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In this paper I thoroughly explore the subject of mass and weight. The last section of the paper is undoubtedly the most interesting and important, since it is there that I apply some simple math to data from the Moon and discover several astonishing things. In doing so, I begin to fill in the last major hole we inherited from Newton.
Implications
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A Redefinition of Gravity
Part VII
Mass and Weight
by Miles Mathis
milesmathis.com
mm@milesmathis.com
I do not see it as any demerit against Newton that he left this hole unfilled, since he did not have the data to address it. If he had been privy to the data I am privy to, no doubt he would have filled the hole himself, perhaps more thoroughly than I am able to.
This being said, the rest of the paper is far from being filler. I present several key ideas of expansion theory and answer many important questions.
To a reader who has patiently followed the first six parts of this theory, a main point may seem to have been ignored. This reader will now present me with two objects that are the same size but of different weights and ask me to explain the difference between them using my new theory. Let us say I am presented with an empty cardboard box 1 meter to a side, and a cubic meter of lead. According to my theory, both objects must be expanding at the same rate. If they did not, we would see one of them getting larger than the other one. At the same time, according to my theory, mass and gravity are supposed to be completely determined by real three-dimensional expansion. These two ideas appear to be impossible to maintain simultaneously.
I admit that it may look very bad for me when stated this way. This argument is, of course, the main argument against all expansion theories, and it is why most of them look so foolish. They cannot answer this question and the theory collapses forthwith. It is the strongest apparent argument against expansion, stronger even than the orbital argument. All expansions theories have so far fallen very quickly to one argument or the other; or I should say both.
However, I have answered the orbital question where no one else could, and you might expect that I have an answer for this one as well. Otherwise, why should I have bothered to write a hundred pages of theory only to end up at a brick wall? The problem is that so far everyone has reached this point and been satisfied with a very cursory analysis. At a glance, it doesn't make sense; in fact it appears absurd. If you are used to thinking about the problem as physics always has, then it is difficult to see that this problem is rather elementary, and rather simple to answer. But because the simple answer is not immediately apparent, everyone has so far retreated back to the shelter of the Standard Model, where they have the safety of numbers if no real explanation for anything.
But remember that the inverse square law also looked like an unclimbable wall. No one had scaled it and no one had thought it was scalable. I leapt it with four lines of high-school algebra (with some help from Relativity). I will make the same quick work of this problem.
In Part I, I hinted at a solution, though I did not provide the full math or theory for it. There I reminded the reader that my spherical acceleration must apply to the material spheres I had postulated. That is, the acceleration of matter applied primarily to the "building block". I said that the idealization I had put forward applied pretty closely to a proton, and that it would apply even more closely to smaller building blocks like quarks or mesons. I cautioned that the simple equation I developed for acceleration, which was basically an acceleration in the length of the radius, would not be applicable to large compound spheres like the Earth or Sun. In these objects, we would be dealing with a summation of internal accelerations, and would have to consider the material make-up of any large object, in order to understand its mass and size.
To better comprehend this, let us analyze closely the empty box and the cube of lead. By one measurement, they must have the same acceleration. They are the same size and stay the same size, therefore the acceleration at the surface of the two objects must be equal, according to my theory. By another measurement, weight, they are vastly different. But these two operations of measurement are not at all equivalent. The relative size of the two objects can be measured in any field. The relative weight of the two objects can be measured only in a gravitational or other force field. Remember that in a zero-gravity field, our two objects would weigh the same—nothing. The difference in weights or masses could be found only by applying a force to them both. But to measure their size in a zero-gravity field, all you need is a ruler: the same ruler you used in a gravitational field.
This means that in large compound objects, the acceleration of the surface of the object is not determined by its mass. The acceleration of the surface is determined only by the size and the structure. Let's look at why this is so before we ask how the total mass is determined by the sum of internal accelerations.
A cardboard box 1 meter to a side is mostly structure. It has nothing on the inside, of course, except air. In the box part, it has a solid latticework of molecules held together by atomic and molecular forces—mainly charge. The size of the box is therefore determined mostly by electromagnetic interactions. Now, if all the protons get bigger, then the atoms get bigger and the molecules get bigger and the latticework gets bigger. The box therefore would be expected to expand proportionally to the protons. So far so good. What about the lead cube? Well, we can use the same argument regarding size. The lead cube is also made up of protons and electrons and so on, and these quanta are assumed to be equivalent to those in the cardboard box. We have a lot more of them and the latticework is more densely packed with them, but any increase in the size of the quark will cause a proportional increase in the size of the cube. Therefore, if the total latticework of the cube was equal to the latticework of the cardboard box before any expansion, then we would expect they would be equal after the expansion.
Most people can accept that, since it is not the conundrum here. The conundrum concerns explaining the difference in weight once I throw out the gravitational field and replace it by acceleration of matter. If gravity is acceleration of matter, it would appear that my cardboard box and cube of lead would create the same gravitational field. The outer surface of the cardboard box creates an outward acceleration vector, a vector I have given a gravity-equivalence. When I created this theory, I turned the g vector around and gave it to a real acceleration. The cardboard box has a real acceleration outward due to its size. Why is its "field" not equivalent to the lead cube? The answer: in some ways it is.
What? A cardboard box is equivalent to a lead cube as a gravitational object? In the first instance, yes. We could say that they are strictly equivalent as gravitational objects, since they are accelerating outward at the same rate. All the differences are in secondary effects of this acceleration.
I know you will be dumbfounded at this assertion, but I stand by it. Bear with me for a moment. Look at the field we have created so far. It is not really a field, is it? It is a pseudo-field. We don't have any forces yet, not with the cardboard box or the lead cube. We just have a centripetal acceleration outward. We have a 360o motion. To generate a real field, we have to have some interaction with other objects. These interactions will give us forces and from these forces we can build a real field with real, physically meaningful, numbers.
So, let's throw a bowling ball into the "fields" of these two objects. Let us say we accept expansion theory and we imagine that we can now see these two objects expanding at some rate. The cardboard box's field looks beautiful until the surface of the box actually meets the bowling ball. When the two objects collide, what happens? The box collapses, of course. We see it wrap around the bowling ball and then it is likely ripped in half and left to expand as two cardboard sheets, flapping in the wind.
But the lead cube meets the bowling ball, causes it to bounce back a few times, and then the bowling ball sticks to the lead cube. Why? Not because the lead cube is attracting it, but because the lead cube is continuing to accelerate toward the bowling ball no matter where it bounces. After some time, the bowling ball will find it easiest just to sit on the surface of the lead cube.
So the difference in the two is not mainly one of gravity or acceleration, it is one of structure. That is to say, it is one of electromagnetic interaction inside the objects. Now, it is the atoms inside the object that create this structure, it is true, but the structure is an E/M structure. The more atoms you have, the stronger the structure, in general (although the particular form of the structure is also important, as we know). Therefore the degree to which an object resists a force is due only indirectly to the amount of matter contained by the object. It is due directly to the strength of the E/M latticework built up by this matter. Another way to put this is that the atoms generate the total E/M field, but it is primarily the E/M field that any object meets, whether it is meeting the E/M field at some distance from the object, or whether it is meeting the primary shell of the E/M field at the surface of the object.
This is only a very partial answer, I know, and I will continue to elaborate. Another major difference between the field created by the cardboard box and the lead cube is the E/M field external to the object. As I have shown in previous papers, this E/M field is an important part of what we currently call the gravitational field of an object. It determines the exclusionary field as well as any tangential torques. This field is generated by all the atoms in the structure of the object, so that it would differ greatly between the cardboard box and the lead cube. The electromagnetic presence of a cardboard box is negligible. The electromagnetic presence of a lead cube is orders of magnitude greater. In small ways, the lead cube would begin resisting the force of the bowling ball before it ever reached the surface. When you get to an object the size of a planet, then you get an electromagnetic presence that can begin to act like an extension of the object. It can hold objects off before they even reach the surface.
So, although the cardboard box and the lead cube have the same acceleration outward, they do not act the same in any event. It is these events that determine the new "gravitational field".
You will say, "That is all fine and good, but what about the weight? Why does the lead weigh more in the gravitational field of the Earth? You have taken away any attractive force from the Earth. In fact, all objects fall at the same rate, whether the Earth is accelerating up or whether the Earth is pulling things down. Why does the lead weigh more?"
This brings us to the Standard Model, which, if you will notice, has not really answered the question above yet either, even after three hundred years. I had my reader put the question in that form so that this would be a bit more obvious. As he said, according to current theory, all objects fall at the same rate. Just like my theory, they have the same acceleration. My theory gives the main part of the acceleration to the Earth and current theory gives it to the falling objects, but the question is the same. Why do the objects have the same acceleration in free fall but have different weights? Current theory skirts the question by giving us a definition instead of an answer. Current theory says that the object that weighs more has more mass. Weight is a force and F = ma. That definition is so ingrained in our minds that we forget to ask, "Yes, but what is it about the mass that causes the force to be different than the acceleration? A force is a mechanical interaction and so is an acceleration. But in adding mass to the equation, you have not shown a new mechanical interaction. All you have done is create a thing called mass." A thing with more weight has more mass. That is just word-juggling. It is like asking "what causes rain?" and being satisfied with the answer "clouds". What we want with rain is not a name for some "rainmaker" but a logical mechanical process. What we want with mass is not another name for weight, but a mechanical process that explains it.
Current theory then gets highly rigorous and calls greater mass "more particles". But this does not explain it either. "More particles" is only half an answer. A whole answer would be in the form of, "More particles doing something." More particles doing nothing is mechanically equivalent to fewer particles doing nothing, so that the current answer is just a ghost. It is once again obscurum per obscurius: explaining the obscure by the use of the more obscure. If you say that more particles resist a force better than fewer particles, this implies that a single particle can resist a force. And so I am allowed to ask, "How does a single particle resist a force? You are saying that the force resistance of a large object is a sum of the force resistance of constituent particles, but you must have something to sum." Current theory is just a summation of a definition. A summation of a definition cannot resist a force. You must have something to sum, and current theory has no motion, action, event, or sub-force to sum. Therefore the current theory is not a mechanical theory, it is just words.
Physics has finally began to recognize this. This is why they are now pursuing so hotly the Higgs boson and other similar mass-creating particles. The Standard Model cannot explain mass. The theoretical underpinning is non-existent. It would have to be non-existent to allow for the creation of new particles. If the current theory had any content at all, no one could insert a new particle at the fundamental level. Mass has so far just been a definition: a definition with no mechanical content. Mainstream theorists are trying to give the content to a new particle. I am giving it to new motion of particles we already have. The matter we already know about, made up of protons and electrons and so on, is accelerating outward spherically. This is what mass is, and no mass-carrying particle will ever be found.
This should have been known even before my theory arrived, since the idea of mass-carrying particles is absurd on the face of it. Logically it immediately creates a reductio, since if the boson gave mass to matter, we would have to ask what gave mass to the boson. An infinite regression is the outcome. It also creates two types of matter: matter with intrinsic mass—bosons—and matter without intrinsic mass—everything else. I can’t imagine what sort of scientist such a theory would satisfy.
As we have seen, my theory gives matter a basic motion and a basic emission at the fundamental level, so that I have something to sum. My theory is not just words. My spherical acceleration outward of quantum particles, combined with the foundational E/M field, gives me both an apparent attraction and a real exclusion. I have shown that the atomic orbit is a combination of basic spherical acceleration, an exclusionary field caused by bombardment (the E/M field), and spin. These three phenomena logically and mechanically create an equilibrium that is capable of resisting a force at the atomic level, and summed over all atoms, they create a solid E/M latticework that is capable of resisting a force at the macro-level. In this way, the summed accelerations of matter create an E/M field, and it is this E/M field that we call the macro-object. It is the elastic strength of this field that determines the resistance to a force, and that thereby determines the mass of the object.
In other words, the mass is not determined simply by counting up atoms. It is determined by summing the fields created by these atoms. The summed field will be at root an E/M field, and this field is the mass. In the final analysis, the field or object more strongly resists a force by being more elastic. Our cardboard box had less mass, but it is the interaction of this mass that determines the field. What we measure is not the "mass" but the "give".
I am trying to explain mass in a non-technical and conversational manner here, but let me pause for a moment to give you links to papers where I address each facet of this complex question with a bit more rigor. In my latest paper on charge, I show that all [spinning] quanta are emitting the foundational E/M field. So it is not just the acceleration of the radius that is creating the mass or the elasticity or the ponderability; it is this bombarding or exclusionary field. This emitted field is a field of real particles, so the force that creates mass is ultimately a force of bombardment. This emitted field (which up to now has been called the charge field, and has been mediated by the virtual "messenger photon") I now call the foundational E/M field, or the B-photon field, and it has a real mass itself. This mass of the field is in fact the charge. In another recent paper, I have even found the radius of the B-photon. In my Unified Field paper I show how this field fits into Newton's gravitational equation. Since it has mass, it must do so, and it does so in a quite simple way, without any difficult math. In short, the mass variables in Newton's equation can be written as a volume times a density. The volume we give to the gravitational field; the density we give to the B-photon field. G then acts as a transform between to the two fields. In this way, mass actually becomes a compound effect. Without the emission of the foundational E/M field, we could not calculate a mass. The force of exclusion is created by this emitted field, and mass is more directly an attribute of the emitted field than it is of the gravitating particle.
You will say that I am replacing inertial mass with some form of rigidity or impermeability, and that is mostly true. Once I replaced the idea of "ponderable mass" with the idea of a spherical acceleration and emission, all need for weight, mass, heaviness, etc. was over. We will keep these ideas because they are human feelings, just like sourness and blue and so on. But heaviness is no longer a physical parameter, not even in the form of inertial mass. You don't have to think of each atom tied to the space fabric in order to imagine inertia anymore. We already accept that acceleration causes force. If we give each atom acceleration outward, we have automatically given it a force in every direction. We no longer have to go on beyond this and continue to make up explanations like inertia or a connection to the space fabric or any of that nonsense. Some will have difficulty knocking those ideas loose from their brain, but there it is.
The only way an atom or a proton or an electron with a spherical acceleration would not have a force in every direction is if the shell of the atom were infinitely collapsible, if it had zero rigidity. As long as it is radiating, it cannot have zero rigidity, since radiation acts as both exclusion and a sort of impermeability. Expansion and radiation therefore take the place of ponderable mass. This is a great trade, since mass has no measurable parameters, except mass itself. But rigidity can be measured as a function of radiation. It is a real mechanical property that we can easily give numbers to. Rigidity is not just a word, it is now a physical concept that can be expressed with physical variables like time and length (which give us field density). When two objects meet and one hits harder than the other, we no longer need to beg the question by giving the reductive argument that one contained more particles. We can explain the mechanics of the difference, since we can now show that more particles emit more and stronger E/M fields and these E/M fields create greater rigidity.
Weight in my theory is still a force, but force in my theory is not mass times acceleration. It is acceleration times expansion/exclusion. If I weigh more than you, it only means that my E/M field is denser than yours. My acceleration will therefore hurt you more than your acceleration will hurt me. You will bend sooner than I will, and you will break sooner than I will.
Other questions arise, so I will continue. A reader may say, "Go lie on the floor with the cubic meter of lead resting on your stomach. Now tell me that the difference between that lead and an empty cardboard box is only one of rigidity." A very colorful example, but I have to maintain my position, painful as it now is for both of us. I feel a lot of pain for two reasons: 1) I am accelerating toward the cube and it is accelerating toward me and the earth is accelerating up behind me and I am accelerating toward the earth. Even though the earth's acceleration mainly trumps all the others, I am caught in quite a pinch, literally. 2) The floor is rigid and the cube is rigid, but I am not. I may weigh quite a bit, but most of that is squishy water, and so I feel a lot of pressure. Humans are not great weight bearers, and everyone knows this. Rest the lead cube on the floor, and the floor bears it with much less implied pain. Why? Not because dead wood feels no pain, but because dead wood is more rigid than I am. It isn't squashed as much and therefore wouldn't feel much pain even if it could feel pain. Weight-bearing is a matter of force-bearing which is of course a matter of the type of rigidity I am talking about.
"OK," says my reader, "Now put the cardboard box on your stomach. It has the same acceleration down as the cube, according to your theory, and you and the Earth have the same acceleration up as before. But you feel nothing. Are you telling me this is all a matter of rigidity? Put the lead cube on a scale and you are telling me that the scale is measuring rigidity?" Amazingly, yes. Lack of rigidity is like a sponge that soaks up force. The Earth, rushing up, attempts to deform me, and it does so—part of its force being lost in compressing my molecules and thereby counteracting their E/M forces. I therefore press back on the Earth less forcibly than it does on me. That is why 1) it squashes me, 2) it moves me through space. Expended force of Earth up = force of me down + force lost to squashing me + force required to move me through space.
Therefore if I were more rigid, I would weigh more. But to become more rigid I would need stronger E/M bonds. To have stronger E/M bonds I would need more molecules and atoms. If I had more atoms, I would have more mass, by current theory. So you can see that my theory is not so strange after all. It is only giving mass a real motion that mechanically creates an E/M field that mechanically creates rigidity.
The lead cube weighs more than most other cubes the same size because it dissipates less of the acceleration up of the Earth into its atoms and molecules. It returns almost all the force back to the Earth. This is what the scale measures. Another way to put this is that the lead cube hits back. What does it hit back with, specifically? Its own foundational E/M field, which it is emitting all the time. This B-photon field is much much denser than the field emitted by the cardboard box, and that is the mathematical difference.
A reader may still not be satisfied. He will say, "So if the lead cube weighs a thousand pounds and the cardboard box weighs 1 pound, you are saying that box is dissipating 999 pounds into its molecules?" No, not at all. The box meeting the earth never has 1000 pounds of force to deal with or to dissipate. Weight and force are not created by adding up accelerations alone. Weight and force are created by a combination of acceleration and rigidity. The box does not have the rigidity of the lead cube, so that if we multiply that rigidity by the acceleration, we come up with something much less than 1000 pounds. In fact, we come up with just 1 pound. In this sense, weight is not the force, it is the resistance to the force. Both weight and mass are resistance to a force. Mass and weight have an even greater equivalence than was always thought, since mass, like weight, cannot be measured without a force. It can be calculated without a force, if we know an E/M field density, but it cannot be measured without a force.
Another reason my reader's analysis is faulty is that it assumes that both the box and the lead cube must resist all the acceleration and rigidity of the earth. But of course they don't. That would be a huge force to dissipate for either object. Either object encounters only the amount of rigidity in the earth that the earth encounters in them. The rest of the force goes into moving them, or it is never encountered at all. Remember that we don't have an equilibrium of forces here, not with the box or the cube. An equilibrium would imply that the box is pushing on the earth as hard as the earth is pushing on the box. But this is not so. The earth is accelerating the box through space, and the box is just along for the ride. The box is not moving the earth at all, obviously. So the box does not have to dissipate the mass of the earth, nor does it have to dissipate a mass equivalent to the lead cube. It only deals with "dissipating" the effects of its own mass or rigidity.
The force felt is determined by the object's own mass or rigidity, not by the earth's. That is true in current theory as well. The force on a cardboard box is not F = Ma, where M is the mass of the earth. An object does not encounter all the mass or rigidity of another object; it encounters only, at most, its own mass or rigidity, and the total acceleration at the point of contact. What I mean by "at most" is that if the force exceeds its own rigidity, it will deform. If it deforms, then some of the force is "dissipated." This, too, is already known and accepted. It is why we don't weigh things on pillows or deformable platforms. Some of the weight gets dissipated in that instance.
One final subtlety. My argumentative reader will say that I still seem to be basing weight on volume. What if we let my box collapse, so that it is lying flat on the ground? Its mass and weight should be the same, but its volume is near-zero. How does my theory explain this? Or, what if we take the box and crumple it up into a ball? We have the same weight and mass, but a smaller volume.
I will take the second question first, since it is a bit easier to explain. I have defined mass as a mechanical function of E/M rigidity and acceleration due to expansion, therefore let us assign some variables.
AE = acceleration due to expansion
R = rigidity
This means that force would now be expressed
F = gRAE
If weight is this force in a gravitational field, then the crumpled up box is pretty easy to explain. We have a smaller volume (or average radius, say) than the original box, which gives us a smaller acceleration due to expansion. But we have greater rigidity, since we have forced more of the same E/M field of the box into a smaller volume. So that the quotient of RAE stays pretty much the same.
The same general analysis applies to the box lying flat on the ground. What we lose in volume we gain in rigidity. Our box will not have zero volume from a microscopic point of view, even if we can no longer put anything in the box. Its usable volume, from a packer's point of view, may be zero. But its physical volume is not zero. The average radius has not changed as much as you might think. Therefore we do not have to imagine that the box gains a near-infinite amount of rigidity. But obviously it gains some rigidity, since it can no longer collapse vertically on a macroscopic scale.
The analysis above forces us into another consideration. I have given g in the last equation to an actual spherical motion of the earth. In my theory, the earth is really accelerating outward at that rate. But this means that g must be studied a bit more carefully now. I have shown that force depends on rigidity as well as acceleration, so that the equation becomes even more complex. We have taken into account the rigidity of our object, but we have not taken into account the rigidity of the earth.
If we are talking about the gravitational field of the earth at some distance, then the rigidity at the surface of the earth obviously doesn't matter. We can look at the "rigidity" of the E/M field of a massive object at some distance from that object, but that is an entirely different subject than the one we are addressing here. For now we are only concerned with the rigidity at the physical surface of the object. And if we are using the variable g at the surface of the earth, then we must take into account the rigidity of the earth's surface or the rigidity of the prepared surface where the smaller object meets the earth. We have always known this. The only difference is that I am showing that this rigidity is not an effect external to the mass of the object in question, whether that mass is a cardboard box or the earth itself. A softer object will act like a less massive object in the transmission of force, and this fact is critical. If we re-define mass as expansion times rigidity, then a softer object does not just act less massive, it is less massive. If mass is resistance to a force, then it is the strength of the bonds we are measuring, not the number of atoms.
Of course, by "softness", I don't just mean the elasticity of the surface. I mean the elasticity of the object all the way down to the quark.
You can see that my theory mirrors current theory in many ways. But there are major differences. Several of these are already apparent, but there are a few that I haven't yet touched on. Some readers will have noticed that my theory implies that mass can vary due strictly to molecular composition or atomic composition, and this is true. It is one of the basic predictions of my theory, in this regard. Meaning that I predict that it will eventually be proved that mass is not just a summation of atomic masses—an adding up of all the masses of the protons and neutrons and electrons. Rather, mass is a summation of bond strengths at all levels. These bond strengths directly create the rigidity I have been talking about. In general, more atoms will create more bonds which will create greater rigidity. But in some circumstances this will not be true. In some cases an object with fewer molecules or fewer atoms would actually have more mass and weight than an object with more molecules or atoms. Since some molecular structures are more rigid than others, these micro-structures must create better weight-bearing and weight-causing macro-structures.
This will be offensive to many, I know, since it initially seems counter-empirical. For example, we can take a block of wood that weighs very little and show that it can bear a lot of weight. It is therefore very rigid. So we have high rigidity and low weight, which would seem to tell very strongly against my theory.
But this only means we must differentiate between cell structure, molecular bonds and atomic bonds. We already know that there are many levels of structure in any material object. We have so far seen examples of macro-structure, molecular structure, atomic structure, and sub-atomic structure. There must also be many levels of rigidity. The block of wood is rigid in its outer structure, but no so much in its inner structure. It therefore resists low forces well, but resists higher forces very poorly. Once the block of wood is physically crushed, it loses most of its rigidity. Whereas a piece of lead might initially be softer, but once you compress its molecular structure, its atomic structure is very rigid.
My difficult reader will reply, "Well, if the wood is more rigid at first, why does it not weigh more at first? How does the earth know? Mass and weight should be measurable from the first touch. But how does the earth know how well the wood will resist the force once it has been crushed before it has been crushed?"
A logical question, but one that is once again easy to answer. The earth knows because the earth is applying a gigantic force—from its direction—from the first touch. The force down on the scale is determined by the expansion and rigidity of the smaller object. But the force up is near-infinite in comparison.
Or I should say the potential force. Think of it this way. You press your hand against the wall. It presses back equally, since nothing moves. You press harder and harder, and the wall's force keeps climbing, too. It has a potential force determined by its structure. Compared to your hand pressing, this potential force is gigantic. You are no good measurement of it, but it is a good measurement of you. It knows you from the first touch. Why? Because it really can "measure" you all the way down to the atomic level from the first touch. It does this because the amount of force it returns is determined by your rigidity all the way down to the atomic level (and perhaps below). For instance, the amount your palm compresses is determined not only by the strength of the wall, but also by the strength of your atomic forces. In a very real sense, your molecular forces are caught between the molecular forces of the wall and the atomic forces of your hand. The atomic forces in your hand are stronger than the molecular forces. This determines that all the compression must go to the molecules (and cells) in your hand. Only if your molecules are compressed to some limit, will any atomic compression begin. Therefore, your atomic compression is known by the wall from the first touch. Every level of both objects knows every level of the other object from the beginning, since all these levels determine the interaction and the resultant force. You immediately obtain a resultant force, therefore the measurement of all forces must also be immediate, even when no actual compression is taking place. Your atoms may not be compressing at all when you press up against the wall. But their strength is known, because this strength in part determines the real compression of your molecules.
In the same way, the earth knows the rigidity of the wood at all levels from the beginning. The rigidity at all levels of both objects (earth and wood) determines the resultant force, the amount of force lost to compression, and the weight. The rigidity at all levels therefore also determines the mass. My reader is confused only because he has taken rigidity to mean only cellular or molecular rigidity, or the sort of rigidity that wood exhibits in the first instance. My use of rigidity goes beyond this meaning, and includes all possible levels of rigidity.
For instance, we know about cellular and molecular rigidity, since that is what we have been mainly talking about. But there is also atomic rigidity, determined by the atomic E/M field. I have given a mechanical explanation of the strength of that field as a function of nuclear expansion and bombardment by radiation. The nucleus keeps the electron at a distance by bombarding it with photons, in the simplest possible visualization.
All these exclusions at all levels determine the mass of the macro-object. These exclusions are certainly a function of the number of ultimate quanta in the objects in question. But they are also a function of the strengths of the fields and bonds at work, so that we can no longer say that mass is strictly a measure of the material present or of ponderable matter. Mass is not a sum of quantum masses or something like that. Mass is a sum of bond strengths and exclusionary strengths at every possible level.
Gravity on the Moon
as proof of Expansion Theory
Now let us move on to another question. My reader will have noticed that I have stated outright that the gravitational field is not a function of mass. Of course this contradicts both Newton and Einstein. Furthermore, it contradicts all current models of exotic celestial objects like pulsars, white dwarfs, and black holes. I stand by my statement, however, and back it up with data much closer to home. I remind the reader that all estimates of gravitational fields of exotic objects are based on speculation and calculation from various assumptions. Obviously, no one has dropped an object in the vicinity of a black hole or a white dwarf and directly measured the acceleration in freefall. We have, however, traveled to the Moon and dropped objects there. What we have found is that the Moon's gravity at the surface is about 1/6th that of the gravity at the surface of the Earth. I now offer you the fact that the Moon's mass is 1/81 that of the Earth and that the diameter of the Moon is .272 or 1/3.672 that of the Earth. It does not take a genius to see which of these numbers is closer to the other.
Current theory explains the discrepancy by measuring the two field strengths at the same radius. Therefore we must imagine an object at 3.672 Moon-radii. If acceleration is proportional to the inverse square of the distance from the center, then:
9.8/6//x = (3.672)2/12
x = .12 m/s2
Thus, the acceleration at that distance from the Moon would be 1/81 that of the Earth. Current theory might justly add: "We never said gravity was a function of mass by itself. According to the equation a = Gm/r2, the acceleration is a function of radius and mass at the same time, which means that it is a function of density. Your object freefalling at 3.672 Moon-radii must be compared to an object falling near the Earth's surface. Only then are both the mass and the radius taken into account."
All very true, except that current theory fails to notice two things. In my paper on the Universal Gravitational Constant I showed that quantum masses were determined by a fundamental acceleration of the radius, by the equation
m = 4Δr3/GΔt2. By substitution we find:
a = Gm/r2
m = 4Δr3/GΔt2
a = 4Δr3/r2Δt2
After any appreciable amount of time, r becomes negligible relative to Δr, so that we may ignore it.
a = 4Δr3/Δt2
This means that the acceleration is a function of the radius alone. Current theory keeps using density to explain proportional differences, but it has not yet recognized that the complex equation it is using can be simplified. The data from the Moon should have been a flag announcing this, but no one got the signal. Because 1/3.67 and 1/6 did not match exactly, everyone has assumed that the two numbers are a coincidence. The Standard Model has assumed that gravitational acceleration cannot just be a function of radius, since none of the equations said so, and the data did not show it incontrovertibly. Newton's equations have always contained a variable called mass that got most of our attention. It got our attention despite being a variable with no mechanical definition, as I have shown. The concept of mass has been a sort of highly successful fudge-factor that no one ever recognized as chocolate. It has such a prominent place in our equations that no one ever considered that it might be superfluous. No one wants to lose it, so no one even tries to lose it. We are in love with our own heuristics.
But if you look closer, you will find that there is a closer mathematical connection between the radius of the Moon and its gravity. If you re-analyze the whole situation using my theory, you find that the discrepancy between 1/6 and 1/3.67 can easily be explained. That is, we assume, using the postulates of my theory, that the Moon and Earth are expanding at the same relative rate. Otherwise we would see the Moon getting bigger. If the Earth is expanding at 9.8 m/s2, then it will double in radius every 19 minutes:
r = at2/2
6,378,000 m = (9.8 m/s2)t2/2
t = 1141s
The Moon must also double in radius in this time.
1,737,000 m = a(1141s)2/2
a = 2.668 m/s2
According to my theory, the Moon should have an acceleration that is 1/3.67 that of the Earth, just like the radius. Why doesn't it?
My answer is: it does. If you could measure the acceleration of the surface of the Moon outward, independent of all other forces and fields, that acceleration would be 1/3.67 that of the Earth. But you can't measure that number independently. Both the Earth and the Moon have complex fields that are a mixing of the "gravitational" field and the E/M field. The Moon, like the Earth, is both moving toward and repelling any object in its field. The weight of an object in this dual field must take into account both the gravitational field and the E/M field. The Standard Model knows that the E/M field must apply an exclusive force at any distance, but in calculating weight this model ignores the E/M field completely. The Standard Model acts as if weight is only a function of gravity. But it is not. An object, whether it is sitting on the Earth, sitting on the Moon, or is in freefall over either body, must be receiving a force from an E/M field. This force is not negligible when we are talking about a body the size of the Earth or Moon. Current theory recognizes that the E/M field totally swamps the gravitational field at the quantum level, but it assumes for some reason that it is near-zero at the macro-level, even in the vicinity of very large masses. Actually, it knows very well that this is not true, but in calculating weight, which is a force, current theory forgets to include the E/M field in adding up that force.
Think of it this way: let us say you have a one-kilogram weight on a scale on the surface of the Earth. In either my theory or current theory, that weight is simultaneously in two fields—the gravitational field and the E/M field. But current theory doesn't really understand the mechanics of the E/M field. It assumes (correctly) that the field is created by photons, but it thinks that these photons are "messenger particles" that can somehow give different messages to differently charged particles. Conversely, I have postulated that the E/M field is simply an exclusionary field. It is created by the (statistically) unidirectional emission of a huge number of photons. These photons bombard all particles indiscriminately, electrons and nuclei alike. But because electrons are so much smaller, they collide with far fewer photons. The E/M field therefore drives the free electrons much more slowly than the nuclei and the other larger particles. The field is not like a stream that floats all micro-particles at the same rate. It is a field of discrete quanta, which means that the electrons can dodge large parts of the stream. This makes them appear to move against the stream, in a negative direction to the stream of larger particles.
If this is true, then the major effect of the E/M field is a force outwards from the center of the large body. The gravitational field attracts (or, in my theory, appears to attract); the E/M field repels. In Part II of this series I have already used the exclusionary power of the E/M field to repel the Moon away from an Earth that is accelerating toward it in order to create the orbit, and I now use the E/M field to explain why the Moon fails to have .272 the weight-causing force of the Earth. If we go back to the 1 kg weight on the scale, we see that the weight is being bombarded with millions of photons. The Earth is repelling it with this bombardment, and the repulsion is not negligible compared to the gravitational "pull". If you took away the E/M field of the Earth, the weight would actually weigh fractionally more than 1kg. The weight we actually measure is the force due to gravity minus the force due to E/M. The weight is a result of the summation of both fields.
You will say, "Even if this is so, you still haven't shown why the Moon fails to have 1/3.67 the gravity of the Earth. Surely both the Moon and the Earth are bombarding, so that if we have failed to recognize it, we have failed to recognize it for both objects. Nor can you explain it by arguing that the Moon has little or no magnetic field. This would make your problem worse, in fact. What you need is more bombardment by the Moon, not less."
Precisely. I would never argue that the Moon is not bombarding because it shows less magnetism than the Earth. Lack of magnetism does not imply lack of an E/M field. An object can be strongly electrical without being strongly magnetic. I have already shown that Venus acts like an object with a large E/M field, even though it has very low magnetism. Mars, also. Plasma research has shown us that massive objects have active E/M fields even in the near absence of magnetism or magnetospheres. I could not have explained the orbit of Mar's moon without giving Mars an E/M torque just like all the rest. You are also quite correct to point out that I must show that the Moon has greater E/M bombardment than the Earth. It does so for this reason: I have shown that the gravitational field varies as the radius. This means it varies ONLY as the radius, which means it no longer obeys the inverse square law. Yes, Newton's equation still obeys the inverse square law, but it does so only because I have shown that it includes the E/M field. In my Unified Field paper I have shown that Newton's equation is a compound equation. There, I have actually done the math to express my claims in these Third Wave papers. I have separated the equation into two fields. I have un-unified what was already unified, so that I could show exactly how it is put together. It is the E/M field that obeys the inverse square law, and it does so simply because the field diminishes with distance from the center. The E/M field is a true field, made up of sub-particles, and in calculating a force, one must sum the field. This summation requires a field density, and this is where the inverse square law enters Newton's equation. Beyond that, I have shown in part 3 of this series that Relativity provides a second, separate, inverse square law, due to the time differential. This second inverse square law is not really applicable to Newton's equation, since Newton's equation is not relative, but whenever we apply force equations to any large system, we must include both inverse square effects. Since we want to keep gravity dependent on radius alone, we apply both inverse square effects to the E/M field.
You may also think of E/M field like this: in a spherical E/M field, the field lines do not remain parallel. The distance between the field lines increases as the radius increases. This is because, if the field is mechanically expressed by photons, and photons are discrete particles, the field must become less dense as the radius increases. If the object is producing x-number of E/M expressing photons during each dt, then the distance between photons must increase at greater radii. If the photon has any characteristics of a discrete particle—and we know that it does—then a spherical E/M field must become less dense at increasing radii. How fast would a bombarding field created by a discrete particle be expected to drop off if the field is spherical? Well, the surface area of a sphere is 4πr2. Which means if you double the radius, you quadruple the surface area and quarter the density. Hence the inverse square law.
You will say that if the gravitational field is expressed by the graviton, then that field must also become less dense at greater radii. But I deny that the acceleration field is expressed by the graviton. I deny it not only because it contradicts my theory, but also because it now contradicts the inverse square law. I have shown that the inverse square law is caused by Relativity [see Part III]. If you add a drop off in the field due to the introduction of a discrete mediating particle like the graviton, then you destroy the inverse square law. You would then have two drop-offs, one caused by Relativity and one caused by the considerations I have just shown. The field would no longer vary with the inverse square, it would vary at some other rate. But we know that it does not vary in any way but with the inverse square. Therefore the graviton cannot exist. I have just shown one more fatal argument against the graviton.
I have shown two inverse square laws. If we apply them both to the E/M part of the compound field (as the only real field present) this makes the force of the E/M field proportional to 1/r4.
Now we can return to the Moon. I have developed a theoretical mechanism to explain a proportionally stronger E/M field for the Moon than for the Earth. The question is, is it enough? The problem is that my theory as stated here hasn't offered a number for the strength of the E/M field of the Earth or Moon. All I have stated is a method for determining relative strength. But if we don't know how much of the Earth's force field is determined by E/M, then it is hard to say how much of the Moon's is. Let us see if we can develop some equations to express it.
We know that the total field of the Earth at its surface creates an acceleration of 9.8 m/s2 and we hypothesize that this is the gravitational field minus the E/M field. And we know the same for the Moon.
gE - EE = 9.8 m/s2
gM - EM = 1.62 m/s2
I have also postulated that the gravitational part of this acceleration should be proportional to the radii.
gE / gM = 3.672
gM = .2723 gE
And I have just postulated that the E/M field is proportional to 1/r4.
EE /EM = 1/3.6724 = .0055
EM = 181.81 EE
But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density.
DE /DM = 5.52/3.344 = 1.6507 = 1/.6057
EM = 110.12 EE
So, we just substitute:
.2723 gE - 110.12 EE = 1.62 m/s2
gE - EE = 9.8 m/s2
.2723gE - .2723EE = 2.6685 m/s2
[subtract the two equations]
-109.85EE = -1.0485 m/s2
EE = .009545 m/s2
EM = 1.051 m/s2
gM - EM = 1.62 m/s2
gM = 2.671 m/s2
Once we correct for the presence of the E/M field, the Earth and the Moon have gravitational fields that are exactly proportional to their radii. We did not get an exact match in the third decimal place only because we used 9.8 m/s2 for gE in the first equation. We must now add .009545 to that, and if we do we get 2.671 m/s2 in the first equation as well.
You will say that we have tested the fields on the Moon already and found them to be quite small. There are two problems here. One, our tests were designed to measure local fluctuations in the E/M field, and especially the magnetic component of that field. This is not the same thing as measuring the strength of the entire field at a distance. Two, the tests of the E/M field are compromised just like all our tests of the gravitational field have been. In neither case have we been successful in separating the effects of the two fields. Whether we are measuring a gravitational field or an E/M field, we must measure a force on a body. But the force on the body is a composite of the two. A differential. If we do not take this into account (and we don't) there is no way we can know what the strength of each field is alone. We would have to block one field or the other in our measurements, and we have never done this. According to my theory, you cannot block the field of gravity, since it just a real acceleration. You cannot block an acceleration. But the E/M should be blockable. If it is the radiation of photons, we should be able to block this radiation. It is unclear how successful this blocking might be expected to be. If you block off a small area of an E/M field, it is doubtful that you can thoroughly block the effects of that field. However, it should be possible to design a simple experiment that would test my theory. Dropping ball bearings above a large sheet of lead would be a beginning. It may be that an experiment that direct would yield an acceleration measurably above 9.8 m/s2.
Notice that the number I have arrived at for the E/M field of the Earth is quite small. This explains why it has always been neglected. Physicists have correctly assumed that it was negligible in most cases, and they went on to assume the same for the Moon. Why, they thought, would the Moon have an E/M field that was more active at the surface of the Moon than the Earth's E/M field is at its surface? The idea was counterintuitive, so no one has ever done any math to show it one way or another. I have just shown, using postulates that are not too difficult for even the mainstream to accept, that the Moon's E/M field should be expected to offset its gravitational field quite strongly. This finding applies regardless of whether you accept expansion theory or not.
You will say that if it applies regardless of expansion theory, it can hardly be strong evidence in favor of it. But this is not true. You do not need to accept expansion theory to see the logic of my postulates and math, but once you see how the calculations prove that the gravitational field is proportional to the radius, you have to re-examine the Standard Model. It would be a giant coincidence if the gravitational field just happened to vary as the radius. And it would be a giant coincidence if the numbers from the Moon fit this proportionality and other celestial bodies do not.
The math above implies very strongly that all celestial bodies, including exotics like black holes and white dwarfs, also have gravitational fields that vary as their radii vary. It suggests in the strongest possible way that the huge additional forces hypothesized for exotics are mainly a function of a super-strong E/M field, and have nothing to do with gravity per se. This means we must reconsider all our theories for exotics, and indeed for non-exotics. Our theory has existed with a very large hole in it and now we must re-calculate many things. I think it would be an extraordinary coincidence if the Moon and the Earth had gravitational fields that varied as their radii and other objects did not. It cannot be because they are of similar origin, since we know that their densities differ. If density is not a factor between the Earth and the Moon, then why should it be a factor in other objects? The only reason we have finally discovered this secret with the Moon is that its statistics are entirely more settled and complete. We have been perfecting this data for centuries, and the data finally bears fruit.
Some will say, "What do you mean density is not a factor? You had to correct for it, it must be a factor!" Yes, it is a factor in the E/M field. A denser object creates a stronger E/M field and I had to correct for the fact that the Moon is not as dense as the Earth. But this correction did not affect the gravitational fields. That is what the final number shows. I predicted that the gravitational field was a straight outcome of the radius, with no other factors involved. That is where we got the first number 2.671 above. That first equation has no correction for density. It is radius and nothing else.
If my postulates here are true, the implications are, of course, beyond number. I could not begin to address them here, even as a list. I begin to address them in other papers, but it will take physics decades to come to terms with the full import of this discovery. Those who have claimed that physics is nearly over will be glad to discover that they have something left to do.
[March 25, 2008: Go to An Update on Weight for more on this topic.]
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