A variation on the twin paradox

[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

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

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:
$\Delta \tau_{A,A} > \Delta \tau_{B,B}$ or
$\Delta \tau_{A,A} < \Delta \tau_{B,B}$ 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 $\Delta \tau_{A,B} > \Delta \tau_{B,B}$ 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

$$\int dt < \int \sqrt (1 - \frac{g_{11}}{g_{00}}v^2) g_{00}dt$$

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

 

[pervect]

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:
$\Delta \tau_{A,A} > \Delta \tau_{B,B}$ or
$\Delta \tau_{A,A} < \Delta \tau_{B,B}$ 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.

Intuition isn't terribly reliable - I intuitively guessed the third option a while back, so my intuition was different than yours. And it was just as wrong, because after doing the calculation I'm convinced any of the above cases can be true depending on the density profile of the planet.

However that would also mean $\Delta \tau_{A,B} > \Delta \tau_{B,B}$ 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!

This is perfectly possible. I'm having a bit of a problem untangling your notation or why you think there is a difficulty.

For any observer, using a +--- sign convention, we can write

$$d\tau^2 = g_{00} dt^2 + 2 g_{01} dt dx + g_{11} dx^2$$

which can be re-written as

$$d\tau = \sqrt{g_{00} + 2 g_{01} \frac{dx}{dt} + g_{11} (\frac{dx}{dt})^2 } dt$$The numerical values of the g_ij will depend on the coordinate system used.

Let's look at it from the viewpoint of the stationary observer at the center of the planet, which we will call obsever A.

You are assuming, I believe, that we adopt coordinates so that g_00=1, g_01=0, and g_11 = 1 for the stationary observer. (We don't have to adopt such coordinates, but that's what I'm getting from what you wrote). Then the above intergal reduces to $$\tau_{A,A} = t$$.

Now let's consider the calculation of $$\tau_{B,A}$$

The moving obserer will have g_00, g_01, and g_11 as functions of time. We do not expect the metric coefficeints for observer B to represent a Minkowskian metric, because observer B is far away from A in a gravity field.

g_00 may be greater than 1 for the moving obsever. In fact, we expect g_00 to be greater than 1. Well, I expect it to be greater than 1.

Therfore it is possible in principle for dtau > dt.

The moving observer will have g_11 approximately equal to -1. Thus it is possible in principle for dtau < dt.

To decide which is the case, we need more information.

In english:

The moving observer is higher in a gravity well. Thus it's clock is ticking faster because of it's height. The moving obsever is also moving. Thus it's clock is ticking slower due to relativistic time dilation. Which effect dominates is not clear - it could be either one.

 

[Garth]

Yes thank you, I simply took the first option as an example to chew on!

And I certainly agree with your analysis from A's POV, I would be interested if it could also be reworked from B's POV, with the metric expressed in B's coordinate system.

What I am chewing over is the comparison of two inertial observers traveling between two close encounter events via separate geodesics, seeing the situation from each POV.

In fact that these two inertial observer's are not equivalent because the gravitational field is stationary for A but not for B.

In other words any difference between them is imparted by the distribution of the matter in the rest of the universe, which in this case is the planet Earth.

I suppose the significance of any issues raised simply depends on whether you think that difference is itself significant or not.

I'll revert back to thinking of the comsological closed space scenario!

Garth

 

[yogi]

Something i am not relating to - why is the bobbing clock a better inertial system than a free-floating clock in circular orbit about a spherically symmetrical mass? The latter is much easier to deal with since from the perspective of a clock at the center of mass the relative velocity and the gravitational potential of the orbiting clock is constant.

 

[Garth]

It was simply that 'relative simultaneity' does not present itself as a problem at the close encounters at the centre of the Earth. If the two clocks are close enough then there is no ambiguity in their meaurements of t₁, t'₁ and t₂, t'₂. the two geometric 'objects' are uniquely determined as
$\Delta \tau_{A,A} = t_2 - t_1$ and
$\Delta \tau_{B,B} = t'_2 - t'_1$.

Garth

 

[yogi]

Grath ..So basically, you were contriving a situation that allowed the clocks to obtain nearly instantaneous measurement of the other during passby. What if the Earth centered clock is synchronized with a clock on top of a large tower, e.g. at the North Pole, which has a height equal to the altitude of a polar orbiting clock - except for the difference in gravitational potential (for which we can adjust) we now have one clock fixed in the Earth centered frame, a second clock in the same frame but at different but constant gravitational potential (The clock on the North Pole tower) and an orbiting clock. Once the North Pole clock is adjusted to account for the difference in the G potential, it will remain in sync with the Earth centered clock - each time the satellite clock (SC) passes by the North Pole clock (NPC) they will read each other. The result is that the NPC will be found to accumulate more time between orbit encounters than the SC clock. In this experiment, which I see as analogous to your oscillating clock, the measurements will confirm that the proper rate of the orbiting clock (SC) is less than than the proper rate of the NPC. I also see it as a local model of the cosmological twin paradox - but it is not a paradox for the same reason, namely since you should not apply the mutually contradictory statement that "two clocks in relative motion each measure the other clock to be running slow" to a situation involving actual time dilation - that statement is only true when the apparent times are measured using the standard two clock technique in one frame. In the case of the oscillating clock and the orbiting clock, the experiment measures actual time difference rather than apparent time dilation - so the commonly heard statement about reciprocal observations of clocks in relative motion show the other to be slow, while true in the sense that that the other clock appears to be running slow, is not true in the sense that the other clock is actually running slow - the SC will be slower than the NPC because your experiment (and my alternative) measure the difference in the proper rate of the clocks in question

 

[Hurkyl]

The point of the bobbing observer is that we don't have to contrive some means of comparing their clocks -- when two clocks have a flyby, there's a "god-given" way to compare them.

What if the Earth centered clock is synchronized with a clock on top of a large tower,

We've discussed before that the word "synchronized" by itself is entirely meaningless in SR.

In GR, the case is far, far worse. In SR, you at least had the notion of global inertial coordinate charts -- in GR, you don't even have that.

For example, if two observers, A and B, have worldlines that never intersect, or only intersect once, then there is a coordinate chart in which A's clock is always running faster than B's clock, and there is another coordinate chart in which B's clock is always running faster than A's clock.

You seem to be doing the same thing you've done in SR: you pick your favorite coordinate chart, and decree that everything is supposed to be done relative to that.

But this obscures one of the most important facts about GR: that any coordinate chart is good enough to do physics! (meaning that all of the laws of physics take an identical form in any coordinate chart)

is not true in the sense that the other clock is actually running slow

In particular, that this sentence has absolutely no physical meaning. To say that a clock is running "slow" at any point in time, you need to have some "standard" time to which you can compare it and take derivatives.

When two clocks are co-located, we have two physical notions of time we can use, one for each clock. It then makes sense to ask about the rate of one clock with respect to the other clock. Mathematically, this question is nothing more than asking for the inner product of their velocity vectors.

When two clocks are not co-located, you cannot physically compare them. You have to invent some sufficiently global notion of time coordinate, and compare your clocks to that time coordinate. Since this requires the invention of some notion of coordinates, it's an unphysical concept.

If the clocks have meetings at which their clock readings can be compared, it does make sense to say that one clock accumulated more time than the other between meetings, but that still isn't enough to give any meaning to the differential rate of one clock with respect to another.

The result is that the NPC will be found to accumulate more time between orbit encounters than the SC clock.

All that being said, I thought the exact opposite would be true in the coordinate chart I think you're using.

Your "clock" at the NPC isn't a clock at all: it doesn't measure proper time. Because you've somehow "synchronized" it with the clock at the Earth's center (ECC), that means that your NPC will be running slower than any ordinary clock that is also located in the tower at the north pole.

I think that this further means that your NPC will accumulate less time between flybys with the SC.

 

[yogi]

Hurkyl - what was intended - perhaps not explained clearly enough - is we have an Earth centered clock (ECC) which we synchronze with another clock. The latter is moved to a tower on the North pole. If no correction is put in, this NPC will run faster than the Earth centered clock. Or if you introduce a scaling factor you can make sure it runs at the same rate as the Earth centered clock. Either way, the clock in orbit (SC) will run slower than the NPC if the NPC and SC have been synchronized at the top of the tower and the SC then put into circular polar orbit.

I am saying the utilization of a simple geometry created by a circular orbit makes it clear there is no paradox - as long as their is an unbroken chronological chain of synchronization between the ECC, NPC and SC, then each time the SC passes by the NPC and the readings are exchanged, they can be related to the ECC.

You state: "If the clocks have meetings at which their clock readings can be compared, it does make sense to say that one clock accumulated more time than the other between meetings, but that still isn't enough to give any meaning to the differential rate of one clock with respect to another."

And I say it does - if at every meeting the SC clock is 7ns less than the reading on the NPC clock - and the NPC clock is always at the same distance from the ECC we have a direct way to compare the SC clock with the ECC because the NPC can send a radio signal to the ECC and the ECC can send a radio signal to the NPC.

Now its true that if the NPC is height corrected so that it runs in sync with the ECC, it will not read the proper time for a clock in that position - but it is a clock - running at a constant offset rate - but that does not negate the impact of what it measures - we know how to adjust for the offset in arriving at the result

 

[yogi]

Hurkyl- one more point which we discussed in another thread - in actuality, we need not even use a clock at the top of the tower - we can simply have the SC clock transmit a signal indicative of the SC clock reading whenever it passes w/i ten meters of the tower - this is picked up by the ECC and compared to the ECC time - the distance is always the same - when you subtract out the difference produced by the different G potential between the ECC and the SC, what is left is an ongoing confirmation updated each time the SC passes the tower, of the actual difference in the proper rate of the ECC and the SC clocks.

 

[Hurkyl]

Or if you introduce a scaling factor you can make sure it runs at the same rate as the Earth centered clock. Either way, the clock in orbit (SC) will run slower than the NPC if the NPC and SC have been synchronized at the top of the tower and the SC then put into circular polar orbit.

In the usual coordinates we use for near-Earth, I'm quite sure that this is wrong. Because you've tuned the NPC to run at the rate of the ECC (according to the specified coordinate chart), the NPC should accumulate less time than the SC between flybys.

And I say it does - if at every meeting the SC clock is 7ns less than the reading on the NPC clock - and the NPC clock is always at the same distance from the ECC we have a direct way to compare the SC clock with the ECC because the NPC can send a radio signal to the ECC and the ECC can send a radio signal to the NPC.

Ok fine.

To the best of my knowledge, to compute a differential rate, you need to take a derivative of something with respect to something else.

So, I don't know how I can take a derivative when all I have is a discrete sequence of values.

I know how to take the derivative of 2t with respect to t.

I do not know how to take the derivative of {2, 4, 6, 8, ...} with respect to {1, 2, 3, 4, ...}. I don't even know what that would mean.

So how do you plan on defining the rate of a clock?

 

[yogi]

If we assume a transmission is sent once each time the SC clock passes by the tower in my post 128, the rate of the SC clock relative to the ECC clock is the time between consecutive transmissions sent by the SC clock as recorded on the SC clock divided by the time between two consecutive signals arriving at the ECC clock. If there is no altitude correction, the time between SC transmissions as measured on the SC clock will be less than the time recorded between receptions as measured by the ECC clock - but since we know the altitude and it is constant, we also know how fast the SC will run if it is corrected to account for the difference in G potential - in fact we can always put two clocks in orbit - one uncorrected for altitude, the other not. Depending upon your objective you can use one or the other.

Note - I do not claim to define a universal clock rate - I only conclude from this simple arrangement that one clock runs fast and the other slow -and there is no ambiguity about which is which - and I claim this is no different than the cosmological twin paradox in compact space - there is no uncertainty as to which clock is running slower. Here we have a well defined set of initial conditions - a central ECC and a way to adjust other clocks so that we can keep track of what has taken place vis a vis synchronization, off setting rates to compensate for either motion of G potential - whatever we want - we actually have a do-able experiment and a lot of confirming date from GPS. In the cosmological twin paradox, an ambiguity is created because it is assumed the two clocks in passing one another are each running at the same proper rate - but that is not the general case. As Garth observes - two clocks following the same geodesic would not in general log the same amount of time between successive passes - one needs to know more about how they got in relative motion to explain why they run at different proper rates - and once this is known, there is no paradox.

 

[Garth]

My musings above were over the ramifications of Hurkl's statement:

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.

I am now satisified, thank you!

Is not the difference between the bobbing/COM (or ECC and SC examples) and the cosmological twin case that of the difference between local and global symmetries?

i.e. The local (non-global) symmetry of the Earth's gravitaitonal field removes the ambiguity (described above #130) whereas in a totally homogeneous and isotropic FRW universe - or even a flat one with a non-trivial topology - that ambiguity can only be resolved by the global topology?

Of course - wearing my Machian hat - I have to ask: "What is it that determines that topology?"

Garth

 

[George Jones]

Is not the difference between the bobbing/COM (or ECC and SC examples) and the cosmological twin case that of the difference between local and global symmetries? i.e. The local (non-global) symmetry of the Earth's gravitaitonal field removes the ambiguity (described above #130) whereas in a totally homogeneous and isotropic FRW universe - or even a flat one with a non-trivial topology - that ambiguity can only be resolved by the global topology?

I wouldn't phrase it this way. All the cases that you mention are examples of a more general situation.

Two obsersvers are coincident at distinct spacetime events A and B, and, between A and B, the worldlines of the 2 observers are different timelike geodesics. To parametrize each worlline by proper time, an integral involving the metric along each worldlne must be performed. The metric is a tensor field, and, in general, assumes different values at the events along the different worldlines. This is the fundamental asymmetry that causes the 2 elapsed proper times to differ. It may sometimes be the case that topological quantities can easily be used to point out the asymmetry in worldlines, but it's the differing set of events, and the thus differing values of the metric tensor that is at the heart of the issue.

General statements like "moving clocks run slow" and "clocks in a gravitational field ..." are often misleading, because they don't always apply, and because it is so easy to apply them incorrectly. These statements can sometimes be useful if applied very, very carefully, but I find that I am too easily led astray by them.

The metric tells all!

Of course - wearing my Machian hat - I have to ask: "What is it that determines that topology?"

And what determines the metric.

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.

Is this also possible when both connections are metric-compatible and torsion-free?

I really have no idea whatsoever if the Einstein field equations would prevent this phenomenon.

Here's my thinking that caused my initial objection. If the distribution of energy, matter, and momentum (i.e, T) in the universe is specified, then, in a chart, Einstein's equation is a set of 10 coupled partial differential equations for the components of the metric. I wondered whether existence and uniquess theorems applied, thus pinning down the metric.

I worried, though, about stuff like: a single chart doesn't cover the entire universe; the "hole argument; etc.

Hawking and Ellis says "Thus the field equations really provide only six independent differential equations for the metric. This is in fact the correct number of equations to determine the spacetime, since four of the ten components of the metric can be givem arbitrary values by use of the four degrees of freedom to make coordinate transformations. ... Therefore the field equations should define the metric only up to am equivalence class under diffeomorphisms, and there are four degrees of freedom to make diffemorphisms."

Regards,
George

 

[Hurkyl]

Is this also possible when both connections are metric-compatible and torsion-free?

First off, I was under the impression that for any connection, there is a metric with which it is compatable. Is that true?

Anyways, according to Spivak, if you don't allow the geodesics to be reparametrized, then there is a unique torsion-free connection with those geodesics.

If you allow geodesics to be reparametrized, then for any connection $\nabla$ and one form $\omega$, the connection defined by $\bar{\nabla}_X Y := \nabla_X Y + \omega(X) Y + \omega(Y) X$ is a connection with the same torsion and geodesics as $\nabla$. (And all such connections are of this form)

 

[selfAdjoint]

First off, I was under the impression that for any connection, there is a metric with which it is compatable. Is that true?

Connections are more general than metrics. You can have a principle bundle over a non-metric space and define a connection on it. This in fact happens in many approaches to LQG.