A COMPLETE RE-DERIVATION of E=MC2 or HOW CORRECTED TRANSFORMS in SPECIAL RELATIVITY affect MASS, MOMENTUM and ENERGY EQUATIONS Einstein hiding behind his paper by Miles Mathis First written October 2004 Introduction 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, Does the Inertia of a Body Depend upon its Energy Content? Fully half of my paper is devoted to analyzing, critiquing and expanding this thought problem and its math. The rest of the paper is devoted to a variant thought problem I devised to clarify Einstein’s variable assignments and conceptual assumptions. This thought problem yields new equations that answer many of the embedded mysteries of relativity and mass transformation.        Einstein’s paper Does the Inertia of a Body Depend upon its Energy Content? has long been a source of confusion. It’s brevity and opacity have made its underlying concepts quite difficult to unravel. As with the time and length transforms of Special Relativity, the mass transforms that this paper yielded have never been corrected. They have been confirmed to the satisfaction of most experimental scientists and therefore the math to derive them has become a moot point. It was long ago swallowed up by much more complex math, including hyperbolic fields, imaginary numbers, Hilbert spaces, Hamiltonians, Lagrangians, and the tensor calculus. Although thousands of papers have been written on the mass transforms, no one has so far offered a crystal clear explanation of Einstein’s algebraic variables and equations. In the past half-century, no famous physicists or mathematicians have even attempted to do so. Some have glossed the derivation as presented by Einstein, but none who accepted his final equations have provided a superior groundwork for them.         Now, a century later, only those who do not accept the final equations spend time on the mass transforms. And they do not attempt to clarify Einstein’s mistakes. Rather, they present a variant math that makes more sense to them. Some of these variant maths have a certain validity, but I believe that none will be looked at seriously until Einstein’s math is proven to be false. That is what my paper does. A falsification of Einstein’s algebra will be a falsification of all the higher maths that rest upon it. Einstein’s paper is a compound—and sometimes a compensation—of several basic algrebraic errors. Although in the body of the paper I will prove these errors exhaustively, here I will just gloss them. Firstly, he incorrectly applies his time transform gamma to the planes of light. Secondly, he misapplies the term m0c2 at the end of the derivation, giving it to the body rather than to the planes of light. This is difficult to understand, since the final equation contains the variable L, which he has explicitly given to the light. Despite these two errors Einstein arrives at a transform that is very nearly correct. That transform is again gamma.         Einstein then solves down from this energy transform to find a mass transform, which is likewise gamma. But in this case he is wholly mistaken: his misassignment of variables has cost him needed clarity, and gamma is not even an approximation of the correct mass transform. This mistake has rarely been seen, since in experimental situations mass is always calculated from energy equations. In working with subatomic particles, for instance, the naked mass transform equation is never used. Values are arrived at from energy equations. As I said, Einstein's energy equation is almost correct. The term for gamma is γ =          1     .       √1 - (v2/c2)]        I will prove that by correcting the math, the energy transform for Einstein's problem is actually kappaκ = 1 + [v'2/(2c2 + cv' - v'2)] or ET = m0c2[1 + (v’/2c)]                  [1 – (v’2/c2)]       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 ≠ mc2.       If we assume that the rest energy is given by the rest mass—as in Er = mrc2—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 mc2 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 = κmrc2 - mrc2 = mv2/2       Simply 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 readt = 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. I have been answered that Relativity is symmetrical around v, but it isn't. It could be symmetrical around velocity only if system S was “me measuring you,” while S' was “you measuring me.” But since in Relativity, S' is “me measuring you,” while S is “you measuring you,” the equations cannot be symmetrical. Without this symmetry, we must find two values for velocity from the beginning. 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. Part OneEinstein's Solutions Now let us proceed to the mass and energy transforms. The best place to start is with Einstein's second paper of 1905, Does the Inertia of a Body Depend upon its Energy Content? In this paper he has a body at rest emit two planes of light in opposite directions. The two planes of light have equal energies; therefore the body remains at rest after the emission. He then asks how the energy of this body before and after the emission would look to an observer moving directly away from the body at velocity v. To be precise, he never specifies that the observer is moving away from the body (in the positive x direction, with the body at the origin) but it is implied by analogy to his previous paper. I will say, in passing, that his failure to specify a direction in this paper has had far-reaching consequences, since it has been assumed (without much argument one way or the other) that the direction is not important. That is, all the transforms of Special Relativity are now assumed to be non-specific regarding direction. This is too bad, since I have shown (and will show again, below) that Relativity must be specific regarding direction.       Einstein lets the two planes of light emit from the body at angles to the x-axis, and therefore to the observer. Let us call B the system of the observer and A the system of the body. Using his nomenclature,       E0 = the initial energy in A. This is not kinetic energy (from the point of view of the emitting body) since he states that the body is not moving in A. It is unclear what E0 is at this point. But from the outcome of the equations, E0 must be what he calls the initial rest energy, as in E0 = m0c2. Since the body is at rest in A, E0 is both the rest energy and the total energy.        E1 is the energy in A after the emission of the two planes of light.H0 is the initial energy of the body as seen from B. That is, it is the initial rest energy plus the kinetic energy.       H1 is the final total energy of the body from B, being the final rest energy plus the final kinetic energy.        L/2 = the energy of each plane of light, as measured from A.E0 = E1 + L/2 + L/2       This is the equation as calculated from AH0 = H1 + aL/2 + bL/2      This is the equation from B, where a is the negative angle transform and b is the positive angle transforma = γ[1 + (v/c)cosφ]b = γ[1 - (v/c)cosφ] where γ = gamma =1/√[1 - (v2/c2)]Now, Einstein says the initial kinetic energy of the body is represented by the equationK0 = H0 - E0And the final kinetic energy is represented byK1 = H1 - E1So that the change in kinetic energy isK0 - K1 = L{       1             - 1} = γL - L                    √[1 - (v2/c2)]That is the whole paper. It takes up less than three pages in Annalen der Physik. It will take me somewhat longer to show all the mistakes in it.       The cardinal error in this whole derivation is in the final two steps. At the end Einstein mixes up the last equation with the next to the last equation, treating them as the same thing. But one expresses the final kinetic energy and the other expresses the change in kinetic energy. They are not the same in this problem, since the body has an initial kinetic energy (from the point of view of the observer). Einstein assigns the term γL to H1 and the term L to E1. He assumes that H1 is mc2 and E1 is m0c2. But look back up the series of steps: L ≠ E1 H1 ≠ γL This is because K1 ≠ K0 - K1.Once you have digested the enormity of that, notice that in the final step Einstein has subtracted the final kinetic energy from the initial. This is backwards. It is standard practice to subtract the initial energy from the final to find a change in energy. Corrected, the equation should read, K1 - K0 = L(1 - γ)       An even greater error is made in assigning values to the light angle transforms a and b.       Notice that the magnitudes of a and b are not equal. The observer in B would therefore expect Einstein's body to change course, since one of the planes of light would have more energy than the other, measured from B. Einstein ignores this. The body must not change velocity, because then the change in kinetic energy would be due to that velocity change and not to a change in mass—which is of course what he is trying to prove. By a mathematical trick Einstein gets the two planes of light to add to unity in both systems, but in B the two light planes do not have equal energies.        Another crucial error in this thought problem is that Einstein applies his transform γ only to the planes of light, L/2 and L/2. He does not transform the mass, velocity, or energy of the body directly. Those transforms are implications of the thought experiment, but they are calculated indirectly, as results of these very energy equations. In truth, the masses are applied to the energies somewhat willy-nilly, and a rigorous explanation has never yet been provided.        The problem can be solved down from the energy equations, of course, but it is a curious method, especially as it stood (and still stands, until the publication of this paper) as the first and only method. To solve from the energy equations one must be extremely careful to keep all the hidden variables in order. Einstein does not do this, as I show in the paragraphs that follow. But the greater problem is that solving by this method keeps those variables in the dark. In solving a problem for the first time, a scientist or mathematician should put all the variables in plain view, showing how they are transformed directly. He should not derive them indirectly by a compact but impenetrable method. This problem is the perfect example of that. Einstein has not been corrected for a century due to the obtuseness of his proofs. In my opinion, it would have been more helpful to do transforms on the basic variables, those being mass and velocity, and then to build energy equation from those. As it is the conceptual basis for relativistic mass, momentum and kinetic energy has been keep under a cloud from the beginning.       As a first example of this cloud, notice that if you insert m0c2 into the last equation above, as Einstein did later and as history still does, this implies that L = m0c2. Not E0 but L. In the beginning of the equations, E0 is assumed to be the rest energy of the particle. At the end, Einstein and history have assigned m0c2 to E0. But according to these equations, L = m0c2. That is, m0c2 is not the rest energy before or after the emission of the light, it is the change in rest energy. It is the energy equivalence of the planes of light. You may say that the situation is different when Einstein expressly assigns m0c2* to the rest energy. In that problem ("Dynamics of the Slowly Accelerated Electron," last part of section 10 of On the Electrodynamics of Moving Bodies, 1905) he applies a force from an electrostatic field, taking the electron from rest to v. There is no L involved.       No, there is not. But the situation is directly analogous, otherwise how could it yield the exact same equation? In it, the electron starts at rest with a given energy. Let us call it E0 again, as above. If we apply all the electric force at the first instant, to complete the analogy to the light planes being emitted, then we can follow the problem in the same way, without calculus.** The body reaches v instantaneously, and we want to know how much energy it has gained from the force. Einstein has his electron accelerate slowly, but that is only to avoid giving off radiation. That is, it is an experimental concern, not a theoretical concern.D = the energy gained from the electrical forceE0 = E1 - D       Einstein says the field imparts a velocity to the electron. So the electron is now the moving body. Let us assign it to B, the observer being at rest in A. It is the electron that is moving, not us. It would be even more precise to say that the electron is B. It is not moving in B; it is the system B itself.H0 = H1 - bD where b is the transformation term.       But, the electron starts at rest relative to A and B, therefore H0 = E0K1 - K0 = H1 - E1 - (H0 - E0) = H1 - E1       But K0 = 0 since the electron has no kinetic energy at rest in both systems. So:K1 = H1 - E1        = H0 + bD - (E0 + D)K1 = bD - DThe kinetic energy is equal to the total energy measured from a distance minus the total energy measured from the body. And this is the energy taken from the field as measured from A minus the energy taken from the field as measured by B. This is precisely equivalent to the example with the light planes—substituting D for L—except that in one the body (the electron) is the moving system and is gaining energy, and in the other the body is the at-rest system and is losing energy.        Here again, though, if you insert m0c2 as it has been historically into the last equation, you find that it is equal to D, not to E0. D is the energy gained from the field, by Einstein's own variable assignments. E0 ≠ m0c2*Einstein assigns the variable in question (my D above) to mc2 not m0c2. He assumes that m stands for the rest mass here, since the electron starts from rest. Unfortunately, the rest mass changes during the acceleration (which is what he is trying to prove) so that m is not the final rest mass. It is the initial rest mass. This conflicts with later interpretations and assignments of the kinetic energy equation. Regardless, D is not assignable to any mass of the electron.**Einstein actually uses calculus, and provides us with a single equation: K = ∫ εXdx = m∫ β3vdv. This kind of math is not helpful in creating a new theory, since precisely none of the concepts are enumerated. Also notice that, just as in the proofs of Special Relativity, Einstein has failed to assign v to either system A or B. This must affect his calculations. Nor does he consider that kinetic energy can be calculated from either system, A or B. If A can calculate a velocity relative to B, then A can also calculate a kinetic energy. He does not specify where K is measured from. The form of the equations implies that K is measured from B, but this is not a necessity. The fact that Einstein does not carry into this problem a v', as I do, has had long-reaching consequences.        And finally, gamma is an incorrect transform, as I have proven elsewhere. It is incorrect mainly because Einstein never saw the existence of v', from the beginning. And, even if gamma had been correct as a transform for mass and time in Special Relativity, it still should not have been applied to the light rays here. Physics already had a transform for frequency that had nothing to do with Special Relativity. This transform always has been equivalent to my basic transform alpha. Because he does not have enough variables or coordinate systems, Einstein has once again been forced to finesse his math. He has done so in several places, in fact. In the first thought problem, the initial trick is letting L/2 stand for half the emitted light. Splitting his variable so that it yields a two-term equation is done only to ensure that it cancels properly. The second trick is using a transformation term that has a 1+ and a 1- in the numerator that also cancels out. This is not just luck. Nor is it necessity. As I will show, it is much more convenient to choose the send the planes of light straight ahead and straight back, since then they are all in the same line as the given v. Everything is then in the x-direction. Why does Einstein choose an angle? He chooses it because it is the best way to finesse this equation. If he lets the planes of light be emitted in a line, he gets into all kinds of trouble. His split equations won't cancel out in that case, according to his own faulty theory, since Einstein's transforms are the same regardless of direction. In using the angles, as he has, he ensures they cancel, but only at the cost of theoretical consistency. Mathematically they cancel. Conceptually they do not, as I have shown. The observer in B should see the body change direction, and Einstein cannot explain why this does not happen.        Einstein's equations do not distinguish between movement toward and movement away. He says that moving things act the same, no matter the direction. Therefore the energy of both planes of light should increase from the point of view of B. If the body had sent out electrons instead of planes of light, Einstein would have found both the electron receding and the electron approaching the observer in B to be slowed and mass-increased. But this is false, as I will show.        Finally, Einstein finesses the equations by assigning (in the last equation of either problem) the first term to the kinetic energy and the second term to the rest energy. Like this,K0 - K1 =           m0c2       -        m0c2}                   √[1 - (v2/c2)]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 K0 - K1 = γ L - L       If L = m0c2 , then what is the value of E0? Let us see. E0 = E1 + 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 m0c2 is the mass equivalent of the light, not of the body. But this is not what m0c2 means in current energy equations. Now let's look at the calculus derivation of E = mc2 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 ≠ mv2/2 . 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 - ∫ pdvbut 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 = mv2 - ∫       m0v     dv                      √[1 - (v2/c2)]     = mv2 + mc2[1 - (v2/c2)] - m0c2 and so onNotice 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 m0c2 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 mc2 the total energy of the body, and we see that the total energy equals the rest energy plus the kinetic energy." mc2 = m0c2 + K       But 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, m0c2 is the change in rest energy, not the initial rest energy or the rest energy at zero. And notice that mc2 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 E0. Part TwoA Correction for Einstein's Thought Problem 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 return to the light-plane problem. I will get rid of the angles of emission, leaving the light to travel only along the x-axis. One plane of light travels directly toward the observer in B, and one plane directly away. Since with light the energy is dependent on the frequency, not the speed, we need linear transforms for frequency, not velocity. The light moving in the +direction of v will be red-shifted, since although it is moving toward the observer in B, the observer is moving away from it. As regards the other plane of light, the case is a bit more subtle. That light is not moving toward the observer at all. It is wrong to say that an object moving away from an observer has a kinetic energy, since that object cannot possibly do any work for the observer. To be even more precise, light moving away from an observer cannot be known to exist at all. However, we can measure the energy of the incoming light, and we can see—or we are given—that the emitting body has not changed speed or direction. Therefore, the receding light must have an equal but opposite energy to the incoming light. This is only an inference though, and may not be measured or seen directly. Let us see if we can express this in equations. I am assuming the given velocity is B as measured from A.         Einstein's nomenclature is (purposefully) confusing so I am going to call the L/2 incoming F0 and the L/2 receding G0.        If F1 is the energy of the light as measured by B, thenαF1 = F0     since F1 < F0F0 = -G0       The energy of a plane of light is dependent only upon its frequency, since its velocity is always c. E = hf, where h is Planck's constant and f is frequency. The transform for frequency isf' = αf       Which makes the transform for the energy of a light planeE' = αE       Amazingly, this is the current transform for frequency, as used by scientists for decades. Richard Feynman used it in his explanations of Special Relativity, at the same time that he was corroborating gamma and other mathematical falsehoods. So my alpha has been a common transform in optics for several generations. But until now it has not been properly tied to Special Relativity and the mass transforms.       The magnitude of the energy of G1 must equal F1, otherwise the observer in B would see the body change velocity or direction after emission. We are told that it does not change velocity. It stays at rest in A, and keeps velocity v in B. We could express the direction of the planes of light as angles of 0 and 180, to mirror Einstein, but notice that it is completely unnecessary in this sort of problem. We are only interested in vectors, not in angles. Both planes of light end up being subtracted from the mass of the body. Einstein's use of 1 + cos and 1 - cos, etc. was just false bombast. This is the way the equations should go:E0 = E1 + F0 - G0E0 - E1 = 2F0 H0 = H1 + F1 - G1      [We are dealing with energy as a vector, remember!]H0 - H1 = 2F0/α        You may say, shouldn't the light plane traveling in the -x direction have a blue shift, and a transform that is the inverse of the red-shift transform? No. Light is blue shifted if it is traveling toward the observer and the observer is traveling toward it (or if the point of emission is traveling toward the observer-which you see is the same thing). A light plane traveling in the -x direction is neither blue-shifted nor red-shifted, nor subject to any possible transformation. It is invisible and undetectable, except by inference.        Now, Einstein says the initial kinetic energy of the body is represented by the equation K0 = H0 - E0       And the final kinetic energy is represented by K1 = H1 - E1       So that the change in kinetic energy isK1 - K0 = H1 - E1 - (H0 - E0)K1 - K0 = -2F0/α + 2F0       Now, if we want to put L back in, and solve, we getL = 2F0ΔK = L[1 - (1/α)]ΔK = L(v/c)The body lost the mass equivalent of the light but gained kinetic energy. This is simply because the body had a negative kinetic energy to start with. It was moving away from the observer and therefore could do no work. Its loss of the energy of the light gave it a smaller negative kinetic energy, which is of course a positive vector change.       Einstein had to finesse his equations to get a positive number at the end. I have shown how to analyze the vectors correctly. You can see that I have done a lot of housecleaning. The way I dealt with the planes of light was quite different than Einstein. Notice, for one thing, that I would never let the planes of light be emitted in any other way than the way I did. Why? Because any other planes of light, emitted at any angle to the x-axis, will be undetectable from B. Einstein assumes that B can perform transformation equations on light that never even comes to B. Light emitted at any angle will never reach B, and is therefore not a source of possible calculation. In that case all energy changes will be inferences; none will be measurements. Which would make Einstein's thought problem a fantasy from beginning to end, rather than a meaningful potential experiment.       Furthermore, if m0c2 is inserted into the equation, then L = m0c2. In this problem, according to Einstein's own assignments, this is the mass equivalence of the emitted planes of light, not the rest mass of the object.        Finally, my corrections make it clear that L/α cannot be assigned to mc2. Currently, Einstein's theory assigns m0c2 to L and mc2 to γL (which is my L/α). He says that ET = γL. But this is false, according to his own variable assignments. From the equations above and current theory, we have ET = H1 + K = H1 + γL - LH1 ≠ L Therefore, ET ≠ γL        Above I showed that ET ≠ H1. Here I have shown that ET ≠ γL. The truth is that ET is not singly assignable to any of Einstein's energy variables; nor is it assignable to mc2. Since I have thrown out Einstein's method for deriving mass increase equations, I must now derive them on my own, using my own thought problems. Part ThreeThought Experiment 3 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. We run the experiment twice. The first time the device in the ball emits a photon toward the mirror at T' = 0s, and then receives the same photon upon its return from the mirror (it does not re-absorb the photon, it simply measures it with an instrument). T' is the time on the ball's clock. At the beginning of the experiment, just before the emission of the photons, T' = T. That is, the clocks of the zero point and the ball are synchronized.        The second time, the device emits a photon at T' = 0s, another at T' = 1s, and then another at T' = 2s. The observer at the zero-point intercepts the second and third photons from the ball in order to calculate where the ball is after T' = T = 0. This observer also intercepts the first photon returning from the mirror.        By the conservation of momentum, the ball must recoil in the opposite direction from the emission of the photon. When the photon returns, the distance the ball has traveled may be measured, and the inertial mass of the ball may be determined. Question: will the mass of the ball as calculated from the ball be equivalent to the mass of the ball as calculated from the zero point of the experiment? If not, how will they differ?L = distance from zero-point to mirrorE = energy of photon = 1 x 10-19J (say)m' = mass of ball, measured by the ballv' = velocity of ball, measured from the ball       Let us calculate from the ball, first of all. In this case, the ball is the measurer, and the system of the ball is therefore the S' system—the primed system (I make the local system the primed system simply to be consistent with my other paper). What does the ball see?        The simplest thing to do is to let the photon return all the way to the ball. We could let the photon return to the zero-point and then let a signal be triggered, but that seems redundant, since the signal would have to be a light signal.       Let the ball be very tiny, to be sure it travels a nice long distance. But do not assume it reaches velocity instantaneously (this will be important later). When the photon arrives back at the ball, the ball looks at its clock and discovers that 2.5 seconds have elapsed. The ball thinks, "This is very easy. The light took one second to get over to the mirror and one second to get back, and half a second to reach me. If the mirror is 300,000km from the zero-point, then I am 150,000km from the zero-point. I went that far in 2.5s, therefore my average velocity (relative to the system of the zero-point) is:v'av = 150,000km/2.5s = 60,000km/s       By the conservation of momentum, the momentum of the light must be equal to the momentum of the ball:E/c = m'v'avm' = E/cv'av = 1 x 10-19J/(3 x 108m/s)(6 x 107m/s)       = 5.55 x 10-36kgNow let us calculate from the zero-point. The first photon arrives at the zero-point in 2 seconds, according to the clock at the zero-point. The observer at the zero-point then must measure the distance the ball has appeared to travel. The observer does this by receiving the other photons from the ball. We could use the t-equation from my previous paper, to calculate the difference between the period of the ball and the period of the zero-point. This would be the most direct way to calculate, since the only data the zero-point is receiving from the ball is ticks. [The zero-point is able to calculate velocity simply from receiving ticks, since the zero-point knows the local period of the ball. When the ball was at rest at the zero-point, at the beginning of the experiment, it's period was 1s.] However, since we have already calculated the velocity of the ball according to the ball, I am going to skip this step and use my velocity transformation equation instead. If the ball calculates its own velocity to be 60,000km/s, then the observer at the zero point will calculate (by receiving ticks) the velocity to be: t = t' + x'/ct' is a given as 1s. t is incoming data. Therefore x' and v' and v may be calculated.v =        v'                1 + (v'/c)vav =        v'av                 1 + (2v'av/c)vav =        6 x 104km/s               1 + (12 x 104/3 x 105)     = 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/cvav       = 7.77 x 10-36kgThe 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). mrB = rest mass of the ball, before emissionmrAB = rest mass of the ball calculated by the ball, after emissionmrAZ = rest mass of the ball calculated by the zero-point, after emissionm0B = mass equivalence of the photon, as measured by the ballm0Z = 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 = m0Zc2 = the energy of the photon relative to the zero-point E' = m0Bc2 = the energy of the photon relative to the ball mrAB= mrB - m0B       We can solve since we also know that mvav = m'vav' = E/c m = m0Zc/vavm0Z = m(vav/c)m0Z = (m - m')/2       = 1.11 x 10-36kg       which agrees with our given value for its energy.m0Z = αm0B m0B = 7.9 x 10-37kg       Now 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 T0. Emission took place at T0 , 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'. mrAZ = m' - m0Z = 4.44 x 10-36kg        The rest mass before emission is just the photon added back in:mrB = 5.55 x 10-36kg = m'       Whenever I speak of rest mass from now on (concerning this problem with the ball) I will be talking about mrAZ, but I will simplify the notation, taking it back to mr. 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 m0B 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 10-36kg, before the experiment. It then would have calculated the moving mass to be 7.77 x 10-36kg, 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 mr, to find that the ball also miscalculated its velocity. p = mrvr. 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: mvav = 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]. Part Four New Mass Transforms These then are the new mass transform equations, for one degree of relativity, if the object is moving away from the measurer. [alpha must be modified if the object is moving toward the measurer—see below for modification process; or see paper on velocity transforms for full proof of modification.]mvav = m'v'avmv'av /[1 + (2v'av/c)] = m'v'avm = m'[1 + (2v'av/c)] = m'α where m' is local mass and m is measured from a distanceWhat if we want to use vav instead of v'av?m = m'/[1 - (2vav/c)] = m'αHowever, these equations tell only part of the story, as the above thought problem made clear. The observer at the zero-point would calculate the ball to have a moving mass of m'/[1 - (2vav/c)]but if the ball subsequently came to rest relative to that observer and was weighed by him, it would weigh mr = m' - m0Z m0Z = m(vav/c) mr = m' - m(vav/c)m' = m[1 - (2vav/c)] mr = m[1 - (2vav/c)] - m(vav/c)               = m[1 - (2vav/c) - (vav/c)]mr = m[1 - (3vav/c)]m = mr/[1 - (3vav/c)]I will call this transform beta.beta = β = 1/[1 - (3vav/c)] This is a very important equation, since it mirrors many experimental situations. Already you can see that there are many equations involved with mass increase, and the correct one must be chosen for the situation. Just as with velocity, we must take into account the direction of relative motion. In addition, we must take into account which mass we are seeking, which mass or momentum we are given, and precisely what we are transforming to and from.In the thought problem we have just solved, the mass changed twice, for two reasons: firstly, it changed because the ball emitted a photon. This changed the mass even from the point of view of the ball, of course. So this is not a consideration of Relativity. Secondly, it changed from the point of view of the observer, since a velocity was involved. This second change required a mass transform due to Relativity.        The first change of mass was not a concern of Special Relativity, meaning it was a mass change that could be (and was) calculated without Relativity Transforms. Part FiveMass Transforms from one velocity to another 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 v1' and ends with a local velocity of v2'. Will its mass appear to increase from a distance? Let us assume (at first) that its local mass will not change, since no particle is being emitted in order to accomplish a higher velocity, as with the photon emission above. First we must specify a direction. Let us say it is moving directly toward an observer or a zero-point. In this case we will not have to make the velocity or the momentum negative. For notice that once we start talking about momentum and kinetic energy, we must think in terms of vectors. Objects moving away will have negative momenta and negative kinetic energies.       Now let us take a closer look at these givens. Are they possible? Is it possible for a ball to change velocity without changing its total energy? Of course not. But can it change total energy without changing its local mass? That is a subtler question. As we saw above, the ball gained a velocity by emitting a photon. Its rest mass therefore changed. In many other situations, especially in particle physics, the local or rest mass of the body in question will be affected by a field or by bombardment, since photons or positrons or neutrinos or other small particles will be emitted or absorbed. It may be that no transfer of energy is possible, even on the macro-level, without a change in mass. However, we will assume that some transfers are totally elastic (nothing sticks or is emitted). At the macro-level this will always be an approximation (although often negligible); at the micro-level it will likely always be a falsification. But for this part of the problem, we will assume that the ball changes velocity without changing its rest mass or local mass.        The initial momentum of the ball as measured by the ball is given by the equation m'v1' and its final momentum by m'v2'.       But in an experiment where energy or momentum is the yield, then the mass will be calculated down from the momentum equation. In this case, the velocity will be measured from a distance, obviously. Scientists do not measure the local velocity of quanta, or anything else. So these scientists will be using these equations for the initial momentum and the final momentum:pi = mivi       where the i stands for initialpf = mfvf      " final       Since there is only one energy output at collision, no matter where it is measured frommivi = m'v1'mfvf = m'v2'v' = v/(1 + v/c)v1' = vi/(1 + vi/c)v2' = vf/(1 + vf/c)m' = mfvf/v2'mivi = v1'mfvf /v2'mi/mf = vf/(1 + vi/c)//vf/(1 + vf/c)              = (1 + vf/c)/(1 + vi/c)mf = mi (1 + vi/c)/(1 + vf/c) = mi(c + vi)/(c + vf)If the final velocity is greater than the initial velocity, the final mass must be less than the initial mass. For an approaching object, there is an apparent mass decrease. Obviously this is just to keep the momentum the same. If you are measuring its velocity and getting a number that is too high (compared to the real value) then you must measure the mass to be too low, so that when it hits you, the real momentum and your calculated momentum are the same thing. If the object were moving away, then you would once again calculate a mass increase.       And there is your mass transform. It has two v's, unlike Einstein's equation; and this is very convenient, since it allows us to calculate from initial to final.       Now let's see if my term causes more change than Einstein or less.If vi = c/4 and vf = c/2 thengamma = 1.03my term = 1.2Somewhat greater change in mass.What if the initial velocity is zero? If vi = 0, then mf = mi /[1 + (vf/c)] = mi/(c + vf)Of course, in the same way we can derive a transformation from local velocities, if we want.vi = v1'/[1 - (v1'/c)]vf = v2'/[1 - (v2'/c)]mivi = v1'mfvf /v2'mf = mi(1 -v2'/c)         1 - v1'/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. mi = m0. We know this not from the momentum equations, but by definition. In which casemf = mr/(1 + vf/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) mr = mf[1 - (vf/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) mr = m[1 - (3vav/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. In all these equations, we see a limit for the unprimed velocity relative to c. The mass goes to infinity as v goes to c/2 or c/3. One thing that makes this easier to understand 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. In all my equations and theories, the real mass is inviolable. Part SixEinstein's Momentum Transformation Equation 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.p' = pm'vav' = mvav        We could not have found a mass transform otherwise. Notice that Einstein, despite never making this assumption, arrives at the same basic substitution I do. His transform for mass is the same as his transform for time and length, gamma. My transform is also unchanged. My transform for mass is the same as it was for time, distance and velocity: alpha. But Einstein does not work in the direction I do. I used my transform for velocity to find the mass transform. Einstein, who assumes he has no velocity transform in the same situation, must instead develop an energy transform first. Remember that in the light plane problem, he had no v'. So he finds an energy equation and solves down from there to find mass transforms and then a momentum equation.        Like me, Einstein does not have a momentum transform equation. For Einstein, momentum can only be calculated by an observer (since he failed to remember that an object can calculate its own velocity). For Einstein,m'vav' cannot equal mvav, since Einstein has no v'.Instead, he finds that p = mv = γm0v        This is the current equation. In it gamma is understood to be transforming the mass. There is no v' to transform. This is the major problem with the current momentum equation. It proposes to transform from one coordinate system to another, but it does so without transforming the velocity. That is to say, this equation assumes that v is correct-that it is unaffected by relativity. Einstein is transforming m0 (which is in the coordinate system that is going v) to the coordinate system of the observer (which is the unprimed system here). The unprimed system is the system of the scientist measuring the particle whizzing by. But Einstein does not transform the velocity. He finds a velocity transform in Special Relativity, but he does not use it in the momentum equation. Why? One must suppose it is because the velocity transform he finds there is for two degrees of relativity, and he does not think it applies in this situation. I have shown in my previous paper that it does apply. The given velocity v is affected by relativity and must be transformed. It is affected by the speed of light. Why would the speed of light affect mass but not velocity, requiring a mass transform but no velocity transform?        Einstein's m0 is equivalent in math and theory to my mr. Therefore his equation for momentum is equivalent to thisp = xmrvavwhat does x equal, using my transformation terms? p = mvav x = m/mr = α p = αmrvav       In my theory, this last equation is not a momentum transformation. It is not transforming from one coordinate system to another. It is simply expressing the momentum in terms of a rest mass. The relativity transforms are between m and m'. Technically you cannot calculate a momentum from a rest mass, since a rest mass is not moving. But if, for some theoretical reason, you want to express momentum in terms of rest mass, this is the equation you should use. Part SevenEnergy Transformation Equations 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.        First, notice that Einstein's thought problem is analogous to mine except for one thing. Upon emission of the planes of light, his body does not change position in system A or velocity in B. My ball, however, does change velocity. It goes from rest at the zero-point to a final velocity of v' as measured from the ball or v measured from the zero-point. Einstein's two planes of light cancel out. My one photon has no twin in the opposite direction, therefore the ball is given a push and it achieves a velocity. In this way my ball is more like Einstein's slowly accelerated electron. So we only need to return to Einstein's equations to make the proper corrections.       E0 = the initial energy of the ball (measured by the ball) before emission of the photon.        E1 = the total energy of the ball measured by the ball after the emission of the photon.         H0 = the initial total energy of the ball as seen from the zero-point.        H1 = the final total energy of the ball as seen from the zero-point.       F0 = the energy of the photon as measured by the ball        F1 = the energy of the photon as measured by the zero-pointF1 = αF0     since F1 > F0E1 = E0 - F0 H1 = H0 - F1 E0 = H0      since the ball is initially at rest in both systems, A and BH1 - H0 = -F1 = -αF0And the final kinetic energy is represented byK = H1 - E1     = H1 - (E0 - F0)     = H1 - (H0 - F0)     = -αF0 + F0     = (1 - α)F0 (The initial kinetic energy was zero.)K = F0{1- [1 + (v'/c)] }     = -F0(v'/c)My kinetic energy is negative. It is negative because the ball is moving away from the zero-point. It can do no work upon a body positioned at the zero-point. To do +K amount of work on the zero-point, a force would have to be applied to the ball creating energy in the amount of 2K. In other words, a force sufficient to turn the ball around and give it v' in the opposite direction.        Now that I have brought Einstein's problem into line with my own thought problem, I may use F1 as the energy of my photon. F1 = m0Zc2 (though I will drop the "z" after this).         We do not need Einstein's derivation of m0c2 here, nor the textbook's simplified calculus derivation. I have shown that both are false. All we need is the equation we have already usedE/c = pL        which says that the momentum of a photon is expressed by E/c. This equation comes from previous theory and has nothing to do with relativity. If we assume that light can have a mass equivalence, then we haveE/c = m0vE = m0c2        where m0 is the mass equivalence of the light. My photon has a mass equivalence of m0 in this particular problem. Putting this into my equation above yieldsK = -F0(v'/c)αF0 = F1K = -m0c2(v'/αc)         But we want kinetic energy in terms of m' not m0.m0 = m'v'/cK = -(m'v'/2c )c2(v'/αc)K = -(1/α) m'v'2/2 To put in some hypothethical numbers: If vav' =.2c, then v' =.4c, and K = .714 x 5.55 x 10-36kg x (1.2 x 108m/s)2/2 = -2.85 x 10-20JNow let me calculate equations from the zero-point K = (1 - α)F0         {For α we will use 1/[1 - (v/c)] instead of 1 + (v'/c)]}K = -(v/c)m0c2m0 = (m - m')/2mr = m' - m0m0 = [m - (mr + m0)]/23m0 = m - mrK = -(m - mr)(v/3c)c2 -3cK/v = mc2 - mrc2 -K ≠ mc2 - mrc2         Which means that ifET = K + mrc2 ET ≠ mc2 ET = mrc2 - (v/c)m0c2= mc2/β - (v/2c)c2 [m - (m/α)] = mc2/β - (v/2c)[mc2 - (mc2/α)]= mc2[(1/β) - (v/2c) + (v/2αc)]ET = mc2[1 - (3v/2c) - (v2/2c2)]= 3.70 x 10-19JNow let us find ET in terms of mr, so that we can compare the transform to gamma.ET = mrc2 - (v/c)m0c2mr = m' - m0m0 = mrβ/α - mrET = mrc2 - {mr(v/c)c2[(β/α) - 1}        = mrc2 - {mr(v/c)c2[v/(2c - 3v)ET = mrc2{1 - [v2/(2c2 - 3cv)]}K = {mrc2{1 - [v2/(2c2 - 3cv)]} - mrc2 I check this against my previous numbers and find that indeed this also is-2.85 x 10-20JI will call this transformation term kappa2, κ2.κ2 = 1 - [v2/(2c2- 3cv)]Notice that it is not the equivalent of either gamma or beta (although it is very close in output to gamma).@ v = .286c, κ2 = {1 - [v2/(2c2 - 3cv)]} = .929        If we had been in an experimental situation where the kinetic energy had been positive, then we would have found the inverse of this number using kappa1, which isκ1 = 1 + [v2/(2c2 - 3cv)]         = 1.07        To show you how close we are to current experimental values, if we had used the average velocity in this equation, we would have found kappa1 to be 1.01gamma (@ v = .143c) = 1.01An exact match. Astonishing, considering all the mathematical and conceptual changes I made in Einstein's derivations. But he was not able to derive the classical equation from his thought problem, and I can:-3cK/v = mc2 - mrc2 (from above)mc2 - m[1 - (3v/2c)]c2 = -3cK/v multiply both sides by v2/c2mv2 - m[1 - (3v/2c)]v2 = -3Kv/c (3v/2c)]mv2 = -3Kv/c K = - mv2/2         = -2.85 x 10-20J Absolutely incredible! Once Einstein's variable assignments are corrected it turns out that the classical equation is precisely correct. Einstein and current wisdom both treat the classical equation as an approximation at slow speeds relative to c. As supposed proof of this, they expand the square root in gamma using the binomial expansion, the first uncancelled term being v2/2c2. But this is once again a fortuitous collision of luck and bad math. I have shown that gamma is an incorrect transformation term, so that expanding the square root of the term is pointless. If there is no gamma, there can be no expansion of the square root and no proof of the approximation of mv2/2. Besides, this expansion proposes to find that K ≈ mrv2/2         Which is absurd. What should have been intended is to show that K ≈ mv2/2 at slow speeds        This latter equation is the classical expression of kinetic energy. As I have shown, expressing kinetic energy in terms of a rest mass isn't even sensible, once it is understood what the different terms mean. The relativistic equation would have to resolve to either mv2/2 or m'v'2/2 at slow speeds, even if gamma and Einstein's theory and the binomial expansion were all completely correct. Having it resolve to mrv2/2 is just further proof that no one knew what was going on with the math and the variable assignments. You cannot have a rest mass in a kinetic energy equation because a mass at rest has no kinetic energy. Or, to be more precise, you cannot express mass as rest mass in your central and fundamental kinetic energy equation. K = mv2/2 is not some lead-up equation. It is the basic expression (and definition) of kinetic energy. It is therefore illogical to use a rest mass as your variable. Let me now clear up some rough spots. I have used Einstein's thought problem to find my energy equations, after a good bit of scouring. But in his thought problem the transform is done on the frequency of the light. This makes sense except for one thing: I explicitly said in my own thought problem that the ball does not use E' in its equations on the photon. How can I reconcile the two statements?         In the momentum equation m'v' = E/c, I say that the ball does not use E'. And this is true. In this equation, the equality applies to two numbers that are both generated by the same field, that field being the field of the zero-point. m'v' is relative to the zero-point, therefore E/c must also be relative to the zero-point. E'/c is not relative to the zero-point; it is relative to the coordinate system of the ball.         But in Einstein's thought problem, we are not creating a momentum equality, or conserving momentum. We are transforming from one system to another, A to B. We are transforming the energy of the light from E to E' at the same time that we are transforming masses and total energies. We must therefore include in the derivation E', which is the energy of the light measured from the ball. E' is a necessary variable in his problem. In my initial thought problem it is not. The second rough spot concerns the variable v'. In my mass transform equations, vav was the average velocity. But in using Einstein's thought problem, I show that v must be the final velocity of the body. Some may ask why I did not simply let v be the final velocity in my own thought problem. It was foresight that made me do it (and the Work-Energy Theorem). Remember that the ball, when it is calculating its own velocity, is in possession of only two pieces of data. It has an elapsed time and a distance. It also knows it started from rest. It therefore must assume an acceleration over one part of the distance or all of it. According to the Work-Energy Theorem, the ball may not assume that a final velocity was achieved instantaneously or over no distance. There is a kinetic energy because there is work (and vice versa) and there is work because there is time and distance involved. A force cannot be exerted over zero time or distance. Since I knew that kinetic energy would be both the end product and the driving "force" of my thought problem, I was astute enough to let v be what it must be under the situation—an average velocity. Of course this gives us just one more thing to be very careful about. Each problem has its own specific variable assignments, which have to be written out in full and kept track of. Even α does not always contain the same variables. In my mass transforms, the velocity variables in α are average velocities. In my time, distance and velocity transforms it does not matter, since no acceleration is involved. In the light frequency or light energy transforms, the velocity variables are commonly assumed to be final velocities. The last rough spot concerns the use of both v and vav in the derivation of the energy equations. When I am transforming the masses within the energy equations, I am using α with vav. But the final equation is expressed in terms of v. Isn't this an illegal mathematical substitution? No. It is perfectly legal to use v and vav in the same equations, as long as you keep track of them. As you can see, α has the same value whether you use v or vav, as long as you use the correct form of alpha each time. Therefore canceling alphas from one equation to the next is not a problem. ~~~~~~~~~~ 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.So let us return to Einstein's thought problem. His problem is different than my ball problem in that he has both a final and an initial kinetic energy. My ball started at rest, so that its initial kinetic energy was zero. These should have been Einstein's final equations, according to my corrections:L = 2F0ΔK = L[1 - (1/α)] ΔK = L(v/c)Whereas my final equations for the ball were:ΔK = (1 - α)F0 ΔK = -F1(v/c)So let us find the equations for Einstein's problem ΔK = 2F0(v/c)In his problem, the body measured the normal frequency for the light, so that F0 = m0c2ΔK = 2m0c2(v/c)        But we want kinetic energy in terms of m not m0.2m0 = mv/2c      (I used my equation m0 = mvav/c from above; but we must use the total mass of the light planes, which is 2m0.) ΔK = 2(mv/2c)c2(v/c) ΔK = mv2/2ET = mrc2 + 2(v/c)m0c2mr = m' - 2m02m0 = mrβ/α - mrET = mrc2 + {mr(v/c)c2[(β/α) - 1}ET = mrc2{1 + [v2/(2c2– 3cv)]}K = {mrc2{1 + [v2/(2c2– 3cv)]} – mrc2 ET = mrc2 + 2(v/c)m0c2ET = (1/β)mc2 + (v/c)[(m/α) - m/β)]c2ET = mc2 [(1/β) + (v/αc) - (v/cβ)]ET = mc2 [1 – (3v/2c) + (v2/2c2)] We have proven that the classical equation also applies to Einstein’s thought problem, and that ET ≠ mc2 there either.        But we are finally in a position to show that Einstein chose his thought problem carefully. He wanted avoid an acceleration and the use of average velocity that my problem entailed. So he chose a problem with no velocity change at all. Kinetic energy changes only because the mass has changed. But this has the effect of oversimplifying the problem of energy transformation due to Relativity. Notice that it is difficult to understand where to apply the Work-Energy Theorem in Einstein’s problem, since it is unclear where there is any force. You can’t have a force without an acceleration, and there is no acceleration here. What has happened is that the two forces from the two light planes have cancelled eachother out. You have forces, but they have added to zero. Some may say that is the beauty of the problem. It sidesteps all non-critical issues. But by sidestepping them it has cloaked them, historically. Einstein’s problem was too subtle by half. It was so subtle that it confused Einstein himself. What is more, by generalizing his findings from this one very unique thought problem (which was not at all general—it was not a standard problem of mass increase or energy transformation) he hid all the variations of mass and energy in Relativity. My thought problem is both more standard and more complex, so that it shows all the issues involved in solving problems of this nature. Now, what if we have a thought problem where the velocity is toward an observer? The kinetic energy will be positive, but the masses will show a decrease. Let us return to our ball and our photon. We will imagine two white lines drawn on the ground now, instead of one. The ball starts at the first white line, as before, but this time it is propelled toward the second white line, where we put an observer. Everything else is the same as in the first experiment. The momentum equalities will be the same, except that alpha for both the velocities and the masses will be inverted. Alpha in this problem will take the value α2 = 1 – (v’/c) = 1/[1 + (v/c)]       Alpha in the energy transform of the photon must stay in the original form, however, since the ball will still be measuring a redshift. The observer will be measuring a normal frequency, just as in the first problem. We must be very careful here, since we have two values for alpha. Not just two equal constructions, as in the first problem; now we actually have different values. We will call the original alpha α1, and the new alpha α2. E1 = E0 - F0 H1 = H0 - F1       [Some will want to add F1 to H0 here, since we have changed directions. But if we subtract the photon’s energy in one system, we must do so in the other system as well. The body cannot lose the mass equivalence of the photon in one system and gain it in the other.]K = H1 – E1F1 = α1F0 K = H0 - α1F0 -(H0 - F0)= (1 - α1)F0 = -(v/c)F1 = -(v/c)m0c2K = -mv2/2        What happened? We still got a negative kinetic energy. Everything worked out just like our first problem. But we know that the body must have a positive kinetic energy relative to the observer, since it is now moving toward it (it was at rest to start, of course). The reason we got a negative value at the end is that the equations don't know the difference between one white line in our system and the other. The equations only know the difference between one system and another. This series of equations gives us values relative to the zero-point of the experiment, which has not changed. We moved our observer, but we did not change the point of emission. The equation does not recognize that our observer has moved, since we did not add any pertinent information to the equation. The movement of the observer took place only in our heads, not in our math. The point of emission is behind the ball, therefore the kinetic energy, as a vector, is still negative. In order to get the right direction, we must make the change by hand. Anyone who has done a lot of vector equations knows that this kind of thing is a common feature of directional problems. The equations don’t always yield the desired information, since it is often impossible to include the pertinent postulates into the math. That is why it is so critical to be able to visualize vector problems and other geometric problems. Juggling equations is not sufficient. No problem in history has made this clearer than Special Relativity. K = mv2/2ET = mrc2 + (v/c)m0c2 mr = m’ - m0 m0 = mrβ2/α2 - mrET = mrc2 + {mr(v/c)c2[(β2/α2) – 1}β2 = 1/[1 + (v/2c)]ET = mrc2{1 + [v2/(2c2+ cv)]}K = {mrc2{1 + [v2/(2c2+ cv)]} – mrc2ET = mrc2 + (v/c)m0c2ET = (1/β2)mc2 + (v/c)[(m/α2) - m/β2)]c2ET = mc2[1 + (v/2c) + (v2/2c2)] ~~~~~~~~~~ We have now found three different transforms for three different problems. The only thing that has remained constant is that K = ±mv2/2        which can be considered ironic in that this was the one equation that was thought to be an approximation; also the one equation that was thought to have been superceded. One very important thing is different from classical theory, though, and that is that I have shown that v has a limit at .5c. Using gamma, physicists now think that v (the velocity as measured from a distance) can approach very close to c—since this is the value for v that gamma gives them. I have shown that v’—which is the true velocity of the body—may approach c; but v may not. This is not a great difference in theory—in that currently the variable v is thought to be the real velocity of the object. But it does take some getting used to, experimentally. An experimental physicist must now use either one of the equations K = mv2/2      orK = {mrc2{1 + [v2/(2c2– 3cv)]} – mrc2         since in an experimental situation he or she will always be dealing with mass as measured from a distance. That is to say, m’ is the mass as measured by the proton or electron itself, for instance, so it will not be part of our data. Therefore v’ cannot be calculated directly, by using one of the primed equations. The scientist will first discover v and then calculate v’ from that. Part EightThe Accelerator Problem (Why 108?) 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. Thought Experiment 4: Let us reverse the situation of our 3rd thought experiment—where a body at rest emitted a photon—and ask what would happen if the body instead absorbed a photon. Let us call our body a proton, so that we can assign it a known rest mass (mr =1.67 x 10-27kg). Now, we discovered an equation for mass increase with the emission problem, but this equation implies that we cannot increase the mass by more than 4 times, even if we take the proton all the way to c. m = mr/[1 - (3v/2c)]        Remember that v cannot exceed c/2. In my math it is the variable v’ that has a limit at c. However, we know from experiments in particle accelerators that the mass of the proton hits a limit at 108mr. We imagine this means that the proton in the accelerator is accelerating by absorbing energy from the acceleration field. To see what I mean by this, notice that both my emission problem and Einstein’s various thought problems all imply that when a body emits a photon, it not only gains an acceleration from the emission, it also loses mass or mass equivalence by losing the “body” of the photon. In other words, the photon leaves a hole. The rest mass of the body decreases after the emission. That is what Einstein’s variable assignments tell us (E1 = E0 - F0). This would be expected, since a body can hardly emit a smaller body, no matter whether that body is a particle of light or not, and expect to keep the same amount of rest energy.        This means that if we reverse the process, the body must gain an acceleration and gain rest mass from the absorbed photon. It gains a sort of double energy increase. Let us use our math from previous papers to express this.        In a real accelerator, the proton is taken to speed in a series of accelerations. This is an experimental concern, however, not a mathematical concern. Scientists do not use one super-field to accelerate since they 1) cannot create it, 2) cannot keep it from destroying the proton if they did create it. But we can simplify the math by allowing ourselves to imagine a super-high frequency photon with which we will bombard our proton in a single go. The proton will absorb this giant photon and we will see if the math we achieve from this absorption can explain the number 108. If it can, then we will have taken a decisive step in proving these corrections to Special Relativity. No one has yet been able to derive this number, and there is currently no theory to explain why there is a limit. The accepted term gamma implies an infinite mass increase capability; nor has the math of quantum theory predicted the existence of a limit or the number 108. First we must differentiate between our different masses and mass-equivalences. m0 = mass equivalence of the photonmri = rest mass of proton before absorption = 1.67 x 10-27kgmrf = rest mass of proton after absorption, measured from Bm = moving mass of proton, measured by an observerm’ = moving mass of proton, measured by the proton, relative to the observerBy the conservation of momentum, the momentum of the proton+photon after the absorption must equal the momentum of the photon before. mv/2 = E/c [remember that we must use the average velocity]E = m0c2m = 2m0c/vm0 = mv/2c1/α =1 – (v/c)v/c = 1 – (1/α)mv/c = m – m’ m0 = (m – m’) /2mrf = mri + m0m’ = mri + 2m0m’ = mrf - m0 + 2m0 = mrf + m0m/α = mrf + m0 = mrf + mv/2c mrf = m[1 – (3v/2c)]       Still the term beta. But let us find m in terms of mri and mri in terms of m0.m/α = mri + mv/cmri = m[1 – (2v/c)]m0 = mv/2cmri = 2m0[(c/v) – 2]So we only need to return to Einstein’s equations to make the proper corrections.       E0 = the initial energy of the proton before absorption of the photon (A as background).        E1 = the total energy of the proton after the absorption of the photon (A)       H0 = the initial total energy of the proton as seen from the zero-point (B)       H1 = the final total energy of the proton as seen from the zero-point (B)       F0 = the energy of the photon in A        F1 = the energy of the photon in BF1 = F0α since F1 > F0E1 = E0 + F0 H1 = H0 + F1 E0 = H0 since the proton is initially at rest in both systems, A and BH1 – H0 = F1 = αF0And the final kinetic energy is represented byK = H1 – E1= H1 – (E0 + F0) = H1 – (H0 + F0) = αF0 - F0 = (α - 1)F0 = (v/c)F1K = (v/c)m0c2m0 = (m – m’)/2mri = m’ - 2m0m0 = [m – (mri + 2m0)]/24m0 = m – mriK = (m – mri)(v/4c)c24cK/v = mc2 – mric2 K ≠ mc2 – mriic2 4cK/v = mc2 – mric2 mc2 – m[1 – (2v/c)]c2 = 4cK/v multiply both sides by v2/c2mv2 – m[1 – (2v/c)]v2 = 4Kv/c (2v/c)]mv2 = 4Kv/c K = mv2/2 Which means that ifET = K + mrfc2 mrf = mri + m0ET = K + mric2 + m0c2ET = mric2[1 + (v’/2c)]                  [1 – (v’2/c2)]ET = mric2{1 + [(v2 + cv)/(2c2– 4cv)]}ET = mc2 [1 – (3v/2c) + (v2/2c2)]ET = mc2 [1 + (v’/2c)]              [1 + (2v’/c) + (v’2/c2)]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)m0c2 + m0c2 + mric2 = 108mric2(v/c)m0c2 + m0c2 = 107mric2       This last step was allowed since mri is the same in both theories. [(v/c) + 1]m0 = 107mrimri = 2m0[(c/v) – 2][(v/c) + 1]/[(c/v) - 2] = 214v = .4982558c v’ = .9930474c = 2.97708 x 108m/sc = 2.99792458 x 108m/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 104m/s less than c. By my theory the gap is a bit larger: 2.1 x 106m/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 10-12m/s2.    Here is a gloss of that math.  Given two equal spheres of radius r touching at a point, we haveF = Gmm/(2r)2ma = 2Gmm/(2r)2 a = 2Gm/4r2    a/2 = 2Δr/2Δt2         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/Δt2 = Gm/r2           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 + Δrm = Δr3/GΔt2a = 2Δr/Δt2a = 2mG/Δr2That 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/Δr2      If the proton has a radius of 10-13m, this yieldsa = 8.88 x 10-12m/s2          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 104m/s        =    85 million years   8.88 x 10-12m/s2          My corrected numbers give an age of the proton of about 15 billion years.v = at/2 = (8.88 x 10-12m/s2 )(4.73 x 1017s)/2 = 2.1 x 106m/s         My 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. *I have been asked where the 2 comes from when I write 2v'av/c in the denominator of my velocity transform above. In answer, consider that the fraction v/c is the real transform here, so we have to be careful to get it right.  We are comparing v to c.  So we can't compare v'av to c.  If we did that, we would be implying that the object accelerated to speed over some interval while the light did not.  But since we assume light is material just like our object, we cannot imply that.  The light may accelerate to speed extremely fast, but it still must accelerate.  Since our given number for c is obviously a final velocity, we must also use a final velocity for our object.   You will say that we have no indication light accelerates to speed when emitted, but that empirical fact is beside the point in this particular thought problem.  Since we are ultimately calculating momenta and energies from this thought problem, we have to let the photon accelerate at emission, otherwise it wouldn't be capable of creating a force back on the object.  If we let light be emitted with no acceleration, it couldn't create an equal and opposite force back on the object, you see.  It would just slip out of the object with a whisper, and no force back would be created.  So to compare v to c and obtain any sort of momentum—using the conservation of momentum—we have to assume light accelerates at emission.  If we assume that, then we have to use a v final in that position to make the equations work.   If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. 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