Dear Jack,

I think what you are missing is the information of the reference system. "Slow" is not to be meant in an absolute sense, but relative to some other reference system. Therefore the question is: the interior clock is "slow" relative to what? To another clock which is located on the outside surface of the shell? If so, then your scenario is correct, but neither clock needs "to know" about the other, of course. When the shell expands or collapses, the clock located on the outside surface undergoes changes of the gravitational field, which is sufficient to discuss the time dilation effects.

Best regards

Oliver

Thank you, Oliver Tennet, for your very thoughtful reply.

I meant slow relative to a clock very far away from the shell, where the shell's gravitational time dilation is negligible. Ultimately, slow relative to a clock in the the comoving frame. Cosmic time in the comoving frame, in which the cosmic background radiation appears isotropic (and far removed from local gravitators), is in a sense absolute. In an oscillating universe, a clock in the comoving frame measures the greatest elapsed time from the Big Bang to the Big Crunch. Other clocks can measure a shorter elapsed time form the Big Bang to the Big Crunch, but no other clock can measure a longer elapsed time. (Newton's statement about "true, absolute, and mathematical time" seems not wrong if applied to cosmic time.) A clock in the comoving frame always measures cosmic time regardless of the radius of the shell. So if the shell contracts (expands), how does the clock inside "know" that its slowdown relative to the cosmic-time clock must increase (decrease) --- even though by Birkhoff's Theorem there is no gravitational field inside the shell for the clock inside the shell to interact with?

Alternatively, if the spherical shell is all that exists, slow relative to a clock extremely far away from the spherical shell, where its gravitational influence is negligible and spacetime is Minkowskian. A clock extremely far away from the spherical shell always measures undilated time regardless of the radius of the shell. But there is also no gravitational field inside the shell, so spacetime is also Minkowskian inside the shell. If the shell contracts (expands), a clock riding on the outside of the surface of the shell senses an increase (decrease) in the shell's gravitational field and thus "knows" to slow down more (less) relative to the clock extremely far away. But by Birkhoff's Theorem spacetime inside the shell remains Minkowskian and hence the clock inside the shell senses nothing, so how does it "know" to slow down more (less) relative to the clock extremely far away? Even if the shell contracts through its Schwarzchild radius, by Birkhoff's Theorem spacetime inside the shell still remains Minkowskian and hence the clock inside still senses nothing until the shell collapses onto it --- yet it must stop completely relative to the clock extremely far away --- but how does it "know" this?

Dear Jack,

I think you might have misunderstood some points in general relativity. Or maybe I am misunderstanding you.

"I meant slow relative to a clock very far away from the shell, where the shell's gravitational time dilation is negligible."

Aha! So you mean an asymptotic observer? But no: then there will be no difference between a clock within a shell of either smaller or larger radius, because the interior gravitational field is zero (in either case), whereas the outer gravitational field is completely unchanged by changing the radius of the shell. (By the way, this is what Birkhoff's theorem actually says.)

"[...]

So if the shell contracts (expands), how does the clock inside "know" that its slowdown relative to the cosmic-time clock must increase (decrease) --- even though by Birkhoff's Theorem there is no gravitational field inside the shell for the clock inside the shell to interact with?"

As there is no change in the gravitational field, there is no change in the time delay.

"A clock extremely far away from the spherical shell always measures undilated time regardless of the radius of the shell."

So now you are saying this yourself, which is correct. But -- by the way -- for Birkhoff's theorem to apply, you need not be asymptotic at all. It applies to the complete outer region of the shell, close or far. Reminder: Birkhioff's theorem applies to the outer region and says that the only spherically symmetric vacuum solution of the Einstein field equations is Schwarzschild's.

Best regards

Oliver

Dear Oliver,

Thank you very much for another very thoughtful reply. I’m sorry if there was any misunderstanding.

Of course, so long as contractions or expansions of a spherical shell are purely radial, Birkhoff’s Theorem holds everywhere, not just at asymptotic (very large) distances from the shell. Indeed, to the best of my understanding, it holds inside as well as outside of the shell. Indeed, since spacetime inside the shell always remains Minkowskian with the metric coefficients never changing from their Minkowskian values, how can Birkhoff's Theorem not hold inside the shell as well as outside of it?

The difference in clock rates between an asymptotic clock very far removed from a spherical shell and a clock inside the shell depends on the difference in potential, not on the field. An asymptotic clock experiences (essentially) both zero field and zero potential regardless of the radius of the shell. By contrast, a clock inside the shell experiences zero field, but not zero potential, regardless of the radius of the shell. If the shell radially contracts, the potential inside becomes more strongly negative, and hence the clock inside the shell ticks more slowly relative to the asymptotic clock. Thus, despite General Relativity being a field theory rather than an action-at-a-distance theory, it seems that: if the clock inside the shell responds to changes in the field, it must respond at a distance rather than locally, but if it responds to changes in the potential, it can respond locally.

An analogy might be the Aharonov-Bohm effect, whereby an electron (or other electrically-charged particle) is deflected by the (vector) potential due to a magnetic field localized within a solenoid, even though the electron is outside of the solenoid and the magnetic field cannot reach the electron. Yet perhaps the Aharonov-Bohm effect can be explained in terms of fields. Although the magnetic field of the solenoid is localized within the solenoid and cannot reach the electron, the electron’s magnetic field, which exists in the solenoid's reference frame because because the electron is moving relative to the solenoid, penetrates to within the solenoid, so there might be interaction between the electron’s electromagnetic field and the magnetic field within the solenoid. (The electron's electric field may not be able to penetrate to within the solenoid, because the solenoid is made of conductive material.) Likewise, although the gravitational field within a spherical shell is zero regardless of the radius of the shell, the gravitational field of a clock that is inside the shell penetrates to the outside of the shell, so there might be interaction between the clock’s gravitational field and that of the shell.

All the best,

Jack

Dear Jack!

Ah! I got it now what you mean! Thanks for the clarification!

Interesting indeed. Yes I think you are right: essentially what we are looking at is a scenario the analysis of which needs the equivalence principle only, not general relativity.

I must admit I have not yet thought it through, nevertheless my first line of thinking is: how will the shell be able to expand/collapse on a scale that will lead to a significant change in the gravitational potential? For this, surely, it either is a semi-classical "source" that somehow changes by some external mechanism, and is not to be treated as part of some sort of self-consistent solution of the Einstein equations. Also, as the expanding/collapsing is spherical, there is no gravitational radiation whatsoever.

Hence your comparison to the AB effect might be correct, in fact it is surely to be considered as a Berry-phase of sorts. On the other hand, there is no cyclic motion in your scenario. But the analogy might fit in general.

But let me think this through, I think your last post seems to be on the right track.

And also, I think I was wrong in saying: there is no change in the gravitational field. Because indeed overall there is: the spacetime region which is flat changes as the shell changes its radius, and so does the outer Schwarzschild region.

Cheers

Oliver

Dear Oliver,

Thank you very much for another very helpful and thoughtful answer.

A spherical shell has zero gravitational field corresponding to TWO values of the potential, zero potential at extremely large distances from the shell and negative potential within the shell. Concerning employing the equivalence principle, I haven’t been able to conceive of ANY accelerational field (construed as gravitational by an observer at rest in the accelerating frame) that can duplicate this property. Two examples: (i) Within an elevator undergoing constant acceleration, the gravitational field of a very large flat disk, near the center of the very large flat disk (distance from center of disk

Dear Ahmida,

first of all, there is no such thing as a global observer in relativity. That actually is the meaning of relativity.

Second, Jack Denur's scenario is the following: a spherical shell with an expanding/collapsing/oscillating radius (for whatever reasons) leads to the following situation:

- there is no gravity and therefore a constant potential inside the shell

- there is no change in gravity to a remote (not asymptotic, but far away from the shell's surface) observer, due to Birkhoff's theorem

- the time dilation for the remote observer's clock relative to the clock inside the shell is changing, according to the changing radius of the shell

The latter occurs, even though (naive and wrong interpretation!) neither clock somehow changes its state of motion or its location!

The correct interpretation is: the change in time dilation occurs because spacetime is changing overall when the spherical shell changes its radius.

Jack Denur: I am not yet 100% sure if you can interpret this as a Aharomov-Bohm-like phenomenon or not. But I am not done thinking about it either...;)

Also, the equivalence principle of course is a local principle, not a global one, and in a strict sense cannot even be extended to any finite volume. No acceleration can lead to a radial gravitational field as given outside the shell. So the answer to your question above is NO.

Best regards

Oliver

Dear Ahmida and Oliver,

Thank you for your interest and replies.

Especially, thank you Oliver for your reply concerning the equivalence principle.

One clarification: If the shell expands/contracts/oscillates, there is a constant (zero) gravitational field inside the shell but not a constant gravitational potential. The potential inside the shell becomes less (more) strongly negative if the shell expands (contracts).

Dear Jack,

if you plot the function of the gravitational potential versus r, the potential is constant within the shell, and then increases linearly from r_S (radius of the shell) on. The variability of the potential due to the expansion/contraction is then given by the varying point r_S from which on the linear increase starts, by the same derivative. Hence the potential difference is dependent on r_S.

Best regards

Oliver

Dear Oliver,

The potential is indeed constant within the shell, as it must be since the field is zero within the shell. But the value of this constant depends of the radius of the shell. The smaller the radius of the shell, the more strongly negative is the constant potential within the shell (assuming the mass of the shell to be fixed).

All the best.

Jack

Yes, correct. If you define your reference potential e.g. as the one at spatial infinity, or at any other well-defined radius r_ref.

Best regards

Oliver

Dear Oliver,

Thank you very much for another helpful reply.

The reference potential at spatial infinity is of course most naturally defined at zero, because at spatial infinity it is as if the shell didn't exist. If the shell was the only thing in the universe, if it didn't exist there would be no source of gravitation, and hence the potential everywhere in such a Minkowskian universe would be most naturally set at zero.

All the best.

Jack

Yeah, it's probably "1" then...

Dear Gentlemen,

I am sorry, I noticed this discussion now, with a large delay. Perhaps, it can still go on.

It was mentioned that the metrics inside the spherically symmetric material shell is the Minkowski metrics. I would like to ask: why? Because I look for a reason, several years, and I have found no actual reason.

I perfectly understand that the net gravitational acceleration in the shell is zero if the Newton law is considered. However, in general relativity, it seems that the Minkowski metrics is just postulated.

Dear L. Nielusan,

If the gravitational field is ZERO, isn't it true that the metric MUST be Minkowskian? Isn't the Minkowski metric the ONLY metric consistent with ZERO gravitational field?

Sincerely yours,

Jack Denur

Deak Jack Denur,

of course, if the gravitational field is zero, then the metrics in general relativity (GR) is the Minkowski metrics. However, field equations say that a particle in the shell, in a position aside the center, is accelerated outward the center. This is what the field equations and equation of geodesic generaly imply, in fact. It would be hard to explain this conclusion, here, with only the plain text. I outlined the derivation of outward oriented gravitational action in shell in GR in the 3rd section of my paper published in Modern Physics Letters A (2019, Vol. 34, id. 1950244-356; it can also be downloaded from my personal web page: https://www.astro.sk/~ne/neslusanMPLA.pdf). If you are interested in my reasoning, please see the section.

I suspect that the Minkowski metrics inside the shell was postulated to avoid an (artificial) problem in modeling of compact objects like neutron stars. Namely, if one integrates the field equations to obtain a model, and the integration starts in a finite distance from the object's center, whereby it is performed in two etapes, stepping outward and inward, then also the inward processed integration ends with zero pressure and energy density in a finite object-centric distance. This implies an existence of inner surface of the object and a vacuum void inside. For a given set of constituting particles, there is only one combination of initial conditions resulting in the full-sphere (zero inner radius) object and infinite number of combinations of initial conditions producing an object in form of hollow sphere. In GR, the hollow-sphere-object solutions are (currently) forbidden. (I do not know the reason why, but are.) So, if we want to obtain and explain the full-sphere solutions, there must be postulated the Minkowski metrics inside a thin spherically symmetric material shell. To establish the postulate, the field equations, equation of geodesic, and discontinuity of metrics in the inner surface of the shell are simply ignored.

Dear L. Neslusan,

Thank you very much for your very thoughtful and very detailed reply, and for your paper. It seems to imply that the gravitational acceleration is NOT zero, but directed radially outwards, inside a spherical shell, except at the center. This seems amazing. I need to think about it. Once again, thank you very much.

Dear L. Neslusan and Oliver Tennert,

I thank both of you for your very helpful comments and suggestions.

I think that I understand the issue of when the gravitational field does and does not vanish within a spherical shell. The gravitational field vanishes within a spherical shell if and only if the gravitational force diminishes as the inverse square of the ruler distance from a point source, or more generally from the center of and outside of any spherically-symmetrical source, of gravitation. If the gravitational force diminishes less rapidly than as the inverse square of the ruler distance from such a source, then a test mass within but not at the center of a spherical shell experiences greater attraction from the far side of the shell than from the near side, and hence is attracted towards the center. At the center of a spherical shell it is in stable equilibrium. (In the absence of friction it can oscillate about the center.) If the gravitational force diminishes more rapidly than as the inverse square of the ruler distance from such a source, then a test mass within but not at the center of a spherical shell experiences greater attraction from the near side of the shell than from the far side, and hence is attracted away the center, until it rests at (or in the absence of friction bounces from) the nearest point on the inner surface of the spherical shell. At the center it is in unstable equilibrium: the slightest disturbance (unavoidable owing to the uncertainty principle) will displace it from the center. The latter situation (gravitational force diminishes more rapidly than as the inverse square of the ruler distance) obtains in General Relativity, owing to the departure from Euclidean geometry of space engendered by gravity, and of the consequent steepening of the gravitational potential from the Newtonian. Gravity stretches space from the Euclidean in the direction of a gravitator. [See Relativity: Special, General, and Cosmological (Second Edition), by Wolfgang Rindler (Oxford University Press, Oxford, UK, 2006), Chapter 11, especially Sections 11.1, 11.2A, and 11.5.].

So a clock within a spherical shell DOES have a gravitational field with which it can interact LOCALLY. This seems to resolve the spherical-shell clock paradox in General Relativity. The gravitational potential is most strongly negative at the inner surface of the spherical shell, and slightly less negative at the center. So a clock will tick most slowly at the inner surface of the spherical shell, and slightly faster at the center.

Dear Jack Denur,

thank you for the detailed explanation of the gravitational force inside the spherical shell and pointing out that a non-zero gravitational attraction resolves the spherical-shell clock paradox in the GR. I have no objection. I would however like to state a warning. I am afraid that the community dealing with the GR, especially with the compact objects, would not accept any analysis based on a non-zero net force inside the shell. Other than the Minkowski metrics inside the shell is forbidden to be supposed in reality. I do not know a reason of this prohibition. The GR people however always claim that the metrics is the Minkowski metrics and net gravity in the shell, aside the center, is zero. (They do so despite the fact that this generates several problems; not only the spherical-shell clock paradox.)

Dear L. Neslusan,

Thank you very much for your very thoughtful reply.

All the best.

Sincerely yours,

Jack Denur

In the context of non-zero gravity inside the spherical shell, I would like to add a remark on the aspect, I realized recently. In the case of spherical symmetry, there should always be zero net gravity. However, in the general relativity, the concept of the spherical symmetry is not so simple as in the Euclidean geometry of Newtonian physics. In the Newtonian physics, a distribution of matter, which is observed as spherically symmetric by an observer in the center of the symmetry (center of shell), is also observed as spherically symmetric by an observer in whatever else position. However, in the general relativity, we can find three independent Killing vectors on a sphere in the coordinate frame with the origin identical with the center of the sphere, but this task cannot be solved in the coordinate frame with other center, I am afraid. So, while the distribution of matter is observed to be spherically symmetric by the observer in the center, it is no longer observed as spherically symmetric by the observer in the center of other frame (observer aside the center). I think that just this asymmetry is the cause of non-zero gravity in the shell, in the general relativity.

Dear L. Neslusan,

Thank you for another informative reply, even though I am not familiar with the higher technical-mathematical aspects of general relativity.

So I rely on a more physical, intuitive explanation. If the gravitational force as a function of ruler distance from a point source or from the center of a spherical source is other than inverse-square, gravity within a spherical shell is non-zero (except at the center). In General Relativity it is steeper than inverse square (sometimes called the pit in the potential).

All the best.

Sincerely yours,

Jack Denur