# Shell Theorem & False Centers

1. Jun 11, 2008

### jonmtkisco

The Newtonian Shell Theorem makes important predictions about the gravitational force experienced by an object located inside a spherically symetrical massive structure, such as a hollow or solid ball. The same predictions are made by Gauss' Law. According to http://en.wikipedia.org/wiki/Shell_theorem" [Broken]:

1. If the body is a spherically symmetric shell (i.e. a hollow ball), no gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

2. Inside a solid sphere of constant density the gravitational force varies linearly with distance from the center, becoming zero at the center of mass.

It bears emphasizing that the Shell Theorem applies only with respect to mass arrangements which are (a) actually centered on the center of the sphere, (b) are symetrically distributed around that center, and (c) have constant density in the non-hollow portion. In this thread I want to discuss some examples that don't seem to meet all three criteria, although the issue can become interestingly subtle.

The most obvious example is that a solid or hollow shell of matter cannot act as a "gravity shield" to avoid the gravity of other nearby objects. For example a dense sphere buried just under the earth's surface is not shielded from the earth's gravity to any significant degree. This is so of course because the earth is not arranged spherically around the sphere, and the center of the sphere is not the "true" center of the earth-sphere system. I attached a simple diagram which illustrates this example.

I'll post a more interesting example later.

Jon

#### Attached Files:

• ###### Sphere A inside Sphere B.pdf
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2. Jun 12, 2008

### jonmtkisco

Infinite homogeneous dust

OK, so here's a subtle example.

Let's consider an infinite, spatially flat, Lambda = 0 universe filled homogeneously with "dust" (matter particles). Let's arbitrarily define an origin coordinate and a metric sphere centered on that origin. Let's say this universe is expanding slowly with a low deceleration factor (relative to the proper size of the sphere). Let's place a massless test particle inside this sphere; it is at proper rest with respect to our origin coordinate. Can we correctly apply the shell theorem to this test particle?

Sure, as long as the center of our sphere represents the true center of the dust cloud. Uh oh... this dust cloud is infinite so it does not have a gravitational center. We can establish an arbitrary center by defining any coordinate system we wish. But it is not a gravitational center; a universe of this type has no true center.

Therefore I hypothesize that the test particle will NOT move towards the center of the sphere in the manner described in the shell theorem. It will not move at all because it feels no directional gravity vector.

Let's test this hypothesis by arbitrarily defining three very large, symetrically arranged, overlapping metric spheres, each with its own arbitrarily defined origin coordinate. An observer is located at each origin point. We place the massless test particle equidistant between the three origin points. In what direction will the test particle move? A diagram is attached.

Jon

#### Attached Files:

• ###### 3 spheres & test particle 6-12-08.pdf
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3. Jun 12, 2008

### Haelfix

Write down the metric for your test universe, and try to derive gauss's law. I suspect you will find a problem (assuming the metric is what I think you think it is).

4. Jun 12, 2008

### Wallace

??? What has 'the true center' of the dust cloud got to do with whether Gauss's law is valid?? You might want to find some basic undergraduate texts to look this up, I think you have an misconceived view of what the shell thereom is.

The answer is very simple, the particle will be accelerated towards the center of all three spheres. Where they actually move in the next instant however depends on their current velocity of course, not the acceleration.Think about what this represents, it's a deccelerating homogenous universe which means that the rate that any pair of particles is moving apart is slowing down. Thus the rate that the chosen test particle is receeding from the centre of all three spheres is slowing. Thus the shell theorem clearly works correctly and gives you precisely the result you expect.

5. Jun 12, 2008

### jonmtkisco

Hi Haelfix,

I'm not sure what point you are trying to get at. Maybe you can explain more.

As I understand it, Gauss' Law is based on the fact that inverse square forces (such as gravity) produce equal flux across a spherical surface regardless of the radius of the sphere.

Since I defined the test universe to be flat, I think that Gauss's law will apply consistent with my description. If it were spatially curved however (e.g. a closed universe) Gauss' Law would not calculate a correct result unless it were corrected for the curvature.

Jon

Last edited: Jun 13, 2008
6. Jun 13, 2008

### jonmtkisco

Hi Wallace,
We agree on this point! The test particle will accelerate towards all three spheres equally. And in the same way, if we define an infinite number of symmetrical surrounding spheres (in three dimensions), the test particle will accelerate equally towards all of them. In other words, the test particle gains no directional vector at all.

So, the test particle will not actually move at all (in proper distance) towards the center of any individual sphere arbitrarily selected from among the multitude of surrounding spheres.

EDIT: If it wasn't clear already, I want to make it clear that the origins of the three spheres ( or infinite number of spheres) are not in proper motion with respect to the test particle at the start of the test.

Jon

Last edited: Jun 13, 2008
7. Jun 13, 2008

### Chronos

That is absurd. The universe has no shell-like observational features.

8. Jun 13, 2008

### Wallace

What is a 'directional vector'? Presumably something different from a non-directional vector, which is a clearly nonsensical concept.

I think you are trying to say that in a homogenous and isotropic universe there is no prefferred direction, which is obviously true by definition. Every direction from the particle is the same so of course it feels no net acceleration in any direction.

Wrong. The particle will move towards the center of all of the other spheres.

I really fail to see the point of this thread? You seem to be attempting to demonstrate that you get different physical results depending on how you choose to define co-ordinates (in this case the centers of spherical Gaussian surfaces). The only way this could possibly occur is if Gauss's law fails. Clearly the resolution is to correctly understand Gausses law instead of claiming you have discovered some interesting paradox.

9. Jun 13, 2008

### Wallace

To Jon's defence on this point, the 'shells' he is reffering to are Gaussian surfaces, which are mathematical constructs to help solve a problem more easily. He is not refferring to any physical shell like features, the situation under consideration is a homogenous universe.

10. Jun 13, 2008

### jonmtkisco

Hi Wallace,
I admit I'm dense sometimes, but I don't understand this statement. In my diagram, from a tops-down two dimensional perspective, the stationary centers of the three spheres are set at 120 degree azimuthal angles relative to each other, so of course a test particle equidistant between them cannot move at three different angles at once.

Any physical movement of a test particle can be defined as a vector in proper coordinates, with a magnitude and a single coordinate direction. Yet you agree that the movement here is non-directional and that it is nonsense to speak of nondirectional vectors. Therefore by first principles the test particle does not move.

I can only interpret your statement to mean that the test particle does not move, but rather the three sphere origin points will gravitationally collapse towards the test particle. That outcome cannot occur in this scenario. The three sphere origins are mere proper coordinate points, they are not massive objects; the gravity of the background dust cloud does not cause their proper distances relative to each other to change.

Yes as you say I am attempting to demonstrate that use of different coordinates changes the physical result. However, I disagree that this means Gauss' law fails. The mathematics of Gauss' law is just fine. What I am suggesting is that Gauss' law simply does not apply to mass distributions in which the center of the sphere is not the gravitational center of the entire mass distribution. You have already characterized this suggestion as nonsense, but I think my example demonstrates that the problem is legitimately subtle.

Jon

Last edited: Jun 13, 2008
11. Jun 14, 2008

### Wallace

Movement, i.e. having a non-zero velocity, means that a particles co-ordinates are changing. In order to know if a particles co-ordinates are changing one needs to know how the co-ordinate system has been defined. It is also crucial to realise that the behaviour of something in a particular co-ordinate system is not universal, so something can be moving on one co-ordinate system but not in another.

Now, if you construct a spherical co-ordinate system centered on the point in the middle of your diagram then clearly the point does not move in these co-ordinates. What you will find however, is that clearly the three other particles all move towards the origin, due to the gravitational attraction of the mass enclosed by a sphere centered on the origin with a radius extending to the other test particle. Your test particles will be accelerated by gravity, so clearly they must be accelerated towards the centre.

Now, if you instead construct a spherical co-ordiant system centred on another particle then we can consider that the particle in the 'middle' which is no longer at the origin will be accelerated towards the new origin by the same reasoning. Which particles move and which do not depends on the co-ordinate system. What does not depend on the co-ordinate system however is the distance between each particle as a function of time. It doesn't matter how you define your co-ordinates you will find that the particles will move together due to gravity at exactly the same rate.

It is obvious what will happen in this situation, that the universe will collapse. It is also clear that this is consistent with the shell theorem.

Yes, absolutely for the reasons described above.

I don't know what you are saying here? You said in post #2 that 'an observer is placed at each origin point' this seems to imply that there are massless test particles at these points. If so then these test particles will be accelerated by gravity towards every other particle in the Universe as described above. They most definately will all move towards the middle particle. If these points are not massless test particles and merely arbitrarily defined points in the space then this whole discussion is meaningless. Co-ordinates may do whatever we define them to in a way unconstrained by physics, we could having expanding co-ordinates in a contracting universe if we wanted.

In terms of physics (i.e. what does the actual matter do?) the situation here is clear. The universe collapsing and we can understand this by considering any one of the four spheres in your picture.

I really don't think this thread belongs in cosmology. You will probably get a better explanation of the fundamentals of Gauss's law in some of the other sub forums. I suggest you ask this question in one of those, since you clearly have misconceptions about it. Even reading the Wiki article might be a good start. Don't take this as a criticism, I'm trying help and it is clear that the confusion here lies in a fundamental misunderstanding of this physical law. Trying to understand this in the context of cosmology is then like trying to run before crawling. Learn to crawl first and you will get to running properly much faster.

12. Jun 14, 2008

### chronon

If the universe is expanding at a rate greater than the 'escape velocity' then it won't recollapse.

If the universe is static then the question is whether it can remain so. This question has a long history.

Einstein said no - he had to introduce the cosmological constant to try to fudge it.

Newton said yes. Each object in the universe will be attracted equally from all directions and so will have no reason to move.

However, even in Newtonian gravity, one can argue that if we divide the universe up into spherical shells, then those outside a given radius will have no effect and so the ball inside that radius will collapse. Applying this to the universe as a whole you get it collapsing to a point.

Certainly this worried those who followed Newton. For instance, Boskovich mentioned the possibility of a form of cosmological constant in 1763, and I've read that Newton had a similar idea (although I haven't been able to find it in his works)

13. Jun 14, 2008

### jonmtkisco

Wallace, your line of logic here is not helpful in understanding the problem. The coordinate system in my example is defined solely by the 3 center points of the 3 spheres, not by the test particle. In the "thought exeriment" way that these scenarios are usually described... we send fictional spaceships to each of the 3 sphere centers. Spaceship #1 drops a small marker at an arbitrary location. The 3 spaceships then use radar ranging and fictional rulers laid end to end (like Barnes & Francis do) to define the 3 sphere center locations with respect to the single marker, and to verify that they remain at a fixed proper distance from each other throughout the exercise.

You err in assuming that the spaceships will be spontaneously pulled by gravity towards the test particle. Why would a homogeneous, directionless dust cloud around the spaceships cause them to move an a single arbitrary direction? Why not the opposite direction? This is another obvious microcosm of the false center problem.

Even if the spaceships did feel a unique directional gravitational pull in one arbitrary direction (which they don't), it would in no way undermine the validity of the simple coordinate system which is defined by points mapped in space, not by the spaceships.

All of the "Tethered Galaxy" experiments define an arbitrary origin, so of course arbitrary origins are meaningful. Using 3 arbitrary origins should be just as meaningful, but my example doesn't need to rely on that, since 2 of the 3 sphere center origins are defined with reference to a single initial arbitrarily placed marker.

My scenario defined the universe to be spacially flat and expanding at the decelerating Einstein-de Sitter rate. The universe I defined will never collapse. Please try to keep the facts straight.

Of course I read the Wikipedia article and much more on this subject. It was reading textbooks which alerted me to the fact that it is a common misconception that the shell theorem and Gauss' Law apply when the center of the sphere is not the center of the mass distribution being examined. May I respectfully suggest that you brush up a bit?

Jon

Last edited: Jun 14, 2008
14. Jun 14, 2008

### jonmtkisco

Hi chronon,
Yes as I understand it the problem with a static universe with a cosmological constant is that it is inherently unstable. Even the most microscopic change in mass will cause the universe to accelerate into an expansion or collapse mode. So it is commonly said that the universe must be either expanding or contracting, and not static.

The only models I can think of for a stable static universe are the "black hole inhibitors" I mentioned in another thread. Virial motion in theory ought to be able to stabilize a gravitationally overdense universe; but over the long term virial motion is perturbed by collisions and near collisions, as well as by gravity waves, so it eventually will collapse. Adding some amount of cosmological constant would help stabilize it, I'm not sure if it can entirely stabilize. The maximum size of a virial universe is also limited by causality, is there enough time in the history of the universe for it to coordinate its virialization?

The other alternative is an electrostatically charged universe, such as one filled with electron plasma. But seems to be a fairly useless model as a practical matter because if regular electrically neutral matter is added to it, the neutral matter will gravitationally collapse even though the electron plasma won't. Perhaps adjusting the exact mix of neutral and charged matter would help stabilize it, and adding a cosmological constant ought to help.

Jon

Last edited: Jun 14, 2008
15. Jun 14, 2008

### jonmtkisco

Hi Wallace,
Here's a slightly simplified example (attachment) that may help clarify this. In this version, 3 very massive objects are attached to a strong but lightweight (insignificant mass) support ring. A "Shell Theorem sphere" is measured around each massive object. Again a test particle is dropped equidistant from the 3 massive objects. Which way will it move?

Of course, it will not move at all.

One more thought about "arbitrary" origins: The earth, the center of our Local Group, or any other point in the universe that one wishes to use use as an "origin" of a coordinate system is equally arbitrary.

Jon

#### Attached Files:

• ###### Sphere with 3 masses.pdf
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Last edited: Jun 14, 2008
16. Jun 14, 2008

### Haelfix

Jon, please write the metric down for what you have in mind and the coordinates you want to use. We can't talk about this quantitatively if you do not. I don't understand what you have in mind, and in particular this can alter the shell theorem (Gauss's theorem is not mantained unambigously bc of coordinate subleties if its not exactly flat).

The reason I ask, is b/c you seem to think this acts like something static, even though elsewhere you assumed a FRW universe with k = 0 and lambda = 0. Alternatively I think there is a confusion about global issues vs local ones.

To wit, in the limit where things are very flat (or when spatial curvature/cc is very small relative to the density term), the universe is in a approximately newtonian limit and the shell theorem *must* be valid.

Last edited: Jun 14, 2008
17. Jun 14, 2008

### jonmtkisco

Hi Haelfix,
First, check out my most recent post where I attached a slightly revised version of the example, it may resolve your question.
I'm sorry but I don't understand your question. The universe in this model is a very simple straightforward one, a homogeneous, isotropic, spatially flat, matter dominated, dust-filled Einstein-de Sitter universe with Lambda = 0. It is expanding and the deceleration factor q is positive. The standard RW line element defines the metric.

As I said, an arbitrary origin point #1 is selected, and then two additional origin points are mapped out at equal distances from origin #1 and separated by 60 degrees of azimuth angle (from the perspective of origin #1).
I agree completely, the shell theorem and Gauss' Law *are* valid. But they are valid *only* when applied to a proper set of facts. The facts must be such that the hypothetical point mass at the center of the sphere is the *gravitational center* of the entire mass distribution.

Jon

Last edited: Jun 14, 2008
18. Jun 14, 2008

### jonmtkisco

Well, on further review the last example I posted (Sphere with 3 masses) doesn't accomplish what I hoped. It serves as only as an example of how the Shell Theorem works with respect to *external* test objects.

My discussion has been about test objects located *inside* a solid or homogeneous mass (e.g., dust filled sphere.) For that discussion, we'll need to revert back to the original "3 spheres" example. Sorry about that.

Jon

19. Jun 15, 2008

### jonmtkisco

Hi Wallace,
These statements of yours may be key to sorting out our discussion. If the particles in the universe were to move closer together in the manner you describe, then by definition the average matter density increases as a function of time. Yet every elementary textbook tells us that the defining characteristic of a homogeneous, expanding FLRW universe is that average matter density relentlessly decreases as a function of time. The matter density in any one (and in all 3) of the spheres will decrease over time, not increase. (Think of density in terms of the average proper distance between dust particles.) Surely you must acknowledge the direct conflict here with your interpretation of the shell theorem. If density were to increase in one spherical region of the universe, then it would need to decrease faster than the FLRW rate in other regions of the universe. That is inconsistent with the very concept of homogeneity.

(Of course for the purposes of this theoretical discussion we are assuming perfect homogeneity and ignoring any local gravitational perturbations caused by primordial overdensities, etc.)

Jon

Last edited: Jun 15, 2008
20. Jun 15, 2008

### jonmtkisco

Sorry to keep coming back to this, but here's one more variation on the thought experiment which I think illustrates the absurdity of permitting only one sphere at a time to measure the valid shell theorem results.

Let's say we build a very large rigid ring using advanced materials which are so astoundingly lightweight that the mass of the ring is insignificant compared to the homogeneous dust density of the region of the universe it occupies. We'll mount 360 center-facing numbered video cameras on the ring, exactly 1 degree of azimuth apart. We will consider each video camera to be the center point of its own Shell Thereom sphere, all spheres being equally sized and just large enough to overlap at the point in space defining the radial center of the ring.

Before the start of each test round, we will write down on a slip of paper which of the 360 cameras we have randomly selected to be the origin of our coordinate system for that round. A different camera is designated for each test round. We will not disclose this information to the spaceship pilot, who will release the massless test particle at the exact centerpoint and depart.

Does the shell theorem predict that in each test round, the test particle will move only towards the particular video camera whose number is written on on the current slip of paper? Need I say more?

Jon

Last edited: Jun 15, 2008
21. Jun 15, 2008

### chronon

Your origin is moving with respect to the surrounding dust which rather confuses things.

If you released a test particle from the ring then it would accelerate towards the centre.

22. Jun 15, 2008

### jonmtkisco

Hi chronon,
This subject can get confusing. The origin can be any one of the cameras. In an FLRW universe, no matter where an observer is located, she sees the dust in her immediate local area not to be moving at all; she observers that objects further away from her are moving directly away from her, at velocities proportional to their distance.

EDIT: I confused myself on my first try at answering this. I'll try again: Only one point at a time on the ring can be aligned with its local Hubble flow. All other points on the ring will observe the background dust cloud moving with respect to them. However, this should have no bearing on the movement of a local test particle.

I've thought a lot about whether the proper motion (or the acceleration) of the dust cloud has any direct bearing on a proper-distance application of the shell theorem. I don't think it does. Acceleration has an indirect bearing in the sense that it *results from* the existence of a gravitational field.
In my scenario the test particle is dropped at the radial center of the ring, not from the ring itself. However, even if the test particle were dropped next to the ring, I am sure the test particle would not move relative to the ring. The background dust cloud is directionless so there is nothing to cause it to impart a unique direction of movement to the test particle. We've defined the mass of the ring itself to be insignificant.

Jon

Last edited: Jun 15, 2008
23. Jun 15, 2008

### jonmtkisco

Another easy example:

A very long rigid ruler is constructed in space out of advanced materials which are so astoundingly lightweight that the mass of the ruler is insignificant compared to the homogeneous dust cloud in the region. The ruler is marked with 4 units of length. "Shell Theorem spheres" are measured, the smaller sphere having radius = 1 unit and its center located 1 unit from the left edge of the ruler; the larger sphere having radius = 4 units and its center located 4 units from the left end of the ruler. A test particle is released at the left end of the ruler, just at the edge of both spheres. Will the test particle initially accelerate at the rate determined by the mass of the r=1 sphere or the r=4 sphere? Diagram is attached.

Note that the total mass of the r=4 sphere is 64 times larger than the mass of the r=1 sphere, proportional to $$r^{3}$$. If the Shell Theorem applied, the initial gravitational acceleration rate imparted to the test particle by the the r=4 sphere would be 4 times greater than the acceleration imparted from the r=1 sphere, due to the inverse square law as measured from each sphere's center.

Obviously, if either sphere satisfies the Shell Theorem here, then the other one violates it. So the Shell Theorem can't be applicable. In fact, the test particle will not move at all relative to the ruler.

Jon

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• ###### 2 spheres & ruler.pdf
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Last edited: Jun 15, 2008
24. Jun 16, 2008

### chronon

You need to specify how the ruler moves with respect to the surrounding dust. I assume that you want it to be stationary, but the dust cloud is expanding, so that only one part of the ruler can be stationary with respect to its surroundings. Is it the left end, the right end or the middle.

25. Jun 16, 2008

### chronon

Actually, for any of the options I suggested in the previous post, the particle will accelerate to the right with respect to the ruler.