A variation on the twin paradox

[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

 

[pervect]

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?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

Consider again my example of two points separated by a mountain. Or better yet, consider two points on an egg:

         A
    xxxxxxxxx
xxxxxxxxxxxxxxx
    xxxxxxxxx
         B

There exists two path on the egg from A to B which are both geodesics but which have different lengths.

You are apparently still confusing "local maximum" with "global maximum". A geodesic path has the property that it's a local maximum (of time in space-time), or a local minimum (of distance in Euclidean space). A given geodesic is not necessary a global maximum (of time in space-time) or a global minimum (of distance). As Hurkyl points out, it is only for paths "near" the target path that you are guaranteed to have an extreme value. "Nearness" of paths is a somewhat abstract concept, while I could write down the intergal, it's probably not worthwhile to apply rigor at the moment.

Instead, it may be simpler and illuminating to consider a simpler problem - finding the minimum of a function.

Consider finding the minimum of a simple function f(x). If df/dx = 0, then the function is locally an extreme point of f(x), which could be a local minimum or a local maximum depending on the sign of the second derivative.

However the existence of a local minimum does not guarantee that that minimum is a global minimum. The local minimum is only a minimum for points "near" x. You actually need to compute f(x) at all points where df/dx = 0 to determine the global minimum. (You also need to address what happens at the two limits as x approaches +/- infinity, for that matter.)

Satisfying the Euler-Lagrange equations (or the geodesic equations) is equivalent to saying that df/dx = 0 in the function minimizing problem. It generates a local extreme point, and we can say that curves "near" the solution of the Euler-Lagrange equations will be local extreme points, but it does not guarantee that a specific solution will be the global maximum (or minimum) anymore than saying that df/dx = 0 will guarantee that a local max or min is a global max or min.

To guarantee that one has the absolute best maximum (or minimum), one must consider all of the extreme points (functions).

In flat space there is only one straight line connecting two points, but this is not true in general. When there is more than one straight line connecting two points, one must evaluate all of them before making any global statements.

This is probably not the only issue bothering you - your other issue is related to Mach's principle, IMO. But that's a separate issue. I think we need to take care of this one, first.

 

[Hurkyl]

"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?

If by "hold" you mean that it's applicable, I would say no. The absurd conclusion is a clear demonstration that it is not applicable.

But that aside, the center-of-Earth observer's path doesn't look anything at all like the oscillating observer's path! So how could the quoted statement possibly apply?


You seem to be arguing something different -- that the problem is so infinitessimally small, that we ought to be able to treat the whole thing special relativistically.

But there's a huge, giant clue that such a treatment must be inherently flawed: you have a pair of straight lines that intersect twice.


In the heuristic sense you seem to be trying to use, the problem is that if you set your scale so that the near-Earth system becomes infinitessimally small, the amount of time spanned by the experiment must be considered infinitely big -- the observers cannot possibly meet twice during any finite interval of time. So the whole thing blows up and our attempt at approximation foiled.

 

[Hurkyl]

Whoops, I was a bit hasty -- I assumed the velocities involved were finite!

Suppose we posit that the field is infinitessimally weak, and the timespan of the experiment is finite (but not infinitessimal), in an attempt to apply SR.

It follows that the relative velocity of the two observers must be infinitessimal. (Otherwise the infinitessimally weak field could never pull them back together)

We can conclude, via SR, that the difference of the readings of their clocks must be infinitessimal.

But our approximation can only be correct up to an infinitessimal error.

Thus, the amount of error in our approximation is on par with whatever result SR might have told us: in other words, we can conclude nothing.

Or, to put it vividly, each observer can say:

"I aged more than the other guy... plus or minus some amount of error that may be greater than the difference in how much we aged."

 

[Garth]

It seems you have no idea what I have been arguing about.

First I agree with everything you said about a local extremum. It is on that fact that there are multiple geodesic paths through curved space-time that the whole paradox depends.

SR has nothing to do with it.

My example of using the Earth's field is all about the weak field approximation of GR. Satellites orbit the Earth, and objects fall towards its centre, quite successfully and GR effects are able to be observed experimentally in Earth orbit, the field and the observer's relative velocities are not infinitesimal.

You seem capable of seeing the situation only from one frame of reference.

On top of that as you stated: "the center-of-Earth observer's path doesn't look anything at all like the oscillating observer's path! " you seem unable to visualise the two geodesics in the perspective of 4D space-time:

One is a straight line about 80 light minutes long.

The other is a very stretched out (rectilinear) spiral, a sine curve, with a pitch, or wavelength, of the same 80 light minutes and amplitude of around 1/10 light second, ie. a sine curve 50,000 times longer in wavelength than amplitude.

Visually that would be almost indistinguishable from a straight line and the other geodesic.

The question is: "How does the oscillating (in the Earth's frame of reference) observer, B, see it?"

In B's inertial frame of reference it is the Earth, and A, that is oscillating.

And as I have said, if anything is moving in B's frame of reference, (and the elongated sinusoidal geodesic now belongs to A,) then the integral of that moving object's proper time interval along its geodesic in B's frame of reference is necessarily shorter in 4D space-time duration than B's geodesic between the two encounter events, because of the signature of the (GR) metric.

Now the resolution of the paradox is that in fact at the second encounter A will find her understanding that \Delta \tau_A > \Delta \tau_B is correct and B's understanding that \Delta \tau_B > \Delta \tau_A is incorrect.

However, at that second encounter as both inertial observers are freely falling and temporarily in a locally small enough region, if we apply the Equivalence Principle we must conclude that both statements \Delta \tau_A > \Delta \tau_B & \Delta \tau_B > \Delta \tau_A are true for their respective observers.

This is the paradox.

But, if we do not want to live with this paradox and we need to resolve it, because of the experimental comparison of the two time durations, how will this selection of the correct observer statement be made?

I would argue only by the geometry of the set up, which determines the geometric gravitational field that is described in either observer's metric. Just as in the cosmological twin paradox, as Barrow and Levin state:

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

so too here the resolution hinges on the existence of a preferred frame introduced by the geometry of the mass distribution, i.e. the geometry of the geometric object, which is the Earth's field.

And that, I would argue, is an application of Mach's Principle.

Garth

 

[yogi]

Garth - Is it necessry that your explanation of the local temporal inconsistency between the A and B clock depend from Mach's principle - From my jaded perspective, the preferred frame seems to be more easily explained by the notion that local matter conditions local space in a way that makes light velocity locally isotropic - i.e., as in Lorentz Relativity (not Lorentz ether theory or MLET). To dissect the thought experiment further, suppose there is only one mass in the universe - e.g the earth. Would you exclude the possibility that the A and B clock would move any differently in the absence of distant concentrations of matter? - I guess I know the answer - but let me have it anyway if you would.

Yogi

 

[Garth]

Thank you Yogi,
1. I'm not quite sure I understand your first point, Is it not that "light velocity" is "locally isotropic" for both observers?

2. Well this is the interesting question! An otherwise empty universe. Would inertial masses be the same or do they depend on that "distant concentrations of matter"?

One answer is that it depends on how you measure Mass, length and time. What are your standards of measurement and how are they affected? In http://en.wikipedia.org/wiki/Self_creation_cosmology inertial masses do depend on the cosmological evolution of matter and the gravitational & scalar fields. So basically in your scenario it would seem that G is much larger (m_i is much smaller).

Garth

 

[Hurkyl]

You seem capable of seeing the situation only from one frame of reference.

Then you're not getting me at all. When I'm trying to explain twin paradox-type problems, I strongly prefer to do it without appealing to any frame of reference at all.

Coordinates and frames of reference are unphysical things -- they're mathematical tools we use to describe the universe. They're not actual elements of reality. Twin pseudoparadoxes generally involve some sort of mistake involving coordinate charts.

IMHO the most important step for correcting these mistakes is to understand that there's a reality out there that doesn't care one whit about coordinates, and the same is true of our mathematical models. Coordinates are no more important to Minowski space (SR), or to a pseudoRiemannian manifold (GR) than they are to high-school Euclidean geometry.

However, at that second encounter as both inertial observers are freely falling and temporarily in a locally small enough region, if we apply the Equivalence Principle we must conclude that both statements \triangleup \tau_A > \triangleup \tau_B & \triangleup \tau_B > \triangleup \tau_A are true for their respective observers.

Why do they have that understanding?

A will observe that \partial \tau_B / \partial t_A < \partial \tau_A / \partial t_A, and B will observe that \partial \tau_B / \partial t_B > \partial \tau_A / \partial t_B. This is what the equivalence principle tells us during each meeting. (t is for an observers local "coordinate time". More precisely, \partial /\partial t_A is just the tangent vector to A's worldline)

But why should B think that \Delta \tau_B > \Delta \tau_A, and why should A think that \Delta \tau_A > \Delta \tau_B?

This is why I thought you were trying to apply SR (i.e. extend the equivalence principle beyond its domain of applicability) -- I thought you were arguing that A can say \partial \tau_B / \partial t_A < \partial \tau_A / \partial t_A for the entire trip, and thus should conclude \Delta \tau_B < \Delta \tau_A, as is done in SR, and vice versa.

If you aren't trying to extend SR this far, then how are you arriving at those final inequalities?

I would argue only by the geometry of the set up, which determines the geometric gravitational field that is described in either observer's metric.

There is only one metric. It's an inherent quality of the universe. (Or, at least of the pseudoRiemannian manifold) The different observers are both using the same metric -- they're just writing it with respect to different bases.


I have more to say, and hopefully I'll remember. But I've got to go. (I should have left half an hour ago. )

 

[Garth]

First I have just realized that LaTex hasn't been working properly in <itex> in my posts, I have corrected them now. I use \Delta \tau_{A,B} to mean the time elapse between consecutive encounters as measured by A or B. It had come out as just \tau_{A,B} I am sorry for any confusion that has arisen.

Secondly Hurkl I agree there is only one geometric 'object' which you describe as one 'metric'. I was using the term 'metric' to mean the coordinate based g_{\mu \nu} expression as measured by each observer.

I agree that the geometric object is an inherent property of the space-time of the universe. However it is experienced by observers in particular frames of reference and expressed in particular coordinate systems in which measurements have to be made.

The paradox arises when the expected time elapses of the two observers are compared, it is resolved by the experimental measurements of the actual time delays being compared.

Garth

 

[pervect]

It seems you have no idea what I have been arguing about.

First I agree with everything you said about a local extremum. It is on that fact that there are multiple geodesic paths through curved space-time that the whole paradox depends.

OK, I was also suspecting communications failure, but now that now that we've hopefuly killed and buried the "local vs global" maximum issue, let's move on a bit.

I think, mainly based on past discussions, that what's bothering you ultimately is how you go about explaining the different observations of different observers in terms of Mach's principle.

Personally I don't actually believe in Mach's principle, so I don't explain things in terms of it. But I can make some general comments anyway.

To calculate the proper times of each observer, we just have to integrate dtau along their paths as has been discussed at length.

To calculate dtau, we only need the metric of space-time. So the means to answer the question of "which observer ages the most" is contained purely in the geometry of space-time.

So the route to answer your question is to ask - how is the geometry of space-time (i.e. the metric) determined by the distribution of matter via "Mach's principle"? Knowing how the metric is determined, we can then determine which observer has the greatest elapsed time, which is a purely geometric question about space-time.

The problem with this formulation is that its difficult, except in a very general way, to describe the distribution of matter in space-time without a metric. (And if you have a metric, it pre-determines the distribution of matter!).

For very simple problems one can get a lot of mileage out of symmetry, and an "equation of state". One assumes, for instance, a spherically symmetrical planet, and then some "equation of state" for the matter in it.

In standard GR, we need only pressure vs density as the "equation of state". I'm really not sure what SCC needs in addition (if anything) - I mention it mainly because I'm pretty sure it's on your mind, and needed for you to answer the question.

So given symmetry (a spherically symmetric planet) and an equation of state, what else do we need to solve Einstein's equations? We need boundary conditions - usually asymptotic flatness of space-time at infinity. This also gets rid of "gravitational radiation" issues, any such gravitational raditionis included in the boundary condtions. For the spherically symmetric planet case, we already know that the boundary conditions we want (asymptotic flatness, no gravitational radiation) are going to yield a Schwarzschild solution outside the planet.

This is a very messy problem, but you only want to do it in "weak fields". Without SCC, we can ignore the pressure terms and think of gravity as only being due to density. This let's us get rid of pressure altogether, except as it affects the density profile. We basically get Newtonian gravity. Following through, we get the calculation I did much earlier in the thread

You'll have to add in additional terms in SCC as needed, or higher order terms.

This gives a rather interesting result that the times are almost exactly the same. The result is so close that we start to have to worry about the equation of state to be able to answer the question - i.e. we actually need to know the density vs depth profile of our planet to be able to tell which observer has the longest time.

If the planet were of totally uniform density, the effects all cancel out to the first order. Since I haven't consistenly included all second order effects, I can't say for sure what happens to the second order.

I'm not totally confident, but based on the Virial theorem analysis, I believe that if the planet has a dense core, the observer that "bobs" will wind up with a longer proper time than the observer who stays still, because he will gain more by leaving the potential well than he will lose from his velocity.

I suspect you may be looking for "short-cuts" from the above rather involved approach, but I don't see how you are going to pull it off and guarantee accurate answers. Given the first-order cancellation of effects, I think the problem really is as hard as I'm making it out to be.

.

 

[pervect]

Another point I wanted to make, I don't know if it matters to you.

If the lines of constant coordinates are taken to be geodesics, only the observer at the center of the planet will have a coordinate system that covers all of space-time.

The geodesics will start to intersect for the "bobbing" obsever, so his coordinate system (if it is based on geodesics) will not cover all of space-time.

 

[Hurkyl]

The paradox arises when the expected time elapses of the two observers are compared

Both observers compute exactly the same value for \Delta \tau_A. Both observers compute exactly the same value for \Delta \tau_B.

I guess, in your new notation, that means \Delta \tau_{A,A} = \Delta \tau_{A,B} and \Delta \tau_{B,A} = \Delta \tau_{B,B}

Do we agree on this much? If so, then what is the paradox?


I think we don't agree on this, and that's a problem. The proper time along a worldline (i.e. \Delta \tau_A) is a geometric quantity; it is entirely independent of coordinates. If the two observers don't agree on its value, then one of them did something wrong.

 

[pervect]

Let me unify & reorganize my comments about the virial theorem.

To the first order, we can represent a Newtonian gravitational field by g_00 = 1 + 2V, where V is the potential energy, and g_11=1.

Then to the first order

<br /> d\tau = \int \sqrt{(1+2V(t)) - v^2(t)}dt \approx 1 + \int V(t)dt - \frac{1}{2} \int v^2(t)dt = 1 + \int V(t)dt - \int T(t) dt<br />

where V(t) = potential energy per unit mass (as a function of time) [capital V], v(t) = velocity as a function of time [small v], T(t) = kinetic energy per unit mass (.5 v^2(t))

Now the virial theorem

http://en.wikipedia.org/wiki/Virial_theorem

allows us to relate \int V(t) dt to \int T(t)[/tex]. We realize that the above intergals are proportional to \bar{V} and \bar{T}, where the overbar represents taking the time average.

For a power-law force, where F = dV/dr = a r^n, we can say that

\bar{T} = \frac{n+1}{2} \bar{V}

by the Virial theorem (see above link).

For a uniform density planet, the force law is linear, n=1, and the two terms are equal, resulting in no effect to the first order. A much more thorough analysis would have to be done to include all second order effects consistently to tell us what happens.

If we want a force that represents a planet that is more dense in the center than it is on the surface, we want the force law to be sub-linear, which we can represent as a power-law force with n<1.

When n<1, the two terms above are not equal, and we see from the virial equation for a power-law force that the kinetic energy term will be lower than the potential enregy term, meaning that the "bobbing" clock will read higher than the "stationary" clock. This is because the "bobbing" clock spends more time at an altitude, where it ticks faster. The positive contribution due to \bar{V} will be greater than the negative contribution due to \bar{T}, the later representing the slowing effects of velocity on the clock.

 

[Garth]

Both observers compute exactly the same value for \Delta \tau_A. Both observers compute exactly the same value for \Delta \tau_B.

I guess, in your new notation, that means \Delta \tau_{A,A} = \Delta \tau_{B,A} and \Delta \tau_{A,B} = \Delta \tau_{B,B}
(Where \Delta \tau_{1,2} is the proper interval between two encounters of observer 1 as measured or computed by observer 2)
Do we agree on this much? If so, then what is the paradox?


I think we don't agree on this, and that's a problem. The proper time along a worldline (i.e. \Delta \tau_A) is a geometric quantity; it is entirely independent of coordinates. If the two observers don't agree on its value, then one of them did something wrong.

Okay, it is that last statement that has given me food for thought.
It means the proper interval between two events is independent of the geodesic route taken between them. So \Delta \tau_{A,A} = \Delta \tau_{A,B} and as measured by B: \Delta \tau_{B,A} = \Delta \tau_{B,B}.

I had not considered this before being so use to thinking that a moving observer's proper interval was necessarily less than that of a stationary one.

Thank you, I shall sleep on it!

Garth

 

[pervect]

Okay, it is that last statement that has given me food for thought.
It means the proper interval between two events is independent of the geodesic route taken between them.

Eeek! No.

A proper time interval is measured between two points along a specific curve, which may or may not be a geodesic.

The proper interval depends on the specific curve. There can be many different curves connecting the same two points - the proper interval along each curve can in general be different. The proper interval is defined for both geodesic and non-geodesic curves.

The proper interval along two curves which both are geodesics and which both connect the same two points in space-time does not have to be the same.

The proper interval does not, however, depend on a choice of coordinates. No physical quantity can depend on the choice of coordinates. Coordinates are just markings on a map, and have no physical significance whatsoever.

 

[Garth]

That is, of course, exactly what normally I understand!

However, in this case, where A is the COM observer and B the 'bobbing' observer and using a notation in which the time elapse between encounters is written, we have:

\Delta \tau_{A,A} being A's time elapse as measured by A,
\Delta \tau_{A,B} being A's time elapse as computed by B,
\Delta \tau_{B,B} being B's time elapse as measured by B,
\Delta \tau_{B,A} being B's time elapse as computed by A,

so \Delta \tau_{A,A} & \Delta \tau_{B,B} are measured by the identical clocks carried by A and B in their inertial frames of reference respectively.

We have said that:
\Delta \tau_{A,A} = \Delta \tau_{A,B} and
\Delta \tau_{B,B} = \Delta \tau_{B,A} as they are geometric objects independent of the coordinate systems in which they may be calculated.

Now either:
\Delta \tau_{A,A} &gt; \Delta \tau_{B,B} or
\Delta \tau_{A,A} &lt; \Delta \tau_{B,B} or
\Delta \tau_{A,A} = \Delta \tau_{B,B} and intuition tells you it is the first of these options that would be proved correct in an actual experiment.

However that would also mean \Delta \tau_{A,B} &gt; \Delta \tau_{B,B} as \Delta \tau_{A,A} = \Delta \tau_{A,B} would it not?

In which case in B's inertial frame of reference, A's space-time interval is computed by B to be greater than B's space-time interval, even though it is A that is moving and B that is stationary in that inertial frame of reference, against all space-time intuition about relative moving clock time dilation.

The moving clock is 'ticking' faster than the stationary one, in other words, the moving observer A is aging more quickly than the stationary B!

As the two intervals measured and computed by B, in r,t coordinates, (suppressing the other two spatial dimensions as B is on a rectilinear orbit,) are given by

\int d\tau = \int dt

and

\int d\tau = \int \sqrt (1 - v^2)dt

We are saying that

\int dt &lt; \int \sqrt (1 - \frac{g_{11}}{g_{00}}v^2) g_{00}dt

where v is A's velocity in B's frame of reference.

I cannot see how this can be, I therefore chose the third option:

\Delta \tau_{A,A} = \Delta \tau_{B,B}

Perhaps you can enlighten me?

Garth