# Curvature inside a cavity

1. Jun 4, 2007

### 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?

2. Jun 4, 2007

Pete

3. Jun 4, 2007

### pervect

Staff Emeritus
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 $T_{ab}[/tex], the stress-energy tensor, is non-zero at the center of the Earth. The problem can be traced to the problems of translating physics into sloppy popular langauge and the ambiguity of the term "gravitational field". Depending on one's interpretation of what popular langauge means, one can conclude that either the "curvature" being talked about when one says "curvature causes gravity" is something other than the "curvature" as defined by the curvature tensors, or that what is measured by the "curvature tensors" is something other than the Newtonian gravitational field, or both. It is specifically correct and helpful to note that the Riemann curvature tensor, [itex]R_{abcd}$, 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.

4. Jun 4, 2007

### lightarrow

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

5. Jun 4, 2007

### pervect

Staff Emeritus
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.

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.

Last edited: Jun 4, 2007
6. Jun 4, 2007

### Chris Hillman

A few pointers

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 possess 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.)

Last edited: Jun 4, 2007
7. Jun 5, 2007

### lightarrow

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

8. Jun 5, 2007

### pmb_phy

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

9. Jun 5, 2007

### lightarrow

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)?

Last edited: Jun 5, 2007
10. Jun 5, 2007

### Chris Hillman

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.]

Last edited: Jun 5, 2007
11. Jun 5, 2007

### Mentz114

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.

12. Jun 5, 2007

### Chris Hillman

Thanks

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.

13. Jun 6, 2007

### lightarrow

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.

Last edited: Jun 6, 2007
14. Jun 6, 2007

### Chris Hillman

Elastic spherical body with cavity deformed under its own weight

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

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.)

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 neccessary 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

15. Jun 6, 2007

### pervect

Staff Emeritus
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.

16. Jun 6, 2007

### Chris Hillman

Lense-Thirring effect

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 alot trickier to discuss!)

Last edited: Jun 6, 2007
17. Jun 6, 2007

### pervect

Staff Emeritus
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.

18. Jun 6, 2007

### lightarrow

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).

Last edited: Jun 6, 2007
19. Jun 6, 2007

### Chris Hillman

Intrinsic gravitomagnetism, anyone?

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.

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.

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.

20. Jun 6, 2007

### pmb_phy

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

Last edited: Jun 6, 2007
21. Jun 6, 2007

### Chris Hillman

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!).

22. Jun 6, 2007

### pmb_phy

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

23. Jun 6, 2007

### pervect

Staff Emeritus
Do you have a reference for that statement?

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

24. Jun 7, 2007

### pmb_phy

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

Last edited: Jun 7, 2007
25. Jun 7, 2007

### 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