Is Space-Time Curvature Zero Inside a Spherical Cavity at Earth's Center?

[lightarrow]

Inside a spherical cavity centered at the Earth's centre, the space-time curvature is 0 or =/= 0?

I know Newtonian gravitational field is omogeneously 0, so no field variation, but does GR give a different answer?

 

[pmb_phy]

Inside a spherical cavity centered at the Earth's centre, the space-time curvature is 0 or =/= 0?

I know Newtonian gravitational field is omogeneously 0, so no field variation, but does GR give a different answer?

GR gives the same answer.

Pete

 

[pervect]

All three of the following statements are true:

1) The Newtonian gravitational field at the center of the Earth is zero.
2) GR gives the same prediction as Newtonian gravity for this case.
3) The curvature, as measured by the curvature tensors $G_{ab}$ (and also $R_{abcd}$) is nonzero at the center of the Earth, as well), because $G_{ab} = 8 \pi T_{ab}$ and [itex]T_{ab}[/tex], the stress-energy tensor, is non-zero at the center of the Earth.

The problem can be traced to the problems of translating physics into sloppy popular language and the ambiguity of the term "gravitational field".

Depending on one's interpretation of what popular language means, one can conclude that either the "curvature" being talked about when one says "curvature causes gravity" is something other than the "curvature" as defined by the curvature tensors, or that what is measured by the "curvature tensors" is something other than the Newtonian gravitational field, or both.

It is specifically correct and helpful to note that the Riemann curvature tensor, $R_{abcd}$, measures Newtonian tidal gravity, rather than "gravity" itself, and it's probably also helpful to know that $R_{abcd}$ gives us more information than $G_{ab}$ because you can compute the later from the former, but not vica-versa.

 

[lightarrow]

All three of the following statements are true:

1) The Newtonian gravitational field at the center of the Earth is zero.
2) GR gives the same prediction as Newtonian gravity for this case.
3) The curvature, as measured by the curvature tensors $G_{ab}$ (and also $R_{abcd}$) is nonzero at the center of the Earth, as well), because $G_{ab} = 8 \pi T_{ab}$ and [itex]T_{ab}[/tex], the stress-energy tensor, is non-zero at the center of the Earth.

The problem can be traced to the problems of translating physics into sloppy popular language and the ambiguity of the term "gravitational field".

Depending on one's interpretation of what popular language means, one can conclude that either the "curvature" being talked about when one says "curvature causes gravity" is something other than the "curvature" as defined by the curvature tensors, or that what is measured by the "curvature tensors" is something other than the Newtonian gravitational field, or both.

It is specifically correct and helpful to note that the Riemann curvature tensor, $R_{abcd}$, measures Newtonian tidal gravity, rather than "gravity" itself, and it's probably also helpful to know that $R_{abcd}$ gives us more information than $G_{ab}$ because you can compute the later from the former, but not vica-versa.

Forgive me pervect, but I don't understand: shouldn't tidal forces be zero inside such a cavity, since the field is uniform there?

 

[pervect]

Let's work this out in Newtonian terms and see. Tidal force is force / meter.

For an object some distance d away from the center, there is no force due to the spherical cavity that the object is inside, but there is a force due to the mass M enclosed in a sphere of radius d around the center.

i.e. the net force is due entirely to that matter which is closer to the center of the Earth than the distance 'd' our test observer is.So if we are some distance d away from the center of the Earth, we will experience a force GM / d^2 = G V rho / d^2 towards the center of the Earth, where V= (4/3) pi d^3 is the volume.

The net result is that we see that force is proportional to distance, and the proportionality constant is (4/3) pi G rho.

Note that this is "Hooke's law force" for a constant density planet, it acts basically in the same manner as a spring.

[add]
I should probably read more carefully - I'm not going to delete my post above, but I'll point out that it does relate to the question of how to describe the "gravitational field" at the actual center of the solid Earth, not what the gravity would be if the Earth somehow had a cavity inside it.

 

[Chris Hillman]

A few pointers

Inside a spherical cavity centered at the Earth's centre, the space-time curvature is 0 or =/= 0?

I know Newtonian gravitational field is omogeneously 0, so no field variation, but does GR give a different answer?

This was going to be an exercise in the "What is the Theory of Elasticity?" thread, which seems to have died for lack of interest. Some brief pointers:

1. The case of a thin uniform spherical shell in the weak-field approximation is treated in many places, e.g. the problem book by Lightman et al. The field vanishes inside such a shell, but this gets quite tricky if the shell is "rotating".

2. The interior of a massive object is most often modeled in gtr as a perfect fluid, and the static spherically symmetric perfect fluids are not only known but a rare example of a well understood portion of the solution space of the EFE! The simplest model is the Schwarzschild constant density fluid ball; but the Tolman IV fluid is probably even better example. Qualitatively:

a. The pressure and density are maximal at the center and decrease monotonically as radius increases, with pressure falling to zero at the surface of the fluid ball (where we can match to a Schwarzschild vacuum exterior solution).

b. The acceleration of bits of fluid vanishes at the center (as must happen by symmetry), and depending on boundary conditions may reach a maximum under the surface (this happens for parameters appropriate for neutron star models).

c. The tidal tensor components are maximal (and positive) at the center and the tensor is in fact diagonal there (wrt any reasonable frame field). The components fall of as r increases, but $E_{11}$ decays faster than $E_{22}, \, E_{33}$ and may even go negative before reaching the surface (again, this happens for parameters appropriate for neutron star models).

d. Similar remarks hold for the three-dimensional Riemann tensor of the spatial hyperslices orthogonal to the world lines of the fluid elements. Put more vividly, near the center, the orthogonal hyperslice always resembles a three-spherical "cap" (see the pictures in MTW, which illustrate the Schwarzschild fluid, where each slice is locally isometric to S^3 everywhere).

e. The minimal radius for a static spherically symmetric fluid ball is $r=9/4 \, m$, i.e. larger than $r=2m$ (Buchdahl's theorem).

3. A perfect fluid cannot sustain a cavity at the center of a spherically symmetric body. However, an elastic solid can do so. A slight modification of an example I gave in the "What is the Theory of Elasticity?" thread gives the displacements and stresses for the exact solution in Newtonian elastostatics which models an spherically symmetric body with a spherical cavity at the ceenter which is made of an isotropic homogeneous material, such as steel. If I ever take up that thread again, at some point I'd get to weak-field approximation and then fully nonlinear elasticity. Static spherically symmetric elastic bodies are one of the few examples which can be treated fairly readily. The basic features of the stress tensor are similar to the above.

4. It is possible to create simple models of collapsing dusts with (shrinking) spherical cavities using the FRW dusts, or more generally the LTB dusts. The Carter-Penrose diagrams exhibiting the conformal structrure of such solutions possesses some interesting features.

Even in Newtonian theory, it is not so easy to generalize this to an oblate spheroid with centrifugal forces. In fact, it is not so easy to find the interior or exterior potentials of a homogeneous ellipsoidal solid. (The exact solutions can be given in terms of elliptic functions, but this apparently wasn't done explicitly until long after Newton! However, Newton did know the potential inside a homogeneous density thin ellipsoidal shell, which I'll let curious students puzzle over.) Nor is it trivial to investigate the stability of a rotating perfect fluid body even in Newtonian theory, with hydrostatic forces in static equilibrium with centrifugal "forces". (McLaurin and Jacobi discovered that there is an interesting sequence of sudden changes as the angular velocity increases.)

 

[lightarrow]

This was going to be an exercise in the "What is the Theory of Elasticity?" thread, which seems to have died for lack of interest. Some brief pointers:
[...]

Unfortunately I don't have enough knowledge to understand your answer. Thank you, anyway.

 

[pmb_phy]

Forgive me pervect, but I don't understand: shouldn't tidal forces be zero inside such a cavity, since the field is uniform there?

If the cavity is spherical and the body also spherically symetric and the center of the cavity is at the same location as the center of the sphere itself then there will be no gravitational field as measured by an observer at rest inside the cavity. This also means that the curvature will also be zero. If the cavities center is not at the same location as the center of the sphere then, in the weak field approximation (nobody I know has ever calculated the field exactly) there will be a uniform gravitational field inside the cavity as measured by an observer at rest inside the cavity. This means that the gravitational field is present but there is no spacetime curvature.

Pete

 

[lightarrow]

If the cavity is spherical and the body also spherically symetric and the center of the cavity is at the same location as the center of the sphere itself then there will be no gravitational field as measured by an observer at rest inside the cavity. This also means that the curvature will also be zero. If the cavities center is not at the same location as the center of the sphere then, in the weak field approximation (nobody I know has ever calculated the field exactly) there will be a uniform gravitational field inside the cavity as measured by an observer at rest inside the cavity. This means that the gravitational field is present but there is no spacetime curvature.

Pete

Thank you for your answer.
My original question was about a cavity concentrical with the earth, but since you have mentioned, I would like to ask you: in the second case you have discussed, when the cavity is not concentric with the earth, you make the assumption that the cavity is very little with respect the Earth (r << R)?

 

[Chris Hillman]

Unfortunately I don't have enough knowledge to understand your answer.

What didn't you understand? Perhaps I can clarify.

[EDIT: I just reread your exchange with pervect. Both pervect and I tried to sketch some of the relevant background. In particular, I think you should try to understand what Newtonian gravitation says before you tackle gtr, and pervect and I agree, I think, that you should try to understand the simplest models of a cavity-free ball deformed under its own weight before you try to understand a ball with a concentric cavity. As you can see from the posts by pervect and myself, there is more background to master than you may be ready to tackle, unfortunately.]

[EDIT: In case you are willing to try to understand the math, I fleshed out the Newtonian analogue, which is easier, in an addendum to "What is the Theory of Elasticity?", a tutorial in progress.]

 

[Mentz114]

Chris Hillman wrote :

This was going to be an exercise in the "What is the Theory of Elasticity?" thread, which seems to have died for lack of interest.

Perhaps not so much lack of interest as lack of time. Alas, many of us are constrained to earn a crust in between learning physics. Anyhow, it's a worthwhile resource as it stands, thank you.

 

[Chris Hillman]

Thanks

Chris Hillman wrote :

Perhaps not so much lack of interest as lack of time. Alas, many of us are constrained to earn a crust in between learning physics. Anyhow, it's a worthwhile resource as it stands, thank you.

You're very welcome!

I didn't express myself very clearly--- I meant that my own interest has been flagging, in part because the thread doesn't look very nice because I couldn't rearrange material, make corrections, additions, etc., except in an awkward way. Also, it probably should have been in the "tutorials" subforum I recently noticed someplace at PF.

Hmm...inspired by the missing example which should help the OP make the transition appreciate the relavistic analog, I'll make one last attempt to revive it.

 

[lightarrow]

What didn't you understand? Perhaps I can clarify.

[EDIT: I just reread your exchange with pervect. Both pervect and I tried to sketch some of the relevant background. In particular, I think you should try to understand what Newtonian gravitation says before you tackle gtr, and pervect and I agree, I think, that you should try to understand the simplest models of a cavity-free ball deformed under its own weight before you try to understand a ball with a concentric cavity. As you can see from the posts by pervect and myself, there is more background to master than you may be ready to tackle, unfortunately.]

You mean it's not true that the Newtonian field is uniform inside a cavity concentric with earth?
(The fact this field is 0 was not relevant, since we were discussing tidal forces, that is, field gradients).
P.S. I didn't mean to tackle GTR (I have no problems in admitting I'm not ready to do it, excepting the simplest concepts); I only wished to know if it gives a different answer from NT.

 

[Chris Hillman]

Elastic spherical body with cavity deformed under its own weight

You mean it's not true that the Newtonian field is uniform inside a cavity concentric with earth?

I don't know how you concluded that anything I wrote implies this! See my post # 30 in the "What is the Theory of Elasticity?" thread https://www.physicsforums.com/showthread.php?t=171079&page=2

(The fact this field is 0 was not relevant, since we were discussing tidal forces, that is, field gradients).

I am utterly baffled by this claim. I hope you noticed that pervect tried rather hard to explain that reference to "the gravitational field" in this context generally requires careful qualification, and that I routinely urge everyone to try hard to express themselves unambiguously.

(EDIT: the fifth time I read this, it occurred to me that perhaps you meant that if a Newtonian potential is constant, possibly a nonzero constant, the corresponding tidal tensor must vanish. That is true, and in fact in the post I cited I normalized the potential to vanish as spatial infinity, which means that it assumes a nonzero constant value in the cavity.)

P.S. I didn't mean to tackle GTR (I have no problems in admitting I'm not ready to do it, excepting the simplest concepts); I only wished to know if it gives a different answer from NT.

As pervect very carefully explained, you need to compare quantities which make sense in both theories, such as the tidal tensor. In the case of an isolated spherically symmetrical elastic body (homogeneous isotropic material) with a cavity at the center, the tidal tensor vanishes inside the cavity, in the case of both Newtonian gravitation and gtr. I carried out the Newtonian analysis in great detail in the above cited post, and if I continue writing the Elastic thread, eventually I'll give gtr treatments (but at present I am quite some ways from providing the necessary background).

(Strictly speaking, in gtr the tidal tensor is always defined wrt some congruence (family of world lines of ideal observers), but in this case that doesn't affect the conclusion since in this situation it will vanish for all observers.)

HTH

 

[pervect]

You mean it's not true that the Newtonian field is uniform inside a cavity concentric with earth?
(The fact this field is 0 was not relevant, since we were discussing tidal forces, that is, field gradients).
P.S. I didn't mean to tackle GTR (I have no problems in admitting I'm not ready to do it, excepting the simplest concepts); I only wished to know if it gives a different answer from NT.

If we hypothesize a hollow cavity inside a spherically symmetrical rotating body, there may be some gravitomagnetic fields inside the body, which would make it different from Newtonian theory.

 

[Chris Hillman]

Lense-Thirring effect

If we hypothesize a hollow cavity inside a spherically symmetrical rotating body, there may be some gravitomagnetic fields inside the body, which would make it different from Newtonian theory.

Right, that's a good point--- in the Newtonian case, a slowly and rigidly rotating uniform mass density spherical shell produces the same field as a nonrotating spherical uniform mass density shell with the same density and dimensions, but in gtr, the fields are different, due to gravitomagnetism. In gtr, the Lense-Thirring effect says that a gryostablized frame inside a slowly rotating spherical shell will slowly rotate wrt the fixed stars. (Since said stars are outside the shell, this takes some further explanation before it makes sense, but with sufficient care, it makes sense, and the shell rotates in a tricky but definable sense; the gyrostabilized frames inside rotate at a different but definite angular velocity depending on the angular velocity, dimensions, and mass of the shell.) Some authors interpret this effect as "Machian"; others argue with equal persuasiveness that it is anti-Machian. (Just one more indication of the amazing variety of interpretations of what a "Mach Principle" might mean in gtr.)

Why did I say "rigidly rotating"? Well, in either Newtonian theory or gtr, an isolated steadily rotating shell made of an elastic material will be deformed into a roughly oblate spheroidal shape due to centrifugal forces opposing its gravitational self-attraction. (Except that in gtr, "shape" is a lot trickier to discuss!)

 

[pervect]

I haven't worked it out in any great detail, but I think it's true that inside a rotating hollow sphere you can't make the curvature tensor vanish because of the gravitomagnetic effects.

This is rather similar to the way a hollow sphere of charge behaves. If the sphere of charge is non-rotating, a void inside the sphere of charge has no forces (neither electric or magnetic). If the sphere of charge rotates, a void inside the sphere has a magnetic field

You can transform away the magnetic part of the Riemann that I was talking about - but I think you'll wind up with an electric part when you do so.

 

[lightarrow]

If we hypothesize a hollow cavity inside a spherically symmetrical rotating body, there may be some gravitomagnetic fields inside the body, which would make it different from Newtonian theory.

Ok. In my initial question I intended just "hollow" cavity (I thought "cavity" only was enough) and "non rotating". From your answer I deduce GTR predicts no curvature there (as pmb_phy wrote, and as Chris wrote in his post # 14).
Thank you, pervect (and thank you Chris).

 

[Chris Hillman]

Intrinsic gravitomagnetism, anyone?

I haven't worked it out in any great detail, but I think it's true that inside a rotating hollow sphere you can't make the curvature tensor vanish because of the gravitomagnetic effects.

That is correct (that's what Lense and Thirring did for a rotating spherical shell in weak-field gtr in 1917). Reproducing their result is a problem in the book by Lightman et al. On at least two previous occasions I have gone through that in great detail, so Googling might uncover those efforts.

This is rather similar to the way a hollow sphere of charge behaves. If the sphere of charge is non-rotating, a void inside the sphere of charge has no forces (neither electric or magnetic). If the sphere of charge rotates, a void inside the sphere has a magnetic field.

Again, a good point. I emphasized this when I worked the computations in the expository posts I just mentioned. This computation is indeed very instructive because of the close formal analogies.

You can transform away the magnetic part of the Riemann that I was talking about - but I think you'll wind up with an electric part when you do so.

This gets a bit tricky, but roughly speaking one of the drawbacks of weak-field theory is that it is ultimately not really self-consistent. In the fully nonlinear gtr, you could appeal to invariants like
$$R_{abcd} {R^\ast}^{abcd}$$
which is analogous to
$$F_{ab} {F^\ast}^{ab}$$
So there's a notion of "intrinsic gravitomagnetism" (analogous to "intrinsic magnetism"), meaning that gravitomagnetic effects which cannot be entirely transformed away. Applying this test to the Kerr vacuum and similar exact solutions, we confirm that these exhibit intrinsic gravitomagnetism. There's a nice discussion in Cuifolini and Wheeler, Gravitation and Inertia.

 

[pmb_phy]

Thank you for your answer.
My original question was about a cavity concentrical with the earth, but since you have mentioned, I would like to ask you: in the second case you have discussed, when the cavity is not concentric with the earth, you make the assumption that the cavity is very little with respect the Earth (r << R)?

This is not an assumption. It was the result of a calculation using GR in the weak field limit. I'm trying to find the web page I created to demonstrate this calulation result. In full GR there may be a very small amount of curvature but for a planet the size of the Earth I can't imagine that it would be detectable with our best equipment. So long that the cavity is spherical and completely contained within the sphere then it is valid.

I failed to mention that I was assuming a uniform mass density throughout the sphere before the cavity is cut out of it.

Edit: Found it. The web page is at
http://www.geocities.com/physics_world/gr/grav_cavity.htm

After reading it I now see I should have made it clearer because I assumed that the reader was familiar with the details of the weak field approximation. I may change it in the future. Thanks.

Pete

 

[Chris Hillman]

From your answer I deduce GTR predicts no curvature there

If the cavity is concentric and if the body is nonrotating. We call a region of spacetime where the curvature tensor vanishes locally flat, and we say that it is locally isometric to Minkowski spacetime (but certainly not globally isometric to Minkowski spacetime, in this discussion!).

 

[pmb_phy]

If the cavity is concentric and if the body is nonrotating. We call a region of spacetime where the curvature tensor vanishes locally flat, and we say that it is locally isometric to Minkowski spacetime (but certainly not globally isometric to Minkowski spacetime, in this discussion!).

I disagree. The term locally flat refers to a system of coordinates for which the metric takes on the same value it does in Lorentz coordinates (i.e. locally inertial frame (ct, x, y, z)).

Pete

 

[pervect]

I disagree. The term locally flat refers to a system of coordinates for which the metric takes on the same value it does in Lorentz coordinates (i.e. locally inertial frame (ct, x, y, z)).

Pete

Do you have a reference for that statement?

We call a region of spacetime where the curvature tensor vanishes locally flat.

sounds like a perfectly reasonable definition of locally flat spacetime to me.

 

[pmb_phy]

Do you have a reference for that statement?

Yes. See A first course in general relativity by Bernhard F. Schutz, page 6.2. The section is called The metic and local flatness. Do you know of a reference for Hillman's statement?

Pete

 

[Garth]

Can we provide examples to illustrate the difference?

i.e. A system of coordinates for which the metric takes on the same value it does in Lorentz coordinates (i.e. locally inertial frame (ct, x, y, z)) when the curvature tensor does not vanish?

Or a situation where the curvature tensor vanishes and a system of coordinates for which the metric does not take on the same value it does in Lorentz coordinates (i.e. locally inertial frame (ct, x, y, z))?

Garth

 

[pmb_phy]

Can we provide examples to illustrate the difference?

i.e. A system of coordinates for which the metric takes on the same value it does in Lorentz coordinates (i.e. locally inertial frame (ct, x, y, z)) when the curvature tensor does not vanish?

Or a situation where the curvature tensor vanishes and a system of coordinates for which the metric does not take on the same value it does in Lorentz coordinates (i.e. locally inertial frame (ct, x, y, z))?

Garth

If you have a flat spacetime then the curvature tensor will vanish, regardless of the system of coordinates.

Pete

 

[pervect]

Yes. See A first course in general relativity by Bernhard F. Schutz, page 6.2. The section is called The metic and local flatness. Do you know of a reference for Hillman's statement?

Pete

I was able to borrow a copy of Schutz, but the only thing I could find in it was a discussion of locally flat coordinates, not a discussion or definition of a locally flat spacetime. For example, on pg 158 (section 6.2, not page 6.2)

This is a very useful result. It is also conceptually an important result
because it is the first example of a kind of argument we will frequently
employ, an argument that uses locally flat coordinates to generalize our
flat-space concepts to analogous ones in curved space.

One of the reasons I'm suspicious about your claims is that it doesn't make a lot of sense for the definition of a "locally flat spacetime" to be dependent on the coordinates one choses. One would expect that the local flatness of a spacetime would be independent of the choice of coordinates.

There is also a good chance, BTW, that mathematicians and physicists have slightly differing definitions here, in particular different definitions of what it means to be "local".

At the moment I don't have a definitive definition to offer, though I continue to think the definition offered by Chris Hillman makes sense. Since you were so emphatic about your claims, I thought it appropriate to ask you for a reference. Unfortunately your reference hasn't convinced me that Chris Hillman is wrong, it appears that he's talking about oranges (locally flat spacetimes) while you're talking about apples (locally flat coordinates).

 

[MeJennifer]

Looks like a confusion about terminologies, in this case the same term for different things

Conform Shutz: "The existence of local Lorentz frames is merely the statement that any curved space has a flat space "tangent" to it at any point." So in that respect every small region on a differentiable curved Riemannian is locally flat.

But at the same time a curved manifold could have extended flat regions, one could also call such regions locally flat.

 

[pmb_phy]

I was able to borrow a copy of Schutz, but the only thing I could find in it was a discussion of locally flat coordinates, not a discussion or definition of a locally flat spacetime. For example, on pg 158 (section 6.2, not page 6.2)

I don't see how that is a problem since I clearly stated The term locally flat refers to a system of coordinates ..., i.e. that was my entire point. And I made this point because the presence of "local flatness" has nothing to do with spacetime curvature or the Riemann tensor etc. It has to do with a locally inertial coordinate system. Personally I think "Locally flat" should never have been given this meaning. But I go but well established terminology so that I don't have to change it everytime somebody new pops in and starts using non-standard terminology then it would every confusing. As far as I know the terms "Locally flat spacetime" has never been defined. I'm still waiting for Hillman to provide a source for this term as he uses it.

pervect - I'm curious about something. Back in 05-01-2005 you used the term "locally flat coordinates". See - https://www.physicsforums.com/showpost.php?p=551833&postcount=25

Do you recall what you meant by that term at that time??

There is one last think that I wanted to point out and that is the fact that all manifolds 'look' flat when you look really close to them. I can see this as being very wel defined and what Hillman may have been speaking of.

Pete

 

[mendocino]

All three of the following statements are true:

1) The Newtonian gravitational field at the center of the Earth is zero.
2) GR gives the same prediction as Newtonian gravity for this case.
3) The curvature, as measured by the curvature tensors $G_{ab}$ (and also $R_{abcd}$) is nonzero at the center of the Earth, as well), because $G_{ab} = 8 \pi T_{ab}$ and [itex]T_{ab}[/tex], the stress-energy tensor, is non-zero at the center of the Earth.

The problem can be traced to the problems of translating physics into sloppy popular language and the ambiguity of the term "gravitational field".

Let's forget about gravitational field for a moment and talk about measured quantity? For example, will the clock runs slowly at the center and will someone experienced a force inside the cavity?

 

[Chris Hillman]

You need to be more precise

Let's forget about gravitational field for a moment and talk about measured quantity? For example, will the clock runs slowly at the center and will someone experienced a force inside the cavity?

Hi, mendocino, this question actually is a very good illustration of why I keep insisting that saying "the clock runs slow" somewhere or even that "time slows" somewhere is Not A Good Idea. Rather, the world lines of radially outgoing time signals are modeled by radial null geodesics and these tend to spread apart in a Schwarzschild vacuum. IOW, an ideal clock can only run slow compared to another ideal clock, and the way in which this comparison is carried out is crucially important.

To keep things manageable, we typically seek a simple operationally significant way of effecting the comparison; one of the simplest is to assume that both clocks A,B lie on the same radius and to look at radial null geodesics from A to B and vice versa. In this case, we have the added complication that the homogeneous isotropic material of which our spherical object with concentric cavity is made, is presumably not transparent to light, and light (or radio waves) presumably travels more slowly inside this material than in vacuo.

Be this as it may, you more or less said what A you have in mind (unless you don't mean, as I guess, the static observer at the very center of the cavity), but not what B, so you need to clarify that before we can compute the answer you seek. You also need to clarify whether you are willing to accept a weak-field approximation or not. And if the answer is "not", you should wait until I get around to generalizing the Newtonian theory in "What is the Theory of Elasticity?" Or else look up previous posts by myself from many years ago in sci.physics.relativity in which I worked this out for the case of a simple model of a perfect fluid ball (no cavity), the Schwarzschild fluid.

However, it should be clear just from looking at the Newtonian potential (see my plot in "What is the Theory of Elasticity?") that this scenario involves the behavior of geodesics "in the large". That is, we should expect a "time dilation" effect (the magnitude depends upon the details you haven't specified), even though the cavity is locally isomorphic to Minkowski spacetime, i.e. locally flat.

(Re what pervect said, this is a completely different concept from the concept of "locally geodesic" or sometimes "locally flat" charts, among which Riemann normal charts are particularly important. The terminology in this subject can sometimes be confusing, and is sometimes incompletely standardarized! If you stay alert to the possibility of confusion, you can probably avoid trouble. In particular, note that "locally flat spacetime" uses "local" in the standard sense of manifold theory, meaning "local neighborhood", whereas "locally flat charts" are defined at the level of the metric tensor and its derivatives at a single event. The latter usage is deprecated because it conflicts with the mathematical terminology required for much modern "geometric" physics, e.g. for fiber bundles. For more about normal charts, see Poisson, A Relativist's Toolkit.)

As for "force", if you were asking whether a stationary observer at the very center of the cavity in a Newtonian model experiences a force, the answer is self-evident from symmetry. (In what direction would the force vector point?) In gtr, there is no notion of "gravitational force", but you can ask whether the stationary observer at the center of the cavity is an inertial observer or not. If not, he must use a rocket engine (or taut cables) to maintain his position, and thus must feel a nongravitational force, corresponding geometrically to the path curvature of his world line. (In what direction would the force vector point?)