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HOW NEW TRANSFORMS First written October 2004
In this paper I will derive new transformation equations for mass, momentum and energy. I will show that Einstein, despite using a thought problem that was useful and mostly correct in variable assignments, made several crucial errors that compromised his final equations. The thought problem I am mainly concerned with here is in his short paper of 1905, [1 – (v’^{2}/c^{2})] You can see that the difference is very small in most situations*, and might pass for decades without final experimental confirmation, especially in a milieu that considered Relativity a settled question. Physical Review Letters, the primary publication of record in the US, doesn't even have a category for Special Relativity. A scientist could not present a finding if he had one. *In comparing kappa and gamma, it is also important to note that I prove below that Einstein's thought problem is not directly analogous to the more common experimental problem of a sub-atomic particle in an accelerator. I show that the energy equations must vary from problem to problem, depending upon the physical situation. In deriving this new transform I also discovered several other facts of great interest. One of these is that E ≠ mc ^{2}.If we assume that the rest energy is given by the rest mass—as in E _{r} = m_{r}c^{2}—then the moving energy cannot be given by the moving mass, in a straight equation. A transform is required here as well, and it is not gamma. This is a consequence of Einstein's own variable assignments. Einstein assumed, with no theoretical or mathematical backup, that mc^{2} must be the term that is applied to the final energy E. It turns out that this is not the case. Even more astonishing is that using my new derivation, where all the variables are rigorously assigned, I am able to prove that the classical equation is precisely equivalent to the relativistic equation. In other words, K = κm _{r}c^{2} - m_{r}c^{2} = mv^{2}/2Simply by correcting the math of Einstein's own thought problem, I arrive at a new energy transform κ that is nearly equivalent to γ. This new transform allows me to derive the classical equation directly, by a straight substitution. In doing so, I prove that the classical equation is not an approximation at low speeds, as has always been assumed. It is an exact equation. The binomial expansion of the differential in gamma is a manufactured proof, since gamma itself does not exist as a correct transform in any part of Special Relativity.
In an earlier paper, I derived new transformation equations for time, distance and velocity. My central transform there was α = alpha = 1/[1-(v/c)] = 1 + (v'/c), which replaced gamma. Interestingly, the term that I call alpha is commonly used in optics to transform the frequency of light. I recently found Richard Feynman using it in a proof of Relativity (Feynman Lectures on Gravitation, lecture 7). So even the status quo should have been surprised to find Einstein using gamma to transform light frequency as he does. No one, apparently, has ever seen the contradiction in this until now. Before I get to Einstein's thought problem, I must first gloss the findings of my earlier paper, since they are crucial to understanding this paper. In that paper I showed that Einstein misunderstood his initial coordinate system and variable assignments, so that his transforms end up being unassignable. He applies gamma to his time and distance transforms, in this way:t = γt' and x = γx'. Unfortunately, these transforms are not correct. In his various thought problems—the most famous of which is the man on the train—Einstein has three coordinate systems. He has the man's system, the train's system, and the system of the platform, for example. But he tries to solve from only two systems. In his equations, he has only a primed system and an unprimed system, but no double-primed system. At the end, when he finds t = γt', he has mistaken a transform from the man to the platform for a transform from the train to the platform. Einstein completely ignores the direct transform from the platform to the train. His given velocity v is the velocity of the train relative to the platform, he tells us. But he does not say whether this is the velocity as measured from the train or from the platform. The two measurements must be different, but Einstein never includes this difference in his calculations. t = γt' therefore applies to a transform from the platform to the man, which is in fact a transform of two degrees of relativity. He never provides transforms for one degree of relativity. t = γt' should read t = γt'' However, I show that gamma is incorrect for two degrees of relativity as well. I was the first to demonstrate first-degree relativity, as well as the first to offer transforms for it. t = αt' and x = x'/α I was also the first to offer corrected second-degree transforms, although these do not enter into mass transform solutions. My first-degree transforms are in inverse proportion between x and t, whereas Einstein's were in direct proportion. His mistake came about by borrowing the light equations of Lorentz, x = ct and x' = ct', which I have shown are incorrect. Finally, my discovery of first-degree transforms allowed me to derive a first-degree transform for velocity, which Einstein never derived. His transform for velocity is for two degrees of relativity, v to v'', as he admitted, and as has never been questioned. Until my paper, there had been no v', nor any idea that it was necessary to the solution. In this current paper, I show that the correct mass transform must be derived from one degree of relativity, using v'. Einstein was not capable of this solution, since he did not have a v' in his choice of variables.
Now let us proceed to the mass and energy transforms. The best place to start is with Einstein's second paper of 1905, _{0}c^{2}}√[1 - (v ^{2}/c^{2})]Einstein says that the second term (the term on the right) applies to the rest energy of the particle. There is no reason to do this. It is an equation with two terms, but the terms are not divisible or singly assignable without a very compelling reason and a full explanation. I have shown that the two terms are simply the outcome of a finessed equation. There is no necessary physical reality to either term. To clarify this, let's look again at the light-plane problem. At the end Einstein finds that K _{0} - K_{1} = γ L - LIf L = m _{0}c^{2} , then what is the value of E_{0}? Let us see. E _{0} = E_{1} + L The initial rest energy = the final rest energy + the rest energy? This only makes sense if the final term is understood to be the change in rest energy. L is actually the mass equivalent of the planes of light. Even if Einstein's final equation were correct in form (it isn't), it would imply that m _{0}c^{2} is the mass equivalent of the light, not of the body. But this is not what m_{0}c^{2} means in current energy equations. Now let's look at the calculus derivation of E = mc ^{2} from a current textbook. It follows Einstein pretty closely—meaning it makes all the mistakes he makes, and then adds a few of its own. The problem for the textbooks is that they try to clarify some of the things that Einstein purposely kept in the shadows. They try to apply real math to things that Einstein simply glossed over. Unfortunately, they are no more thorough than he was.The first thing they do is state that they take the Work-energy Theorem as still valid in Relativity. This is false. It is valid, but not with the same equations. Remember that Einstein throws out the classical equation for kinetic energy. In fact, the textbook finds, at the end of these very equations, that E ≠ mv ^{2}/2 . But it assumes, for some reason, that the integral based on this equation is valid!W = ∫ Fdx = ∫ dpdx/dt = ∫ vdp But this is absurd, since according to Einstein, energy is not expressed in the same terms in relativity as it is in classical mechanics. The integral ∫ Fdx works because F = ma and v = x/t , etc. From this they get W = ∫ dpv - ∫ pdv but p ≠ mv in relativity. You can't just juggle the same old variables for a few steps and then suddenly introduce a mass substitution to make it all right. But this is what is done. W = mv ^{2} - ∫ mdv_{0}v √[1 - (v ^{2}/c^{2})]= mv ^{2} + mc^{2}[1 - (v^{2}/c^{2})] - m_{0}c^{2} and so on Notice that if we so much as lose the square root of the gamma term, then the integration is ruined. You have no third term from v = 0 as you do in the current integration. So you have no m_{0}c^{2} term. But it does not matter since the integration was compromised long before that.The textbook I have in hand says this (which is typical): "We call mc ^{2} the total energy of the body, and we see that the total energy equals the rest energy plus the kinetic energy." mc ^{2} = m_{0}c^{2} + KBut this assigning of physical realities to the terms in the equations is completely ad hoc, whether it is done by Einstein or the current textbooks. As I have shown above, the equation works in the same way whether the particle starts from rest or not. In these equations, m_{0}c^{2} is the change in rest energy, not the initial rest energy or the rest energy at zero. And notice that mc^{2} has been assigned to the total energy with no mathematical or theoretical proof whatsoever. With the givens we have in Einstein's thought problems, the real mass of the body or electron is not calculable or assignable, beyond the given E_{0}.
The first thing to do, before I derive new equations for mass increase and energy, is to correct the thought problem I have just critiqued. If I have asserted that Einstein has made mistakes, I should rerun the problem in the right way.
Let us say that we have a tiny ball containing a device that emits light. It is able to emit light one photon at a time, with a known energy. At a distance of 300,000km from this ball is a mirror that reflects directly back to the ball. This distance has been measured locally (by walking it, say). It is a given, not a measurement by the ball after emission. Also, the zero-point of the experiment is marked on the ground with a white line, so that an observer may be placed there. 1 + (2v _{av}/c)v _{av} = 6 x 10^{4}km/s 1 + (12 x 10 ^{4}/3 x 10^{5})= 42,857km/s (Again, the zero-point could have arrived at this number without knowing v' _{av}. This is of some importance below.)m = E/cv _{av}= 7.77 x 10 ^{-36}kgThe mass of the ball has appeared to increase, if measured from the zero-point, as compared to measurement from the ball itself. This much is consistent with the findings of Einstein: mass appears to increase as time dilates. But the transformation term is obviously different. I have used a variation of my velocity term alpha rather than gamma.Now, one may ask, which mass is correct? The mass measured from the zero-point or the mass measured from the ball? Either mass conserves momentum, as long as we keep it in its own equation. But you can see that the mass as calculated by the ball itself must be the correct moving mass, since it is connected to the correct velocity. The zero-point calculates a larger mass only because it has used an incorrect velocity. Its visual data has been skewed by time dilation, making the velocity wrong and then the mass. Next, one may ask, what was the rest mass in this problem? Well, there must be three calculable rest masses: the rest mass before the emission and two rest masses after (the ball and the zero-point will calculate different rest masses, unfortunately). m _{rB} = rest mass of the ball, before emissionm _{rAB} = rest mass of the ball calculated by the ball, after emissionm _{rAZ} = rest mass of the ball calculated by the zero-point, after emissionm _{0B} = mass equivalence of the photon, as measured by the ballm _{0Z} = mass equivalence of the photon, as measured by the zero-point The photon will have two mass equivalents, since the photon will have a different energy relative to the ball than it will have relative to the zero-point. The ball is moving away from the photon when the photon returns, so that its energy will be redshifted. E'< E. E = m _{0Z}c^{2} = the energy of the photon relative to the zero-point E' = m _{0B}c^{2} = the energy of the photon relative to the ball m _{rAB}= mr_{B} - m_{0B}We can solve since we also know that mv _{av} = m'v_{av}' = E/c m = m _{0Z}c/v_{av}m _{0Z} = m(v_{av}/c)m _{0Z} = (m - m')/2= 1.11 x 10 ^{-36}kgwhich agrees with our given value for its energy. m _{0Z} = αm_{0B} m _{0B} = 7.9 x 10^{-37}kgNow all we need is the rest mass. Some will think that is just the mass measured by the ball, since that is the only mass that is truly at rest with regard to its background. But the ball, using the equation above, is calculating with redshifted light. This means that its value for the mass equivalence of the photon is incorrect. In this way my thought problem is not like that of Einstein. In the light planes problem, the body is at rest and the observer is moving away. Therefore the body measures the normal frequency and the observer sees a redshift. But in my thought problem, the observer at the zero-point sees the normal frequency and the ball sees the redshift. Upon emission the ball lost a certain amount of energy. This amount of energy is expressed by E, not by E'. Therefore, the rest mass relative to the zero-point must be calculated with E. This is simply because we must imagine that the ball was not moving at the instant of emission. The ball did not start moving until the instant after T _{0}. Emission took place at T_{0} , therefore the light has its normal frequency relative to the zero-point.This seems somewhat strange at first, since the ball is not at rest relative to the zero-point. How can we calculate a rest mass for it relative to the zero-point; and what is more, why would we want to? We want to in order to get the correct energy equations. If we work with the wrong rest mass, we will get the wrong equations. We must use the rest mass that is at rest relative to the light. That is the only true rest mass, in any problem whatsoever. You will say, "doesn't Relativity imply that all bodies are at rest relative to light, since light travels c relative to all bodies?" Relativity does say that light travels c relative to all bodies, and it is correct to do so; however, it is quite obvious that a body that is measuring red or blueshifted light is not at rest relative to that light. The true rest mass of any body will be calculated from unshifted light—that is to say, light with a normal frequency (see a full definition of "normal frequency" below). And so, in this particular problem, I must seek the rest mass relative to the zero-point. How can I find it, since I don't yet have an equation for it? All I need is a mass after emission from which to subtract my photon from. I have two masses, m and m', but m' is the correct mass since it is connected to the correct velocity. The variable v' was measured locally, meaning that the t variable did not need to be transformed. That makes m' a reliable mass. But it is not the rest mass itself. It is a moving mass. To find the rest mass, we simply subtract the mass equivalence of the photon from m'. m _{rAZ} = m' - m_{0Z} = 4.44 x 10^{-36}kg The rest mass before emission is just the photon added back in: m _{rB} = 5.55 x 10^{-36}kg = m'Whenever I speak of rest mass from now on (concerning this problem with the ball) I will be talking about m _{rAZ}, but I will simplify the notation, taking it back to m_{r}. We should take note that all these masses were calculated from an average velocity over the interval of acceleration up to a final velocity. If we had used a final velocity, we would have found a mass equivalence for the photon that was twice too little. This final velocity is not used in the mass or momentum transforms; but it will be used in the energy transforms, simply so that I may be sure to derive equations that are analogous to the ones that are currently used. The current energy equations are used given a final velocity. In many experimental situations, the scientist does not know or is not concerned with how the particle reached velocity. His or her only data is a final velocity. Some might complain that the ball must use m _{0B} since that is the mass it would calculate from the frequency of light it actually sees. But since we have as one of our givens the fact that the ball knows it is moving and is already calculating a velocity for itself relative to the background of the zero-point, it is not difficult to require that the ball notice that the normal frequency of light is f rather than f'. Both the observer at the zero-point and the ball itself are calculating a moving mass when they use a momentum equation, since the momentum equation includes a velocity. The variable m' could hardly be understood as a rest mass, since it was calculated from an equation that describes movement. Using other methods than this experiment (such as a gravitational method), the zero-point would have found the rest mass of the ball to be 5.55 x 10 ^{-36}kg, before the experiment. It then would have calculated the moving mass to be 7.77 x 10^{-36}kg, from an experiment like this one—a mass that would appear to be confirmed by any subsequent collision of the ball, since the momentum equation used by the zero-point would be assumed to be correct. The momentum would in fact be correct, but neither the velocity nor the mass would be.Some may want to calculate a momentum using m _{r}, to find that the ball also miscalculated its velocity. p = m_{r}v_{r}. But this cannot be done. A rest mass is at rest, by definition, and can have neither velocity nor momentum nor kinetic energy. The rest mass is defined as the mass at rest relative to the normal frequency of light. As you can see, the momentum is the same measured from either the ball or the zero-point, which is just as it must be: mv _{av} = m'v'_{av}. It could hardly be otherwise, since the masses were calculated from a momentum equation in the first place. All we have had to do is keep our variables in order, so that we understand precisely what we have been given and what we are seeking in each event and with each solution.
Finally, let me address the comment that E/c = m'v' _{av} cannot be the correct equation describing the initial situation, since the ball will not receive the photon back from the mirror at energy E. It is true that when the first photon returns to the ball its frequency will have changed, due to the movement of the ball. Because E = hf, the ball will receive the photon at E', not at E, and E'< E. However, we do not use E' in this equation for this reason: we are not concerned with the energy the photon has when it returns to the ball, not from any vantage; we are concerned with the energy the photon has when it leaves the ball. The equation E/c = m'v' _{av} describes an equality of numbers, when all the numbers are relative to the same background. This background is the background of the zero-point, or the background of the ball before it gained a velocity. You may say, no, the variables as measured against that background are unprimed variables, by definition. The primed variables I have said are measured from the ball. However, if you think this, you are not being rigorous enough in your variable assignments. Just as in my first paper, the variable assignments here are very subtle, and we now must write them out in full, to avoid confusion.v' is the velocity of the ball relative to the zero-point, as measured from the ball. v is the velocity of the ball relative to the zero-point, as measured from the zero-point. There is no velocity of the ball measured by the ball, relative to the ball. In the same way, E' is not just the energy as measured by the ball, it is the energy of the returning photon relative to the ball. It is not the energy we want for any of our equations. Both velocity measurements above have the same background. Therefore in the equation E/c = m'v' _{av}, E must also be measured against this background. E must be the original given energy of the photon.
Before we continue, I wish to make one final comment regarding this problem. We have just seen that light may have a different frequency depending upon who measures it. Of course this is not news: we have known of redshifts for decades. But our experiment above has shown us that frequency may be privileged just as I have privileged certain measurements of velocity and mass. What is the privileged measurement of light? The measured frequency of light is normal, and therefore privileged, when the system that measures the ray or photon is at rest relative to the point of emission. That is fairly straightforward, I think, besides appealing to common sense. This effectively privileges the point of emission of light regarding measurement of the light's frequency. Notice it is just the opposite of the privileging of time, velocity and mass to the local system. Local time cannot be wrong. But the measurement of the frequency of light can be wrong, from what we have heretofore called a local system. The ball was the local system above, but it would have measured f', which is not the normal frequency. If you say, we can't privilege certain fields like that—how can we know if we are moving relative to the light source? Well, I say, we can't always know. But it is possible to know in certain situations, from spectra shifts. The fact that it is possible to know means that there is a pre-existing fact. Light does have a normal frequency. For instance, we know, due to stability, that the sun is not moving relative to us. It is neither approaching the earth, nor fleeing it. Therefore measurements of the frequencies of sunlight from the earth are privileged. Notice, however, that measurements of sunlight from the sun are not privileged, since the sun is moving through space. You will say, it doesn't matter, since the sunlight is moving away from the sun, and is therefore undetectable from the sun. But sunlight reflected back to the sun could be measured from the sun. [See my paper on the mirror experiment to replace Michelson/Morley].
These then are the new mass transform equations, for one degree of relativity, if the object is moving away from the measurer. [
Now let us find equations for a velocity change that is not from zero. Let us imagine an even simpler situation. Let us say that a ball of local mass m' starts out with a local velocity of v 1 - v _{1}'/cYou may be surprised to find that the body can calculate its own mass increase due to velocity. But if it can calculate its own velocity, it can calculate its own mass increase. The body itself would of course interpret this not as a real change in mass, but as a change in mass equivalence relative to its background. The body, for itself, has not gained mass but kinetic energy. The classical interpretation would be that this is kinetic energy and nothing else. The modern interpretation is that mass is a sort of energy, especially in a momentum equation, so that they may be lumped together. I prefer to think of the measurement of mass from the object itself as the moving mass. The object must then do further calculations to obtain its own rest mass. The question is, can we also use these equations to transform from a local mass at rest to a relative mass at velocity? Let us set the initial velocity to zero, in which case the initial mass in the relative system should equal to the local mass or rest mass. m _{i} = m_{0}. We know this not from the momentum equations, but by definition. In which casem _{f} = m_{r}/(1 + v_{f}/c)This is only if the object is moving toward the observer, since we simplified an equation from that problem. The mass variables would switch if the object were moving away: (Eq. 1) m _{r} = m_{f}[1 - (v_{f}/c)]However, we now have two equations for the same situation, and they don't match. Even if we switch directions, the equation we found above isn't equivalent: (Eq. 2) m _{r} = m[1 - (3v_{av}/c)] How can we explain this? It is because the experimental situations aren't the same. In the first thought problem, the ball emits a photon in order to reach velocity. In the second, it doesn't. Notice that the ball has borrowed the energy of the photon in the first experiment. A scientist wouldn't necessarily know this, if he came upon the ball after emission, but it is an important fact of the equations. In the second experiment we are just imagining that the object goes from rest to a final velocity, and we calculate the mass increase due to that velocity. But again it might be asked, is this possible? Can an object gain or lose velocity without borrowing the energy of another object, by collision, emission, or other method? I don't think so. In any experimental situation, we must assume that any object under consideration—that is not at rest relative to our field—gained its velocity by some means external to our initial measurement. We may postulate emission, collision, or the influence of a field, but we may not postulate a relative velocity that was gained without energy transfer. Therefore, I hypothesize that eq. 2 is always the correct one. This equation implies some rather shocking things, of course. The most important being that there is a limit at .5c for v. The mass m becomes infinite when v = .5c. But this is the same limit for v that I found with my velocity transforms. [There is not really a limit for v' in my velocity transforms, but when v' = c, v = .5c. When v' is infinite, v = c.] So at least my equations are consistent. Whether I can reconcile this with the findings of particle physics is yet to be seen (below). One thing that makes it slightly easier to accept is that I am not postulating a real mass approaching infinity. m is not a real mass. It is a measurement. I am postulating a measurement to approach infinity. Therefore, there is a limit to measurement; but the variable m does not apply to the real mass at all.
I said that according to my equations, momentum does not need to be transformed. In order to find our initial transforms for mass, we had to assume that the momentum of our object from the zero-point was equal to the momentum measured from the object itself.
Let us now return to my correction of Einstein's energy equations and see if we can apply them to my problem with the ball and the photon. ~~~~~~~~~~
The equations and terms we found above apply only to the thought problem with the ball. I have shown, both in this paper and in my paper on the time and velocity transforms, that there is no one problem in Special Relativity. Trajectories must always be taken into account. Which means that if we want to generalize the mass and energy transforms we must do a bit more work. We must be sure that what seems to be true, is true. In other words, we must run the equations for 1) Einstein's problem with the light planes—in which there is a mass change but no velocity change-and also for 2) A problem in which the relative velocity is toward an observer. Only then will we fully understand the mechanics of mass and energy in Relativity. ~~~~~~~~~~ We have now found three different transforms for three different problems. The only thing that has remained constant is that
K = ±mv
The one variation of this problem we have not solved is the one that will prove this theory of mine beyond any doubt. That variation is the real problem of subatomic particles that achieve high velocities in accelerators. Notice that this problem is not like any of those we have solved. We have found transforms when 1) a body with an initial velocity emits a smaller body, but does not change velocity, 2) a body initially at rest emits a smaller body and moves away from an observer, and 3) a body initially at rest emits a smaller body and moves toward an observer. We now seek the situation when 4) a body with an initial velocity is bombarded by smaller bodies and achieves a final velocity. This case is obviously closer to Einstein’s slowly accelerated electron, except that in this case the subatomic particles are not slowly accelerated. [1 – (v’^{2}/c^{2})] E _{T} = m_{ri}c^{2}{1 + [(v^{2} + cv)/(2c^{2}– 4cv)]}E _{T} = mc^{2} [1 – (3v/2c) + (v^{2}/2c^{2})]E _{T} = mc^{2} [1 + (v’/2c)][1 + (2v’/c) + (v’ ^{2}/c^{2})]Notice the last bolded equation above tells us why gamma works so well in accelerators despite being slightly incorrect and being derived with so many mistakes.In accelerators we are finding a limit at 108. Therefore, we set my equation equal to 108 and see what velocity the proton is really achieving. (v/c)m _{0}c^{2} + m_{0}c^{2} + m_{ri}c^{2} = 108m_{ri}c^{2}(v/c)m _{0}c^{2} + m_{0}c^{2} = 107m_{ri}c^{2}This last step was allowed since m _{ri} is the same in both theories. [(v/c) + 1]m _{0} = 107m_{ri}m _{ri} = 2m_{0}[(c/v) – 2][(v/c) + 1]/[(c/v) - 2] = 214 v = .4982558c v’ = .9930474c = 2.97708 x 10^{8}m/sc = 2.99792458 x 10 ^{8}m/sAccording to current theory, gamma is equal to 108 at v = .999957c. The v variable in gamma is equivalent to my v’, since current theory has no v’, and since I have defined my v’ as the true velocity of the object.
So, we now have all our numbers in hand. How am I going to explain the number 108? Notice that we have an unexplained velocity differential in both current theory and my theory. By current theory the limit in velocity for the proton is 1.2 x 10 ^{4}m/s less than c. By my theory the gap is a bit larger: 2.1 x 10^{6}m/s. What causes this gap? And which gap is correct? If I can answer these questions, then I can show where the number 108 comes from.Let’s say that the proton already has a velocity or velocity equivalent due to some motion or force or other unexplained phenomenon. Let’s say that the proton’s total velocity cannot exceed c, and that this other unexplained motion or force makes up the difference. That is precisely what I have done in my paper on the Universal Gravitational Constant. Using a hint of Maxwell and the dimensions of G, I showed that the proton can be shown to have a constant acceleration in any direction of 8.88 x 10 ^{-12}m/s^{2}. Here is a gloss of that math. Given two equal spheres of radius r touching at a point, we haveF = Gmm/(2r) ^{2}ma = 2Gmm/(2r) ^{2} a = 2Gm/4r ^{2} a/2 = 2Δr/2Δt ^{2}We now let the spheres expand at a constant and equal rate. We assign Δr to a change in the radius instead of a change in the distance between the spheres, and this allows us to calculate even when the spheres are touching. Δr/Δt ^{2} = Gm/r^{2} After time Δt, the radius will be r + Δr. After any appreciable amount of time, r will be negligible in relation to Δr, so that Δr ≈ r + Δr m = Δr ^{3}/GΔt^{2}a = 2Δr/Δt ^{2}a = 2mG/Δr ^{2}That is the acceleration of each of two equal masses in a gravitational situation. But if we want to give all the acceleration to one of them, holding the other one steady for experimental purposes, then we simply double the value. a = 4mG/Δr ^{2}If the proton has a radius of 10 ^{-13}m, this yieldsa = 8.88 x 10 ^{-12}m/s^{2} If we allow the proton to accelerate at this pace over its entire lifetime up until the current moment, then we can achieve a number for its present velocity due to mass. My velocity is a much better fit. Using this acceleration due to mass and gamma, we get an age of the proton of only 85 million years. v = at/2 = 2 x 1.2 x 10 = 85 million years^{4}m/s 8.88 x 10 ^{-12}m/s^{2}My corrected numbers give an age of the proton of about 15 billion years. v = at/2 = (8.88 x 10 ^{-12}m/s^{2} )(4.73 x 10^{17}s)/2 = 2.1 x 10^{6}m/sMy number is therefore a match to current estimates, as you see. Current theory based on gamma is clearly wrong, since the proton cannot be as young as 85 million years. That would make protons 50x younger than the earth.[To see a shorter way to derive the number 108, you may now visit my more recent paper called Redefining the Photon. There, I use the density of the charge field to calculate the number.] In conclusion, my mathematical connection of this paper with my other papers therefore does several very important things. 1) I have explained the velocity limit of the proton in the accelerator. It cannot achieve c due to its mass. This was assumed by all. But I have shown precisely how and why the mass limits the velocity. 2) The mass has a calculable velocity equivalent and I have provided the math to achieve this velocity. In doing so I have dismissed the mass dimension altogether, showing that mass can and must be expressed with the dimensions of length and time. I have given the dimensions of G to the mass, so that G is now just a number. This means that the kilogram must be redefined in terms of the meter and the second. 3) I have provided further mathematical proof of my corrections to Special Relativity. I have shown one more instance in which gamma fails to give us correct numbers. Findings in particle accelerators could not be tied to other theory for two reasons: we didn’t have the correct theory to tie it to, and we didn’t have the correct velocity of the particle. My corrections from both ends allow us to tie up in the middle in a very satisfying way.4) The explaining of mass as motion is a huge step in the quest for a unification theory. One important implication of my new theory is that gravity doesn’t even exist at the atomic level. We don’t have to call the motion I have given to mass gravity. We can continue to call it gravity at the macrolevel if we like, but we can let the motion explain one of the other “forces” at the atomic level. Gravity is not a force at all. According to the new theory, you can assign mass, gravity and inertia to the same basic motion. Mass, gravity and inertia are not three different things, they are three different expressions of the same thing. And all three resolve to length over time. If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |