clocks.

I am not sure what you mean here. By "identical action on matter" being extended to physical clocks, do you mean the two clocks should record identical time durations between consecutive encounters? Surely this is not the situation we are discussing in this paradox.

In fact, as we have established, in the situation posited above the clocks record different time durations; the paradox lies in the fact that at the second encounter, in an arbitrarily small enough region, both the two clocks have remained in inertial frames of reference, on geodesics, throughout, and therefore, as you said, are equivalent.

So on what basis do you choose between them?

Garth

 

[JesseM]

Because, in their own inertial coordinate systems, they both can take themselves to be stationary. Therefore, if as you say the two observers are equivalent, why should one have the 'right' answer and the other the 'wrong' one?

Are you talking about the situation of one observer at the center of the Earth and the other oscillating up and down? How can either have a non-local "inertial coordinate system" when they are in curved spacetime?

 

[Garth]

Are you talking about the situation of one observer at the center of the Earth and the other oscillating up and down?

Yes

How can either have a non-local "inertial coordinate system" when they are in curved spacetime?

Let A be the observer at the center of the Earth be A
and the other oscillating up and down be B

They are both inertial observers and can base a coordinate system with themsleves as the origin. Times and distances are measured by standard clocks and rulers kept in inertial frames of reference at their respective origins, and radar may be used to measure distance.

A's metric is that of the Schwarzschild solution. B's is very complicated and I don't know of anyone who has looked at it, but perhaps others do.

Nevertheless, although we cannot easily calculate the exact time durations between consecutive encounters, measured by A as $\Delta \tau_A$ and B as $\Delta \tau_B$, as the signature of the metric is (-+++) , $d\tau^2 = - g_{\mu\nu}dx^{\mu}dx^{\nu}$, we can be sure that:

A thinks that $\Delta \tau_A > \Delta \tau_B$
and B thinks that $\Delta \tau_B > \Delta \tau_A$

So, which is correct when $\Delta \tau_A$ and $\Delta \tau_B$ are compared at the second encounter, and how is that observer selected by the physical setup?

Garth

 

[JesseM]

Let A be the observer at the center of the Earth be A
and the other oscillating up and down be B

They are both inertial observers and can base a coordinate system with themsleves as the origin. Times and distances are measured by standard clocks and rulers kept in inertial frames of reference at their respective origins, and radar may be used to measure distance.

A's metric is that of the Schwarzschild solution.

What do you mean when you say the clocks and rulers would be "kept in inertial frames of reference at their respective origins"? If a given clock or ruler was at a constant distance from A's position the center of the earth, then it would not be following a geodesic path and would therefore be moving non-inertially, right?

 

[Garth]

What do you mean when you say the clocks and rulers would be "kept in inertial frames of reference at their respective origins"? If a given clock or ruler was at a constant distance from A's position the center of the earth, then it would not be following a geodesic path and would therefore be moving non-inertially, right?

Right. The standards of measurement, a regular clock and a fixed ruler, are kept by A and B with them, at their respective origins. To make measurements away from their origins they would have to construct some kind of Schild's ladder, or use radar, to establish a metric around them. Remember the space-time around them has geometric properties that are independent of the coordinate systems used to describe them, but you do have to work out how distant measurements are made in anybodies frame of reference.

For the sake of this gedanken the actual values of the measurements are not important, only whether
$\Delta \tau_A > \Delta \tau_B$
or $\Delta \tau_B > \Delta \tau_A$.

Garth

 

[JesseM]

Right. The standards of measurement, a regular clock and a fixed ruler, are kept by A and B with them, at their respective origins. To make measurements away from their origins they would have to construct some kind of Schild's ladder, or use radar, to establish a metric around them. Remember the space-time around them has geometric properties that are independent of the coordinate systems used to describe them, but you do have to work out how distant measurements are made in anybodies frame of reference.

OK, so my argument is that in curved spacetime there's really nothing corresponding to the difference between inertial and non-inertial coordinate systems, at least not when dealing with global coordinate systems as opposed to purely local ones--no global coordinate system can be called "inertial" in curved spacetime. And in SR, when you use a non-inertial coordinate system and express the laws of physics in that system without using tensors or a metric (just writing down the equations of motion for objects in terms of the space and time coordinates of that coordinate system, for example), then the equations for the laws of physics will not be the same ones used in inertial coordinate systems--in particular, you can't assume that the time dilation of a clock moving in a non-inertial coordinate system is just a function of its coordinate velocity, or that the clock will slow down by $$\sqrt{1 - v^2/c^2}$$. So in curved spacetime, if no global coordinate system is inertial then presumably the same is true, time dilation will not just be a function of coordinate velocity, and each coordinate system will have to have its own expression for time dilation. So as long as you correctly express the equations for time dilation in whatever global coordinate system you choose for A and B (and there should be multiple possible choices for A and for B, it is only for inertial observers in SR that there is a standard canonical way to construct each observer's own rest coordinate system--no reason A must use Schwarzschild coordinates, for example), then each should make the same prediction about how much time elapses on each clock between the time they depart and reunite, there's no reason to expect the kind of symmetry that you see between the time dilation of inertial observers in SR.

 

[Hurkyl]

I don't understand this. A solution to Einstein's equation includes a pair (g, t), where g is the metric and T is the energy-momentum tensor.

Are you saying that it's possible for (g, T) and (h, T), where g =/= h (but T is the same), to both be solutions to Einstein's equation, including the same boundary conditions?

Could you give an example?

What I know (just recently learned!) is that it's possible for there to be two different connections with the property that a curve is a geodesic of the first connection if and only if it is a geodesic of the second connection. (Am I correct in concluding from this the corresponding statement if I replace connection with metric?) And the statement is also true if I allow the curve to be reparametrized.

I really have no idea whatsoever if the Einstein field equations would prevent this phenomenon. (And this is a question to which I would like to know the answer! But I doubt I have the sophistication to follow the details of an answer either way)

 

[Hurkyl]

Because, in their own inertial coordinate systems, they both can take themselves to be stationary. Therefore, if as you say the two observers are equivalent, why should one have the 'right' answer and the other the 'wrong' one?

To reiterate, they both think that their clock, and not the other clock, should have recorded the greater time elapse.

Ah, that's the problem. This argument is invalid in both frames!

Now, you can say something like:

"If I'm traveling inertially, then my clock reads the greatest amongst all clocks that have been traveling roughly the same path as me."

but this is clearly inapplicable to the situations at hand.

In agreement with what Bernard said later, what is it that determines the metric if it is not the distribution of mass and energy?

For an analogy, consider electrodynamics. The corresponding claim would be "What is it that determines the electromagnetic field if it is not the distribution of charge and current?" But in EM, the distribution of charge and current is clearly not the entire story! For example, you can have all sorts of electromagnetic waves when there isn't a lick of charge or current anywhere in the entire universe.

I am a bit worried about pushing analogies too far, but it seems obvious to me that the distribution of mass and energy is insufficient to completely determine the metric.

 

[Garth]

Ah, that's the problem. This argument is invalid in both frames!

Now, you can say something like:

"If I'm traveling inertially, then my clock reads the greatest amongst all clocks that have been traveling roughly the same path as me."

but this is clearly inapplicable to the situations at hand.

Why? I think it is quite applicable.

We have only been talking about weak fields $\frac{GM}{rc^2} \sim 10^{-9}$ and the oscillating observer only needs to travel less than an Earth radius:
~ 10^-2 light secs, or perhaps ~ 10^-1 light secs, where the orbital period ~ 80 minutes, and we are not talking about a weird topology induced by one observer zipping around a Black Hole or something, so doesn't your statement: "If I'm traveling inertially, then my clock reads the greatest amongst all clocks that have been traveling roughly the same path as me" hold?

For an analogy, consider electrodynamics. The corresponding claim would be "What is it that determines the electromagnetic field if it is not the distribution of charge and current?" But in EM, the distribution of charge and current is clearly not the entire story! For example, you can have all sorts of electromagnetic waves when there isn't a lick of charge or current anywhere in the entire universe.

I am a bit worried about pushing analogies too far, but it seems obvious to me that the distribution of mass and energy is insufficient to completely determine the metric.

Yes, gravitational waves would introduce a slight modificaton to the metric, but basically the 'meat' of $T_{\mu\nu}$ is in the distribution of mass in the Schwarzschild solution.

Garth