and the Proton Mass is just its Radius Squared

Please note that this paper is a simplification by me of a paper or papers written and copyrighted by Miles Mathis on his site. I have replaced "I" and "my" with "MM" to show that he is talking. All links within the papers, not yet simplified, are linked directly to the Miles Mathis site and will appear in another tab. (It will be clear which of these are Miles Mathis originals because they will be still contain "I" and "my".) The original papers on his site are the ultimate and correct source. All contributions to his papers and ordering of his books should be made on his site. (This paper incorporates Miles Mathis' proton paper and proton3 paper). |

Newton Coulomb Rutherford

By importing my Unified Field from Newton's law is a Unified Field of Gravity and E/M into Rutherford's scattering equations Miles Mathis is able to show that the diameter of the nucleus has been underestimated. Specifically,MM will show that the equation for the impact parameter is compromised by importing a mass into a charge field without transforming it in any way. Using simple math MM was able to develop this transform, proving that the historical method has been both incomplete and (in many ways) false. This recalibration of the scattering equations changes the size estimate of the nucleus, as well as all other atomic and quantum sizes. It also gives us an elegant method for showing where the so-called fine structure constant comes from, solving what Feynman dubbed one the most important problems of the 20th century.

The atomic diameter is currently thought to be on the order of one angstrom or 10^{-10 }meters. The proton radius is thought to be 100,000 times smaller. Both numbers were arrived at by scattering experiments. The first important experiments in scattering of this sort were done by Rutherford and his students Geiger and Marsden in 1909. Rutherford developed an equation at that time to estimate the size of the gold nucleus (which nucleus was doing the scattering in his experiments), and his assumptions have never been questioned. Both his equations and his size estimates remain in use today. They have been updated but have never been seriously recalibrated, much less reworked. Electron scattering has improved the accuracy of Rutherford's size estimates—since the *beta* particle is smaller than the *alpha* particle Rutherford and his students were using—but the theory of scattering has not changed in a century.

Rutherford's equation is called the Rutherford formula:

N(θ) = NnLZ_{i}^{2}k^{2}e^{4}/4r^{2}K^{2}sin^{4}(θ/2) |

This equation is arrived at by treating the scattering as due to the Coulomb force, with the nucleus as a point-charge Z*e*. This equation matches data up to a certain kinetic energy, but fails after that. Rutherford assumed the failure in his experiment was due to *alpha* particle interaction with the body of the nucleus itself, bringing not only the charge but the mass in to play. Later physicists have assigned this failure to interaction with the strong force. Either way, this "departure from the Rutherford formula" is what creates a method for finding the size of the nucleus and thereby the proton.

However, the math used to analyze scattering is incomplete. It is incomplete rather than false because it is correct as far as it goes. It simply fails to take into account the presence of gravity at the atomic level. This means that although its manipulations are done correctly, its assumptions are faulty. Rutherford assumes that the force can be expressed as the Coulomb force, and that therefore it is solely an electrostatic force, but this will shown to be false.

The intent of this paper is not to return to the Thomson model or to throw into question the usefulness of the Coulomb equation. The Coulomb equation allowed Rutherford to estimate the correct answer, so it must be correct as a heuristic device over some energies, at the very least. MM is not intending to question the experimental findings of the last century, but to accept that the energies and forces have been measured correctly and that the data has been read properly. What is intended to be shown here is that the force field in this scattering problem is not the single field it has been assumed to be for a century. By applying the findings in Newton's law is a Unified Field of Gravity and E/M to the problem of scattering and using that field, the size estimates at the atomic and quantum level clearly need to be reassessed.

In Coulomb's equation is a Unified Field equation in disguise it was shown that the charge field is present in Newton's gravitational equation F = GMm/r^{2}. The charge field, like any other physical field that creates real forces, must be a real field. It cannot be mediated by virtual or messenger photons. It must be mediated by photons with energy and mass equivalence. This being true, this field must be present at all levels of size, not only at the quantum or charge level.

Unlike others who have tinkered with Newton's equation, Miles Mathis has not added the charge field as an extra term in the equation, or done anything that would change the current predictions of the equation. Rather, he has inserted the charge field into the existing terms, so that Newton's equation may stand as-is. In G is the Key to the Secret of Gravity the equation was re-expanded, so that we may see clearly the fields that lay under it and generate it.

This makes Newton's equation a compound equation, with G as the transform between the two fields it contains. MM showed that the two fields are in vector opposition and that Newton's equation gives us only the field differential. That is, his equation tells us the total field but tells us nothing of the constituent fields. We can discover the relative size of these fields only by studying the interactions of real bodies. MM has done this starting with the Earth and Moon and now has firm numbers not only for the relative field strengths of these two bodies, but for the relative field strengths in general. (See The Secrets of the E/M field are revealed at the Moon's surface.) This discovery has also allowed me to find the Solution to the Ellipse problem, the cause of orbital stability on MM's site, the cause of planetary torques on MM's site, and many other things.

A major implication of this expansion of Newton's equation is that gravity is a function of radius and nothing else. If we write the masses in Newton's equation as density times volume, the density describes a quality of the charge field (or what MM calls the foundational E/M field) and has nothing to do with gravity at all. Gravity is shown to be an acceleration only, and whether that acceleration vector points in or out from the center it has nothing at all to do with density.

This being so, the size of the field must change at a different rate, depending on the size of the gravitating object. In the end, this means that the size of the gravitational field at the quantum level must be some 10^{22} times larger than we have thought. Although it is still small compared to the E/M field at that level, it is no longer negligible. It plays a very real and measurable part in quantum and atomic motions and forces.

This also gives us an easy way to calculate gravity in QED. If the charge field is present in Newton's equation, then the gravitational field must be present in Coulomb's equation. MM has shown that the two equations are really the same equation, with one hiding the charge field and the other hiding the gravitational field. When we re-expand Newton's equation, we find:

the charge field F = GMm/r ^{2} = H - E = [m(A + a)] - E |

When we re-expand Coulomb's equation, we find:

the gravitational field F = kQq/r |

H is the gravitational field and it is found by the same equation at all levels of size. The variable "m" is the mass of the smaller of the two objects, being the gravitating object in Newton's equation and the scattered particle in Rutherford's experiment. "A" is the gravitational acceleration of the larger object and "a" of the smaller.

As you see, this must impact the findings of Rutherford and all scattering experiments. It will not change the data, of course, but it must change the mechanical assumptions. If the force is not only electrostatic, then the mechanics cannot be what we have assumed it is. If we mis-assign forces, we end up with wrong numbers when we start calculating down from those forces. In these scattering equations, we are not calculating energies or forces from lengths or times, we are calculating lengths or times from energies or forces. Logically, that is upside down, and it is a dangerous mathematical manipulation. It requires an assumption of complete knowledge of the field mechanics, so that we can solve down in the correct way. Having shown that we do not have a complete knowledge of the field mechanics, we should not be able to solve down with such complete assurance. In fact, Rutherford and those who followed him have solved down using false assumptions, and have thereby gotten the wrong numbers for their lengths.

To find the diameter of the nucleus, current math seeks what is called the "impact parameter", which is the perpendicular distance to closest approach during scattering. Once the scattering departs from the Rutherford formula, it is assumed that this impact parameter is equal to the nuclear radius, since the *alpha* particle must now be impacting the body of the nucleus. This is the equation currently used, with "b" as the impact parameter:

b = √
[1 + cosθ
/1 - cosθ
] kQq/mv^{2}

As you can see, the impact parameter is calculated from both the Coulomb force and the kinetic energy. **The problem in this equation is with the mass variable in the denominator. **

In the numerator we have the constant k transforming the charges into usable numbers. In Newton's equation, G has the job of turning the masses into usable numbers: numbers that will give us the correct force. We need a constant or a transform like G because the mass variables are not really standing for a single measurable field parameter. In G is the Key to the Secret of Gravity MM has shown, the mass variables in Newton's equation are standing for two separate qualities in two separate fields. The mass variable should be written as DV (density times volume), and we should give the D variable to the E/M field and the V variable to the solo gravitational field. Since we don't do this, and since we have had no knowledge of the two fields involved, we have simply let G make all the corrections for us. G turns our misunderstood and mis-defined mass variables into numbers that work.

The constant k does the very same thing in Coulomb's equation. Charge has never been defined mechanically, so the charge variables here are physically meaningless. They are little more than place fillers. We have never given "charge" a definition, not even to the extent we have given "mass" a definition. The definition of "mass" is pretty slippery, from a mechanical point of view, but the definition of "charge" is even more slippery. Charge is now defined as a trading of virtual particles, which is about as non-mechanical as you can get. No, all the mathematical content is in the constant k. It does all the numerical work in Coulomb's equation. It is quite easy to see this, since the value of charge is just 1 or -1. The number "1" does nothing in an equation, even when it is given an exponent. These charge variables are just ghosts. We could remove them from the equation altogether and it wouldn't matter.

In Coulomb's equation, the constant k expresses the unified field directly, compressing both the gravitational field and the E/M field into a single number that gives us the right force at this level of size. But it does even more work than G does in Newton's equation, since it not only acts as a transform between the two fields, it also acts to transform those two 1's into real numbers. At least in Newton's equation the mass variables have given numbers. We don't give every body in the universe a mass of 1, do we? But in Coulomb's equation we give every important body a charge of 1. Pretty silly, if you think about it.

Anyway, k is basically transforming radius and density once again, the radius giving us the gravitational field and the density giving us the density of the *B*-photon field (and therefore the strength of the foundational E/M field). Density and radius are real measurable field parameters, and they are the real working variables in both Newton's and Coulomb's equation. Newton's equation and Coulomb's equation are really the same equation, expressing the same fields. Historically we have just assigned different variable disguises to them, making them difficult to interpret or collate.

This being so, we have a problem in the impact parameter equation, since we have transformed the charges into usable numbers with the constant k, but we have not transformed the mass into a usable number. In this equation, we aren't just dealing with kinetic energies, we are dealing with compound force fields. Once you insert a mass or a kinetic energy into a field, you need to transform it so that it gives the right number. But here we have a mass being used in a field equation without the transform G, k, or any other transform.

It would appear to be impossible to develop a transform. You can't use k as the transform, since k transforms charges, taking them from the number 1 to a usable number. You can't use G, since G transforms two intersecting fields of two bodies into one number, given two masses. Here you just have one mass, entering fields that have already been transformed in the numerator by k.

To accomplish this let us start with an expression of k that isn't used very much:

k = 10^{-7}c^{2}

The c^{2} looks very similar to c^{2 }in Einstein's famous equation and it also has a mass in it, which is convenient here:

E = mc^{2}

Setting the two c^{2 }'s equal to each other gives us:

E/m = k/10^{-7}

m = 10^{-7}E/k

In the Rutherford experiment, departure from the Rutherford formula began at about 27.5MeV.

1eV = 1.6 x 10^{-19 }kg m^{2}/s^{2}

m = 10^{-7}(27.5)(10^{6})(1.6 x 10^{-19 }kgm^{2}/s^{2}) /k

m = 4.4 x 10^{-19 }/k

According to that equation, the mass in our field equation should be

m = 4.4 x 10^{-19 }/k = 4.89 x 10^{-29 }kg

But we already know that the mass of our *alpha* particle is 6.64 x 10^{-27 }kg. This makes our transform in this equation .00736

η_{m} = m_{k}/m_{α}
= .00736

b = √
[1 + cosθ
/1 - cosθ
] kQq/η
_{m} mv^{2}

We needed to find a transform for our mass, and we have done that. Look at the ease in which this was developed. There are longer and more complex ways to derive this transform, but this is by far the most elegant.

There are other ways we could write that last equation, too. We could just write it in terms of m_{k}

b = √ [1 + cosθ/1 - cosθ] kQq/m_{k}v^{2}

Or we can take the mass out of the equation altogether

b = √
[1 + cosθ
/1 - cosθ
] k^{2}Qq/ξ
v^{2}

Where ξ
is a new constant

ξ = 4.4 x 10^{-19 }J

Perhaps not surprisingly, that is the energy of a photon. We multiply our mass m_{α}
by ξ/km_{α}
in order to bring it into the field equation. Ultimately, that makes perfect sense, because we are trying to match our field in the denominator to our field in the numerator, so that the ratio will work properly. The numerator expresses a charge field and a charge field is mediated by photons. Therefore we can't simply import a raw mass into the denominator. We have to ask how it is acting as an electrostatic object. Since k is already in the equation, and since k is most directly expressed as a function of the speed of the photon, c, it makes perfect sense that we transform the mass into the mass equivalence of a photon with energy equal to our *alpha* particle. That is what we did with E=mc^{2}, since in E ≠ mc2 (Gamma is Kappa) MM showed that Einstein's equation—written baldly like that without *gamma* or another transform—applies only to photons. The variable m_{k} is the mass equivalence of a photon with the same energy as our *alpha* particle. [The *alpha* particle isn't going c, so it must be bigger to create the same energy.] And once we put m_{k} next to v^{2} in the equation, that velocity transforms it back into the *alpha* particle (photons don't go v, they go c). But now that *alpha* particle is acting like a charge in the equation, instead of a mass. That is why and how that beautifully short derivation worked.

1/.00736 = 136, so our new value for b must be about 136 times the current value. Which means that the nucleus is 136 times larger than the current estimate.

You may want to compare that number to the fine structure constant. MM just showed that you have to multiply a mass by .00736 to turn it into charge, at the quantum level. The fine structure constant is about .0073. Therefore, MM has just showed you where the fine structure constant comes from, answering one of Feynman's top questions, a question he put on his blackboard every morning for 20 years.

In G is the Key to the Secret of Gravity, MM used 10^{-13}m instead of 10^{-15}m for the radius of the proton, and here is the answer. That paper may now stand as confirmation or proof of this paper. In that paper, MM use the number 10^{-13}m to confirm the age of the universe as about 15 billion years, by a pretty transparent method. Since that number comes from observables and equations that have nothing to do with our observables or equations here, that number stands as experimental evidence in favor of this paper and this finding. In other words, MM has shown that 15 billion years confirms 10^{-13}m.

Everything in the atomic and quantum world—including the atom, the nucleus, the proton, and the electron—is about 100 times larger than you were taught.

For more on the fine structure constant, see Planck's Constant and the Fine Structure Constant, where the current fine structure constant defining equations are torn apart, showing how they are overwritten and redundant. Also in that paper where he shows that the fine structure constant is a mass to charge transform, and tie it to other important charge numbers.

In G is the Key to the Secret of Gravity, MM showed that according to Newton's own equations, the mass of any spherical particle could be written as an acceleration of its radius. Then, in Bohr's Three Mistakes, MM showed that by correcting the angular momentum equations, we could correct both the Bohr radius and the proton radius. MM found a proton radius of about 4.09 x 10^{-14}m. Of course that is very close to the square root of the mass, so we must now ask if that is a coincidence. Is the proton mass just its radius squared, and if so, why?

To answer this, we return to the first paper on G. One of my readers suggested that the mass was the radius squared because the radius squared gave us a sort of area. The area of a circle is πr^{2}, so perhaps mass is a two-dimensional entity. We must shoot this down for several reasons. Firstly, by the quantum equations in π (pi) is 4 not 3.14...., so it is unlikely we are dealing with area. Secondly, this would create a scaling problem, since mass would be less than radius for all particles less than 1m and more than radius for particles more than 1m. An area is normally greater than its own radius, but here we find the opposite. The mass is much less, as a number, than the radius. That is counterintuitive as well as paradoxical. Thirdly, MM has shown that mass acts like a 3-dimensional motion, not a 2-dimensional motion. Using Newton's equations, mass reduces to three lengths over two times, and three lengths implies three dimensions. For this reason, mass should be analogous to volume, if anything. It shouldn't be the same as area, it should be the same as volume. But if we start talking about volume, we have r cubed, not r squared.

It turns out that the square doesn't come from area, it comes from acceleration. As MM has already shown, the mass of any spherical particle can be written as the acceleration of the radius. In other words, it is the time squared in the denominator that is important here, not anything in the numerator. A gravitational acceleration is always written with a time squared in the denominator, and that is what we should be studying.

If, as Maxwell showed, mass can be written as length cubed over time squared, then with a spherical particle, mass is the acceleration of the radius. By this analysis, mass is not operationally a measure of ponderability, it is a measure of motion. It is the motion of the proton shell over a given time. This being the case, we just have to ask how this affects the math and the definitions. To answer this, we have to ask why the mass isn't the same number as the radius. Why can't all the radius express itself as mass over the time period? Why only the square root?

Well, it is because Newton's equations, as written, define mass as the motion of the radius, not as the radius. That is, mass is a change in the radius, not the radius itself. And not just as a velocity, but as an acceleration, as we have seen. Velocity is a simple change, and acceleration is a change of a change. A second order change. In this way, we see that it is a matter of exponents. Mass is tied not to the radius or the velocity of the radius, it is tied to the acceleration of the radius. Since acceleration is a squared change, mass must be a function of that exponent of 2. Logically, mass cannot be greater than the radius, because that would imply a change that was greater than what was changing. Therefore, mass must be smaller than the radius, and that means that the exponent of 2 must be working as a square here, rather than a square root. A square root would give us a mass greater than the original radius, since we are working below the number 1 here.

In other words, the square comes from the exponent of 2, and the exponent of 2 comes straight out of the definition of acceleration. In this way, we have unified mass and gravity at the foundational level. It is thought that Newton defined gravity as a function of mass, or that he defined mass as a function of gravity. But that is not the whole story. When we look really closely, we find that neither is a function of the other, since they aren't mechanically separable. As with time and distance, gravity and mass are two names for the same thing.

In A Revaluation of Time and Velocity, MM showed that time is really a measure of distance. We simply relabel it to suit ourselves. It is a second measure of distance, so that we can compare the two distances in a ratio, obtaining a velocity. Operationally, it turns out that the same is true of gravity and mass. Both are defined and determined by the acceleration of the radius, but we give that measurement two different names in order to calculate force because Newton did it, and we do as he did.

All this also applies to the electron. The mass of the electron is 9.109 x 10^{-31}kg. The square root of that is 9.544 x 10^{-16}kg. The only problem there is that MM has already calculated the radius of the electron to be 2.24 x 10^{-17}m. This is off by a factor of 42.6. Fortunately, 42.6 is roughly the square root of the Dalton, 1822. So that we get this equation:

m_{e} = Dr_{e}^{2}

But that begs the question, “Why can we go from proton mass to radius with no transform, but not do so with the electron? Why the transform D?” The reason that MM has shown is that the Dalton is basically an outcome of spin. The electron is four spin levels below the proton. Mechanically, the Dalton is not really an expression of the atomic mass unit or anything abstract like that, it is the number relation between the velocity on the shell of the spinning electron and the velocity on the shell of the spinning proton.

D = m_{e} /r_{e}^{2} = r_{p} /r_{e}

m_{e} = r_{e} r_{p}

As you can see, mass once again expressed as distance. Mass is not a function of distance, mass is distance. That last equation is amazing not only because it expresses mass as distance, but because it defines the electron mass in terms of the proton radius. Because the proton mass is the straight square of its radius, the proton must be more primary in this math. The electron mass is derived, and depends on the presence of the proton. But again, why? It turns out to be another instance of a sort of relativity. The proton is primary here because matter is composed of protons. It is the same reason that protons are given the plus sign in E/M and electrons are given the minus sign.

MM has shown that mechanically this is because protons act as the baseline and electrons are measured relative to that baseline. In other words, electrons are not attracted to protons, they are repulsed less than the baseline repulsion of proton/proton. Which makes them negative in the math. They are mathematically negative, not mechanically negative. Another way to see this is to remember that the proton creates the bulk of the charge field, by recycling and emitting far more photons than the electron. This is why the proton is the defining particle in the E/M field. For the same reason, the proton is again the baseline in this study of mass and radius, simply because protons are the primary constituent of matter. Electrons are a secondary constituent. Mass, as we measure it, is a unified field number, so it is never beside the point that the proton emits and defines the E/M field. Beyond that, due to the spins, we cannot define both the electron and the proton as primary. For reasons of relativity, we have to pick a point of view. Because matter is composed of protons, we naturally and inevitably measure from the point of view of the proton. The actual material structure we measure in is composed of protons, and the charge field we exist in is emitted mainly by protons. We live and measure in the proton field.

We can see this just by looking at motion. We are not living primarily in the photon field, because photons are moving c relative to us. We are not living in the electron field, because electrons are either moving very fast or are orbiting relative to us. But nearby protons are nearly at rest, as an average, relative to us, since our “at rest” is determined by their “at rest.” This is why the proton is the defining particle in both the E/M field and the gravity field. This is why we require no transform, just a square, when we go from proton mass to radius, and why we require the transform D in going from electron mass to radius.

But let us return to that last equation, for a bit more analysis. MM has just shown you why the electron's numbers are dependent on the proton's numbers. It is because for us the proton is the baseline. That being so, we see that the mass of the electron is determined by the motion of the electron's shell in the field of the proton. We monitor the expansion of the electron relative to the expansion of the proton. That is what the multiplication is about. We have two simultaneous motions, so we integrate or multiply them. The two motions give us an acceleration over some interval, and that acceleration is what we call mass. In that sense, we could write the last equation more rigorously as

m_{e} = Δr_{e} Δr_{p}

It takes two expansions to give us a mass, since mass is meaningless except as a relationship. A single particle in the void has no mass, strictly speaking, since MM has shown that mass is motion, not ponderability. But motion is relative. Therefore mass is relative. A lone particle has only a size. It has a potential mass. Of course we must postulate that our particle is not penetrable, which gives it a structure, which structure might be called ponderability or mass. But if we can and do write mass as motion, as in L/T, then strictly speaking mass must be relative, like motion.

MM has shown that mass is really a function of motion, like everything else. My analysis implies that we must give the shell of the proton an impenetrability, yes, but other than that mass is just motion. Mathematically, mass can be calculated straight from rectilinear motion. This is a great improvement on current theory, since it reduces the mass variable to distance and time. It is also much preferable to all the talk we hear of bosons. We don't need a reductio like the boson to explain mass, we just need a bit of clear thinking.