return to homepage The standard model tells us that gravity is on the order of 10^{36}  10^{39} weaker than E/M at the quantum level. To find this number, they scale the forces they find at the macrolevel to the forces they find at the quantum level. Seems pretty straightforward, but it turns out to be spectacularly wrong. The problem is that the standard model is scaling force numbers, but gravity is not a force, it is an acceleration. Gravity can cause a force, but as a defined variable it is an acceleration. The standard model knows this, but it tends to forget it at crucial times. They "forget it" on purpose in this case, simply because they can't figure out how to scale accelerations. You can't really scale accelerations, since accelerations don't tell us velocities or distances or times over a single interval, an infinitesimal interval, or what is now called an instant, unless we have an initial state. When comparing quantum particles to planets and moons, it is not clear what these initial velocities might represent; nor is it clear how to find them. Given an acceleration of gravity, how do you develop a velocity or distance that you can compare? The only velocity belongs to an object in the field, not to the large central object. So you see the problem: given an acceleration of gravity, you need a velocity or distance that belongs to the quanta or planet, so that you can then compare those velocities—in order to scale them. But there doesn't seem to be any way to develop those distances or velocities, since the quanta and planets aren't moving to create the accelerations. This is why the standard model chooses to scale the forces instead of the accelerations. Unfortunately, forces don't scale like accelerations. In fact, the two diverge very quickly, as we will see below. The only way to scale properly between the quantum level and the macrolevel is to scale distances or velocities. I will now show you the simple way to do that. It was in my paper The Moon gives up a Secret that I first discovered a simple method of separating the E/M field from the gravity field, and finding the sizes of each field. I assumed that gravity was dependent on radius only, then ran the equations using the current numbers of the Earth and Moon. I discovered that, according to this analysis, the Earth must have an E/M field acceleration of .009545m/s^{2}, and that the Moon must have one of 1.051m/s^{2}. Although I developed these equations in 2005, I have not yet applied them to quanta, although it is very easy to do so. I have been busy with many other projects, as anyone who has visited my website will see. But now that I have shown a new mechanics for quantum particles, one that overthrows QED and QCD, it is time to do the full calculations. This ties into my most recent papers in two ways. It makes good on a promise I made in my paper on mesons to show how my margin of error is caused by the gravitational field at the quantum level. But it also will prove in a more direct fashion that the current Coulomb equation is terribly misused. I showed in my latest paper on Coulomb that textbooks were running the equation with bad numbers, and getting forces that were off by many exponents. This paper will add warehouses of ammunition to that claim. In my Cavendish paper I used my unified field equations to find the separated forces of E/M and gravity on Cavendish’s balls. Those who have read that paper will understand that when I say unified field equations, I am not talking about difficult or esoteric equations. I am talking, in the first instance, about an equation that scales down directly from the numbers of the Earth, based only on size. All I need is a radius to perform this unified field calculation. All we need here is a radius for the proton. In my paper on the Bohr magneton, I calculated a precise radius for the proton of 4.09 x 10^{14 }m. That is about a hundred times larger than the current estimate, but I show why the current estimate is wrong in my paper on the scattering experiments. We know that the Earth’s gravitational acceleration is 9.809545m/s^{2}, and that its radius is 6,378,100 m. That is all we need for the gravitational part of this unified field equation. We just make a proportionality equation: 9.809545/6,378,100 = g_{P}/4.09 x 10^{14 } g_{P} = 6.29 x 10^{20 }m/s^{2} That is the gravitational acceleration of the proton. In my paper on Electrical Charge, I show that the permittivity of free space ε_{0} actually stands for gravity at the quantum level. The value I found there, 2.95 x 10^{20 }m/s^{2}, differs from my number here only because the gravity at the quantum level is also caused by the electron and neutron. The neutron, being nearly the same size as the proton, will not affect the number much, but since the electron is smaller, it must pull the total number down by a fraction. The gravity of the electron is 3.43 x 10^{23 }m/s^{2}, but, being smaller, the electron is a lesser partner in the total field. If we add the two fields and divide by two, we get 3.147 x 10^{20 }m/s^{2}. That is still above my number for ε_{0}. This is probably telling us how many electrons we have relative to nucleons, but I will save that calculation for another paper.
The acceleration we just found is an acceleration relative to the acceleration of the Earth. The form of the equation alone tells us that, since we are comparing the proton to the Earth. This means that we have found an acceleration for the proton as measured from the size of the Earth. As an example of the relative nature of this number, I will return for a moment to my first paper on G, where I found a different number for the gravitational acceleration of the proton. In that paper I found an acceleration about 7 exponents greater than this acceleration. It was greater because it was an acceleration relative to the meter itself, not to the radius of the Earth. The radius of the Earth is almost 7 exponents higher than 1 meter, hence the difference. At any rate, the number I just found is an interesting number, but it is not really the number we want here, in this paper. We want to know how the proton would measure its own acceleration. Forces are transmitted and felt locally, and the relative field strengths must be calculated locally. To do this, we must compare velocities instead of accelerations. Velocities are always comparable, because they happen over one time interval. To get a velocity from this acceleration, we just use my trick from many other papers: we reverse the gravity vector, a la Einstein’s equivalence principle. We let it point out, and we let the surface of our object move with it. We look at how far the surface of the proton moves over one time interval, say one second. At that acceleration above, it moves: x = at^{2}/2 = 3.145 x 10^{20 }m The Earth moves 4.91 m during the same second. If we compare these velocities to the radii, we find something very interesting. 3.145 x 10^{20}/4.09 x 10^{14 } = 4.905/6,378,100 You will say that is just doing math in circles, and in a way it is. But I have done it this way to show that the speed of the proton’s surface relative to its radius is the same as the speed of the Earth’s surface relative to its radius, at any given time. And this means that the speed of the proton’s surface relative to its SIZE is the same as the Earth. And this means that if the proton measures it’s own acceleration due to gravity, it will find the number 9.81 m/s^{2}! It would have to, wouldn’t it, or the proton would be getting larger or smaller relative to the Earth. Of course this is a direct outcome of making the gravity field a function of radius alone, but it will be an unexpected outcome for most regardless, I think, even those who had already accepted my postulate. Once we apply the acceleration of gravity to a real motion, by reversing the vector like this, we see things we never could see before. We see that the acceleration of gravity is itself a relative number. It is relative not in an Einsteinian way, but in a completely new way. Acceleration depends on size, and who is measuring it. Einstein’s relativity was due to speed or distance. This relativity is due to size. So, we have found two new gravitational fields for the proton, one as measured from the size of the Earth and one as measured by the proton itself. Now let us look at the E/M field. The E/M or charge part of the equation is only slightly more difficult. As I showed in my paper on the Moon, the E/M field is proportional to 1/r^{4}. This is because the E/M field dissipates as it moves out from the surface, so we get one inverse square law right out of the surface area equation. But the E/M field also dissipates within the gravity field, so it picks up that inverse square law as well, just from the acceleration field. The radius of the Earth is 1.55 x 10^{20} greater than the radius of the proton, so, E_{E} /E_{P} = 1/(1.55 x 10^{20})^{4} = 1.73 x 10^{81} E_{P} = 5.52 x 10^{78}m/s^{2} But we must make the same corrections to that number that we did to the gravity number. The E/M field is an emission field, so those photons will be traveling inside the gravity field. Which is to say, if we let the gravity vector point out, the photons will be emitted while the surface of our object is moving. The E/M field is a velocity inside a velocity, so anything that is happening to gravity will also be happening to E/M. As it turns out, we can represent all this by simply squaring our gravity correction. We have the fourth power here, which is square the power of the gravity field. Another way to look at it is that the size differential applies here twice, since we have two fields that are both smaller: one field inside the other. I have already found a size differential between the Earth and proton of 1.55 x 10^{20}, so we just square that, to get 2.4 x 10^{40}. E_{P} = 5.52 x 10^{78}/2.4 x 10^{40} = 2.3 x 10^{38 }m/s^{2} That is the number the standard model is finding for the strength of the E/M field in quantum interactions. 9.8 is 4.26 x 10^{38} weaker than that, so that is where the numbers you read about in textbooks are coming from. Problem is, the standard model has only done the first part of the math. Let us complete it now. E_{P} = 2.3 x 10^{38}/1.55 x 10^{20} = 1.48 x 10^{18 }m/s^{2 } Now, I could have ignored all the mechanics and just told you to divide once by 1.55 x 10^{20}, instead of going to the fourth power and then dividing by a cube. But although that would have gotten the right answer in much less time, in this case efficiency is not my primary concern. My primary concern is uncovering the mechanics, and showing them to you in a transparent manner. Let us look at one more outcome of our new numbers. I went to an inverse fourth power and then cubed to get the number 1.48 x 10^{18}, as you have seen. I might just as easily have started with the proposal that the E/M field at the quantum level acts like the gravity field at the quantum level: that is, it is the same number as the Earth. If we can propose that the proton would get 9.81 for its own gravity, why not propose that the proton would get .009545 for its own charge field, just like the Earth? Well, actually, we could, and we would get the right answer, provided we made the right corrections. All we have to do is scale down using the radius differential again. In this case, the radius differential is representing the fact that the photon is that much larger relative to the proton. We have scaled the momentum of the photon up, relative to the proton, so that it has the right energy in the field. 1.55 x 10^{20 }(.009545) = 1.48 x 10^{18} The photon is going c whether it is taking part in quantum charge interaction or taking part in the E/M field of the Earth. But the photon will seem much bigger and more powerful to the proton. Let us see what the current unified field acceleration of the proton is, using Coulomb’s equation. Yes, I have shown that Coulomb's equation, like Newton's equation, is already a Unified Field Equation. We can compare it to my acceleration above if we use the current charge on the proton and the current quantum force, but apply them to a motion of the proton instead of to the field. a = F/m = 8.2 x 10^{8 } N/1.67 x 10^{27 }kg = 4.9 x 10^{19 }m/s^{2 } My acceleration is about 33 times less than that of the standard model. That makes sense, since my quanta are larger than standard model quanta. Being larger, they don’t need as much acceleration to cause the same field effects. And we can find a force using my new acceleration: F = ma = 1.48 x 10^{18 }m/s^{2 }(1.67 x 10^{27 }kg) = 2 x 10^{9 } N So you can see that my force is not far off the proposed standard model force, despite all the corrections to the math I did. According to textbooks, using the Coulomb equation, the force between the hydrogen nucleus and the orbiting electron is 8.2 x 10^{8 } N. This means that the Coulomb equation is not hopelessly off course, in the first instance. At this point it is less than 100x wrong, and most of that is because the Bohr radius is wrong. Unfortunately, I am far from finished in my corrections. We have a very important step outstanding. Before we do that, let us look at our new fields. I have shown that the local charge field at the quantum level is E_{P} = 1.48 x 10^{18 }m/s^{2 }. The local gravity field is g_{P} = 9.81 ^{ }m/s^{2}. From this we see that the gravity field is still negligible, but not nearly as negligible as QED thinks. It is about 10^{22} times more powerful than we are told. We are told it is about 10^{39 }times weaker than E/M, but, from the point of view of the proton, it is only 10^{17 }weaker. I have been able to strengthen the gravity field by a large margin, but it still appears to be negligible. If it is still about 10^{17 }smaller than the E/M field, it is difficult to see how it will involve itself in quantum interactions. In various papers I have used gravity to explain the quantum orbit, one of the quantum waves, the apparent attraction of the electron (instead of its real attraction), and the margin of error in my meson states. None of these things can be explained with a gravity field that is still so weak. That would be true if my math and theory stopped here, but we still have to consider the drop off of charge due to the spherical shape of the field. This problem has already been encountered by QED in (one of) the famous zerocharge paradoxes. Although the measured charge is now said to be 1.6 x 10^{19} C, according to the original math of QED the effective charge at any distance from the electron or proton would seem to be zero. This problem is at the root of renormalization, as I showed in my paper on asymptotic freedom and as most involved in QED know. I won’t get into all that again here, but it is true that the charge would be expected to drop off quickly. In the equations above, I developed an acceleration due to charge, but this acceleration was and is an analog of the charge field of the Earth. My charge field of the Earth is the charge field measured right at the surface. So the charge field I calculated for the proton must also be right at the surface. This would apply only to protons at impact, but that is not what we normally mean by the charge strength at the quantum level. What we mean is the charge strength at an the radius of an orbiting electron (or what I have since proved is actually the radius of capture of the electron, not its orbit). How far are the quanta apart given this situation? I have recently developed a corrected number for the Bohr radius, which is 9 x 10^{9} m. I have already given a number for the proton radius of 4.09 x 10^{14 }m. That is a difference of about 100,000x [using the standard model we get the same factor]. So, using my 1/r^{4 } equation, we get a drop off of charge of 10^{21}. If the local charge is 1.48 x 10^{18}m/s^{2}, then the effective charge at the orbital radius is about .00064 m/s^{2 }. This means—according to the proton’s own measurements—that the surface of the proton is moving toward the electron at 9.81 m/s^{2}, and that the electron is being driven off at .00064 m/s^{2}. This is why we have an apparent attraction. So let us use those new numbers to find a unified field force between the proton and electron in orbit. a = 9.80 m/s^{2}  .00064 m/s^{2 }= 9.80 m/s^{2} But we must ask if we want the force on the electron or the force on the proton. There is no such thing as a generalized force, since the equation must have a mass in it. Let us find the force upon the electron: F = ma = (9.11 x 10^{31} kg) 9.80 m/s^{2 }= 8.93 x 10^{30} N On the proton we get: F = ma = (1.67 x 10^{27} kg) 9.80 m/s^{2 }= 1.64 x 10^{26} N And now we see that the textbook use of Coulomb’s equation was a catastrophe. We are told that the force between the proton and electron is 8.2 x 10^{8 } N, which you can see is off by a factor of 10^{22}. Notice that this is also the factor that the standard model is off with the strength of gravity at the quantum level: 10^{22}. Next time someone tells you that QED is so fabulously exact and precise, send them to this paper and tell them that you know that QED is wrong by 10^{22 }on one of the fundamental questions of the atomic world. How does the current number compare to my number? That is, how is the mistake made, precisely? Watch this: If we take the constant out of Coulomb's equation (when we apply it to quanta) and sum the charges, we get very near my number. I have found separate forces for the proton and electron, but we can find the total force by adding them together and dividing by two. That gives us a sort of average force of 8.2 x 10^{27} N. Coulomb's equation squares the charge, but you don't need to square the charge. The forces and charges don't square, since they are field charges and forces. They shouldn't multiply, they should add. If we were being rigorous, we would have to add them as density fields, so they don't even obey a straight addition, but since the electron force is so much less, we can just divide by two and get the right answer, as you see. Now, since the current equation multiplies both charges by the constant, we have to unmultiply twice to make the correction in the field. In other words, we have to divide by the square of the constant. We are going smaller, so we use the inverse of the constant, which is 1/k = 1.11 x 10^{10}. Squaring that is 1.23 x 10^{20}. So we multiply that by the current figure, to make the correction. (1/k)^{2}(8.2 x 10^{8} N) = (1.23 x 10^{20})(8.2 x 10^{8} N) = 1.01 x 10^{27} N If we take the constant out of Coulomb's equation when we apply it to quanta, then when we seek the force between the electron and proton, we get 1.01 x 10^{27} N. If we use my new math, which gives an acceleration of 9.80 m/s^{2} to both the electron and proton, we get an averaged field force of 8.2 x 10^{27} N. I would call that fairly astonishing, no matter how you look at. The standard model will try to write it off as some sort of mathematical coincidence, but this paper shows that it is not a coincidence. The standard model has vastly overrated the forces between quanta, and it has done so because it doesn't understand how the constant is working in Coulomb's equation. It also doesn't understand how the charge field works mechanically, how the charge field interacts with gravity, or how to scale accelerations. All these mistakes have snowballed on them, creating a fundamental field error of 10^{22}. We can also now see that the force between the proton and electron—the fundamental charge—is numerically almost precisely equal to the mass of the proton. Remember that the mass of the proton is 1.67 x 10^{27} kg. If we take the constant out of Coulomb's equation when we apply it to quanta, then when we seek the force between the electron and proton, we get 1.01 x 10^{27} N. Wow. Another coincidence, right? No. The numbers are nearly the same because one is measured in kilograms and the other is measured in Newtons, but the Newton is based on the kilogram. 1 N = 1 kg m/s^{2}. So this is what we should have expected from the beginning, if we had understood our equations. It makes no sense for the masses of the quanta to be at a scale of 10^{27} and the forces to be at a scale of 10^{8}. Since accelerations scale with the radius, as I just showed, the force must scale with the mass. Please note that I have simultaneously shown a very strong repulsion between protons at close range and an attraction between the proton and electron in orbit. I have done this without having an attractive field of any kind. The charge is always repulsive, and gravity is represented by a real motion, as you see. Of course, by this logic, protons would also be attracted to each other, up to a point. And it explains why free protons are excluded at greater radii than electrons. Free protons would indeed attract each other, but any kind of proton/proton orbit would be well outside the electron orbits. Electrons, being smaller, can get closer to the proton. They encounter a smaller cross section of the charge wind. It may be that electron orbits nullify any possible proton orbits, by making the atom neutral. But it is much more likely that the tendency of matter to form liquids and solids is due to this attraction between distant protons that I have just shown. Liquids and solids do form, and they form very near the quantum level. If protons are so exclusionary, to a power of 38 over gravity at the quantum level, it is not clear that liquids and solids could ever congeal. Even though solids are solid at a molecular level, not an atomic one, the molecular level is not far above the atomic level. Given the current quantum model, it is not at all clear how chemical bonds form, much less solid structures. “Sharing of electrons” and other similar statements have always been long on mystery and short on mechanics. A charge or E/M field that is so gigantically overwhelming at the quantum level can hardly have evaporated at the molecular level, only a few factors of ten above the Bohr radius. And a gravity field that is so attenuated at the quantum level can hardly have become a major player again at the molecular level. Without these two fields acting in normal ways, the standard model is forced to explain molecular bonds and liquid and solid structures in very strange ways. But my theory begins to make some mechanical sense of it. You see that my charge field, though very strong as a bare charge, is weak enough to evaporate beyond the Bohr radius. And my gravity field is always at your service. Conclusion: One very large scaling error has jeopardized the entire standard model explanation of forces at the quantum level. The standard model has historically treated the gravity field as a force field instead of an acceleration field. They have thought that forces could be scaled by a straight comparison of numbers, whereas I have shown that it is the acceleration numbers that must be scaled, not the force numbers. And the acceleration numbers cannot be scaled in a straightforward manner. To scale acceleration fields, you must develop a velocity over one interval, and scale the velocities. This is what I have done, and in doing so I have shown not only a new sort of relativity, based on size, I have shown a very powerful gravity field at the quantum level. In fact, it is precisely the same gravity field we have at our level, and at all levels. Gravity is a constant, at all scales. But it is a constant only if it is a local measurement. If you measure gravity at one scale from another scale, you have to do a transform. I have shown that very simple transform: the radius differential.
