return to homepage This paper is a compression of my full paper showing that Einstein’s energy and mass transform equations are flawed. In my long paper on the mass and energy transforms of relativity, I followed Einstein’s original derivations in his original papers line by line, showing precisely where the mathematical errors were. To exhaustively dig out and solve all these errors took me 41 pages. I believe that history will ultimately forgive me this length, since I did not say anything that did not need to be said. However I can understand that this length—and the inherent difficulty of the subject—made the paper a chore, to say the least. In my defense I can only say that if the problem had been an easier one, Einstein would not have mistaken parts of it—and subsequent physicists would have uncovered his mistakes before me. My longest suit may be tenacity; and the tenacious, when clearing up errors, do not like to stop until they are finished.
We will start with Einstein’s second major paper of 1905, 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, K _{1} - K_{0} = L(1 - γ)If you insert m _{0}c^{2} into Einstein’s last equation (as Einstein did later and as history still does) this implies that L = m_{0}c^{2}. Not E In the beginning of the problem, E_{0} but L._{0} is assumed to be the rest energy of the particle: Einstein and history have assigned m_{0}c^{2} to E_{0}. But according to these equations, L = m_{0}c^{2}. That is, m_{0}c^{2} 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.Another 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. Also, just as in his first paper on Special Relativity, Einstein has failed to assign v to either system A or B. We are told that B is moving v relative to A. But is v measured from A or B? We have two possible numbers for v: B rel A measured from A, or B rel A measured from B. Kinetic energy can also 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. Einstein 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 not the transform to use here. Even if gamma had been correct as a transform for distance and time in Special Relativity, Einstein still should not have applied it to the light rays here. Physics already had a transform for frequency that had nothing to do with Special Relativity, and that should be applicable in this problem. This transform is f’ = f(1 + v/c). Relativity has not overwritten or jettisoned this transform; Richard Feynman actually used it much later as part of his proof of Relativity.* That is, he uses the correct transform to derive the incorrect one. Current theory is built on a simultaneous and inconsistent use of both transforms.
Now let us correct these errors. We will have to change the thought experiment a bit in order to produce all the clear and definable variables we need. Let us start with the body and the observer both at rest together. Let us have the body emit only one light ray in one direction, and let us limit this ray to a single photon. What difference will this make? Einstein rigged his math so that his body did not change position in system A or velocity in B. Our new body, however, does change velocity. It goes from rest to a final velocity of v’ as measured from itself in A, or from rest to v measured from the observer in B. Einstein’s two planes of light cancel out. My one photon has no twin in the opposite direction, therefore our body is given a push and it achieves a velocity. The kinetic energy is negative in my problem because the body is moving away from the observer. It can do no possible work on the observer. Once Einstein’s variable assignments are corrected it turns out that the classical equation is precisely correct, meaning that it gives us exactly the same numbers that relativistic equations give us. 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 v^{2}/2c^{2}. 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 mv^{2}/2. Besides, this expansion proposes to find that K ≈ m _{r}v^{2}/2 Which is absurd. What should have been intended is to show that K ≈ mv ^{2}/2 at slow speedsThis latter equation is the classical expression of kinetic energy. As I have shown, expressing kinetic energy in terms of a rest mass in a classical equation isn’t even sensible, once it is understood what the different terms mean. You can't express kinetic energy in terms of a rest mass, in a classical equation. The mass variable in Newton's equation must be a moving mass. The relativistic equation would have to resolve to either mv^{2}/2 or m’v’^{2}/2 at slow speeds, even if gamma and Einstein’s theory were correct. Having it resolve to m_{r}v^{2}/2 is just further proof that no one knew what was going on with the math and the variable assignments.Now let us derive the new energy transforms. Above we found that -3cK/v = mc ^{2} – m_{r}c^{2} -K ≠ mc ^{2} – m_{r}c^{2}Which means that if the total energy, E _{T} = K + m_{r}c^{2} E_{T} ≠ mc^{2} m’ = m[1 - (2v _{av}/c)] m _{r} = m[1 - (2v_{av}/c)] – m(v_{av}/c) = m[1 - (2v _{av}/c) – (v_{av}/c)]m _{r} = m[1 - (3v_{av}/c)]beta = β = 1/[1 - (3v_{av}/c)]E _{T} = m_{r}c^{2} - (v/c)m_{0}c^{2} = mc^{2}/β - (v/2c)c^{2} [m – (m/α)] = mc ^{2}/β - (v/2c)[mc^{2} – (mc^{2}/α = mc^{2}[(1/β) - (v/2c) + (v/2αc)] E_{T} = mc^{2}[1 – (3v/2c) – (v^{2}/2c^{2})]Now let us find E _{T} in terms of m_{r}, so that we can compare the transform to gamma.E _{T} = m_{r}c^{2} - (v/c)m_{0}c^{2}m _{r} = m’ - m_{0}m _{0} = m_{r}β/α - m_{r}E _{T} = m_{r}c^{2} - {m_{r}(v/c)c^{2}[(β/α) – 1}= m _{r}c^{2} - {m_{r}(v/c)c^{2}[v/(2c – 3v)E _{T} = m_{r}c^{2}{1 – [v^{2}/(2c^{2}– 3cv)]}K = m _{r}c^{2}{1 – [v^{2}/(2c^{2}– 3cv)]} – m_{r}c^{2}The transformation term here is 1 – [v ^{2}/(2c^{2}– 3cv)], which is not gamma. In my long paper I show that there are several variations of this transformation term. For example, the solution to Einstein’s original thought problem gives us the transformation term 1 + [v^{2}/(2c^{2}– 3cv)], which is also not gamma. A body moving toward an observer would have the term 1 + [v^{2}/(2c^{2} + cv)].In my long paper I showed that Einstein’s own thought problem also resolved to the classical equation. All the various problems I solved resolved to K = ±mv ^{2}/2. Amazingly, this was the one constant, no matter what variations of energy transformation I was dealing with.
[1 – (v’^{2}/c^{2})] E _{T} = m_{ri}c^{2}{1 + [(v^{2} + cv)/(2c^{2}– 4cv)]}E _{T} = mc^{2} [1 – (3v/2c) + (v^{2}/2c^{2})]E _{T} = mc^{2} [1 + (v’/2c)][1 + (2v’/c) + (v’ ^{2}/c^{2})]Notice the last bolded equation above tells us why gamma works so well in accelerators despite being slightly incorrect and being derived with so many mistakes.In accelerators we are finding a limit at 108. Therefore, we set my equation equal to 108 and see what velocity the proton is really achieving. (v/c)m _{0}c^{2} + m_{0}c^{2} + m_{ri}c^{2} = 108_{ri}c^{2}(v/c)m _{0}c^{2} + m_{0}c^{2} = 107m_{ri}c^{2}This last step was allowed since m _{ri} is the same in both theories. [(v/c) + 1]m _{0} = 107m_{ri} m _{ri} = 2m_{0}[(c/v) – 2][(v/c) + 1]/[(c/v) - 2] = 214 v = .4982558c c = 2.99792458 x 10 ^{8}m/s v’ = .9930474c = 2.97708 x 10^{8}m/sAccording to current theory, gamma is equal to 108 at v = .999957c. The v variable in gamma is equivalent to my v’, since current theory has no v’, and since I have defined my v’ as the true velocity of the object.So, we now have all our numbers in hand. How am I going to explain the number 108? Notice that we have an unexplained velocity differential in both current theory and my theory. By current theory the limit in velocity for the proton is 1.2 x 10 ^{4}m/s less than c. By my theory the gap is a bit larger: 2.1 x 10^{6}m/s. What causes this gap? And which gap is correct? If I can answer these questions, then I can show where the number 108 comes from.Let’s say that the proton already has a velocity or velocity equivalent due to some motion or force or other unexplained phenomenon. Let’s say that the proton’s total velocity cannot exceed c, and that this other unexplained motion or force makes up the difference. That is precisely what I have done in my paper on the Universal Gravitational Constant. Using a hint of Maxwell and the dimensions of G, I showed that the proton can be shown to have a constant acceleration in any direction of 8.88 x 10 ^{-12}m/s^{2}. Here is a gloss of that math. Given two equal spheres of radius r touching at a point, we haveF = Gmm/(2r) ^{2}ma = 2Gmm/(2r) ^{2} a = 2Gm/4r ^{2} a/2 = 2Δr/2Δt ^{2}We now let the spheres expand at a constant and equal rate. We assign Δr to a change in the radius instead of a change in the distance between the spheres, and this allows us to calculate even when the spheres are touching. Δr/Δt ^{2} = Gm/r^{2} After time Δt, the radius will be r + Δr. After any appreciable amount of time, r will be negligible in relation to Δr, so that Δr ≈ r + Δr m = Δr ^{3}/GΔt^{2}a = 2Δr/Δt ^{2}a = 2mG/Δr ^{2}That is the acceleration of each of two equal masses in a gravitational situation. But if we want to give all the acceleration to one of them, holding the other one steady for experimental purposes, then we simply double the value. a = 4mG/Δr ^{2}If the proton has a radius of 10 ^{-13}m, this yieldsa = 8.88 x 10 ^{-12}m/s^{2} If we allow the proton to accelerate at this pace over its entire lifetime up until the current moment, then we can achieve a number for its present velocity due to mass. My velocity is a much better fit. Using this acceleration due to mass and gamma, we get an age of the proton of only 85 million years. v = at/2 = 2 x 1.2 x 10 = 85 million years^{4}m/s 8.88 x 10 ^{-12}m/s^{2}My corrected numbers give an age of the proton of about 15 billion years. v = at/2 = (8.88 x 10 ^{-12}m/s^{2} )(4.73 x 10^{17}s)/2 = 2.1 x 10^{6}m/sMy number is therefore a match to current estimates, as you see. Current theory based on gamma is clearly wrong, since the proton cannot be as young as 85 million years. That would make protons 50x younger than the earth.[To see a shorter way to derive the number 108, you may now visit my more recent paper called Redefining the Photon. There, I use the density of the charge field to calculate the number.]
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