by Miles Mathis
Both Coulombís experiment and Coulombís equation have suffered various kinds of attack over the years, but my analysis here will be of a different sort than any other you are likely to have seen. I would agree that Coulombís experiment has been oversold historically and presently, but that question, though perhaps interesting in its own right as a question of the politics of science, is not the central question. The central question is one of fact: is the equation correct in math and theory. I will show that it is not.
One crucial thing that is always ignored is the fact that Coulombís spheres are spheres. The experiment and equation could not work if they were not, but we are never told this. It is always a matter of charge, never of shape. But the reason we get an inverse square law is due to the spherical shape of Coulombís objects. Because they are spheres, they must emit a spherical field, and a spherical field must obey the inverse square law.
Current theory ignores this in a most flagrant manner, and they do so because they must ignore that the charge field is an emitted field. The field stops being mysterious once we realize that the density of spherical emission must fall off with the inverse square, but current theory cannot follow this reasoning. Physicists would rather cloak the field in mystery than admit that it acts precisely like a field of emission. If it is an emission field, they have to consider its mechanics. If they consider its mechanics, they have to quit talking about virtual photons or messenger photons and start talking about real photons with real energy. And if they do that, they have to ask how quanta emit quanta without dissolving. It becomes a fundamental question of conservation of energy.
The reason current theory has to block this line of questioning right out of the gates is that QED and QCD have no answers for it. No only that, but the math they have offered us denies that the field is an emission field. To put it baldly, if the E/M field were shown to be a real field with real energy and mass equivalence, all of QED and QCD would fall into a heap. Faraday assumed the ďfieldĒ was non-physical and non-mechanical in the 1830ís and QED still assumes this.
This is why the E/M field has been cloaked in strange units, constants, permittivities, and permeabilities, as I show in another paper. At the level of QED, the math does the cloaking without the need of more obvious blankets, but once we get down to the level of classical electrical theory (the kind taught in high school and undergraduate levels) there is a great need for these more ham-handed misdirections. Physicists canít really import Hamiltonians and tensors and gauge fields and complex operators at this level, so they load down the explanations with jargon instead. Instead of drowning you in mathematical jargon, they drown you in verbal jargon: complex terminology and undefined terms.
But the fact is, the E/M field, like the gravitational field, obeys the inverse square law because the objects in the field are spheres. Coulombís objects were spheres, and electrons and protons and photons are also spheres. The field emitted by the electron is spherical, and the field emitted by Coulombís objects is spherical. This determines all the charges and forces.
The reason Coulombís equation is so simple is because the spherical nature of the field allows us to scale up and down using only a radius or a diameter, as I will show below. Coulomb was measuring a force between two small balls. The force was a force measured at the macro level, between macro (visible) objects. But this force is actually a summed force caused by a field of forces. It is caused by a bombarding field. Coulombís balls are made up of billions of smaller balls (atoms), and the force at the atomic level is a bombardment of these smaller balls by even smaller balls (photons). To find a force on one atom, we need to know the density of the field of photons relative to the size of the atom. How many photons are impacting it each second? To sum up from the quantum field to the macro field requires we sum all these collisions. Since charge in this problem is defined by field density, we sum the field. But there is a problem. Density is a relative term. It is relative to volume, and volume is relative to radius. So a charge measured at the macro level is not the same as a charge measured at the quantum level.
This is where Coulombís constant enters the equation. We now define charge in terms of current, one Coulomb being one Ampere second. This is a crazy way to define charge, since current and charge are two different things, but the operational crux of this definition is that current is a macro-phenomenon. In measuring Amperes, we are NOT measuring quanta: we are measuring changes in macro-objects like batteries and voltage meters and so on.
Therefore, to get a force from charges defined in this way, we have to multiply our charges by a scaling constant. The scaling constant takes the macro-density (the density of the charge field relative to macro-objects) and takes it down to the quantum density (the density of the charge field relative to protons and electrons and atoms). This gives us a correct force, one we can then sum.
If our macro objects are spheres, we can achieve this sum simply by the scaling up of the charge. In this way, you see that the force IS the charge. Force and charge arenít fundamentally different parameters: force is just charge in a given space. All we have to do beyond the scaling is let the field diminish with the inverse square law. That is to say, the field is emitted, and so it must lose density as it expands into space.
In sum, defining charge in terms of current doesnít really give us charge. But it does tell us a field density of the charge, at the macro-level. We canít get a force from this macro-field density, but we can get it from the quantum density. The density of the charge field relative to the size of the atoms will be much greater, and so will the force felt. This is the force we want, since it is the real force. So we simply multiply by a number that will give us the quantum density from the macro-density. This is what Coulombís constant does. And it does it by comparing radiuses only. We need only ask the radius of the atom relative to the radius of our macro-objects: with spherical objects the radius will give us the volume and therefore the density difference.
So, Coulombís constant tells us directly that the charge field at the atomic level is 9 x 109 relatively more dense than the charge field at the macro-level. And this tells us that the atom is 9 x 109 smaller than the macro-world.
But there is still some confusion here. How big is the macro-world, exactly? According to my analysis, the macro-world is defined in Coulombís equation by his balls, which were about 6 mm in diameter.1 But Coulomb was not able to calculate Coulombís constant from his experiment. He could not measure charge and force, so he had no way to find k. He only proved that the repulsion was inversely proportional to the distance of separation. The constant was calculated much later, once we knew the charge on the electron and the Bohr radius and so on. And the diameter of Coulombís balls did not come into that calculation.
You may know that the constant was calculated using the charge on the electron, but we are never told that the constant is related to the Bohr radius. In my paper on the Bohr magneton, I showed that Coulombís constant depends on the Bohr radius. As a number, it is the inverse of the Bohr diameter. But this means that Coulombís constant is scaling up to the size of 1 meter, not 6 mm. That is a difference of 170 times.2
You will say, ďSo what? Normally Coulombís constant has nothing to do with sizes or lengths, but even if we admit that this mathematical side street exists, how is it possibly pertinent? Does the E/M field act differently at 6 mm than at 1 meter?Ē
Yes, it does. Not only have I shown that the field is different at different sizes, but more importantly, the density must be different. We are at a larger radius and a larger volume, which must create a smaller field density.
I will gloss the first assertion first, though it is covered more fully elsewhere. I have shown that both Coulombís equation and Newtonís equation are unified field equations. Both of them contain the E/M and gravitational fields. Newtonís equation hides the E/M field in the mass variables, and Coulombís equation hides the gravitational field in the charge variables. Masses have both volume and density, as do charges, and in each equation the two fields can be separated along with volume and density. Once we do this, gravity varies only with radius, and E/M varies with density. But this must mean that they do not stay the same size, relative to each other. They scale differently. This is why gravity is larger at large scales and E/M is larger at small scales. We already knew this about scaling, we just didnít have a way to show it in a unified field.
Since Coulombís equation is a unified field equation, it matters very much what size we are dealing with. Even a difference of 170x is significant. The relative sizes of the two constituent fields are different, and so the unified field will not act the same. There will be no variance from the inverse square law, but due to the form of the equation, there will appear to be a variance from it. To be specific, the denominator of the equation will follow the inverse square law perfectly, as it must, but the numerator will betray the equation. This is why you get different curves in different experiments: the objects in the experiments are not the same sizes. The curves roughly follow the inverse square law, but a 1 mm charged object will not act like a 1 m charged object or a 100 meter charged object. The reason is that these objects donít act the same as unified field objects.
Beyond that, a 6 mm object is not equivalent to a 1 meter object in another way, a way that strongly impacts the operation of this constant. For Coulombís constant to work correctly, it must transform between the right levels. It must take us from the atomic level to the experimental level. In Coulombís experiment, this would be between the Bohr radius and 6 mm. Using the current Bohr radius [5.29 x 10-11m], that would be 8 levels (taking each factor of ten as a level). But the current constant is the inverse of the Bohr diameter, which, taking us up to 1 meter, takes us through 11 levels. The constant is too big for Coulombís experiment. Either the constant is too big or the Bohr radius is too small, or both.
As I have shown, the Bohr radius is too small, by about 170x. Amazingly, I discovered this in an analysis that had nothing to do with Coulomb. I was re-doing the math for the Bohr magneton, by correcting the angular momentum equation, and found the number 170 that way. Only afterwards did I realize that this connected to the Coulomb equation.
The Bohr radius is 170 larger than we thought, and 1 meter is 170 times larger than Coulombís balls. The standard model pushed the balls up three levels, when they should have pushed the Bohr radius up three levels.
What this means is that Coulombís constant is not a constant. It is a variable. You have to consider what force you are talking about, what force you are seeking, and how it is measured. If you are applying Coulombís equation to quanta, as they effect each other through the charge field, the constant is one. You donít need a density transform, since you are already at the quantum level. If you are applying the equation to quanta as they affect macro-objects (like detectors), then you have to consider the size of your detector. If you are applying the equation to macro-objects, you must consider the sizes of your objects. If you are seeking Coulombís constant relative to 1 meter, then the constant is 1.11 x 108. Current math is wrong. Using the correct math, we find that the Bohr radius is 9 x 10-9 m, which looks an awful lot like Coulombís constant, 9 x 109. But that does not increase our density by 9 x 109. It increases it by 1.11 x 108 [1/(9 x 10-9) = 1.11 x 108]. By the same token, if we apply Coulombís equation to Coulombís balls, we have a density difference of only 6.5 x 105. If we have an equivalent charge on the Earth, then Coulombís constant increases to 7.08 x 1014. As you see, Coulombís constant is not a constant.
The historical fact is that Coulombís equation is hardly ever used in real life, and when it is used it is misused. It isnít used because there is no place in the equation to represent the fact that charge can vary. Charge is not a constant at any size. Coulomb transferred a small static charge to his pith balls with a pinhead, but there is no indication that he could not have transferred more or less charge. A sphere 6 mm in diameter could hold any range of charge, depending on its elemental makeup, its density, and the charge applied. You will say that Coulombís equation represents the charge, but it is not just a matter of the charge size or the separation between charges, it is also the charge density. In other words, the charge size relative to the size of the object charged. If you vary the ďconstantĒ like I am doing, you can represent this density, but not otherwise. With a constant that is constant, the equation cannot work, except by accident (on an 81 meter object).
If we were dealing with magnetism, for instance, we would have to differentiate between H and M, where H is the magnetic field strength and M is the magnetization per unit volume. But Coulombís equation ignores this difference. Charge is said to create both electricity and magnetism, but in the historical and fundamental equation of charge, we have no possible expression of M or its charge equivalent. We know that electricity and magnetism both have density considerations like this, and that both are caused by charge, but we do not express density in the charge equation.
To see how wrong the Coulomb equation is, let us look at an example problem in a textbook. In this problem, the author calculates the force between an orbiting electron and a hydrogen nucleus.
F = (9 x 109 Nm2/C2)(-1.6 x 10-19 C)(1.6 x 10-19 C)/(5.3 x 10-11 m)2
= -8.2 x 10-8 N
Just as a comparison, this is about the same force between Cavendishís balls in his experiment. He had 350 pounds of mass separated by a few centimeters, and found a force of about 7 x 10-8 N. This textbook expects us to believe that there is an equivalent force between a single proton and electron. If, as I have shown, k is a scaling variable, what is it doing here in this equation? We have nothing to scale. If it is not a scaling variable or constant, what is it? Why is it here? Why does no one ever ask this question?
[To see the correct derivation of the force between the electron and proton, you can go here, but get ready to be shocked.]
If the constant k is just a cgs kludge, as has been claimed by many physicists and engineers, why does it happen to be equal to the inverse of the Bohr diameter? They might answer that it is because you can invert it and put it in the denominator, so that we have
F = q2/r3
They will say that the charge has to dissipate into three dimensions, not two, so the cubed radius gives us a volume. But that only appears to work in this one case where the radius is the Bohr radius. The equation is supposed to work for all charges, not just the charge on the hydrogen nucleus. We use the same constant when we calculate larger charges, as in this example from the same textbook:
F = (9 x 109 Nm2/C2)(-4 x 10-6 C)(-3 x 10-6 C)/(.5 m)2 = .43 N
The constant k is not a third dimension there, is it? We canít combine it with the radius, can we?
While the cgs system is preferable in many ways, it does not clear up the mystery here, not in the form of a kludge or in any other way. The cgs system uses the statcoulomb instead of the Coulomb, and it does indeed get rid of the constant, but it does so by defining the charge in terms of the force, which is of course circular.
ďIf two stationary objects each carry a charge of 1 statC and are 1 cm apart, they will electrically repel each other with a force of 1 dyne.Ē So instead of defining charge in terms of current, we define it in terms of force. This means that the mystery is not solved, it is simply swallowed. With electrostatic force, the Coulomb is preferable in one way: it makes the problem a bit more transparent. The Coulomb leaves the door open a bit. I would have been very unlikely to discover the things I have if I had been limited to the cgs system.
So it is not only the newer SI system that blocks questions about the E/M field, it is also the old cgs system. To show once again how far the current standard model will go to block questions, notice this from Wikipedia, under the heading ďstatcoulombĒ:
Performing dimensional analysis on Coulomb's law, the dimension of electrical charge in cgs must be [mass]1/2 [length]3/2 [time]-1 .That is misdirection, since that can be reduced. Maxwell showed that we can easily do a dimensional analysis on mass, using only Newtonís equations. Mass is [length]3[time]-2 . If we substitute that into Wikiís analysis and reduce, we find that electrical charge is equal to mass. Why is Wiki, the mouthpiece of the status quo, so keen to avoid telling you that? Why is Wiki like all standard texts, online and off, in misdirecting you from the truth?
And another question never answered by textbooks: why is the difference between the Coulomb and the statcoulomb ten times the speed of light or 10c? How did c get into an equation of charge and length? The standard model wants you to think it has something to do with vacuum permittivity or one of the other meaningless constants, but it actually comes from the fact that the charge field is material. You can calculate a force from a density only if you know the velocity of the field. In this case, that velocity is c, and that is how it enters this equation. In my use of the constant k above as a scaling constant, we could ignore c, since it was the same at all densities and volumes. But when you get down to the groundwork mechanics, you find that you need the velocity in order to sum the force. To get the force, you have to know how many particles are hitting your object over some time interval. The density at a given volume wonít tell you that. But if you have a velocity and a density, you can calculate the force, since you then have a field strength. You have both the area of impact and the time of impact, you see.
The standard model must hide this, since bringing a velocity into the equation should destroy the virtual nature of the field. If the field has a velocity, it must be a material field. Everything with velocity has energy, and everything with energy has either velocity or mass, or both. So you cannot logically propose a virtual field that has no energy and no mass equivalence. If your field has velocity, it has energy, and if it has energy, it has mass equivalence. If it has mass equivalence, it is not virtual. If it is not virtual, it must take part in all interactions mechanically, and obey all the rules of mechanics. If the charge field becomes real like this, QED and QCD immediately fall.
Conclusion: even if one disregards all my new theory and math, the very fact that Coulombís constant can be shown to be the inverse of the Bohr diameter means there must be a mathematical and mechanical connection between radius and charge. That is to say, a direct, causal connection between length and charge. And not just a connection of the charge separation, as we already knew from the inverse square law. No, the Bohr radius has nothing to do with the charge separation or the inverse square law. The Bohr radius is the atomic radius (of hydrogen), and the only way it could be mechanically connected to Coulombís constant is if the charge field is an emitted field, with a real density and therefore a real mass.
If the charge field had no mass or energy, we could not connect it to any length at all, not the Bohr radius or any other radius. Any length necessarily implies an extension, and things without mass and energy cannot have extension, or be related to things that do. We are not taught this connection of the Bohr radius and Coulombís constant because it is known to be a tip-off. If physicists were taught that the Bohr diameter is the inverse of Coulombís constant, all the questions I begged in my introduction would be begged for everyone. Everyone would immediately begin to see that the E/M field must be mechanical and physical: that it must occupy space and exhibit real density. If they did this, QED would fall overnight and be replaced by a theory with a real field particle.
2The actual size was 5.88 mm, as I found by deduction.