Constant of gravitation

[elas]

“There is presently no theory (general relativity included) that is able to predict the value of G”

 

.....In particular then, the force that two symmetric spheres exert on each other is the same as if each sphere were replaced by its equivalent particle.

 

Extracted from ‘Classical Mechanics’ by R Douglas Gregory. Cambridge University Press,

 

Previously swansont had objected to the use of the classical electron radius, this caused a change to the quantum mechanical Compton radius where the similarity between the electron radius and the gravitational constant started a chain of thoughts that eventually lead to the following:

Table 1 uses the equation:

 

Mass multiplied by radius = constant (Table 1:col.d)

 

Noting that the constant is close to G/2 and as there is no generally accepted measurement of particle radii, the next stage was to construct a table with particle radii determined by G/2 as shown in Table 2. The new particle radii are compatible with the values for e given by McGregor.

This allows the linear force model to explain G as follows:

 

Within the graviton compaction state each body (electromagnetic compaction) is treated as a single particle. Without G the equation for finding the gravitational force applies the inverse square law to two ‘particles’; with G the equation multiplies two particles by twice the linear vacuum force constant (i.e. G) to produce the linear vacuum force (i.e. gravitational force) for the given distance.

 

Note that the G/2 constant is the same for all force carriers proving that there is only one elementary force.

 

Taken together with other submissions this also proves that there is only one elementary particle and that the only true conservation law that applies throughout infinity, is the conservation of particle numbers. Because the ratio of neutrons to protons can vary between cosmic bodies it follows that the Law of gravitation cannot be applied to a system with more than one elementary particle.

 

[mooeypoo]

“There is presently no theory (general relativity included) that is able to predict the value of G”

 

.....In particular then, the force that two symmetric spheres exert on each other is the same as if each sphere were replaced by its equivalent particle.

 

Extracted from ‘Classical Mechanics’ by R Douglas Gregory. Cambridge University Press,

I'm sorry, elas, I might be missing something here, but who said these quotes? swansont? can you cite, please? I am completely lost on context here.

 

~moo

[elas]

I'm sorry, elas, I might be missing something here, but who said these quotes? swansont? can you cite, please? I am completely lost on context here.

 

~moo

 

Douglas Gregory is professor of Mathematics at the University of Manchester. He is a researcher of international standing in the field of elasticity, and has held visiting positions at New York University, the University of British Columbia, and the University of Washington. He is hghly regarded as a teacher of applied mathematics.

'Classical Mechanics' was published in 2006

Going by the text, it seems they are his own statements.

elas

 

PS: http://www.cambridge.org/features/mathematics/gregory/

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Merged post follows:

Consecutive posts merged

I'm sorry, elas, I might be missing something here, but who said these quotes? swansont? can you cite, please? I am completely lost on context here.

 

~moo

 

Another quote that applies to the proposed model comes from the work of Malcom H McGregor in his book 'The Enigmatic Electron' which descibes in great detail the various radii measurements applied to the electron.

 

With respect to the way we regard the electron, the factor of a million disparity between the radii R(QMC) and R(E) is crucial. If the electron has a radius that is comparable to RQMC, then we can quantitatively reproduce its basic properties in a classical context, which demonstrates that classical physics still apply in this domain.

 

The application of G to the electron radius is a classical physics application. The radius found using G falls between R(QMC) and the 'corrected' R(QMC).

Edited May 25, 2009 by elas
additional info.

[swansont]

I'm sorry, elas, I might be missing something here, but who said these quotes? swansont? can you cite, please? I am completely lost on context here.

 

~moo

 

My objection was the use of the classical electron radius as a physical value, as it is a value derived from equating the electrostatic self-energy with the mass. In reality, the data are consistent with the electron being a point particle.

 

The main thrust seems to be that if you take a number and divide it by a larger number, you get a fraction. elas is selectively quoting MacGregor; if you read the entire passage you'll see that he says that experiment confirms that the electron is indeed a million times smaller (at least) than the classical radius, and that as a result classical physics does not apply.

 

The compton radius depends on the mass of the particle, and AFAIK the QM corrections to it are a constant, so it's not surprising that certain ratios give you a constant. 5/10 is the same as 3/6. OMG!

[elas]

My objection was the use of the classical electron radius as a physical value, as it is a value derived from equating the electrostatic self-energy with the mass. In reality, the data are consistent with the electron being a point particle.

 

The main thrust seems to be that if you take a number and divide it by a larger number, you get a fraction. elas is selectively quoting MacGregor; if you read the entire passage you'll see that he says that experiment confirms that the electron is indeed a million times smaller (at least) than the classical radius, and that as a result classical physics does not apply.

 

The compton radius depends on the mass of the particle, and AFAIK the QM corrections to it are a constant, so it's not surprising that certain ratios give you a constant. 5/10 is the same as 3/6. OMG!

 

elas is selectively quoting MacGregor; if you read the entire passage you'll see that he says that experiment confirms that the electron is indeed a million times smaller (at least) than the classical radius, and that as a result classical physics does not apply.

 

My copy of McGregor’s book states:

The electron was the first elementary particle to be discovered, and all its properties have been exhaustively investigated. Hence the question as to it size is one that seemingly should have been decided long ago. And, indeed, in the minds of most present-day physicists, this question has already been decided: the electron is a point like particle – that is, a particle with no measurable dimensions, at least within the limitations of present day instrumentation. However, a rather compelling case can be made for an opposing viewpoint: namely, that the electron is in fact a large particle which contains an embedded point like charge.

 

Emphases are the authors own. swansont has failed to grasp the purpose of the book as a whole.

 

The Compton radius depends on the mass of the particle, and AFAIK the QM corrections to it are a constant, so it's not surprising that certain ratios give you a constant. 5/10 is the same as 3/6. OMG!

 

That is not what I have shown which is:

mr =G/2

And therefore:

The constant of gravitation (G) is twice the constant of the single particle linear force. To support this proposal I have also shown that the wavelength of a photon contains the linear force of two particles; and explained why. Note that the constant of gravitation and photon wavelengths are found by experiment (not by prediction) therefore my ratio is not a matter of chance but can be related to the results of two entirely different types of experiment (measurement of G and measurement of spectral wavelength). Gregory states that QT cannot predict G; the linear force model shows why - it is because particles are not point like entities they only appear to be so in quantum theory just as the law of gravition considers the Earth to be a point-like body; but who believes that the Earth is a point like body.....(OMG surely not swansont!).

Edited May 26, 2009 by elas

[swansont]

swansont has failed to grasp the purpose of the book as a whole.

 

I haven't read the book, nor did I claim to. I was just pointing out that there's more to it than the snippet you quoted. The existence of a book does not make the information contained in it true, however. You need e-v-i-d-e-n-c-e, i.e. you need to come up with a way of testing the claim in a way it can be falsified.

 

 

The Compton radius depends on the mass of the particle, and AFAIK the QM corrections to it are a constant, so it's not surprising that certain ratios give you a constant. 5/10 is the same as 3/6. OMG!

 

That is not what I have shown which is:

mr =G/2

And therefore:

The constant of gravitation (G) is twice the constant of the single particle linear force. To support this proposal I have also shown that the wavelength of a photon contains the linear force of two particles; and explained why. Note that the constant of gravitation and photon wavelengths are found by experiment (not by prediction) therefore my ratio is not a matter of chance but can be related to the results of two entirely different types of experiment (measurement of G and measurement of spectral wavelength). Gregory states that QT cannot predict G; the linear force model shows why - it is because particles are not point like entities they only appear to be so in quantum theory just as the law of gravition considers the Earth to be a point-like body; but who believes that the Earth is a point like body.....(OMG surely not swansont!).

 

Rqmc/Rc will give you a constant because that's all there is left — the QM correction term (which MacGregor gives as sqrt(3))

 

What are the units for mr? What are the units for G?

[elas]

I haven't read the book, nor did I claim to.

 

elas is selectively quoting MacGregor; if you read the entire passage

 

I was just pointing out that there's more to it than the snippet you quoted..

 

What is good enough for the goose is good enough for the gander! (or perhaps one man's snippet is not another man's passage.)

The existence of a book does not make the information contained in it true, however. You need e-v-i-d-e-n-c-e, i.e. you need to come up with a way of testing the claim in a way it can be falsified.

 

This business of falsifying has always baffled me, being unedified; but I might be able to get some help and come back later on this point.

 

Rqmc/Rc will give you a constant because that's all there is left — the QM correction term (which MacGregor gives as sqrt(3)) .

 

I cannot find where I have entered Rqmc/Rc

 

What are the units for mr? What are the units for G?

 

As given in the table. Although experimenters give G in units (because they are only dealing with one force (i.e. compaction state); the value in reality is an arbitrary value that can be used in all compaction states (i.e. for all forces). Recall that in the law of gravity, the mass of stars and planets are treated as one particle then obviously one unit cannot be applied to both cosmic bodies and particles.

 

Perhaps it will help if the diagram showing the relationship between the particle and anti-particle fields of the linear force model is repeated on this thread; the field is shown in logarithmic form to match Fig.1.1 copied from McGregor’s book.

 

 

It shows that the electric radius (R[e]) is indeed point-like. The electromagnetic radius (classical electron radius (R[o]) extends to the high and low force and anti-force concentric. The Hall radius is on the concentric where the forces change over. And the two Compton radii (and the proposed mr = G/2 radii) are the possible field limits. The sections below A and B are the extract from McGregor.

Edited May 26, 2009 by elas

[swansont]

elas is selectively quoting MacGregor; if you read the entire passage

 

 

 

What is good enough for the goose is good enough for the gander! (or perhaps one man's snippet is not another man's passage.)

 

Oh, stop already. A passage is not the same as the whole book. I provided a freaking link to the page in Google books, where you find the context of the passage in question. MacGregor at least provides a context for the discussion, whereas you do not — you made it sound like the electron being treated classically is a perfectly acceptable thing to do, and upon reading the entire quote one realizes that this is not what MacGregor is claiming. He admits that the QM viewpoint is almost universally accepted, and if one were to only look at your quote, one would not get that impression.

 

This business of falsifying has always baffled me, being unedified; but I might be able to get some help and come back later on this point.

 

It's what separates science from nonscience. Rigor vs ad-hocery.

 

 

I cannot find where I have entered Rqmc/Rc

 

Last column of the first table in the first post.

 

 

As given in the table. Although experimenters give G in units (because they are only dealing with one force (i.e. compaction state); the value in reality is an arbitrary value that can be used in all compaction states (i.e. for all forces). Recall that in the law of gravity, the mass of stars and planets are treated as one particle then obviously one unit cannot be applied to both cosmic bodies and particles.

 

And this is where the wheels come off the wagon.

 

No, mass is not treated as one particle — that's missing the point, as it were. They are treated as point objects, because that's a perfectly valid mathematical treatment. Any spherically symmetric mass of radius R can be treated as if all mass were at the center for r>R (Gauss's law, or Newton's shell theorem). And even if not symmetric, if r>>R, it reduces to that case if you expand the equations. That's all math, and nothing more. There are no other claims made as to the physical nature of the bodies.

 

Your units don't match up. Which means that the relationship between the two depends entirely on the unit systems chosen — if you did this in english units, or even cgs, you would not get the relationship you claim. That indicates that it is accidental rather than physical. Numerology.

[elas]

— you made it sound like the electron being treated classically is a perfectly acceptable thing to do, and upon reading the entire quote one realizes that this is not what MacGregor is claiming. He admits that the QM viewpoint is almost universally accepted, and if one were to only look at your quote, one would not get that impression.

 

That is not the way McGregor puts it; he main thrust is that there is a case for an alternative, despite the popularity of the Standard model.

 

This business of falsifying has always baffled me, being unedified; but I might be able to get some help and come back later on this point

It's what separates science from nonscience. Rigor vs ad-hocery.

 

So does:

Explaining the cause of G

Explaining why light waves have the same energy as two particles.

Explaining the anomalous magnetic moment on the muon.

Explaining why all observed elementary particles have the same charge.

Explaining the structure of an atom of each element in a manner that allo0ws the elements to be explained without any of the exceptions common to the current model. (Table of elements).

It’s what separates science from mathematics.

 

I cannot find where I have entered Rqmc/Rc

Last column of the first table in the first post.

 

This column can be omitted without doing any damage to the proposal.

 

And this is where the wheels come off the wagon.

No, mass is not treated as one particle —

 

Since any element of mass isw attracted by a symmetric spere as if the sphere were a particle, it follows that the force a symmetric sphere exerts on any other mass distribution can be calculated by replacing the sphere by a particle of equal mass located at its centre. In particular then, the force that two symmetric spheres exert upon each other is the same as if each sphere were replaced by its equivalent particle. Thus as far as the forces of gravitational attraction are concerned, symmetric spheres behave exactly as if they were particles

Extracted from ‘Classical Mechanics’ by R Douglas Gregory. (Author's emphases in bold.)

 

This means that going from the densest compaction to the least dense compaction one step at a time; each compaction is a 'particle' within a weaker 'particle' compaction state. Because of this, the linear force constant (G/2) is common to all compactions and the force between any two particles (G) is common to all compactions. All that is needed is a measurement of the total number of elementary particles in each compaction and this is achieve by measuring the mass. Hence in all compactions mr = G/2 (the linear force constant) and the force between any two particles (regardless of compactions) is G. Note that what is currently referred to as gravity covers a large number of different compaction states. The gravitational forces around galaxies, stars and planets are different compaction states that share a common constant (G).

 

Your units don't match up. Which means that the relationship between the two depends entirely on the unit systems chosen — if you did this in english units, or even cgs, you would not get the relationship you claim. That indicates that it is accidental rather than physical. Numerology.

 

All one would get is a different arbitrary value for G. Science not mathematical prediction.

Edited May 27, 2009 by swansont
fix tag

[swansont]

— you made it sound like the electron being treated classically is a perfectly acceptable thing to do, and upon reading the entire quote one realizes that this is not what MacGregor is claiming. He admits that the QM viewpoint is almost universally accepted, and if one were to only look at your quote, one would not get that impression.

 

That is not the way McGregor puts it; he main thrust is that there is a case for an alternative, despite the popularity of the Standard model.

 

The way MacGregor put it is easily discerned by reading the linked passage.

 

And this is where the wheels come off the wagon.

No, mass is not treated as one particle —

 

Since any element of mass isw attracted by a symmetric spere as if the sphere were a particle, it follows that the force a symmetric sphere exerts on any other mass distribution can be calculated by replacing the sphere by a particle of equal mass located at its centre. In particular then, the force that two symmetric spheres exert upon each other is the same as if each sphere were replaced by its equivalent particle. Thus as far as the forces of gravitational attraction are concerned, symmetric spheres behave exactly as if they were particles

Extracted from ‘Classical Mechanics’ by R Douglas Gregory. (Author's emphases in bold.)

 

Wrong emphasis

"the force that two symmetric spheres exert upon each other is the same as if each sphere were replaced by its equivalent particle." There is no claim that it is a single particle.

 

Your units don't match up. Which means that the relationship between the two depends entirely on the unit systems chosen — if you did this in english units, or even cgs, you would not get the relationship you claim. That indicates that it is accidental rather than physical. Numerology.

 

All one would get is a different arbitrary value for G. Science not mathematical prediction.

 

You would get a different relationship for the ratio. Truly meaningful relationships between constants that describe nature, like the fine structure constant, are unitless.

 

Mathematical prediction is one thing that separates science from nonscience. There is a current thread on this very topic.

[ajb]

Is it possible that quantum gravity could predict Newton's constant? (Which is probably running).

 

Maybe not. For example QED and QCD do not predict the renormalised parameters in the theories. Then, quantum gravity is also unlikely to provide an answer here to Newton's constant.

[elas]

The way MacGregor put it is easily discerned by reading the linked passage.

 

 

 

Wrong emphasis

"the force that two symmetric spheres exert upon each other is the same as if each sphere were replaced by its equivalent particle." There is no claim that it is a single particle.

 

 

 

You would get a different relationship for the ratio. Truly meaningful relationships between constants that describe nature, like the fine structure constant, are unitless.

 

Mathematical prediction is one thing that separates science from nonscience. There is a current thread on this very topic.

 

And this is where the wheels come off the wagon.

No, mass is not treated as one particle —

 

Since any element of mass is attracted by a symmetric sphere as if the sphere were a particle, it follows that the force a symmetric sphere exerts on any other mass distribution can be calculated by replacing the sphere by a particle of equal mass located at its centre. In particular then, the force that two symmetric spheres exert upon each other is the same as if each sphere were replaced by its equivalent particle. Thus as far as the forces of gravitational attraction are concerned, symmetric spheres behave exactly as if they were particles

Extracted from ‘Classical Mechanics’ by R Douglas Gregory. (Author's emphases in bold.)

 

Wrong emphasis

"the force that two symmetric spheres exert upon each other is the same as if each sphere were replaced by its equivalent particle." There is no claim that it is a single particle

 

REPLY

I made clear (in brackets) that the emphases are Gregory’s not mine. Neither am I claiming that they are single particles. What the G constant proves is that all spheres can be treated as single particles complete with point-like mass – my case is that this can only be so if all particles have the same content.

If the contents of a quark differed from the contents of an electron then the gravitational force between neutron stars and other bodies would require a different G constant than the G constant between a hydrogen star and other bodies.

 

Alternatively it can be observed that stars made from different compositions of the elements have different ratios of neutrons to proton/electron pairs, therefore there would be no G constant if the particles within the neutrons had a different content (i.e. density:volume ratio) from the particles outside the neutrons.

It is the existence of a G constant found by experiment that proves the linear force model to be correct; that is why the linear force model is the only model that predicts the G force constant via the G/2 constant of single particle linear force. Because G/2 is the single particle constant; G/2 is the fundamental constant.

mr = G/2

Using the same units for mass and radius in all experiments will produce a G constant because all the particles that make up a single mass are different states of a single elementary particle and the contents of each single elementary particle is conserved throughout all changes of state.

[swansont]

I made clear (in brackets) that the emphases are Gregory’s not mine. Neither am I claiming that they are single particles. What the G constant proves is that all spheres can be treated as single particles complete with point-like mass – my case is that this can only be so if all particles have the same content.

If the contents of a quark differed from the contents of an electron then the gravitational force between neutron stars and other bodies would require a different G constant than the G constant between a hydrogen star and other bodies.

 

Here, finally, is a prediction, and something that is potentially testable. Mainstream physics predicts that gravity is solely dependent upon energy density (classically, mass). Even our own sun should change behavior as protons fuse into helium, which means the creation of neutrons and the ejection of neutrinos. This should have some effect in when different fusion cycles occur, since that depends on gravity. Now all that there is to do is coming up with some evidence that supports this.

[elas]

Here, finally, is a prediction, and something that is potentially testable. Mainstream physics predicts that gravity is solely dependent upon energy density (classically, mass). Even our own sun should change behavior as protons fuse into helium, which means the creation of neutrons and the ejection of neutrinos. This should have some effect in when different fusion cycles occur, since that depends on gravity. Now all that there is to do is coming up with some evidence that supports this.

 

I have repeatedly said that constructive criticism is the best aid to progress and over the past 20 years I have pestered the forum experts for just such criticism. Starting with integral and the nutcases on the theory development forum, I have met with varying responses, the worst case being Tom Matteson who for some years has refused to read my submissions. My fellow nutcases have long ago dropped out of view, most are silent but one or two can still be found on less critical forums.

So you will appreciate that my thanks to you for staying this far is both genuine and heartfelt. A testable theory is probably the best I could hope to achieve being unsure of how to proceed; not that I am going to stop but clearly there is a whole new area to study and it may be some time before I have anything further to contribute. So this is an appropriate time to say a sincere thank-you for your help this far,

regards

elas

[elas]

Here, finally, is a prediction, and something that is potentially testable. Mainstream physics predicts that gravity is solely dependent upon energy density (classically, mass). Even our own sun should change behavior as protons fuse into helium, which means the creation of neutrons and the ejection of neutrinos. This should have some effect in when different fusion cycles occur, since that depends on gravity. Now all that there is to do is coming up with some evidence that supports this.

 

I wonder if cosmologist could make use of the following observation:

 

The equation for the Schwarzschild radius of black holes can be simplified to: r = m/constant.

The proposed formula for particles is mr = constant (r = constant/m).

I note that if both equations are applied to elementary particles then:

 

( r = m/constant) x (r = constant/m) = 1

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Consecutive posts merged

Furthermore, by referring to McGregor's work on electron radii, it can be sugested that the Black Hole equation applied to particles produces the electric radius.

[Sha31]

 

That is not what I have shown which is:

mr =G/2

And therefore:

The constant of gravitation (G) is twice the constant of the single particle linear force. To support this proposal I have also shown that the wavelength of a photon contains the linear force of two particles; and explained why.

 

In my theory photon is electric dipole. Electron-positron pair, two opposite electric fields trying to stick together, but because of magnetic fields, instead of orbiting, they end up spiraling each other describing double-helix, also known as transverse EM wave. I discovered this by chance when I was simulating electron trajectories in magnetic field, I was curious to see how electron and positron interact and I saw they produce spiraling waves where opposite electric and magnetic fields oscillate around each other, just like in the real-world.

 

 

Can you tell us more about how did you come up with your conclusion?

[elas]

This is an outline of what I am trying to achieve. At present I am trying to do the falsifications mentioned by the administrators and in addition, bring the various submissions together in one article. This quick reply is just that, not a complete work in itself.

 

Go to http://en.wikipedia.org/wiki/Light and scroll down to electromagnetic theory where the diagram describes a light wave as extending from wave peak to wave peak; this I think is an incorrect assumption.

 

I view the photon as shown fig.aa36 below. This shows the electron and positron in there bosonic state. That is to say that in a partial vacuum field (particle) the vacuum field has collapsed into a Zero Point (represented by a black dot, but in reality dimensionless) leaving the positron and electron matter in its wave form. In this form it passes through other particles distorting the vacuum fields of other particles as they pass through (red wave). We observe the red wave as colour and measure the striking force (energy) of the matter wave; hence the energy is always twice the elementary particle energy, but the colour varies according to the element occupied at any give instant. When passing through gravitons (particles weaker than positrons and electrons) photons retain the wave length of the last element with which they interacted.

 

 

In a submission on the work of Malcolm H McGregor (The Enigmatic Electron) I converted the partial vacuum field of an elementary particle into its logarithmic form to show how the various radii given for the electron compare to measurements along the radius to different points of the partial vacuum field (McGregor’s diagram is given in logarithmic form).

 

 

Taken together with my Table of Elements what is being developed is a proposal for an explanation of the structure of particles and atoms using only vacuum and matter. Electricity, electromagnetism, gravity, strong force and weak force apply to the actions of different states of a partial vacuum field (i.e. single elementary particle).

[Sha31]

Ok. But, I can model this by simply simulating electric and magnetic field interaction according to classical electromagnetic equations, so why do you think zero-point energy/forces are necessary, how do you conclude they are more "basic" than magnetic and electric forces?

[elas]

Look again at the photon diagram and consider possible decay states:

1) no change

2) electron and positive neutrino

3) positron and negative neutrino

4) electron and positron

5) no other change is possible

the current model suported by experimement, gives us (1) to (4) it does not give us (5)

 

One of the tables I have been told to 'falsify' is the table that shows that mr = G/2 that is to say that gravity (vacuum) is the fundamental force, electromagnetism etc are names we give to actions arising from different compaction states of both matter and vacuum field (gravity). The actions of all force carriers is determined by the difference in density of partial vacuum fields, not just differences in the density of matter.

 

new readers should also read: http://www.scienceforums.net/forum/showthread.php?p=533307#post533307

Edited December 18, 2009 by elas

[swansont]

If the contents of a quark differed from the contents of an electron then the gravitational force between neutron stars and other bodies would require a different G constant than the G constant between a hydrogen star and other bodies.

 

In light of some recent refresher reading I now see that this statement is in flat contradiction to Einstein's equivalence principle. Local position invariance dictates that gravity is the same, independent of the constituents of the mass. Comparisons of atomic clocks in the varying gravitational potential of the sun (due to our elliptical orbit) using different elements confirms this.

http://prd.aps.org/abstract/PRD/v65/i8/e081101

[elas]

The Cs-H maser comparisons show no correlation with variations in the solar potential, within an uncertainty that is about 30 times smaller than the previous most sensitive comparisons.

http://meetings.aps.org/link/BAPS.2007.MAR.N32.2

 

The experimenters and their equipment are participants in the experiment in that all the force fields associated with planet Earth are enclosed within the solar G field; the enclosed planet Earth force fields expand and contract remaining in balance with the solar field: to observe changes in frequency the experimenters need to be outside the experiment.

[swansont]

The Cs-H maser comparisons show no correlation with variations in the solar potential, within an uncertainty that is about 30 times smaller than the previous most sensitive comparisons.

http://meetings.aps.org/link/BAPS.2007.MAR.N32.2

 

Um, this result confirms LPI, which predicts a null effect. You, on the other hand, predicted a difference based on composition.

 

The experimenters and their equipment are participants in the experiment in that all the force fields associated with planet Earth are enclosed within the solar G field; the enclosed planet Earth force fields expand and contract remaining in balance with the solar field: to observe changes in frequency the experimenters need to be outside the experiment.

 

What?

 

We can measure time dilation effects in gravitational potentials. We don't need to rocket off to Mars to do so.

[elas]

I cannot understand what time dilation has to do with the original point so I will try and explain my case in a different way as follows:

 

Local position invariance dictates that gravity is the same, independent of the constituents of the mass.

 

The challenge is to construct two bodies with the same gravitational force, mass, and volume, but with different number of elementary particles. My case rest on the claim that this is not possible.

 

Perhaps QT can do this because it regards particles as point-like, but I have made a case that shows that particles have volume and justified this by showing how the density of particles in a given volume determines the force carried by the particle,( and in the case of atoms, the nature of the elements). This means that Newton was correct to suggest that the universe is “corpuscular in nature” and Einstein was right to insist on a classical explanation of the structure of nature.

 

QT is brilliant at predicting actions, but it fails to explain cause. QT lacks the simplicity predicted by both Newton and Einstein because the simplicity lies in the structure itself and not in the actions of that structure. Structural simplicity is founded on showing that there is only one elementary particle and only one elementary force coupled with some (incomplete) evidence that suggests that nature repeats the same structural pattern in all compactions states.

[swansont]

The challenge is to construct two bodies with the same gravitational force, mass, and volume, but with different number of elementary particles. My case rest on the claim that this is not possible.

 

That's not the only consequence. For a spherically symmetric object, the volume doesn't matter — that's a ramification of a 1/r^2 force (i.e. it's purely a mathematical phenomenon). Unless, as you claim, that G is different for different particles, which is at odds with LPI.

 

Now, you do have the option to show that the experiments are not sensitive enough to measure the effect you claim.

 

Perhaps QT can do this because it regards particles as point-like, but I have made a case that shows that particles have volume and justified this by showing how the density of particles in a given volume determines the force carried by the particle,( and in the case of atoms, the nature of the elements). This means that Newton was correct to suggest that the universe is “corpuscular in nature” and Einstein was right to insist on a classical explanation of the structure of nature.

 

QT is brilliant at predicting actions, but it fails to explain cause. QT lacks the simplicity predicted by both Newton and Einstein because the simplicity lies in the structure itself and not in the actions of that structure. Structural simplicity is founded on showing that there is only one elementary particle and only one elementary force coupled with some (incomplete) evidence that suggests that nature repeats the same structural pattern in all compactions states.

 

You do realize that gravitation/relativity is not quantum theory, right? And we've already been through the fact that theories ultimately do not explain cause — that's philosophy and metaphysics. It's not a failing of science.

[elas]

That's not the only consequence. For a spherically symmetric object, the volume doesn't matter — that's a ramification of a 1/r^2 force (i.e. it's purely a mathematical phenomenon). Unless, as you claim, that G is different for different particles, which is at odds with LPI.

 

I do not claim that G is different for different particles, I pointed out the possibility that mr=G/2 by which is meant to show that electromagnetic force and strong force are compactions of G force.

 

I do realize that gravitation and relativity theories are classical theories, but I am also aware of attempts to produce a Quantum Gravity theory. By showing that fractions produced by so-called ‘two dimensional’ electromagnetic compression experiments are also produced naturally in three dimensional atomic structure, I am showing that QT (i.e. that part of QT devolved from FQHE) and classical theory (derived from atomic structure) produce the same fractional result, in fact classical theory does it to a greater degree of accuracy. By using compression fractions to show the structure of the elements I am using fractional quantities that we have shown to be both classical and quantum in origin; the intention being to make a start on the underlying structure that must not only describe quantum theory, but must also describe classical theory. Cause is essential to understanding; the lack of cause is not a failing of science; it is a failing of quantum theory.

 

Writing in "Quantum Physics, Illusion or reality" Alastair I.M. RAE of the Department of Physics at the University of Birmingham states that Quantum physics is about "measurement and statistical prediction". It does not describe the underlying structure that is the cause of quantum theory.