return to homepage As a lead-in to this solution I will show some interesting facts, so far uncompiled in the correct way, but related to the solution revealed here. The first, surprisingly enough, is a quote by Jonathan Swift. As a matter of science, Swift is most famous (or notorious) for making strange and precise statements about the moons of Mars, statements that turned out to be very near true. Somewhat less well known in our own time is a statement he made regarding gravity and his friend Newton’s efforts to explain it. In Before I prove this with math, I would like to show some data that pushes us where we are already going. In the 1940’s the Dutch geophysicist and ocean explorer F. A. Vening Meinesz showed that gravity is very slightly stronger over deep oceans. This phenomenon has never been explained, although Vening Meinesz attributed it to continental drift and the standard model how tries to explain it as an outcome of plate tectonics and isostasy. Using the offered mechanics of isostasy and plate tectonics, the solution is both fuzzy and unverifiable. Proof would require measurement of large sections of deep earth that we simply cannot measure. And even then the postulated mechanisms are farfetched and everchanging.
A similar phenomenon is explained in much the same way. In the 1850’s J. H. Pratt showed that the Himalayas do not exert the expected gravitational pull. They do not deflect a plumbline. This result was so surprising that the scientific world has really never gotten over it. They have never explained it either, except by more desperate theorizing. The astronomer G. B. Airy came up with the idea that there are “reverse” Himalayas under the ones we see, buried in the sub-crust magma like a mirror image. There is no way to prove or disprove that, short of a lot more digging than we are prepared to do, but the reverse mountains wouldn’t solve the problem anyway. This was basically the invention of isostasy, but isostasy doesn’t solve the problem of the Himalayas. True, the plumbline would then be affected by both the upper and lower mountains, but the upper mountains should still deflect the plumbline. The whole fix is absurd and counterintuitive, since it was never thought the real mountains were sitting on a void. They were assumed to be sitting on a huge mass already, a mass called the Earth. Putting reverse mountains down there doesn’t solve a thing. Even if the reverse mountains were made of lead, the real mountains would still be expected to affect a plumbline, according to the given theory of gravity. The mountains have a huge mass, and talking about masses underneath is not to the point. The only new mountains that could offset the plumbline would be mountains directly behind the plumbline (assuming the real Himalayas are in front of it). Dr. Airy needed to postulate very heavy ghost mountains behind him no matter what way he turned.
Now that I have shown a couple of experimental proofs of my assertion, let us look at the mathematical proof. We will start with Newton’s equation. Newton’s famous gravity equation is a heuristic equation, and Newton admitted that from the very beginning.
That is the E/M field equation that was buried in Newton’s equation. I could manipulate it into other forms, but I won’t bother with that right now. Notice that we don’t need the larger mass to calculate a gravitational force, but we do need it to calculate an electromagnetic force. This is logical since we assume that both masses are creating a real bombarding field with subparticles, in order to mechanically express the E/M repulsion. We do not assume this with the gravitational field, since we are expressing the gravitational field with motion only. I put the constant G with the larger mass, since that is why it is in the equation to start with. It acts as an electrical field transform from the mass or density of the atomic field to the density of the foundational E/M field, so that the two fields can be compared correctly. Notice that if both masses are very small, G loses much of its power. If we use that equation with quanta, for example, the two acceleration terms will do most of the work, since the mass terms and G will become negligible. Also notice that the gravitational field has nothing to do with distance of separation or with the constant G. These variables enter Newton’s equation only as part of the field E. In fact, I show in another paper that G can be used directly to transform the radius of the E/M photon to the radius of the average atom in the objects. G is a SIZE transform, more than anything else. We would expect the electromagnetic field to diminish with the inverse square of the distance. Why? Simply because our objects are spherical. If, as I have proposed and is already assumed by many, the E/M field is caused by a bombarding field of subparticles (like tiny photons), then this field will disperse simply due to the spherical way it is emitted from the surface of the body. But we would not expect an acceleration field to diminish that way, classically, for the reason I have shown. The distance between the objects makes no possible difference, and it cannot enter the equations in a logical way. Does this equation get the right number? Let's apply it to the Moon, as affected by the Earth. Using the values of A and a that I derive below, it gives us a total force of -9.17 x 10 ^{23}N. If we divide that by the mass of the Moon, we getA _{E} = -12.477 m/s^{2}That offsets the total acceleration A + a, leaving a difference of .00272m/s ^{2}, which is the current acceleration due to the compound field at the distance of the Moon. But if we divide by the combined mass of the Earth and Moon, instead we obtain,A _{E} = -.151m/s^{2}We divide by the combined mass because the E/M field has to repel both bodies. To counteract both accelerations, it has to work in both directions. So we have just found a number for the total acceleration of the E/M field of the Moon and Earth. I will show below that this is in fact the correct number. I have now un-unified Newton’s classical equation of gravity, showing that it is a compound equation of two force fields (or one force field and one acceleration field). But I have some work left to do, since in order to create a completely updated and modern Unified Field Theory, I have to include Relativity. Meaning, I have to express the time differential in my equations above. I will not do this with the tensor calculus. I will do it in the same way that I have solved other major problems of General Relativity: I will solve by keeping that acceleration vector pointing out and by looking at the absolute time separation between events provided by the speed of light. If you don’t know what I mean by that, I recommend you to my papers on bending of starlight, Mercury’s perihelion, and the Metonic Cycle, where I solve GR problems very quickly, without tensors.
That's the new Relativistic E/M field equation. It describes the repulsion between any two objects. It is always in vector opposition to the gravitational force. According to this equation, no two objects attract eachother due to the E/M field, not macro-objects and not quanta. Therefore, all objects have the same charge. Any apparent attraction is only a result due to compound motions or fields. Now let us re-unify the field. F = E + H F = (GmM/R ^{2} )(1 – 2R/ct ) F = (GmM/R
^{2}) – (2GmM/Rct )That is the new Relativistic compound equation, which can replace Einstein's equations. Einstein's field equations are updates of Newton, so his equations are also compound equations. This equation includes both the gravitational field and the E/M field. Therefore it is a Unified Field Equation. Einstein’s field equations are also Unified Field Equations, and it is sad that he never realized it. He spent half his life trying to solve a problem that was mis-defined. At first it looks like if we make R=ct, that equation solves back down to Newton's equation, giving us no new information. But that isn't right. I say R=ct only to create the equation, but in that form it only applies to a photon, which really is going c. To actually use the equation on real particles that aren't photons going c, we have to substitute in their speed here. We do that by recognizing that in that time t, they don't travel r, they travel rv/c. So we substitute rv/c for ct. This helps us get rid of t, which we won't want in many situations.
[I have now used this relativistic unified field equation to solve the problem of galactic rotation, one of the greatest problems of current astrophysics. I develop a velocity equation straight from this UF equation, showing how the problem can be solved without either dark matter or any modified Newtonian dynamics (MOND). The problem is solved with nothing but charge. More recently, I have used this unified field equation to replace the Lagrangian. See my paper Unlocking the Lagrangian to see how my UFE mirrors the Lagrangian while correcting it. See my paper on Schrodinger's Equation to see how replacing the Hamiltonian with my UFE clarifies and corrects many problems there. And see my paper on Lev Landau's orbital proof to see how replacing the Lagrangian with my UFE updates and simplifies the badly compromised textbook proof.] Notice that R/t can be thought of as a velocity. Since the problem is gravitational, we are dealing with accelerations, not velocities. Nonetheless, we get a transform in a familiar form. [1 – (2R/ct)] then becomes [1 – (2v/c)], which should look familiar to all experts on Relativity. My new unified field equation includes the Relativity transform. Before we close, let us look at a couple more things. First let us re-analyze the new E/M field equation. E = (GmM/R ^{2} )(1 – 2R/ct ) – m(a + A + 2AR/ct )
I showed above that the mass could be treated as a sort of acceleration, according to Maxwell, and I used this equation: M = AL ^{2}
That gives us L ^{3}/T^{2}. But let's take it even further. Notice that we have been treating the acceleration due to gravity and the acceleration due to mass as the same thing. What I mean is that the same acceleration can be used to explain the apparent gravitational attraction and the "ponderability" of the object. We don't have two accelerations here, we have one. The only difference is that in the case of mass, we add an L^{2}. So let's combine the two accelerations in the equation, too, and get rid of some of the redundant variables.
E = (GmAL ^{2}/R^{2} )(1 – 2R/ct ) – m(a + A + 2AR/ct )E = mA[(GL ^{2}/R^{2} ) – (2GL^{2}/Rct ) – (a/A) – (2R/ct) – 1]
Looks great, but what does it mean? Since we have taken L/T ^{2} to be the acceleration of M coming right at us (measured from the smaller mass, remember) we must take L^{2} to be the motion in the other two dimensions. If we take the x-dimension as running from m to M, then L^{2} is the yz plane. Since we are defining both mass and gravity as motion, this planar motion must stand for mass in that plane. If so, then it must give us the mass of the field over some infinitesimal interval and over some square "footage." The question then becomes, how big is the square and how small is the interval? That question translates into this one: Can the acceleration give us a mass? If we calculate a gravitational acceleration from our new equation, can we then use it to get a mass directly? It looks possible from here. In that case, we won't just have dimensions for Maxwell's length and time expression of mass, we will have a number.
In pursuing this number, let us first apply the new Relativistic equation to the Earth and Moon. According to Newton's equation, the force between the two should be 2 x 10 ^{20}N. According to my correction, 2GmM/Rct, where we find t in this way:
s = (a + A)t ^{2}/2t = √[2s/(a + A)] = √[2(384,400,000)/(2.671 + 9.78)] = 7855s 2GmM/Rct = 6.49 x 10 ^{16}N
That is a .03% change due to Relativity. That works out perfectly, since, as I said, I previously found a 4% change for Mercury due to Relativity. Mercury’s mass is 4.5 times that of the Moon, its density is 1.62x, and its distance is 390x. .03% x 390 x 1.62/4.5 = 4.2% But now let's find L for the Earth, using this equation, F = (GmM/R ^{2}) – (2GmM/Rct )
But doing the same thing to it we did to the equation for E. M = AL ^{2}F = (GmAL ^{2}/R^{2}) – (2GmAL^{2}/Rct )
To obtain a number for L, we only need the first term and our numbers that we just derived. GmAL ^{2}/R^{2} = 2 x 10^{20}N L = √ [(2 x 10 ^{20}N)(R^{2})/(GmA)]L = 7.84 x 10 ^{11}m
M = AL ^{2}6 x 10 ^{24}kg = (9.8m/s^{2})(7.84 x 10^{11}m)^{2} = 6 x 10^{24}m^{3}/s^{2}1kg = 1m ^{3}/s^{2}
That works out perfectly. But it would be expected to, since we used A = 9.8m/s ^{2}. Problem is, we get that number from experiment, and in experiment we are measuring a compound or resultant force. In every historical experiment to measure g, both the gravitational field and the E/M field are present. We have never tried to isolate one from the other. But we need a number for the gravitational field alone, so we will have to keep working.
In the equations above, A and a stand for raw gravitational accelerations; they are not compound numbers, expressing the resultant acceleration that also includes the E/M field. Therefore we cannot use 9.8 in those equations. This means that trying to find A from the equations we already have appears to be circular. I need another trick in order to find a variance from 9.8. We are looking for a number that is a fraction larger than 9.8, since we need to subtract the E/M field from it in order to get 9.8. That much is clear, I hope. The trick to obtain this number is in another paper of mine from last year, The Moon Gives up a Secret. There I do the math to find the segregated field numbers for the Moon and Earth, simply by looking at the way they are related.
I use several postulates. The first is that the gravitational acceleration is dependent only upon radius. It is not dependent on density. The density affects only the E/M field, not the gravitational field. Perhaps you will have noticed that this is one of the necessary outcomes of my math above, although I did not make it a postulate it in this paper. If we define mass in the way of Maxwell's suggestion, as L ^{3}/T^{2}, then clearly mass is defined only by extension. This has the curious affect of making mass not dependent upon mass. The density of the object must now contain all idea of mass, by the old definition, and density is not necessary to calculate acceleration or gravity. So, in a way, mass is no longer necessary to calculate gravity. You only need a radius. If you also have a time, then the two together will give you the acceleration and therefore the gravitational field numbers.
The second postulate is that the E/M field drops off at 1/R ^{4}.** I have already shown the math for the double drop-off above. We had an inverse square law even before we made the field Relativistic (from the spherical shape of the field), then we added a second drop-off due to the time differential. The transform, as written above, does not make this clear, since I actually have an R in the numerator [1 – (2R/ct)]. But an analysis of the mechanics, as I gave above, shows that there is indeed a double drop-off due to distance.
g _{E} - E_{E} = 9.8 m/s^{2}g _{M} - E_{M} = 1.62 m/s^{2} R _{E}/R_{M} = g_{E} / g_{M} = 3.672g _{M} = .2723 g_{E}
E _{E} /E_{M} = 1/3.672^{4} = .0055E _{M} = 181.81 E_{E}
But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density. Notice we are correcting the E/M field for density, not the gravitational field. D _{E} /D_{M} = 5.52/3.344 = 1.6507 = 1/.6057E _{M} = 110.12 E_{E}
So, we just substitute: .2723 g _{E} - 110.12 E_{E} = 1.62 m/s^{2}g _{E} - E_{E} = 9.8 m/s^{2}.2723g _{E} - .2723E_{E} = 2.6685 m/s^{2} [subtract the two equations]-109.85E _{E} = -1.0485 m/s^{2} E _{E} = .009545 m/s^{2}E _{M} = 1.051 m/s^{2}g _{M} - E_{M} = 1.62 m/s^{2}g _{M} = 2.671 m/s^{2}
We check that against our first postulate, and find that indeed the gravitational field is dependent on radius alone. R _{E}/R_{M} = g_{E} /g_{M} 6378.1/1738.1 = 9.81/ g _{M} g _{M} = 2.673 m/s^{2}
[In a subsequent paper I have confirmed this number .009545 m/s ^{2} for the charge field of the earth, in an unrelated problem with unrelated math. In my paper on atmospheric pressure, I calculated an effective weight of the atmosphere, as a percentage of the gravity field. Using novel but very simple math and diagrams, I found that the force down on any gas semi-contained in the curved field of the Earth would be .00958 m/s^{2}. Since this matches the force up, the atmosphere is effectively weightless. That these two numbers match with such simple math and postulates is one of the outstanding outcomes of my unified field theory, and I highly recommend you take the link, if you haven't already read that paper.] So we have achieved the golden ring. We have found actual numbers for our new fields. We see that the gravitational field of the Earth must be .009545 m/s ^{2} greater than we thought, since the rest must apply to acceleration caused by the E/M field. Even more shocking is the difference we found on the Moon. Why is the Moon's difference so much greater? It is due simply to my second postulate. Because the Moon is smaller, it is nearer in size to the E/M field it is creating. The field doesn’t have as much space in which to dissipate. Because the field is created from the surface of the body, it must be exponentially denser. This is precisely why the quantum field is so strongly electromagnetic, and so weakly gravitational. The Moon, being smaller, is nearer the quantum field.
We must be surprised that the affect is so great, just moving from the Earth to the Moon. The E/M field of the Earth is only a small part of the compound field, which goes a long way to explaining why we have ignored it. But on the Moon, the numbers betray a gigantic secret, very close to home. Physicists have assumed that the Moon's field must be proportionally weaker than the Earth’s, since the Moon is known to be almost non-magnetic, as a whole. But this has turned out to be spectacularly wrong. Even before my paper here, we knew that an E/M field continues to exist in the absence of the expression of its magnetic component. Venus and Mars exclude the Solar Wind just as if they had powerful magnetospheres, even though they do not. To see a fuller explanation of the Moon's E/M field, see my The Solution to Tides. I said above that I would show that my E/M Field Equation got the right answer, and now that I have all my numbers I can show that. We found that the total E/M repulsion created an acceleration of -.151m/s Let us move from the Moon to the Earth. We have famous experimental proof of my number for the Earth's foundational E/M field. The number .009545 m/s Now that we have made some progress in refining the gravitational field, let us return to the E/M field. If we look for equations to compare our new equations to, we don’t find any. I said earlier that I would have more to say about QED, but now the we get here, you will find that it is mostly of a negative sort. [For a more positive answer, you may now go to my paper on gravity at the quantum level and my paper on Coulomb's equation, both of which explain how the unified field works at the quantum level.] I have nothing at all to say about matrix mechanics, which, like Planck, I find "disgusting." But Schrodinger's wave equations don't give us anything either. This is because QED is not interested in describing the field as a whole, and it is especially not interested in describing the field in simple mechanical terms. It is mainly concerned with describing statistical interactions of quanta. And even when it gets beyond statistics, as with Schrodinger, it is concerned with the motions in a given field. QED mainly accepts the field of Maxwell, but adds some novel postulates that allow it to track the motions of quanta. This has many experimental uses and benefits, but theoretically it is a nearly total wash. No, it is even worse than that. QED, looked at theoretically, is worse than Maxwell's equations, since it is even more opaque.
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