Newtonian and GR Friedman equation

[ChrisVer]

Consider an homogeneous spherical universe, with mass density [itex] \mathcal{\rho}_M [/itex]. Then the total energy of some test-mass [itex]m[/itex] at radius [itex]R(t) [/itex] from the center is given by:

[itex]E_{tot} = E_{kin} + E_{pot} = \frac{1}{2} m \dot{R}^2(t) - \frac{4 \pi}{3} Gm \mathcal{\rho}_M R^2(t) [/itex]

Or that:

[itex] \Big( \frac{\dot{R}}{R} \Big)^{2} \equiv H^2= \frac{8 \pi G}{3} \mathcal{\rho}_M + \frac{2E_{tot}}{mR^2}[/itex]

Comparing with the Friedman equation, I have the feeling this derivation , either using the classical Newtonian mechanics (so not taking into consideration the curvature of spacetime and so GR) or the solution of Einstein Equations (Friedman equation) gives the same results by the identification of:
[itex]k= - \frac{2E_{tot}}{m}[/itex]
(connected to the total energy of the system, something to be expected of some "energy" existing in the curvature of space)
where [itex]k[/itex] the curvature in the [itex]g_{rr}^{FRW} = \frac{1}{\sqrt{1-kr^2}}[/itex]

Are these same results, something to be expected or not? To me it is something unexpected because of the difference between Newtonian and GR mechanics/ideas.

 

[PeterDonis]

the total energy of some test-mass ##m## at radius ##R(t)## from the center is given by:

##E_{tot} = E_{kin} + E_{pot} = \frac{1}{2} m \dot{R}^2(t) - \frac{4 \pi}{3} Gm \mathcal{\rho}_M R^2(t)##

The last term should have an ##R^3(t)## in it, shouldn't it? And that messes up the correspondence with the Friedmann equation, doesn't it?

the identification of:

##k= - \frac{2E_{tot}}{m}##

(connected to the total energy of the system, something to be expected of some "energy" existing in the curvature of space)

But ##k## is not determined by the "total energy of the system" (that term doesn't really have a well-defined meaning in the FRW model anyway). It's either 1, 0, or -1, depending on whether the universe is closed, flat, or open. In other words, it can only take on a few discrete values, whereas something determined by the "total energy of the system" should be able to take on a continuous set of values.

Are these same results, something to be expected or not?

As a general question, the answer to this is that you can sometimes find apparent correspondences between Newtonian and GR equations, but if you do, it will turn out to be a coincidence and not tell you anything really meaningful about the physics.

In this particular case, the results aren't the same anyway, so it's a moot point. See above.

 

[ChrisVer]

The last

term should have an R3(t)R^3(t) in it, shouldn't it? And that messes up the correspondence with the Friedmann equation, doesn't it?

No. The grav. potential is [itex]U= G \frac{M m}{R} [/itex]... because the sphere vollume*density is the mass [itex] M= \mathcal{\rho}_M \frac{4}{3} \pi R^3[/itex], the grav. potential gets an [itex]\frac{R^3}{R}=R^2[/itex] dependence.

As for [itex]k[/itex], I think sometimes in bibliography the corresponding term in the Friedman equation is the "curvature energy density". So it's some existing energy because of the curvature of the space. In classical mechanical view, the result for [itex]k=0[/itex] (flat) would be that the total energy would be 0, so the test mass would decelerate and stop asymptotically at infinity.
For [itex]k>0 [/itex] the E<0 and so the universe would be closed - the test mass will be trapped.
For [itex]k<0[/itex] the E>0 and the universe will expand forever.

It's a rather funny coincidence :)

 

[PeterDonis]

The grav. potential is ##U= G \frac{M m}{R}## ...the grav. potential gets an R3R=R2\frac{R^3}{R}=R^2 dependence.

Ah, of course, you're right. Sorry for the confusion on my part.

 

[DrStupid]

Consider an homogeneous spherical universe, with mass density [itex] \mathcal{\rho}_M [/itex]. Then the total energy of some test-mass [itex]m[/itex] at radius [itex]R(t) [/itex] from the center is given by:

[itex]E_{tot} = E_{kin} + E_{pot} = \frac{1}{2} m \dot{R}^2(t) - \frac{4 \pi}{3} Gm \mathcal{\rho}_M R^2(t) [/itex]

Did you use the shell theorem for potential energy? That wouldn't work. With the energies

[tex]E_{pot} = - \frac{3}{5} \cdot \frac{{G \cdot m^2 }}{R}[/tex]

[tex]E_{kin} = \frac{3}{5} \cdot \frac{{m \cdot \dot R^2 }}{2}[/tex]

for the total mass distribution I get

[tex]H^2 = \frac{{8 \cdot \pi \cdot G}}{3} \cdot \rho + \frac{{10}}{3} \cdot \frac{{E_{tot} }}{{m \cdot R^2 }}[/tex]

Are these same results, something to be expected or not? To me it is something unexpected because of the difference between Newtonian and GR mechanics/ideas.

When I got this result for the first time I was also very surprised and I still have no explanation. Maybe it is just a coincidence (similar to the Schwarzschild radius).

 

[ChrisVer]

what is that kinetic and potential energy you've written down?
Also you've used an [itex]m^2[/itex] which doesn't make "sense" I think for the model I'm considering...

 

[DrStupid]

what is that kinetic and potential energy you've written down?

As mentioned above these are the kinetic and potential energy of the total spherical universe.

Also you've used an [itex]m^2[/itex] which doesn't make "sense" I think for the model I'm considering...

Sorry, my m is the mass of the total universe and v the velocity of it's surface. It would have been better to use other symbols to avoid confusions.

 

[ChrisVer]

The potential energy of a test mass [itex]m[/itex] at some position [itex]R(t)[/itex] depends on the mass existing in the shaded "volume" in the attached picture. http://prntscr.com/4whjko
The shaded area will contain a mass: [itex]M(R)= \rho_M \times \frac{4}{3} \pi R^3[/itex] (it's the interior sphere after applying Gauss's law)

So the [itex]E_{pot} = - G \frac{ m M(R) }{R} [/itex] that's why I don't understand your factor [itex]\frac{3}{5}[/itex] and the existence of the same masses [itex]m[/itex] in the numerator. In your case you are setting [itex]m=M(R)[/itex] and I don't understand the reason. So it's not what they denote, but how you used them.

The kinetic energy again makes some sense apart from the [itex]3/5[/itex] again, and the way you define [itex]m[/itex] to be the total mass of the universe. That's because the universe is not moving wrt to itself (??) so there would be no velocity. So indeed your [itex]m[/itex] has to denote a test mass (maybe a galaxy) as in my image

 

[DrStupid]

The potential energy of a test mass [itex]m[/itex] at some position [itex]R(t)[/itex] depends on the mass existing in the shaded "volume" in the attached picture. http://prntscr.com/4whjko

...and on the mass outside the shaded volume. The potential energy of a test mass inside a sphere (not a ball) is equal to its potential energy on the surface of the sphere. To get the total potential energy of the the test mass you need to add the integral of its potential energy over all spheres from R to the surface of the entire mass distribution.

(it's the interior sphere after applying Gauss's law)

It's better to use Gauss's law only. For a homogeneous and isotropic mass distribution it gives the tidal acceleration

[tex]\ddot r = - \frac{4}{3} \cdot \pi \cdot G \cdot \rho \cdot r = - \frac{{4 \cdot \pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{3 \cdot r^2 }}[/tex]

for two test masses with the distance r. Integration results in

[tex]H^2 = \frac{{8 \cdot \pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{3 \cdot r}} + k = \frac{8}{3}\pi \cdot G \cdot \rho + k[/tex]

This method does not allow to specify the integration constant k but it is not limited to finite spherical mass distributions. It also works for an infinite mass distribution.[edit]To get the potential and kinetic energy of the total mass distribution you need to integrate the potential energy and the kinetic energy of any particle over the volume of the total mass distribution. That's where the factor 3/5 and the square of the total mass comes from. But I have to admit that am not sure if I solved all integrals correctly.[/edit]

 

[PeterDonis]

It also works for an infinite mass distribution.

Can you give an example, please? I'm not sure the integral involved for an infinite mass distribution is convergent.

 

[DrStupid]

Can you give an example, please? I'm not sure the integral involved for an infinite mass distribution is convergent.

I'm not sure, what you mean. The formulas in my last post apply to an infinite homogeneous mass distribution and the integral is convergent. I just forgot some r². The correct result should be

[tex]H^2 = \frac{{8 \cdot \pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{3 \cdot r^3 }} + \frac{k}{{r^2 }} = \frac{8}{3} \cdot \pi \cdot G \cdot \rho + \frac{k}{{r^2 }}[/tex]

You are right that the integral of the potential energy is divergent for infinite mass distributions. That's why ChrisVer's approach works for finite mass distributions only (btw: I double checked my corresponding calculations and did not found an error). But that doesn't affect my second approach. It don't uses potential energy or even forces but tidal accelerations only and the tidal accelerations within an infinite mass distribution are always finite and well defined.

 

[PeterDonis]

my second approach. It don't uses potential energy or even forces but tidal accelerations only

I think I'm not understanding your notation. What are ##\rho_0## and ##r_0##?

 

[DrStupid]

I think I'm not understanding your notation. What are ##\rho_0## and ##r_0##?

##\rho_0## is the density of the mass distribution when the distance between two test particles is ##r_0##.
##\rho## is the density of the mass distribution when the distance between the same two test particles is ##r##

 

[PeterDonis]

##\rho_0## is the density of the mass distribution when the distance between two test particles is ##r_0##.

##\rho## is the density of the mass distribution when the distance between the same two test particles is ##r##

Ok, so you're basically adopting coordinates where one test particle is at rest at the origin, and ##r## is the radial coordinate of an arbitrary test particle in this chart. (Angular coordinates can be ignored since this universe is homogeneous and isotropic.) Then ##r## for a given test particle is a function of coordinate time ##t##, and so is the density ##\rho## (but ##\rho## is not a function of ##r## since the universe is homogeneous). ##\rho_0## and ##r_0## are just initial conditions.

That all makes sense, but now I'm not sure how you derived your equation for ##\ddot{r}## (more precisely, the second equality, with ##\rho_0## and ##r_0## in it).

 

[DrStupid]

Ok, so you're basically adopting coordinates where one test particle is at rest at the origin, and ##r## is the radial coordinate of an arbitrary test particle in this chart.

It is not necessary that one particle is at rest at the origin. r is just the difference between the radial coordinates of two arbitrary test particles. Actually it is not even possible to determine if the particles are accelerated or not because the rest system of all particles are free falling.

That all makes sense, but now I'm not sure how you derived your equation for ##\ddot{r}## (more precisely, the second equality, with ##\rho_0## and ##r_0## in it).

I started from the differential form of the gauss law for Newtonian gravity:[tex]\nabla \cdot \vec g = - 4 \cdot \pi \cdot G \cdot \rho[/tex]Due to isotropy the derivation of g should be identical in any direction:[tex]\frac{{dg_r }}{{dr}} = \frac{{dg_x }}{{dx}} = \frac{{dg_y }}{{dy}} = \frac{{dg_x }}{{dx}}[/tex]This results in[tex]\frac{{dg_r }}{{dr}} = - \frac{4}{3} \cdot \pi \cdot G \cdot \rho[/tex]and after integration finally in[tex]g_r = \ddot r = - \frac{4}{3} \cdot \pi \cdot G \cdot \rho \cdot r[/tex]Now I just need to include the variation of the density. Imagine a spherical volume with initial radius ##r_0## and initial density ##\rho_0## within the mass distribution. The mass within this sphere remains constant even when its radius changes:[tex]m = \frac{4}{3} \cdot \pi \cdot \rho \cdot r^3 = \frac{4}{3} \cdot \pi \cdot \rho _0 \cdot r_0^3[/tex]This gives[tex]\ddot r = - \frac{4}{3} \frac{{\pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{r^2 }}[/tex]This works because during collapse or expansion of the mass distribution all distances scale with the same factor. Replacing the radius by the scale factor[tex]a = \frac{r}{{r_0 }}[/tex]results in an equation of the same structure:[tex]\ddot a = - \frac{4}{3} \frac{{\pi \cdot G \cdot \rho _0 }}{{a^2 }}[/tex]

 

[PeterDonis]

Ok, this makes the derivation clearer; and it makes the issue with divergence clearer as well:

after integration

Over what range of integration? What this integral is saying is that the magnitude of ##\vec{g}## is linear in ##r##, so for an infinite universe, ##\vec{g}## can increase without bound as ##r \rightarrow \infty##. The integral does not converge.

Reframing the integral in terms of ##m = (4/3) \pi G \rho_0 r_0^3## doesn't help either, because in an infinite universe, ##m## also increases without bound as ##r_0## increases, i.e., as you consider particles with larger and larger initial separations.

Also, physically, even if we leave out the above issues, the picture given by these equations isn't really consistent. For example, suppose I have particle A which starts out halfway between particles B and C. If I compute particle A's ##\ddot{r}## based on its separation from particle B, I get an acceleration in one direction; but if I compute it based on A's separation from particle C, I get an acceleration in the opposite direction. Both can't be right, since they contradict each other.

More generally, if the universe is homogeneous and isotropic, then a given particle moving freely at any point in the universe should not be accelerated in any particular direction, because that would pick out a particular direction and violate isotropy. But your equations seem to be saying that a given particle will be accelerated in some particular direction.

If your response is that the actual acceleration of a given particle is obtained by summing the results for all possible values of ##r## and all possible directions, then we're right back to the divergence issue. And even if we leave that out again and say that the sum must be zero (by isotropy), that invalidates any useful concept of "potential energy", since ##\vec{g}## is supposed to be the gradient of the potential energy, so if it's zero, the "potential energy" must be the same everywhere, which amounts to saying it's physically meaningless.

 

[DrStupid]

Over what range of integration?

Over the distance between two test particles.

For example, suppose I have particle A which starts out halfway between particles B and C. If I compute particle A's ##\ddot{r}## based on its separation from particle B, I get an acceleration in one direction;

No, you don't get an acceleration of a single particle. You get the second derivation of the distance between two particles with respect to time.

More generally, if the universe is homogeneous and isotropic, then a given particle moving freely at any point in the universe should not be accelerated in any particular direction, because that would pick out a particular direction and violate isotropy.

Homogeneity and isotropy do not require constant escape velocities.

If your response is that the actual acceleration of a given particle is obtained by summing the results for all possible values of ##r## and all possible directions

The actual acceleration of a given particle depends on the frame of reference. Each particle can be assumed to be at rest. Than the acceleration of all other particles result from their distance from this particle according to the equations above. But there is no absolute acceleration.

If you have problems with the equations for the distance between specific test particles, just use the equation with the scale factor instead.

 

[PeterDonis]

you don't get an acceleration of a single particle. You get the second derivation of the distance between two particles with respect to time.

Same problem: any given particle can be paired with multiple other particles, and the results obtained in the individual cases will not be consistent. The fact that you are not using a well-defined coordinate chart (or, if you like, you're switching coordinate charts depending on which pair of particles you are looking at) may be preventing you from seeing this.

Homogeneity and isotropy do not require constant escape velocities.

I would say that homogeneity and isotropy are inconsistent with the concept of escape velocity, just as they are inconsistent with the concept of gravitational potential energy. Perhaps you are using a non-standard definition of these terms: can you state more precisely how you are defining them?

 

[DrStupid]

Same problem: any given particle can be paired with multiple other particles, and the results obtained in the individual cases will not be consistent.

I don't know what you mean. Can you please demonstrate this with a specific example?

I would say that homogeneity and isotropy are inconsistent with the concept of escape velocity

That means all modern cosmological models are inconsistent with the cosmological principle?

 

[PeterDonis]

That means all modern cosmological models are inconsistent with the cosmological principle?

How so? The cosmological principle doesn't say anything about escape velocity.

 

[PeterDonis]

Can you please demonstrate this with a specific example?

First I'd like to make sure I'm not still misunderstanding your model. That's why I asked for your definitions of homogeneity and isotropy. Also I'd really like to see a description of your model in a single coordinate chart, instead of solely in terms of relative distances between different pairs of particles.

 

[haushofer]

Another such "coincidence" is that Newton considered, besides the 1/r2 potential, a k*r potential to give stable orbits. This is what we now would call the cosmological constant term, and it appears e.g. in the same way in an (A)dS black hole solution. I believe the cosmology text by Liddle treats the Newtonian Friedmann equation.

 

[DrStupid]

The cosmological principle doesn't say anything about escape velocity.

The cosmological principle says that the mass distribution in universe is homogeneous and isotropic (on large scales).
All modern cosmological models say that test particles within this mass distribution escape from each other according to Hubble's law (on large scales).
You say "that homogeneity and isotropy are inconsistent with the concept of escape velocity".
Therewith you say that all modern cosmological models are inconsistent with the cosmological principle

 

[DrStupid]

That's why I asked for your definitions of homogeneity and isotropy.

That means the mass distribution and it's dynamics looks the same in any direction for any co-moving observer.

Also I'd really like to see a description of your model in a single coordinate chart, instead of solely in terms of relative distances between different pairs of particles.

Though that makes no sense, you can do that if you want. Just integrate the gauss law in a coordinate chart of your choice. The only difference is the acceleration in the origin:[tex]\ddot r = g_0 - \frac{4}{3} \cdot \pi \cdot G \cdot \rho \cdot r[/tex]

 

[PeterDonis]

All modern cosmological models say that test particles within this mass distribution escape from each other according to Hubble's law (on large scales).

Do you have a reference for this particular interpretation? I've never seen "escape", in the sense of escape velocity, used in this connection. Escape velocity, as I understand it, is only meaningful in an asymptotically flat spacetime, where there is a flat region at infinity to escape to. In a universe (even an infinite one) with a homogeneous and isotropic matter distribution everywhere, there is no flat region at infinity, so there's nowhere to escape to.

 

[PeterDonis]

The only difference is the acceleration in the origin

Shouldn't that just be zero, since by hypothesis the object at the origin is at rest in the coordinate chart?

 

[DrStupid]

Do you have a reference for this particular interpretation?

http://en.wikipedia.org/wiki/Metric_expansion_of_space

The first sentence: "The metric expansion of space is the increase of the distance between two distant parts of the universe with time."

The "increase of the distance between two distant parts of the universe with time" is exactly what we are talking about. Of course in classical mechanics it is not a result of metric expansion of space but of the motion within space but that doesn't matter. The original question is, why the resulting equations can be identical for both cases.

Shouldn't that just be zero, since by hypothesis the object at the origin is at rest in the coordinate chart?

No, there is no reason for an object in the origin to be at rest or even not accelerated. You may choose such a frame of reference but you do not need it.

 

[PeterDonis]

The "increase of the distance between two distant parts of the universe with time" is exactly what we are talking about.

But that's not what "escape velocity" means. I was asking about your claim regarding escape velocity. Saying that distant parts of the universe are moving apart is not a claim about escape velocity in the standard interpretation of that term, which I gave in a previous post.

there is no reason for an object in the origin to be at rest or even not accelerated.

But then it will only be at the origin for one instant. I thought the whole point was to have it at the origin for all time. I'm still confused about exactly what your model says, and I also don't understand why you think a description of it in a standard coordinate chart "makes no sense", since coordinate charts are used all the time in physics, Newtonian as well as relativistic.

Perhaps I can ask another question that will help: consider a particle that, at time ##t = 0##, is at the spatial origin in the chart you used in post #24; i.e., its position ##\vec{r} (t = 0)## is (0, 0, 0). What will its position ##\vec{r} (t)## be as a function of time, in that chart? I realize this seems obvious to you, but it's not obvious to me or I wouldn't be asking. More precisely, the "obvious" answer based on your post #24 makes no sense to me (it would make sense with ##g_0 = 0##, since then the position would just be (0, 0, 0) for all time; but not with ##g_0 \neq 0## ), so I'm trying to understand how you're coming up with it. Are you just generalizing to a non-comoving chart (so ##g_0## would be 0 in a comoving chart but can be nonzero in a non-comoving chart)?

 

[DrStupid]

But that's not what "escape velocity" means.

Yes, you are right. This was a translation error. In German the term "Fluchtgeschwindigkeit" is also used for the increasing distance between objects in the universe according to Hubble's law. What's the correct English term?

But then it will only be at the origin for one instant. I thought the whole point was to have it at the origin for all time.

No, the point is to describe the dynamics of a homogeneous and isotropic mass distribution. An object at rest in the origin makes it much easier but it is not required.

I'm still confused about exactly what your model says, and I also don't understand why you think a description of it in a standard coordinate chart "makes no sense", since coordinate charts are used all the time in physics, Newtonian as well as relativistic.

The description in a coordinate chart makes no sense because it is much more difficult to solve but does not provide any additional information. The main problem is the variable density. It depends on distances but not on positions. I have no idea how to solve the equations without using distances or ratios between distances.

Are you just generalizing to a non-comoving chart (so ##g_0## would be 0 in a comoving chart but can be nonzero in a non-comoving chart)?

Yes, of course. As a limitation to co-moving charts results in the same equations as the for distances I expected this generalization is what you were asking for.

 

[PeterDonis]

In German the term "Fluchtgeschwindigkeit" is also used for the increasing distance between objects in the universe according to Hubble's law. What's the correct English term?

I'm not sure there is a single English term, other than "expansion of the universe" or perhaps "Hubble expansion" (though I haven't really seen that much).

The main problem is the variable density. It depends on distances but not on positions.

I don't understand. Isn't density a function of position? That is, isn't the point to assign a density to each point in space, i.e., each position? In general, the density at a given position can be a function of time, and density can vary with position; in the homogeneous and isotropic case, the density is constant with position (but can still vary with time at a given position).

If you mean that, in order to know the actual function that gives density vs. time, you need to know distances but not positions, I see how that would be true given the solution method you described earlier in this thread, but since you can express distances as differences in positions, you can still do the same thing using a coordinate chart.

 

[PAllen]

I think the most common English scientific term for what Dr. Stupid was referring to is recession velocity. It is, of course, a function of distance (in the standard foliation) for objects with no peculiar velocity; and it is not a relative velocity in the normal sense.

 

[DrStupid]

If you mean that, in order to know the actual function that gives density vs. time

Yes, that's what I mean.

you need to know distances but not positions

And that's why I do not see why I should use positions at all.

 

[PeterDonis]

I do not see why I should use positions at all.

If you're the only person that's going to look at your calculations, of course it makes no difference what method you use as long as your method gives the correct answers for observable quantities. But since coordinate charts are the standard tool for this job, if you want other people to understand what you're doing, it helps to be able to describe it in standard terms.

 

[DrStupid]

if you want other people to understand what you're doing, it helps to be able to describe it in standard terms.

In #24 I provided such a description but it seems that it wasn't very helpful and as I do not know how to integrate this equation I can't provide you with the corresponding solution. In the meanwhile I tried to start from Newton's law of gravitation without using gauss law but I stuck with the same problem:

A test mass i at position r_i within an infinite homogeneous mass distribution can be assumed to be located on the surface of a ball with the centre r₀, surrounded by an infinite large spherical shell. According to the shell theorem the force acting on the test mass depends on the mass M_i of the inner ball only and is independent from the shell outside. The resulting acceleration is

[tex]\ddot r_i = - G \cdot M_i \cdot \frac{{r_i - r_0 }}{{\left| {r_i - r_0 } \right|^3 }}[/tex]

With

[tex]M_i = {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \left| {r_i - r_0 } \right|^3[/tex]

this turns into the same equation as already posted above

[tex]\ddot r_i = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \left( {r_i - r_0 } \right) = g_0 - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot r_i[/tex]

The only difference is that in this case g₀ is not some unknown integration constant but the result of the arbitrarily choice of the centre r₀. As an infinite mass distribution has no centre it is impossible to determine this constant except per definition. There simply is no absolute acceleration in such a configuration.

To solve this problem I need to switch to differences. In this case the unknown constant cancels out:

[tex]\Delta \ddot r_{i,j} = \ddot r_i - \ddot r_j = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \left( {r_i - r_j } \right) = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \Delta r_{i,j}[/tex]

That's why I prefer differences over positions to solve this problem. If you want to use positions you may do this, but then you have to live with unknown or arbitrarily offsets and ugly differential equations. As I don't see any benefits I still don't think that this makes sense.

 

[PAllen]

I find myself doubting the validity of this line of reasoning (post #34). Since the center is arbitrary, both direction and magnitude of force is arbitrary. You then try to remove this via differences. But the if the approach is valid, since the center is arbitrary, it should work if I choose a different center for each point. Since this is obviously not the case, without further justification, I doubt the whole argument.