The Death of the Centre of Mass Theorem

R

Rogue Physicist

Registered Senior Member
Many people believe that the Centre of Mass Theorem is a powerful and useful tool in Newtonian Mechanics. In fact it is a farce.

(1) It is trivially true at distances in which the massive object is virtually a point-mass, such as between distant stars.

(2) It is completely incoherent and self-contradictory at distances in which the size of the object(s) approachs 1/20 the distance between them.

Try it yourself. I'll post the simple proof that it is nonsense after a few people try to guess what is wrong.
 
Warren:

Please copy the derivation of this theorem (as available in any freshman-level college physics textbook), then show us where the limit is.
 
It's not really even a "theorem," Rogue Physicist, it's just an approximation. When people say things like "when d >> a," where a is the size of something and d is the distance from it, they mean that d is much larger than a. In other words, they assume the limit as a goes to zero.

- Warren
 
for anything advanced you would just use calculus anyway, so whats the point?
 
Consider a sphere. The center of mass is located at the Geometrical center. It is a fixed point. Let us place a test-particle a few diameters away from a solid sphere on the right along the x-axis. (make the radius of the sphere 1 unit, and put the sphere at the origin). According to the CMT, the mass acts as if it were concentrated at the centre and the distance will be measured centre to centre for purposes of using Newton's Gmm/d^2 formula.

(1) divide the sphere logically into two halves vertically. Each half will have its own centre of mass, located 3/8 radians down the axis of symmetry. The actual position is n't important, but because it is an average mean of the atoms in the solid, it is fixed relative to the geometric skin of the half-sphere.

(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.

(3) The actual force calculated by summing the vectors separately will be larger than the one calculated treating the sphere as a whole. Conversely, the force calculated by dividing the sphere horizontally will be weaker than the 'whole' force, since vertical components from each vector will cancel.

(4) The Centre of Mass Theorem contradicts itself, and also the Sphere Theorem as well, which is a special case of the CM.

So as it turns out, the failure of the theorem is not directly connected to absolute sizes, or the discrete localization of mass or charge, but simply the error is directly connected to the ratio of the effective radius of the objects versus the distance between them.
 
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I assume you mean to divide your sphere in half by cutting it with a plane perpendicular to your x-axis, Rogue Physicist.

Your point is perfectly valid, and well understood. A complete sphere is a special case in that it always acts as a point mass located at its centre of mass for any object outside it, in terms of gravitational attraction. The same is not true for any other shape, including a half-sphere.
 
Rogue Physicist said:
(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.

(3) The actual force calculated by summing the halves separately will be larger than the one calculated treating the sphere as a whole. Conversely, the force calculated by dividing the sphere horizontally will be weaker than the 'whole' force, since vertical components from each half will cancel.
Click to expand...

Math please
 
Rogue Physicist said:
Consider a sphere....
(1) divide the sphere logically into two halves vertically. Each half will have its own centre of mass, located 3/8 radians down the axis of symmetry. ....
(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.
(3) The actual force calculated by summing the halves separately will be larger than the one calculated treating the sphere as a whole. Conversely, the force calculated by dividing the sphere horizontally will be weaker than the 'whole' force, since vertical components from each half will cancel.
(4) The Centre of Mass Theorem contradicts itself, and also the Sphere Theorem as well, which is a special case of the CM....
Click to expand...
I too would like to see the math. (I.e. sum of the two hemispheres' forces compared to the sphere force with all mass at GC.) For this comparison you will need to pick a numeric location for the test mass (or "field point" as no "test mass" need actually exist unless you were measuring the force,and of course you are not as you "division" is only conceptual, not an actual separation of the two halves.)

You seem to think the "error" is greatest when the field point is not far from the surface of the sphere, so I suggest you pick a field point of 2 or less, but obviously more than 1, the radius of the sphere. Not sure if the CM of the half is 3/8 from the CM of entire sphere, but if it is, then a field point of 1+ 3/8 might be covenient as then the square in the demonators would be 1x1= 1 for the near hemisphere and (1+3/4)x(1+3/4)= 49/16 for the far hemisphere.

Let stick with exact fractions so any difference you show is not due to rounding errors. BTW, distance is not measured in radians and the gravity law is inverse square, not "exponentail" but I am reasonably confident that you intended to state it correctlya nd can do the two calcutions required for the comparison correct, but I want to see your math. (If you do show a difference, then I want to see the derivation of the 3/8, rather than just accept your value for where the CM of the hemisphere is.)

I supported you when talking about fact there is small field inside a hollow sphere because on atomic dimensions mass and charge distributions are never a function of radius onlybut assume in this problem your postulating truely uniform density of mass with in the sphere but think you are completely wrong here.

Let see the math!
 
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Here is the first diagram for this argument:
Horiz-balls3.jpg


( I'll try to explain the diagram in words...a horizontal barbell lies on the x-axis at the origin, with a radius of one. A test-particle is on the x-axis to the right, outside the range of the barbell. The distance between the systems is simply the x-coordinate of the test-particle (centre to centre). A right-angle triangle is formed by marking the same distance d on the y-axis and drawing a line to the nearest end of the barbell from (0,d) to (1,0). The acute angle between the Y-axis and this line is Theta. I hope this helps.)

Case 1: Horizontal Barbell / sphere split vertically
(system A located at origin, test particle on the +ve x-axis.)
For clarity we normalize the equations by making the masses of each system 1 unit, and choose distance units to make the gravitational constant G = 1 as well. This lets us drop the Gmamb part of Newton's formula. Now the formulas simply become 1/ d2 for each part, the d in this case is the distance between the test-mass (system B) and each each end of the barbell. The new equation for the force between the systems is restated as

F = 1 * ( cos theta )2 ( for theta < 45 o )
......d2 ( cos 2theta )2

In this form we can keep the system to system distance constant while spreading our balls. We could normalize d as well, ( i.e., d = 1 ) but we leave it in to show that the formula reduces to the inverse square law. We construct a simple triangle in the diagram to identify theta in the equation. By inspection the angle theta in the diagram is always less than 45o because the right triangle is formed using the distance d , the distance between the geometrical centre ( GC ) of system A and the test-mass.
The force depends upon the spread in system A and becomes Newton’s formula when the two point-masses merge as r = 0 and theta = 0 o.

In this case, spreading the barbell ends to the point of contact with the test-mass is equivalent to moving the test-particle all the way into the sphere to 3/8 radians (of the sphere r = 1), penetrating it moving below the surface. Even at distances outside the sphere the error is massive.
 
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Case 2: Vertical Barbell -
(= Sphere split horizontally on a plane at the x-axis)

Vert-balls3.jpg



Consider a barbell made of two equal uniform spheres, rigidly connected by a rod of negligible mass. Call this system A. Some fixed distance away from the barbell, place another uniform sphere B, having the same mass as the barbell. It should be obvious that the total force on sphere B is just the sum of the pull from each end of the barbell. Each sphere in A exerts a force on B. We can calculate each contribution separately, and then add them. This will be a vector sum however, since each end pulls in a slightly different direction. The total force is the vector sum of the forces for A1 and for A2.

In the diagram, A1 and A2 are the ends making up the barbell, and r is the radius of system A, or the distance from each sphere in A to the geometric centre or centre of mass for system A. mA is the total mass of system A (i.e., of both spheres), and mB is the mass for B. Finally, d is the basic centre-to-centre distance between the two systems. The angle theta shows the direction of pull from each sphere in the barbell, relative to the overall direction of pull, which happens to be toward the geometric centre of system A, due to symmetry.

2. Getting the Net Force for the Barbell Using the Cosine
Any vector can be split up into a vertical and horizontal component. Here the vertical components for each sphere in system A are opposing and equal so they cancel, leaving only the horizontal components. We can ignore the vertical parts. Since we really only need the horizontal components, we can write,

FAB = ( horizFA1 + horizFA2 ) = ( FA1 cos theta + FA2 cos theta )

Multiplying by the Cosine function correctly scales down the force from each end to the net horizontal component only. We then just add these components to get the total force in the X direction.

Substituting the definition of cosine (adj / hyp = d / h), and using Newton,
(each sphere includes only ½ mA, letting us cancel & drop the 2 also.)

FAB = GmAmB * d ...given d & h ( eq 1.1 )
........... h^3

FAB = GmAmB * cos theta ...given angle ( eq 1.2 )
........... h^2

To get it for ( d and theta ) we pick a multiplier and combine it in:

FAB = GmAmB * cos theta *(hd)^2 = GmAmB * cos^3 theta (eq 1.3 )
........... h^2 .......................... *(hd)^2........... d^2

What is the physical difference between equation (1.2) and (1.3) ? In terms of h, if we spread the barbell ends apart in an orbit around sphere B, keeping the distance h to each ball constant, the force is proportional to the cosine. But if we keep d constant (the distance between systems) and just lengthen the barbell, the spheres spread apart vertically, and the force drops as the cube of the cosine!

The model of the barbell is equivalent to the cutting of the sphere horizontally, and expanding the sphere so that the barbell ends coincide with 3/8 of the radius of the sphere at all times.

It should be clear that although the Centre of Mass theorem is an approximation that works when the sphere is negligible in size relative to distance, it will diverge arbitrarily depending upon how we want to chop an object up and sum its components.

The point is, the CM is not the same as the EPM (true Equivalent Point Mass of equal mass position) which could be substituted for the object and exert identical force upon the test-particle.
 
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Rogue Physicist said:
Case 2: Vertical Barbell -
(= Sphere split horizontally on a plane at the x-axis)...
Click to expand...
Whoa! you were talking about two ways to look at the sphere (undivided and still unified but considered as two hemispheres separately.) You claimed the gravitational field calculated by these two differnt methods would not be the same.

Do you now admit you were wrong and want to move on to talking about bar bells? Until you either back up your original claim of admit it was wrong, I for one, will not read anything about your bar bells.

Why should I - you may just drop discussion of them also. Pronounce you possition - Is, or is not, the result of the field from the sphere the same as the added field from the two hemisphers, that are only conceptually divided i.e sill joined on their circular faces?

If you make a claim about sphere, you can't just switch the subject to bar bells.
 
Okay lets go slowly and carefully:

(1) The Centre of Mass theorem makes the claim that an object acts as if all its mass is concentrated at the Centre of Mass, a uniquely defined fixed point for any system of particles rigid or not (a macro-sized object).

(2) The Centre of Mass theorem is admitted to be an approximation, but the margin of error (its quantification) and its accuracy is never specified except vaguely as d >> r.

(3) By the rules of simple vector addition of forces, the force of a system upon a test-particle should be the sum of the forces from the individual components. (at least according to classical gravitational field theory)

(4) The Sphere Theorem is a special case of the Centre of Mass theorem, but is unique because in this case the result of integration is supposed to be exact and actually reflect the field potential, especially inside a hollow or solid uniform sphere, at least with objects large enough to treat as a continuum of mass or charge density.

(5) Treating a macro-sphere (continuous density) by way of the Centre of Mass theorem should give an accurate result.

(6) Treating contribution to the force of the components of the sphere separately should also give a unique and and accurate result, identical to the result when treating the sphere as a whole. (laws of vector addition of forces)

(7) By definition, the sphere halves each have a unique, fixed Centre of Mass defined by the averaged weighted position of all the atoms that compose them. This is located 3/8 of a radian along the axis of symmetry but its position is not important for this discussion, only the fact that it is fixed relative to the hemisphere's position.

Diagram of Equivalent Systems: Sphere/Barbell

(8) The Centre of Mass theorem (not me personally) makes an identity of the sphere halves with a barbell made of equal point-masses with a radius of 3/8 of the sphere's radius placed at the geometric centre, in *any* orientation. (the halving of the sphere will be orthogonal to the barbell shaft).

(9) This identity makes it easy to test the Centre of Mass Theorem for self-consistency, and quantify the error if any in the results.

(10) In two orientations tested, the results of the Centre of Mass Theorem contradict each other, and also the one predicted by the Sphere Theorem. If the Sphere Theorem is accurate in spite of the inaccuracy of the Centre of Mass Theorem, we can then quantify the error accurately and usefully as a formula.

It would surely be worthwhile to quantify the error in the Centre of Mass theorem for a variety of purposes, one being to enable scientists to accurately judge when it can be used, and what margin of error is introduced by using it.
 
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Rogue Physicist,
Nice thread.
Let me see what I understand after a brief one time skim of the thread. The difference of forces measured differently from a model using CM assumptions is only seen at "close" distances. The distance between the solar system and nearest star (approximately 4 ly) is assumed CM measured. A least from the nearest star to here, the solar system should not suffer external perturbations from gravitational variations. As a result of observation it was concluded that the nearest, nor the farthest star exterted any observable forces on the solar system. [Personal observation starring at the summer nightime sky.]

Here ia a question:Assume the sun's motion through space drags the solar system along as an ever developing, or evolving helix. As large numbers confuse me I have scaled down the solar system to a 1cm in diameter earth orbiting the 109cm diameter sun at 330 meters, where Mercury orbits at 88 meters, the outer planet, Neptune, orbits at approximately 9.5 km. Does any expansion of your model account for the unique spatial geometry of the solar system that the inverse square model doesn't provide? The (A) trouble with many solar system origin stories is the cpature of he planets in the helical orbit where at some time, before or after caqpture (?), the planets coalesced into the spherical orbs we witness today.

Your model becomes observationally significant at relative sizes much smaller than the average solar system distances such as the various radii of planetary orbit for instance (so I understand); yet orders of magnuitude closer to "near" than the star to planetary distance ratio. Where do we stand here?. What I am struggling to say is, or ask, can your model be applied to the orbiting distances of the solar system? Does your system fit with some form of Bode's Law? Or in the sense of harmonics does the expansion of the model fit with some harmonic energy stacking structure? (which you may define as you will). I ask the question so from an understanding that the earth/sun radius of orbit is 10[sup]8[/sup] meters magnitude (assumed a much larger distance that the few radii separation of massess demanded by your model. Or is it demanded?);Yet an ordered system, with a high degree of complexity can be expected to use a minimum of energy more efficiently[than inverse square models] to maintain the system, implying, to me at least, that radiant forces are spread efficiently throughout the solar system; the forces that are causally performing the helical geometry of the solar system trajectory must be other than a simple field like model of gravity we are most familiar with. The formation of the solar system suffers if we use the gravity inverse square law to justify the prevailing coalescent of planetary molecules model of planetary formation. Assuming the reality of this then it is unlikely(?) that those formation forces would evolve into the inverse square law we assume modernly. I am speculating here beyond that allowed by my understanding of your model.

Am I close enough (1 meter) to the surface of the earth to have my spatial reality defined by a coherent expansion of your model?

You know Rogue Physics, I have been idly considering a conservation of angular momentum structure as a substitue for a common, "gravity sucks" model. You certainly have given us something to consider. It is easy to say that your model is intuitve, when we recognize that the closer massive objects approach each other the farther they shoulddrift from the gross inverse square law and into an unrecognizable transformation where "extreme" results can be anticipated. Perhaps the result is from the concentration of energy in necessarily confined and limited volumes of space when the exchanges of energy require more predefined channels of force exchanges we are not privy to observe in a gross inverse square law universe.

Geistkiesel​
 
One thing I am willing to say with confidence at this time is that the Gravitational Constant is calculated from careful experiments close to the surface of the earth for the most part, and assume the simple Newtonian inverse square law.

In my view the Gravitational Constant is certainly wrong as a constant to be applied at galaxy-plus distances. What we call the Gravitational Constant is a number that for us matches units of distance with mass, but which has a hidden component of error due to proximity effects. The 'Real' Gravitational Constant will have to be teased out of the working constant by subtracting these effects.

In my view, while the inverse-square law may still be applicable, the Gravitational Constant cannot be saved in this perspective.

The evolution of the solar system over long time periods is something I have not considered, but at a glance I would say that what we see here is clearly as significant as corrections to Mercury's precession are for Relativity. Newtonian Gravitational Theory needs an overhaul.
 
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Rogue Physicist said:
One thing I am willing to say with confidence at this time is that the Gravitational Constant is calculated from careful experiments close to the surface of the earth for the most part, and assume the simple Newtonian inverse square law.....
Click to expand...
I am not sure your are correct in this claim either. It is my understanding that the solar mass, G, the mass of the planets and their moons is determined by astronomical observations, not "near Earth" measurements. I don't have the details, so I will describe how it could be done, but the way it is done is better, and can avoid Earth entirely.

For example the distance to the sun can be measured and so can the orbital velocity of the Earth. By equating the gravitational attraction of the sun to the centrafugal force of the orbiting Earth moon system, one gets an equation envolving G, Ms and Mem and other factors that are known. Ms is solar mass and Mem = Me + Mm where Me is Earth mass and Mm is moon mass. I.e. one equation with four "unknowns" (G, Ms, Me & Mm)

Now one can get a second equation in these same four unknows by considering the moon in orbit. - first approximation only G, Me & Mm are significant, but lets assume you do it completely correct. Now we have two equations with four unknows.

Now suppose that prior to launch into a very distant orbit (so that your "near Earth presummed problem is not important) I carefully measure the mass of the satellite (by measuring its inertia, if you a going to object that the mass measured on a scale is in the same trouble as you are postulating for other means of determining G such as using pendulum of known mass etc.) That is Ms, the mass of the satellite is not a new unknown, I now set up the same equation (balance centrafugal force observed/ calculated from orbit height and speed independently measured etc. against the gravity force of attraction) Again althought the main source of the gravitational attraction is Me, I do it correctly with all the mathematical mess necessary, but still have only four unknows. I now have three equations in four unknows.

I could launch a second satallite, into different distant orbit and get four equations in four unknowns and solve this mathematical mess for G.

In addition, I can measure the mass of a pendulum and get 5 equations in with only four unknowns (over determine the solutions), just to prove that the pendulum mass measured on with a scale (gravitational mass) is the same as the inertial mass I measured for the satellites.

I am losing my initial hope for you as a "promising student." This is the second unsupported assertion you have made that is false. The first false claim, which you are now ignoring or avoiding backing up with a calculation, was that the gravitational attraction computed for a sphere at a point near by the sphere surface would differ form the sum of two calculations that each separately calculated the gravitational attraction of the conceptually divided, but not physically disjoint, hemispherical parts of of the sphere.
 
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The first false claim, which you are now ignoring or avoiding backing up with a calculation, was that the gravitational attraction computed for a sphere at a point near by the sphere surface would differ form the sum of two calculations that each separately calculated the gravitational attraction of the conceptually divided, but not physically disjoint, hemispherical parts of of the sphere.
Click to expand...
I am stunned that you seem to have missed that the easiest way to calculate the force from the two halves of the sphere is to simply use the 'barbell' method, which does *not* require any additional axioms, postulates or premises other than the Centre of Mass theorem itself, which is why it is a perfect test of the accuracy of the theorem. Am I talking to the same person as earlier? Or are you his son, using his account?

Your version of how to go about calculating the Gravitational Constant G is actually fairly close to what has actually happened, although there are discrepancies in its periodic recalculation by independant field workers:

Boys.........1895 = 6.658
Richard..... 1898 = 6.685
Heyl..........1930 = 6.67
Rose..........1969 = 6.674
Facy..........1972 = 6.6714
Luther........1982 = 6.6726

Although experimenal methods have steadily improved, the best we can claim is 3 decimal places. And all of these experiments assume both the validity of the additivity hypothesis of the vector forces for the potential on any particle, and also the validity of the law of universal attraction itself. (an exhaustive list of measurements and the main sources of error can be found in G.T.Gillies (1983).)

In reality the Gmm/r^2 law is *only* very well verified on the scale of the solar system. It is still an open question whether or not it is true at the laboratory scale. (see An Introduction to Relativistic Gravitation by Hakim (1999) pg 26) Further out it is even less clear, especially because of the problem of estimating masses for distant stars and systems. The problem of 'dark matter' et al is well known.

GalaxyFailure.jpg


Most people looking at this evidence and using Occam's Razor would simply admit the failure of Newtonian (and Relativistic) Gravitational Theory rather than desparately try to invent invisible 'matter' and other fantasies to save the Inverse Square Law. Notice below (next message) that 'dark matter' also failed to explain the misbehaviour of the satellites too.

Serious physicists have already formulated a more complex version of the Gravitational Constant as a 'function' in which they can plug in various parameters to try to 'curve-fit' the predictions with realistic observations.
 
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The problems of using Satellites, even with the latest equipment and theoretical advances should be clear to you after reading this:
28 September 1998
Newtonian Physics - All The Wrong Moves
A team of planetary scientists and physicists has identified an unexplained sunward acceleration in the motions of the Pioneer 10, Pioneer 11 and Ulysses spacecraft. The anomalous acceleration was identified after detailed analyses of radio data from the spacecraft.

The research team, led by John Anderson of NASA's Jet Propulsion Laboratory and including Michael Nieto of The Department of Energy's Los Alamos National Laboratory, considered and ruled out many possible causes for the perturbation in the spacecrafts' motions. The team expects the explanation, when found, will involve conventional physics and understanding, but the team has also considered what implication the anomalous motion has for some new physical effect.

The accelerations are so persistent that they could be pointing to some relevant physics that's been overlooked in trying to explain the motions of bodies in the universe.

"In order of decreasing probability the possible causes are some systematic effect associated with the spacecraft; some subtle effect associated with our tracking systems, which would be important to know for space navigation; or some manifestation of 'new physics,' " Nieto said. "By looking at the third possibility we can examine how well 'normal' physics works, which in itself gives you further insight into the universe.

The researchers analyzed signals sent from Earth that were actively reflected by a transponder on the spacecraft. The resulting Doppler shift in the signal was used to calculate the motions. NASA's Deep Space Network sent and received the signals.

Newton's laws of gravity alone - with the sun providing the dominant gravitational force - are good enough for NASA to send spacecraft on planetary rendezvous with near-pinpoint precision. But the anomalous motions of these spacecraft are so small that the researchers had to consider numerous possible causes: perturbations from the gravitational attraction of planets and smaller bodies in the solar system; radiation pressure, the tiny transfer of momentum when photons impact the spacecraft; general relativity; interactions between the solar wind and the spacecraft; possible corruption to the radio Doppler data; wobbles and other changes in Earth's rotation; outgassing or thermal radiation from the spacecraft; and several others.

The researchers have so far not found that any of these effects can account for the size and direction of the anomalous acceleration.

After exhausting the list of possible "normal" explanations, the researchers looked at possible modifications to the attractive force of gravity or the possible influence or non-ordinary matter, or "dark" matter. The dark matter explanation didn't hold up because so much matter would have been required to create the measured spacecraft acceleration it would have affected motions of other bodies in the solar system.

Looking at other mathematical representations for gravitational interactions also "come up against a hard experimental wall," the researchers wrote: namely that the gravitational effect would also be seen in planetary motions, especially for Earth and Mars "If the anomalous radial acceleration acting on spinning spacecraft is gravitational in origin, it is not universal," the researchers concluded. It would have to affect bodies massing a thousand kilograms or so more than bodies the size of planets.

Nieto has long been interested in the possibility that gravity works differently on antimatter than on the familiar matter that makes up our everyday world. This led him to consider how well we understand gravity's influence on normal matter and whether studies of the motions of comets or spacecraft could be used to identify any deviations from the expected influence of gravity.

Meanwhile, John Anderson and his JPL colleagues had for years puzzled over "residual errors" between the calculated and measured positions of the Pioneer spacecraft. Anderson first saw the effect in 1980, but until he had accumulated data over the next 15 years, he could easily dismiss it as systematic errors. "Like a lot of problems in astronomy, many years of observation are needed," Anderson said.

"Clearly, more analysis, observation, and theoretical work are called for," the researchers concluded.
Click to expand...
 
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