Corrections to Newton's Shell theorem

geistkie · May 15, 2009

No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way.

All you've shown is that a half shell does not behave as if all the mass is located at its COM. You are taking a weighted average (as it were) dependent on r^2, so this isn't going to work. The theorem only works for spherical symmetry.

I am merely stating the obvious that is, that the closest half shell contributes the greater share of the total force and this from the law of gravity which is conditiomned on the inverse distance squared. To say that the shell theorem on works for spheres is an arbitrary statement - that the sphere is symmetrical is true but not relative to m, where the mass distribution on the shell is asymmetrical - to assert the 'sphere only' rule requires that the law of gravitation has a separate law for spheres in thagt the closest halh shell on a sphere would contribnute equal force on m as does the farthest half shell, which is nit a tenable conclusion.

Where is it proved about the ;sphere only' law.?

yes I am taking the weighted average of the locations of the centers of gravity. the shell theorem conclusion is that the 'shell really acts as though the mass M of the shell were concentrated at the COM of the shell'. All of this comes from F = GmM/d^2 which states only the force on a test mass m whose COM is located a distance d from m. I have seen the words that the shell theorem only works for spherical symmetry, yet the theorem itself makes no such claims, nor is there any such conclusion inferred from the maths.

SO, if spherical symmetry is the only configuration that the theorem pertains then a sphere stretched out along the x axis with (the mM axis) would provide a different resultwith both masses equal and symmetrically spread on both halves - the halves mirror each other -- would not give a similar result as the shell theorem. This is doubtful.

**swansont** · May 15, 2009

Prove this claim, with math.

Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign.

Your entire argument against the shell theorem is based on this faulty assumption.

I think geistkie is talking about a test mass outside the sphere, but the underlying problem is still there, and so is the need to prove the assumption mathematically.

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I am merely stating the obvious that is, that the closest half shell contributes the greater share of the total force and this from the law of gravity which is conditiomned on the inverse distance squared. To say that the shell theorem on works for spheres is an arbitrary statement - that the sphere is symmetrical is true but not relative to m, where the mass distribution on the shell is asymmetrical - to assert the 'sphere only' rule requires that the law of gravitation has a separate law for spheres in thagt the closest halh shell on a sphere would contribnute equal force on m as does the farthest half shell, which is nit a tenable conclusion.

Where is it proved about the ;sphere only' law.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

geistkie · May 15, 2009

You have an obligation to answer questions put to you. There will be little tolerance for shifting of the burden of proof. That's all that I was trying to say.

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This is not functionally equivalent to the shells. What allows you to say that a
half
-shell should act the same as a point mass located at its center-of-mass?

m will always be subject to the strongest force bearing down on it. If the force vectors are all aligned along the mM axis then using the point as a concentration of the half sphere mass on the mM axis is understood. The force vectors will always point along the mM axis. The only question left unanswered is should the half shell act as if all the mass were concentrated at its center?

In thinking about this I stumbled a tad. Actually the half sphere doesn't act as if the mass were concentrated at the COM of the half sphere. However, using the half sphere as indicated, the cg of the mM system is more accurately located. Slicing the sphere up into 1/4 and 1/8 etc merely increases the accuracy of locating the 'net' CG. for the sphere. To get the exact location this calculatioon using mirror image pairs that are then used to calulate the effective cg of the pair can be determined exactly.

You ask what allows using a point source of mass concentrated at the COM of the shell half? The same allowance for using point sources in performing the calculations in the shell integral.

As stated elsewhere if the shell integral is evaluated from d-R to d and if the differential masses, say differential spheres of mass dM as used in the integral [depending on the shell model used] the limits stated should result in a force greater that when compared with the integral evaluated from d to d + R., Remember, the dM on all rings all show a different distribution of mass to m, whichj was solved by using the differential spheres, where the half closest to m contributes a greater share of force on m than the half shell farther away.

Now don't argue shell theorem regarding the differential mass of spheres used as an essential element in forming the basis for the shell theorem.

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Originally Posted by D H

Prove this claim, with math.

Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign.

I have a serious problem here. If the force on m from the closest half shell were +1 and the force of the farthest half shell was -1, then the total sums to zero. Is this what you meant?

Your entire argument against the shell theorem is based on this faulty assumption.

I think geistkie is talking about a test mass outside the sphere' date=' but the underlying problem is still there, and so is the need to prove the assumption mathematically.

[/quote'] YUes I am talking about a test mass external to the shell.

The shell theorem says the shell acts as if all the mass was concengtrated at the sphere COM. I say the 'shell theorem' does not say the foregoing. The expression F =[ GmM/d^2 says the force acting on m from a shell [sphere] located a distance d from m. The development of the shell integral does not include the calculation of the cg, it calculates the force from a sphershell located at d.

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Originally Posted by geistkie

I am merely stating the obvious that is, that the closest half shell contributes the greater share of the total force and this from the law of gravity which is conditiomned on the inverse distance squared. To say that the shell theorem on works for spheres is an arbitrary statement - that the sphere is symmetrical is true but not relative to m, where the mass distribution on the shell is asymmetrical - to assert the 'sphere only' rule requires that the law of gravitation has a separate law for spheres in thagt the closest halh shell on a sphere would contribnute equal force on m as does the farthest half shell, which is nit a tenable conclusion.

Where is it proved about the ;sphere only' law.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

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Originally Posted by D H

Prove this claim, with math.

Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign.

I have a serious problem here. If the force on m from the closest half shell were +1 and the force of the farthest half shell was -1, then the total sums to zero. Is this what you meant?

Your entire argument against the shell theorem is based on this faulty assumption.

I think geistkie is talking about a test mass outside the sphere' date=' but the underlying problem is still there, and so is the need to prove the assumption mathematically.

[/quote'] YUes I am talking about a test mass external to the shell.

The shell theorem says the shell acts as if all the mass was concengtrated at the sphere COM. I say the 'shell theorem' does not say the foregoing. The expression F =[ GmM/d^2 says the force acting on m from a shell [sphere] located a distance d from m. The development of the shell integral does not include the calculation of the cg, it calculates the force from a sphershell located at d.

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Originally Posted by geistkie

I am merely stating the obvious that is, that the closest half shell contributes the greater share of the total force and this from the law of gravity which is conditiomned on the inverse distance squared. To say that the shell theorem on works for spheres is an arbitrary statement - that the sphere is symmetrical is true but not relative to m, where the mass distribution on the shell is asymmetrical - to assert the 'sphere only' rule requires that the law of gravitation has a separate law for spheres in thagt the closest halh shell on a sphere would contribnute equal force on m as does the farthest half shell, which is nit a tenable conclusion.

Where is it proved about the ;sphere only' law.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

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Where is it proved about the ;sphere only' law.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

Gauss' theorem does not distingjuish between spheres or arbitrary shaped masses, it is concerned with the flux in/out of a a physical body.

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Otiginally by Geistkie -"Where is it proved about the 'sphere only' law.?"
The "sphere only" restriction is in the derivation of gauss's law' date=' where you do the math —

it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.[/quote']

The half-shell was used as a point mass to make a point thaqt there is a contradiction if the sphere is used as a point mass in locating the cg at other than the shell COM.

use of half shell, or 1/4 or 1/8 etc shells are used to demonstrate the increase in accuracy in locating the shell cg that would resolve on 100% accuracy is calculated using the shell integral calculating mirror image pairs together with a runniung acciunt of the cg location, just like the force is a running account of forces until the integral completess.

J.C.MacSwell · May 15, 2009

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself.

The half spheres, not being spheres or spherical shells, cannot be modeled this way.

Spheres and spherical shells can be modeled this way.

Look in particular at the closer half sphere, the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned further from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

geistkie · May 15, 2009

H;491354]This is gobbledygook. For starters, there is only one sphere involved here. Next, before you advance to looking at the sphere as a whole you really need to understand the gravitational field induced by a ring of mass. You do not understand that result.

Each ring contains differential masses dM acting on m a distance d from the center of the shell and where the net vector force is projected along the mM axis.

I merely substituted a differential sphere for the various configuations offerred in various models explaining the shell theorem. When I suggest that the differential sphere must be treated like an asymmetric distribution of mass relative to m I mean that from the inverse distance squared condition of the gravity law, masses of equal mass closer to a test mass contribute more force on m than an equal mass located farther away.

Forces of attraction being confined to mass-mass interactions are not geometrically conditioned such that only spheres can be treated with shell logic and physics - spheres [shells] do not such that the closest mass to m of 1/2 a sphere contributes the same force as all the mass in the farthest half shell. This is saying that a cubic mass M equal to a shell volume acts like the closest half cube to m does contribute a greater share of force to m than the farthest half cube, or that a cylindircal shell of equal M and equal volume [with or without the cylinder hole covered] acts like the cube but the sphere is geometrically such that the shell acts as if the mass of the shell was concentrated at the COM of the shell.

If the shell is so unique in the law of gravity such that for all other shapes the general rule of equal masses closest to a test mass contribute a greater share of the total force on m than equal masses further from m holds rigidly, but that for spherical shells and solid sheres the contribution of force on m ree equal from all segments of the shell - this is saying that distance of physical mass from a test mass is irrelevant and that that the pure

mathematical abstracted condition of mass

[as opposed to actual concentration of mass] concentrated at the shell COM is

physically demonstrated

Who made that claim? If anyone here made it, its wrong.

Split a sphere in half. In fact' date=' split a sphere any way you want into two separate parts. Now compute the gravitational force induced by the two parts of the sphere at some test point inside the sphere. No matter where you place the test point and no matter how you split the sphere into two parts, the forces from these two parts on the test point will be equal in magnitude and opposite in sign.[/quote']

Here, like the paragraph above, an object in some location within a shell will be closer to some mass and farther away from other segments - I recognize the accepted shell theorem that states what was claimed re zero forces on the object will result inside the shell. The statement, however, that, "
these two parts on the test point will be equal in magnitude and opposite in sign
" is hard to accept for two reasons. First, the shell theorem evaluated within the sahell produces a 'zero' result, but this says nothing about the fiorces on the test mass being equal and opposite in sign - the evaluated integral for the force is simply zero.

Take a point on the shell axis a short distance from the midpoint of the sphere. Your statement says that all the force in the shell segment having slightly less than half the mass will nevertheless contribute a force equal in absolute value as that of the shell segment having slightly larger a quantity of mass. Further, the statement says that as the object recedes from the near midpoint along the axis the forces remain balanced in absolute value and differ only in sign. Again, this rule of physics indicates an extreme departure from the F = - GmM/x^2 for objects subject to gravitational forces where the mass is oriented in a thin spherical shell.

Besides the incredulity of the claim, I have never been exposed to a mathematical proof of that claim. I have, of course familiarity with then generally understood claim of the shell theorem regarding the mass ncentrated at the shell COM.

Do you have a reference or can you construct a math model indicating compliance with your claim?

Think of it this way. If a sin(theta) term does creep in' date=' that means half adn each according tlo gravity law are imposing attractive force on the objective ring has negative mass. There is no sin(theta) term.

I have asked you many times now to compute the gravitational force exerted by a thin hoop of mass on a test point mass. You have yet to do this.[/quote']

This thread is clear - the challenge to the statement that "the shell acts as if all the mass was concentrated at the shell COM" is simple. The statement should read that "the shell acts as if the force on m was from a sphere [shell] with COM located at d." Whether I submit to computing the force exerted by a ring of mass on a test mass m has absolutely nothing to do with the thread. All I have to do is get a shell model from the Internet and show it to you. What would be proved?

cAN YOU PROIV

DH would you please indicate to me exactly where the shell theorem jutifies itself in the conclusion people assigned to the result? The resulting exzpression after the integration is F = - GmM/d^2 says nothing about concentrated mass. The vector projection of GmdM/x^2, projects a force onto the mM axis, it does not project mass into the center of the shell. The expression is simply a term for the total force calculated from the shell integral..

Can you rpove your "equal and opposite theory"?

D H · May 15, 2009

Note: I was talking about points interior to the spherical shell. It is a direct consequence of Gauss' law of gravitation and the additivity of Lebesgue integrals for measurable subsets of a space.

geistkie · May 15, 2009

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself.

Using the mass points is of course incorrect, but the result is more accurate

if using as many pairs of points in the shell integral and calculalting the cg of each pair (mirror image pairs). Do not misread me. I am not opposed to concentrating the mass at some point as long as it is the correct point.

I calulated the force on m from each "point" of the half spheres COM. I computed the resulting force of these two points and then determined the location of the cg

The half spheres' date=' not being spheres or spherical shells, cannot be modeled this way.[/quote']

I calculated the force from the sphere M =1 = m located at 10. The Fs = .01' Next the two shell halves were calculated and using R = 1 and the COM of each sphere located at 9 and 11 for a total force of .0103 and a calculated cg of 9.85 F12/Fs = .0103/.01 = 1.03 a three % discrepancy. The location difference for the cg was .15 out of 10 or .015 a 1.5 % discrepancy.

Spheres and spherical shells can be modeled this way.

Look in particular at the closer half sphere, the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned further from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

I arbitrarily put the sphere at 10 and for R = 2, the COM for the closest sphere is located at 9. The farthest sphere at 11. Equal widths of shell strips have equal mass, ergo for R = 2 the COM of the two half shells are located at 9 and 11. If I use 4 equal segments the mini cgs are located at 8.5, 9.5, 10.5 and 11.5 for a force total of .0105 the cg is at 9.75 or for 8 points the cg is 9.74 with a total force of .01052 So the force gets bigger the more segments used as the cg creeps toward m which is consistent with my "off set from the COM in the direction of m".

First, I calculated the force on the sphere of mass M located a distance 10 from m using, for m = M = 1 the force is .01 the lowest of all calculated forces and the cg farthest from m. So using the conmcentrated mass a of the shell at the COM gives the lowest of all forces, which is consiustent with the "closest mass contributes the greater share of the forces."

If the force from the shell were measured at .01 the calculations would apply as stated. F8/Fs = .0105/.01 = 1.052 or a 5% error from the ell calculatioin as a whole sphere, the magnitude of which is dependent on the scale of the variables used.

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Note: I was talking about points interior to the spherical shell. It is a direct consequence of Gauss' law of gravitation and the additivity of Lebesgue integrals for measurable subsets of a space.

Is Gauss' Law valid only for spheres?

I finally got the fact that you were discussing the conditions inside the sphere which I discussed above.

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J.C.McSwell

I don't know if you picked up another 'asymmetry' in the mass distribution but each dM in the shell development has an intrinsic error due to the closest/farthest problem we have been discussing. I eliminiated the gross complexities by assuming each dM was a differential sphere with a unit vector r defined along the x axis and where the resiultant plane bisects the differential sphere and where the plane is always perpendicular to x, duh, that's what thr r does right?. There is of course, the gross closest/farthest conditions we have been discussing.

J.C.MacSwell · May 15, 2009

J.C.McSwell

I don't know if you picked up another 'asymmetry' in the mass distribution but each dM in the shell development has an intrinsic error due to the closest/farthest problem we have been discussing. I eliminiated the gross complexities by assuming each dM was a differential sphere with a unit vector r defined along the x axis and where the resiultant plane bisects the differential sphere and where the plane is always perpendicular to x, duh, that's what thr r does right?. There is of course, the gross closest/farthest conditions we have been discussing.

Not sure if If I follow what you are saying exactly, but by continuously using the COMs along the x axis in your calculations you end up with higher forces, because you are generally picking a point closer to the test mass than the average distance from the test mass of all the points it represents.

If, say, your segments were an infinitesimally thin set of discs perpendicular to the x axis, and you used the COM of each disc, each point chosen to represent the disc would be the closest point on each disc to the test mass and lead to an overestimation of the force, since you continuously underestimate the effective distance.

For some shapes this effective distance is further than the COM, for others it is closer, but for the sphere it is exactly the right distance.

Edited May 15, 2009 by J.C.MacSwell

geistkie · May 15, 2009

Not sure if If I follow what you are saying exactly, but by continuously using the COMs along the x axis in your calculations you end up with higher forces, because you are generally picking a point closer to the test mass than the average distance from the test mass of all the points it represents.

What I do is put the sphere at 10 say. Then arbitrarily I set 9 as the COM of the closest half shell and 11 as the COM of the farthest half shell. Or if I waant to segment the shel in quarters m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. This moves the cg found using two 1/2 masses toward m. The more segments the more accuracy and the cg creeping toward m slows with successive doubling of the segments.

If, say, your segments were an infinitesimally thin set of discs perpendicular to the x axis, and you used the COM of each disc, each point chosen to represent the disc would be the closest point on each disc to the test mass and lead to an overestimation of the force, since you continuously underestimate the effective distance.

I agree.

For some shapes this effective distance is further than the COM, for others it is closer, but for the sphere it is exactly the right distance.

Here there is a slight glitch in logic. Replacing all dM on the rings with spheres halved by the plane

r

unit vector along the x axis. What do you suggest here? Should we use the shell theorem to treat each differential mass as concentrated at the COM of each sphere? or would this be circuitous?

Previously you mentioned something to the effect that my treating shell halves as point masses and performing the calculations as I have been doing might not be "allowed". I am being general here and if I need correcting by all means go for it. However, when calculating the force on m for each dM there is no question of using dM as a point concentration of an infinitesimal chunk of mass. Now as the order of calculating the forces is insignificant then calculating the force due to the mirror image of the nearest shell dM on a mirror image ring in the farthest half sphere poses no perturbation to the total force as resulting from a the sum of single dM calculations at a time.

I am being overly stating things but I want to make a point. When calculating the pair of dMs one dM in 1/2 of the shell, the other dM in the other 1/2 of the shell. Each dM in a shell has an exact counter part in the opposite shell half, and ergo all dM are counted in the pair system.

From this it must be obvious that each dM in the closest shell contributes more force than the dM mate in the mirror image ring in the farthest shell half. Therefore the shell half closest nto m contributes a greater share of the force on m than does the farthest shell half.

If calculating the force in pairs only and continuing with the shell integration the results will be identical to the unmodified integral model hence why bother?

If when calculkating the force on m for two dMs, one force for the dM in the closest shell half, Fc, and one force force for the farthest dM mate, or Fc, the total force of these two dM pairs is Fp = Fc + Ff. Using F = GmdM/r^2 and setting dM = 2dM, Fc + Ff = Fp = Gm(2dM)/r^2, or r = sqrt(Gm(2dM))/Fp and the cg of each pair of forces is determined from which the final cg of the mM system can be calculated by some weighted average. This algorithim, more like numerical analysis determienws the cg exactly.

To repeat, each calculated pair of forces places the

cg

of each pair in the nearest half shell to m, which means that the shell theorem must be read as the physics dictates, that is, the shell acts as if the mass of the shell was concentrated at the

cg

of the mM system and where the force F of the shell whose COM is located a distance d from m.

I know the risk of saying "it is so clear", but it is so clear. I cannot understand the objection to changing, not the shell maths, even without the 'paired force' modification to the shell integral, observation justifies the correction proposed here.

But then when the speed of gravitational forces are integrated into the paradigm I suspect an increase in discussions on the topic of, "conservation of angular mlomentum", that will begin to push "gravity" concepts as generally understood into the file of interesting historical attempts of physical descriptions of stellar phenomena.

J.C.MacSwell · May 16, 2009

What I do is put the sphere at 10 say.
Then arbitrarily I set 9 as the COM of the closest half shell and 11 as the COM of the farthest half shell.
Or if I waant to segment the shel in quarters m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. This moves the cg found using two 1/2 masses toward m. The more segments the more accuracy and the cg creeping toward m slows with successive doubling of the segments.

What makes you arbitrarily do this? It is wrong.

Segmenting further into quarters with m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. is

more wrong

The more segments the

more wrong it gets

even if it converges to limit you error.

geistkie · May 16, 2009

Originally Posted by geistkie

What I do is put the sphere at 10 say. Then arbitrarily I set 9 as the COM of the closest half shell and 11 as the COM of the farthest half shell. Or if I want to segment the shell in quarters m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. This moves the cg found using two 1/2 masses toward m. The more segments the more accuracy and the cg creeping toward m slows with successive doubling of the segments.

What makes you arbitrarily do this? It is wrong.

Segmenting further into quarters with m1 at 8.5' date=' m2 at 9.5 m3 at 10.5 and m4 at 11.5. is more wrong

The more segments the more wrong it gets even if it converges to limit you error. [/quote']

If the error is limited or corrected within some experimental error what is wrong? Why do I do this? I do this because the laws of physics hold that masses closest to a test mass contribute a greater share of the total force on m than equal masses located farther away. You are saying or so it appears to me that the mass in a shell cannot be segmented in a way that selectively isolates the location of the force centers seen by m looking along the mM axis.

Is this it?

What about making the caculations in the shell integral in pairs as I suggested? Does this corrupt the integral, that is to merely vary the order in which various forces attributable to various dMs in the shell is a 'wrong'? Certainly not. I am calculating the total force between paired dMs and calculating the cg of these two dMs. When all pairs are calulated and the totla florce of the shell is determined, the cg has also been determined, which is not the case ion the shell integral. I mean by this that in developing nthe shell integral, even as modleed by DH in an earlier post, does not provide for a claculation iof the shell cg , center of gravity [FORCE] as I emphacise.

What makes you arbitrarily do this? It is wrong.

Segmenting further into quarters with m1 at 8.5' date=' m2 at 9.5 m3 at 10.5 and m4 at 11.5. is [b']more wrong
[/b]

The more segments the
more wrong it gets
even if it converges to limit you error.

What is wrong with it. If it converges to limit an error, theoretically the error converges to zero. What is worng with this?

Let me remind you that monitors are literally breathing down my neck. I am making every effort to answer every post and question, but I have a problem with an unembellished "It is wrong" statement. Do you see my problem? Prove it to me that I am wrong and I will admit to it. I'd rather knoew the truth than to live a life of lies.

**swansont** · May 16, 2009

I am being overly stating things but I want to make a point. When calculating the pair of dMs one dM in 1/2 of the shell, the other dM in the other 1/2 of the shell. Each dM in a shell has an exact counter part in the opposite shell half, and ergo all dM are counted in the pair system.

Counting them is not enough. They contribute different amounts to the total, and not simply because their projected distance along the axis is different.

Your differential mass dm is located r from your point, P. There are two contributions from each dm: one along the vector r, and one perpendicular to it. If we look at the "twin" of this point, projected through the center, the angle to it is smaller, thus, the perpendicular contribution to the force is smaller — it makes a larger radial contribution. Simply pairing them doesn't work and when you do the math you find that the method you are using is flawed.

Doing the on-axis calculation doesn't show this effect. You need to look off-axis to see it.

geistkie · May 16, 2009

No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way.

See below where the error introduced by segmenting the sphere is less than if the shell sphere was not clorrected. So to your statement that the shells do not behave this way needs correcting. So I have assumed only that segmenting the shell results in a more accurate location of the shell center of gravity, than the misinterpreted results of the shell theorem..

All you've shown is that a half shell does not behave as if all the mass is located at its COM. You are taking a weighted average (as it were) dependent on r^2' date=' so this isn't going to work. The theorem only works for spherical symmetry.[/quote']

Not quite. I have shown that using the the half shell can lead to a contradiction of the shell theorem. By this I mean that even though the 1/2 shell considered concentrated at the COM of the half shells has an error, the error is less than that of the uncorrected shell. The shell has an intrinsic error, due, of course, to the asymmetric disribution of the 1/2 shell mass relative to m.

If using 1/4 segments, or 1/8th segments the errors diminish rapidly as the result of the 2, 4 or 8 mass centers are calculated and where the cg slowly creeps in the direction of m as the segmentation lof the shell increases. If such a calculation was performed in the shell integral such that mirror image pairs of forces are calulated the dM closest to m contributes a greater share of force on m than the shell farther from m. All mirrore image pairs places the center of gravity in the half closest to m.

This last statement does not distort the shell into segmented parts and the calculations are performed on a differential level the only differfence with this model is the order of calculating the dM centered forces acting on m. This eliminates all error in calculation of the cg. Your objection vanishes when all dM in the shell are in the form of differential spheres pointed to by x terminating at the differential sphere COM defined by the plane r where r is a unit vector pointing down the x axis, the plane segmenting the differential sphere in two equal spheres where x is always perpendicular to the plane for all dM.

A number of persons have told me in this thread that using of 1/2 shell masses assumed concentrated at their centers of mass that the shell theorem does not work for this configuration. I never said it did. Even though the segmenting the shell into equal halves, quarters, eights etc creates an error, the error is minimized with succeeding increases in the segmentation - just like the differential the error tends to zero as the segmentation increases.

Taking your assessment that the half shell model is not covered by the shell theorem, then I assume you are saying is do nothing - the caclulation results in errors therefore leave the shell alone. I am sure it is because of my poor powers of communication that neither yourself, nor DH (JC MCSwell is starting to grasp the implications of my correction thesis) , see that the segmenting of the shell leads to a test of the validity of the shell theorem.Validity should be replaced by accuracy. How so?

Say a shell is centered at distance d = 10 with a mass M = 10. Calculating this force with m =1 and G = 1. F = 10/100 = .1. When the shell is segmented in two parts one half centered at 9 and the other at 11, the force on the closest segment is 5/81 the farthest segment is 5/121. The total force of the half shells is .1030, which has an error which would decrease if quarter, eights or 16th segments were used and where each succeeding doublin g of the segments increases accuracy considerably. I never said this was 'shell theorem' compatible, but then I've seen no proof that my system is just plain "wrong" to quote JC McSwell.

The ratio of the the segmented system to the shell sphere calculated force is .1030/.1 = 1.03 a 3 % error which has the effect of significantly affecting many thousands of astronomical calculations. I had an inspiration that dark matter was just swept from the astronomical literature with geistkie's correction. Maybe so. I can handle it.

What I have just done above is to use a cal culation assuming the mass of the shell was concentrated at the shell COM. Then by segmenting the shell I show the errors in the assumption that the mass of the shell is located at the COM of the shell. The more the number of segments, not unlike the rings distributed in the shell integral, the more accurate the determination of the cg of the shell.

This just dawned on me. For sure the segmenting of the shell as I have been doing introduces errors, but these errors are less than that if the shell sphere was not corrected for and even for the half shell configuration the error is lessened from using the uncorrected force using the shell theorem constraints.

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I think geistkie is talking about a test mass outside the sphere, but the underlying problem is still there, and so is the need to prove the assumption mathematically.

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The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

First, I have never said that using half shells as I have been doing is error free. The very fact of the geometry of the half shells creates an asymmetric distribution of mass relative to m, but the error resulting in using the segmenting paradigm decreases as the segments double.

------------------------------------------------------------

Here is a form of Gauss' law I got from the Internet. Just as I remember from good ol' school days. the "∂V is any closed surface"

Integral form

The integral form of Gauss' law for gravity states:

Int{[over ∂V] g.dA = -4piGM

where ∂V is any closed surface,

dA is a vector, whose magnitude is the area of an infinitesimal piece of the surface ∂V, and whose direction is the outward-pointing surface normal (see surface integral for more details),

M is the total mass enclosed within the surface ∂V.

The left-hand side of this equation is called the flux of the gravitational field. Note that it is always negative (or zero), and never positive.

J.C.MacSwell · May 16, 2009

Counting them is not enough. They contribute different amounts to the total, and not simply because their projected distance along the axis is different.

Your differential mass dm is located r from your point, P. There are two contributions from each dm: one along the vector r, and one perpendicular to it. If we look at the "twin" of this point, projected through the center, the angle to it is smaller, thus, the perpendicular contribution to the force is smaller — it makes a larger radial contribution. Simply pairing them doesn't work and when you do the math you find that the method you are using is flawed.

Doing the on-axis calculation doesn't show this effect. You need to look off-axis to see it.

I think that is his answer right there if he thinks about it carefully, even without doing the math he should see where he is erring.

In the above the closer "twin" contributes more to the resultant force though, correct? The net force along the axis is greater for the closer twin.

The only reason I ask is that I'm not sure Geistke feels we see this.

**swansont** · May 16, 2009

First, I have never said that using half shells as I have been doing is error free. The very fact of the geometry of the half shells creates an asymmetric distribution of mass relative to m, but the error resulting in using the segmenting paradigm decreases as the segments double.

Then you can't use it! If it's wrong, then you can't get the right answer by using it!

The shell model gives NO error when using it. If you want to show it is in error, you can't used flawed math to do so.

------------------------------------------------------------

Here is a form of Gauss' law I got from the Internet. Just as I remember from good ol' school days. the "∂V is any closed surface"

Integral form

The integral form of Gauss' law for gravity states:

Int{[over ∂V] g.dA = -4piGM

where ∂V is any closed surface,

dA is a vector, whose magnitude is the area of an infinitesimal piece of the surface ∂V, and whose direction is the outward-pointing surface normal (see surface integral for more details),

M is the total mass enclosed within the surface ∂V.

The left-hand side of this equation is called the flux of the gravitational field. Note that it is always negative (or zero), and never positive.

It is any closed surface, but there's a dot product, so you are only looking at the the component normal to the surface. if you want to get the whole (i.e. correct) answer, you must have the flux lines going through it normal to the surface. Otherwise you have to do a rather messy integral.

geistkie · May 16, 2009

Prove this claim, with math.

Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign.

Your entire argument against the shell theorem is based on this faulty assumption.

Are you ta;lking about a test mass inside a shell?

I was discussing a test mass m external to the shell.

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Consecutive posts merged

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself.

The half spheres, not being spheres or spherical shells, cannot be modeled this way.

If I am holding a chunk of a half shell in my hand and I attach a string at the shell point normally closest to a test mass [along the mM axis] and then I attach a string at some point on the shell surface that would be the midpoint of the mass drawn on a line from the shell center to m along the mM axis.I hold the 1/2 shell in the air by the first string. The open end of the 1/2 shell is parallel with the ground facing down. The string would point along the mM axis. Now I hold the 1/2 shell up by the other string. The 1/2 shell circular opening would be perpendicular to the horizon and the string would point to the mM axis directly below. From this I know the COM of the 1/2 shell which I will use as an approximation of the mass concentrated at the COM of the 1/2 shell.

I do the same for the other 1/2 shell. Now I calculate the force on m from each 'point'. Then I find the total force of both points and then find the effective cg which is not in the center of mass of the two 1/2 shells.

Where the Spheres and spherical shells can be modeled this way.

Look in particular at the closer half sphere' date=' the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned [b']further
[/b]
from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

Let us see if you are correct.

If the1/2 shells were molded into spheres then I could use the COM of the sphere. But tnhis would have a different force effect on m due to the differences in the distribution of mass relative to m than the unmodified 1/2 shell. If I find the COM by calculatioin or 'string' I get the same location for the COM. If I use the COM as a point I do so knowing the force affect of the point would be different than the unmodified 1/2 shell. However, the errors can be corrected.

I almost missed your statement "that would bring about an equivalent share of the force". I do not do this. I set the shell down and calulate the force a la the shell theorem for the in tact sphere. Then I segment the shell and

calculate

the two 1/2 shell points. where IOget the contradiction.

First I calculate the the forces of each 1/2 shell where their COM was symetrically arranged around the COM of the shell - each of the two points are located equal distances from the COM of the whole shell. The total force on m from the segmented shell is greater than the force on whose mass is assumed concentrated at the shell center. For M = 10 located at 10 from m the force on m is 10/100 = .1. The force of 1/2 M llocated at 9 is 5/81. The force from 11 is 5/121 or .062 + .041 = 1.03 > .10. I do this for the shell segmented in 1/8ths or 1/16ths the calulations tend to have the errors go to zero.

I don't follow why I would "have to locate the 1/2 shell that would bring about an equivalent share of the force" farther from the COM.

About a matter of old business I would like to bring up what this post hints at. You agree that the closest 1/2 shell contributes the greater share of the total force. This is seen from inspectioon for one.

Try calculating the force from each dM in mirror image pairs. The calculate the total force on m from each dM pair as well as calculating the effective cg of each pair and keep a running total of the cg point of each pair. As each pair has one dM in the closest 1/2 of the shell and its mirror image mate in the farthest 1/2 of the shell integral can spew out the total force frlom all dM on the shell and the exact location of the cg.

This is one method to calculate the cg and forces. The other way is to do the calculations as I have been doling in this thread and in fact this post.

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Consecutive posts merged

Prove this claim, with math.

Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign.

Your entire argument against the shell theorem is based on this faulty assumption.

DH We have been talking abiout different conditions as all my test masses

are external to the shell. I think you are discuyssing confitions inside the shell.

I have a proposition for you as your rank of 'expert' attests and I have no doubt you are probably over qualified for a simple but most important mission which should remove a bulk of our differences.

Using the shell integral as you constructed in your demonstration to me and modify the algorithim slightly that rearranges the order of calculating the dM forces on the shell only and adds one other calculated parameter as follows:

1.When calculating the force attributable to one dM on a ring, place a mirror image ring in the opposite side of the shell and calculate the force from that mirror image dM.

2. Then calculate the total force of the pair and from this

3. calculate the position of the effective center of gravity of this pair on the mM axis, then,

4. Either keep a running total of the effective centers of gravity and make one final calculation or keep an up to date running center of gravity

after calculating each force pair.

If you do this all will be explained to you in a manner best suited to convince you of something without my saying another word.

**swansont** · May 16, 2009

DH We have been talking abiout different conditions as all my test masses

are external to the shell. I think you are discuyssing confitions inside the shell.

I have a proposition for you as your rank of 'expert' attests and I have no doubt you are probably over qualified for a simple but most important mission which should remove a bulk of our differences.

Using the shell integral as you constructed in your demonstration to me and modify the algorithim slightly that rearranges the order of calculating the dM forces on the shell only and adds one other calculated parameter as follows:

1.When calculating the force attributable to one dM on a ring, place a mirror image ring in the opposite side of the shell and calculate the force from that mirror image dM.

2. Then calculate the total force of the pair and from this

3. calculate the position of the effective center of gravity of this pair on the mM axis, then,

4. Either keep a running total of the effective centers of gravity and make one final calculation or keep an up to date running center of gravity

after calculating each force pair.

If you do this all will be explained to you in a manner best suited to convince you of something without my saying another word.

Why can't you do it? This is shifting the burden of proof.

J.C.MacSwell · May 16, 2009

Geiestke.

Imagine a sphere lying with it's COM directly along the x axis from a test mass

Now replace the sphere with a magic stick.

1. The magic stick has the same COM as the sphere. It's COM is positioned in the same place as the COM of the sphere was.

2. The axis of the stick lies along the x axis

3. The magic stick has the same length as the sphere's diameter

4. The magic stick has the same mass as the sphere

5. The magic stick has the same mass distribution as the sphere in the x direction but it is magically concentrated along the x axis

Now the question is: Is the force of gravity between the test mass and the magic sphere greater, lesser, or the same as it was between the test mass and the sphere? Do you see the difference?

geistkie · May 17, 2009

Then you can't use it! If it's wrong, then you can't get the right answer by using it!

It isn 't wrong, it has an error that is, nevertheless an error less than the error associated with mischaracterization of the shell results; also, the methodology takes advantage of the laws of physics that includes the fact that equal masses situated at different distances from a test mass m the closest mass contributes more force on m another equal mass located farther from m. This is obvious is it not?

Some say the sphere can be considered a point of concentrated mass. Well enough, even though everyone realizes the physical impossibility of this condition being realized in nature. I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere.

For G = 1, m = M = 1 and the shell center located at d = 10. Therefore F = 1/100 = .010. No one can complain if I see what occurs when place .5M at 9 and .5M at 11 units distances from m. The closest 1/2 shell force is .5/81, the other force .5/121. Adding these forces gets, .0062 + ,0041 = .0103 = Fcf, and as Fs = ,01, Fcf/Fs = 1.03. Further, for the combined force, r^2 = 97.09, r = 9.85 which is what is expected. I have been screaming that the cg is off set along the axis toward m and here the %ERROR of the two methods has been identified..

What can be pointed to that focusses on the differences in the two systems? The .01 force either resulted from the mass effect of a shell centered at d relative to m, or the .01 force resulted from the placement of the concentrated mass of the shell at d.

There is one argument that resolves the matter conclusively. If m is looking at a sphere centered at d from m, then it must be concluded that my calculations using the segmented concentrated mass in two locations is what is expected from the condition that the closest 1/2 shell contributes most of the total force.

If we assume a condition claimed by shell theorem enthusists that there is no difference in the contributions of force from any two 1/2 shell halves, the calculations just completed show that indeed if a sphere is assumed the shifted cg in the direction of m is consistent with this arrangement.

In the concentrated mass of the shell at the COM arrangement we have only one point to consider, and for this reason alone, can it be said that the concentrated mass, besides being mathematically concentrated, we are told that the sphere must not be subjected to the "equal mass force Fc, closest to m, is greater than equal mass Ff, farther from m and where Fc > Ff". Did Newton's genius extend to the realization of this phenomenon being a fantastic exception to the universal gravitational forces of attraction doctrine.

To accept the stated claim sjustained through the years, from the 17th century to the present, one must accept the supremacy of the mathematical model over physical law regarding "closest equal mass contributes the greater force", and to accept instead, that for spheres only, the mass of the shell can be considered concentrated at the COM;

This is not consistent with the front 1/2 of the sphere contributing more force than the farther 1/2 sphere.

The shell model gives NO error when the COM. using it. If you want to show it is in error' date=' you can't used flawed math to do so. [/quote']

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case. When I segmented the point mass the two segments indicated agreement with the 'closest mass greatest force' paradigm, which means the shell theorem did not concentrate the mass mathematically at d, the center of the shell, the results of the shell theorem places the shell centered at m and the the accuracy in placing the cg is directly proportional to the number of segments chosen for calculation. There has to be some clever looping algorithm to place the cg as accurately as desired.

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont, Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific resposes such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

You thjee should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem, as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as throughly as our scrutinizing processs was applied and from a very unlkely beginmning where imatter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

We may not be listed as the 'number one whizz bang gravitationa subjects research group', but from my perspective we got a huge job completed, we the job done. For those insisting on maintaining the status quo of shell dogma, good luck and whatever personal differences may have been hinted at, no harm no foul.

It is any closed surface' date=' but there's a dot product, so you are only looking at the the component normal to the surface. If you want to get the whole (i.e. correct) answer, you must have the flux lines going through it normal to the surface. Otherwise you have to do a rather messy integral.[/quote']

You say the dot product gives the normal component only and that the correct answer requires the flux lines going through it be normal.

OK. I'll give this one tlo you. [/indent]

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Consecutive posts merged

Geiestke.

Imagine a sphere lying with it's COM directly along the x axis from a test mass

Now replace the sphere with a magic stick.

1. The magic stick has the same COM as the sphere. It's COM is positioned in the same place as the COM of the sphere was.

2. The axis of the stick lies along the x axis

3. The magic stick has the same length as the sphere's diameter

4. The magic stick has the same mass as the sphere

5. The magic stick has the same mass distribution as the sphere in the x direction but it is magically concentrated along the x axis

Now the question is: Is the force of gravity between the test mass and the magic sphere greater, lesser, or the same as it was between the test mass and the sphere? Do you see the difference?

I had to think about it for a spell. All the mass on the stick is equivalent to shrinking the rings to collapse onto the stick, Therefore the vector linking a point on the stick to m is stronger as the mass is the same around the ring, but the distance is greater for each dM on the stick. Priojecting the force from the shell effectively results in losing some of the force due to off axis cancelation of vector forces perpendicular to the axis.

Somewhat repetitive:

Thje mass on the stick is same as the total mass of a ring circling the stick, but the distance to a point on the stick is less than the distance from m to the ring. Thnerefore by distance effects alone the stick should provide a greater force.

When considering that the force from the ring was decreased by the cosine projection, no such loss is seen in the stick arrangement. - all forces are preserved.

For these reasons I pick the stick to contribute more force than the shell.

.

**mooeypoo** · May 17, 2009

It isn 't wrong, it has an error that is, nevertheless an error less than the error associated with mischaracterization of the shell results; also, the methodology takes advantage of the laws of physics that includes the fact that equal masses situated at different distances from a test mass m the closest mass contributes more force on m another equal mass located farther from m. This is obvious is it not?

But the right calculation has no error in it, as swansont has REPEATEDLY told you.

Which is it, geistkie? If it has an error, it is wrong. It can't have an error and still be right. Be serious, please.

Some say the sphere can be considered a point of concentrated mass. Well enough, even though everyone realizes the physical impossibility of this condition being realized in nature. I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere.

Only from a distance is a sphere "concentrated mass". If your distance is considerably larger than the sphere's radius, then you can treat it as a concentrated mass in the calculation. Otherwise, you can't.

It's not about physics consideration, it's about practicality. A planet whose radius is a few thousand kilometers has effects over an object that is a few LIGHT YEARS away that are so close to an estimation of it as a "concentrated mass" dot, that we use just that. It's practically the same in large distances. You can also show that mathematically by calculating the limits when the distance X is extremely larger than the radius R.

It's NOT the same in close distances.

To accept the stated claim sjustained through the years, from the 17th century to the present, one must accept the supremacy of the mathematical model over physical law regarding "closest equal mass contributes the greater force", and to accept instead, that for spheres only, the mass of the shell can be considered concentrated at the COM;

Huh?

Physics and math go hand in hand. If you prove mathematically, and can pose *working* predictable models, then it's physics. They're not mutually exclusive.

Your problem is not the criticism. Your problem is that your math is wrong. Look again at what swansont wrote you. I'm not going to repeat this again for what seems like the 5th time (at least). You have some basic flaws in the construction of your formula, and those result in flawed math.

Flawed math does not lead to working physics.

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont, Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific resposes such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

But you are wrong, you don't understand the math and the shell theorem does only pertain to spheres. You're just ignoring these claims instead of facing them. You're avoiding proving your math when people tell you you're wrong; that will not get people to stop telling you that you're wrong. It will just get people to stop debating with you.

If you have problems in your math, you should fix them. If you don't understand where the problem lies, you should ask for clarification (though, I must say, I think with the amount of time people repeated these counter claims, you should just go over the thread again). But you can ignore this 'till the sun explodes and it still will not make your math right or your conclusion valid.

You thjee should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem, as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as throughly as our scrutinizing processs was applied and from a very unlkely beginmning where imatter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

This is utterly unfair. The people you refer to are professional scientists who work in the field of physics and math. They are not mindless automatons who swallowed what they're schoolbooks told them. The claims they put forth in this thread had EXPLANATIONS of why you're wrong, unlike your counterclaims which seem to be consisting of stomping-the-ground insistence that you're right and they're wrong.

May I remind you that there's nothing better to a scientist than discovering some "new thing" - some revolutionary new idea - in physics (or math). Slightly and marginally less exciting (and yet still extremely desired) is having someone else discover this new revolutionary idea. Revolutionary ideas lead to more opportunity for research, more opportunity to find new stuff that no one has thought of before and stem from that new discovery, and - to the materialistic among us - an opportunity for fame and fortune.

The rejection of your claim has nothing to do with them being stuck in "oldschool" science or being automatons. Their rejection has to do with your inability to provide proper mathematical concepts and face the problems that are raised by your claims.

We may not be listed as the 'number one whizz bang gravitationa subjects research group', but from my perspective we got a huge job completed, we the job done. For those insisting on maintaining the status quo of shell dogma, good luck and whatever personal differences may have been hinted at, no harm no foul.

Oh, knock it off. No one's insisting on dogmas. We would LOVE to have revolution in physics. You're just not supplying the goods.

You say the dot product gives the normal component only and that the correct answer requires the flux lines going through it be normal.
OK. I'll give this one tlo you.

You're not giving anything to him. He provided proper math, not some "whoopass" claim you can "give him" the "win" for.

Seriously, it's as if I claim that 1+10 is 9, and a math expert tells me that it's 11, and I say "Okay, I'll give it to you". Really?

Other than that, please stop using "INDENT" tag on all your posts. There's a purpose for indent, and it's not to get your posts glowey-greeny so that they might attract the eye.

~moo

geistkie · May 17, 2009

Why can't you do it? This is shifting the burden of proof.

I looked at the problem and i will produce an answer.

However, you must be aware that I did not intend to shift the buirden onto DH. I was attempti ng to get him to see the real effect of the calculating the forces in pairs in order to see the effect of distance on mirror imaged dM pairs. The simple re arrn gement starts with adding a cos9neta) term to account for the dM of the mirror imaged dM. asnd to see thagt each pair of calculation places tnhe greater share of the two nbody force onto the closest dM such that when all dM are calculated the force center, i.e. the cg will be located in the nearest segment of the shell. If he refused to do it , fine, I would then provide at least an outline of the algorithm needed for paired force calculation.

Don't you think you might have interferred in my grand plan and misjudged my post as a possible 'shifting of burden'?

Even if i was totally lost and unable to prioduce a working model what does this prove about the necessary corrections required for the shell integral?.

**mooeypoo** · May 17, 2009

geistkie, you are not different than other member of this forum. Your posts are not "more special", and they do not require extra highlighting.

Stop abusing the BB code. Use the "INDENT" tag when the INDENT tag is needed, and not for the entire post.

That said, you have yet to produce actual math, even though it was repeatedly requested throughout the thread.

The other debaters showed you how your theory FAILS in the mathematical aspect. If you disagree, the ONLY WAY to provide evidence that the theory is still valid is by supplying the MATH to show that it does NOT fail.

Pretty explanations will just not cut it. You are the one making the claim, you are the one who has the burden of proof to convince all of us that you have a valid theory. The math is against your theory. Supply alternative math or accept defeat in this mini peer review.

Don't take it hard; a lot of theories fail peer review. You should take it as a chance to improve it, think it over, re-evaluate it, and perhaps post it again when it holds more merit.

~moo

J.C.MacSwell · May 17, 2009

I had to think about it for a spell. All the mass on the stick is equivalent to shrinking the rings to collapse onto the stick, Therefore the vector linking a point on the stick to m is stronger as the mass is the same around the ring, but the distance is greater for each dM on the stick. Priojecting the force from the shell effectively results in losing some of the force due to off axis cancelation of vector forces perpendicular to the axis.

Somewhat repetitive:

Thje mass on the stick is same as the total mass of a ring circling the stick, but the distance to a point on the stick is less than the distance from m to the ring. Thnerefore by distance effects alone the stick should provide a greater force.

When considering that the force from the ring was decreased by the cosine projection, no such loss is seen in the stick arrangement. - all forces are preserved.

For these reasons I pick the stick to contribute more force than the shell.

That is correct. So you see that, yet you are continuously treating the sphere as if the mass was all concentrated along the same axis when you do your calculations. This causes your error.

**swansont** · May 17, 2009

It isn 't wrong, it has an error that is, nevertheless an error less than the error associated with mischaracterization of the shell results; also, the methodology takes advantage of the laws of physics that includes the fact that equal masses situated at different distances from a test mass m the closest mass contributes more force on m another equal mass located farther from m. This is obvious is it not?

Unless you are looking at an approximation, if it has an error associated with it, it's wrong.

Some say the sphere can be considered a point of concentrated mass. Well enough, even though everyone realizes the physical impossibility of this condition being realized in nature. I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere.

Nobody says this is realized in nature. It is a mathematical realization that you get the same answer when you integrate over the volume, and thus is an allowable shortcut. A spherically symmetric mass behaves as if all the mass were concentrated at the center.

For G = 1, m = M = 1 and the shell center located at d = 10. Therefore F = 1/100 = .010. No one can complain if I see what occurs when place .5M at 9 and .5M at 11 units distances from m.

Sure we can. You haven't provided any justification for concentrating the mass at those points. You're just pulling numbers out of some dark, foul region, and to almost nobody's surprise, you get the wrong answer. GIGO.

To accept the stated claim sjustained through the years, from the 17th century to the present, one must accept the supremacy of the mathematical model over physical law regarding "closest equal mass contributes the greater force", and to accept instead, that for spheres only, the mass of the shell can be considered concentrated at the COM;

I hate to break it to you, but you have presented no physical evidence contradicting anything. All you have produced is a mathematical model, and are trying to supplant another mathematical model. However, the huge difference is that one is based on valid math, and the other is based on a supposition that is not true.

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case.

No, because it only works for shells. Do the math.

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont, Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific resposes such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

"You're doing the math wrong" (or worse, "you aren't doing the math at all") and "the shell model only works for spherical systems" are not irrelevant nor are they unscientific.

You thjee should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem, as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as throughly as our scrutinizing processs was applied and from a very unlkely beginmning where imatter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

Yeah, I know the drill all too well. Someone comes along and thinks they have shot down some scientific finding, only it's based on a flawed assumption that they are unwilling to shed. Then they insinuate that scientists have been brainwashed into thinking what they do.

You are actually geistkiesel, right, just re-registered with a slightly shorter name?

Merged post follows:

Consecutive posts merged

Let's say I have two identical masses, separated by some distance, and I am at a point on the line the bisects them. They are each a distance r away from me, at an angle of theta with respect to the line between them

The field contribution from each is Gm/r^2 (if my mass is M, then the force is GmM/r^2), but the sideways (x) contributions cancel, and only the y-component matters.

The y-component force is [math]\frac{GmM sin\theta}{r^2}[/math], making the total force [math]\frac{2GmM sin\theta}{r^2}[/math]

Now, what of I look at the force as if the mass were at their CoM. We will call this m'. The distance away is [math]rsin\theta[/math], which makes the force [math]\frac{Gm'M}{r^2sin^2\theta}[/math]

OMG! m' is not 2m! Why? Because there's no reason to think it will be! It works for spherical symmetry, but not in general.

geistkie · May 17, 2009

Originally Posted by geistkie

It isn 't wrong' date=' it has an error that is, nevertheless an error less than the error associated with mischaracterization of the shell results; also, the methodology takes advantage of the laws of physics that includes the fact that equal masses situated at different distances from a test mass m the closest mass contributes more force on m another equal mass located farther from m. This is obvious is it not?

Unless you are looking at an approximation, if it has an error associated with it, it's wrong. [/quote']

You are confusing the issue. Neither you, Swansont, DH or J.C.McSwell has described with sufficient specificity what was "wrong". You take the phrase out of context then adopt an approach that prevents a rational response.

Steps: 1. Calculate the force from the shell theorem F = GmM/d^2 using convenient unit distances,. masses etc to simplify the calculations..

As the F above is an expression for force, there is no reason to claim that the "shell acts as if all the, mass was concentrated at the shell center". Why don't you just simplify the matter and specify w here in the development of the shell theorem did the concentration of mass occur? The statement says, "the force of the mass on a spherical shell of mass M located a distance d from m."

Step 2. Using the force calculated above segment the shell in equal ½ M amounts of point masses placed symmetrically with respect to the sphere location at the shell COM.. As the F expression represents only one mass entity it cannot be tested for mathematical or physical integrity, hence the need to segment the concentrated mass into two equal concentrated half shells.

Step 3. Calculate the force on m1 of the two shell halves separately.

Step 4. Calculate the total force on m1 from the two ½ shell halves.

Step 5. Calculate the cg of the segmented shell with respect to m1,

Step 6. Compare the two calculated m/1M3forces to the sphere, the segmented to ; which shows the segmented cg

Originally Posted by geistkie

Some say the sphere can be considered a point of concentrated mass. Well enough' date=' even though everyone realizes the physical impossibility of this condition being realized in nature. In any event I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere.

Nobody says this is realized in nature. It is a mathematical realization that you get the same answer when you integrate over the volume, and thus is an allowable shortcut. A spherically symmetric mass behaves as if all the mass were concentrated at the center.

When the spherically symmetric mass is located near m1 this mass sees an asymmetric distribution of mass on the shell. What an observer considers is irrelevant to the matter being considered by us. Surely you can prove that a spherically symmetric mass behaves as claimed. Be as if the mass was concentrated at the center of the shell

Originally Posted by geistkie

For G = 1, m = M = 1 and the shell center located at d = 10. Therefore F = 1/100 = .010. No one can complain if I see what occurs when place .5M at 9 and .5M at 11 units distances from m1

.
Sure we can. You haven't provided any justification for concentrating the mass at those points. You're just pulling numbers out of some dark' date=' foul region, and to almost nobody's surprise, you get the wrong answer. GIGO.[/quote']

Your concentrated mass is not even a pure mathematical abstraction as the force expression F = GmM/d^2 states only the force on m1 from the shell M located at d. However, you take the number as gospel So I simply 'liberate' the concentrated mass and cut it in two and then I place the two ½ M1 symmetrically around the COM of the shell.

Now calculating the forces of the two 1/2 shell segments can calculate the combined force and locate the system center of gravity which is off set from the COM of the sphere in the direction of m1 .

Mooeypoo, you must see that the sphere cannot be considered as a point mass for the reason that the shell has extension, it occupies space and the distribution of mass is asymmetric relative to m1, yet you insist that the forces generated is really other than it actual function – you are treating the shell as not did not obeying the universal gravitational law. You admit the shell isn't concentrated in fact, but it only acts like it is so concentrated which denies on one side of the mouth and agrees on the other side of the mouth.

BTW, the two point masses when arranged in space with extension is seen to behave normaly i.e. that the equal mass closest to m1 contributes more \force on a test particle than a mass farther away.

Posted by geistkie To accept the stated claim sustained through the years, from the 17th century to the present, one must accept the supremacy of the mathematical model over physical reality.

Originally Po

law regarding "closest equal mass contributes the greater force", and to accept instead, that for spheres only, the mass of the shell can be considered concentrated at the COM;

I hate to break it to you' date=' but you have presented no physical evidence contradicting anything. All you have produced is a mathematical model, and are trying to supplant another mathematical model. However, the huge difference is that one is based on valid math, and the other is based on a supposition that is not true.[/quote']

You aren't giving me any news you are sharing an opinion.

Originally Posted by geistkie

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case.

No' date=' because it only works for shells. Do the math.[/quote']

And you have a proof of the claim it only works for shells?

Originally Posted by geistkie

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont' date=' Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific resposes such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

"You're doing the math wrong" (or worse, "you aren't doing the math at all") and "the shell model only works for spherical systems" are not irrelevant nor are they unscientific.

Originally Posted by geistkie

You thjee should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem, as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as throughly as our scrutinizing processs was applied and from a very unlkely beginmning where imatter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

Yeah, I know the drill all too well. Someone comes along and thinks they have shot down some scientific finding, only it's based on a flawed assumption that they are unwilling to shed. Then they insinuate that scientists have been brainwashed into thinking what they do.

You are actually geistkiesel, right, just re-registered with a slightly shorter name?

Yes. Geistkie was input in error tlo which I informed this forum' Oh so now you argue with psychology. We are wasting each others time with this kind of exchange, but you are from the Wizards's Office so you win, right?

Merged post follows:

Let's say I have two identical masses' date=' separated by some distance, and I am at a point on the line the bisects them. They are each a distance r away from me, at an angle of theta with respect to the line between them

The field contribution from each is Gm/r^2 (if my mass is M, then the force is GmM/r^2), but the sideways (x) contributions cancel, and only the y-component matters.

The y-component force is , making the total force

Now, what of I look at the force as if the mass were at their CoM. We will call this m'. The distance away is , which makes the force

OMG! m' is not 2m! Why? Because there's no reason to think it will be! It works for spherical symmetry, but not in general.[/quote']

Nice little expression proving that you haven't been paying attention to my calculations. Just because the calculations contradict your opinion I must be in error.

my calculations are with the two masses in line with the test mass. It isn 't wrong, it has an error that is, nevertheless an error less than the error associated with mischaracterization of the shell results; also, the methodology takes advantage of the laws of physics that includes the fact that equal masses situated at different distances from a test mass m the closest mass contributes more force on m another equal mass located farther from m. This is obvious is it not?

Unless you are looking at an approximation' date=' if it has an error associated with it, it's wrong. [//quote']

But you have no clue of how I make my calculations and why I make them.

Originally Posted by geistkie

Some say the sphere can be considered a point of concentrated mass. Well enough' date=' even though everyone realizes the physical impossibility of this condition being realized in nature. I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere.

Nobody says this is realized in nature. It is a mathematical realization that you get the same answer when you integrate over the volume, and thus is an allowable shortcut. A spherically symmetric mass behaves as if all the mass were concentrated at the center.

Originally Posted by geistkie

For G = 1, m = M = 1 and the shell center located at d = 10. Therefore F = 1/100 = .010. No one can complain if I see what occurs when place .5M at 9 and .5M at 11 units distances from m.

Sure we can. You haven't provided any justification for concentrating the mass at those points. You're just pulling numbers out of some dark, foul region, and to almost nobody's surprise, you get the wrong answer. GIGO.

I use the two (or more_ points) to give extension to the sphere that remains unchecked because it is a single entity and is assumed to act like a concentrated mass at a point .

If I used 100 points 50 in the closest ½ shell the accuracy of my calculation would be effectively perfect..

Originally Posted by geistkie

To accept the stated claim sustained through the years' date=' from the 17th century to the present, one must accept the supremacy of the mathematical model over physical law regarding "closest equal mass contributes the greater force", and to accept instead, that for spheres only, the mass of the shell can be considered concentrated at the COM;

[quote=

I hate to break it to you, but you have presented no physical evidence contradicting anything. All you have produced is a mathematical model, and are trying to supplant another mathematical model. However, the huge difference is that one is based on valid math, and the other is based on a supposition that is not true.[/quote']

What is your supposition that is true? And what is my supposition that is false?

Originally Posted by geistkie

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case.

No' date=' because it only works for shells. Do the math.[//quote'] I have, done the math, you just don't

understand it, and you never will.

Originally Posted by geistkie

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont' date=' Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific responses such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

"You're doing the math wrong" (or worse, "you aren't doing the math at all") and "the shell model only works for spherical systems" are not irrelevant nor are they unscientific.[/quote']

Then show me the proof that they are "scientific"

Originally Posted by geistkie

You three should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem' date=' as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as thoroughly as our scrutinizing process was applied and from a very unlikely beginning where the matter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

Yeah, I know the drill all too well. Someone comes along and thinks they have shot down some scientific finding, only it's based on a flawed assumption that they are unwilling to shed. Then they insinuate that scientists have been brainwashed into thinking what they do.

You are actually geistkiesel, right, just re-registered with a slightly shorter name?

[/quote'] Right Geistkiesel – I pushed the register button before editing the username.

You know I am sure you mean well but you argue in generalities and non-specifics which prevents a rational response from me. I make the simple request that you leave me alone.

Merged post follows:

Let's say I have two identical masses' date=' separated by some distance, and I am at a point on the line the bisects them. They are each a distance r away from me, at an angle of theta with respect to the line between them

The field contribution from each is Gm/r^2 (if my mass is M, then the force is GmM/r^2), but the sideways (x) contributions cancel, and only the y-component matters.

The y-component force is , making the total force

Now, what of I look at the force as if the mass were at their CoM. We will call this m'. The distance away is , which makes the force

OMG! m' is not 2m! Why? Because there's no reason to think it will be! It works for spherical symmetry, but not in general.[/quote']

The y component you should use is ® cos(theta). making the total force

2GmMr(cos(theta))/ r^2.

You are substituting m' for the effective mass located at the midpoint of the two equal masses m. You want to equate the forces to determine m'

2GmM[cos(theta}]/r^2 = Gm /r^2 (cos(theta))^2

m' = 2m[cos^2(theta)]

Your OMG drama is petty –and you have done nothing to this thread except distract from any useful discussion.

I will finish the responses to the unaswered posts at a later time.

[/indent]

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