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All that being said, I have fielded questions for some months and I can now see that a gloss is
necessary. It would be useful to have a short argument, one
that moved quickly and was fairly easy to follow mathematically.
Readers with questions could then refer to sections of the long
paper for clarification. Readers who fully understood and accepted
the short argument could perhaps dispense with whole sections of the long
paper. K = κmrc2 – mrc2 = mv2/2 In short, I derived a new transformation term kappa to replace gamma. I did not pull this term out of thin air or manufacture it from theory or philosophy. I achieved it directly by making simple algebraic corrections to Einstein’s paper of 1905. This new term elegantly showed the equivalence of Einstein’s math and Newton’s, as regards kinetic energy. One final remark before I proceed. I want to stress once again that deriving the classical energy equation from Einstein’s postulates and math does not falsify the major findings of relativity or suggest that science should return to Newton. Relativity is true. Distance, time, velocity, mass, momentum and energy all require transforms from one observer to another. Beyond this, Einstein was correct in his other postulates, such as the fundamental postulate of energy transformation—that energy and mass are equivalent and transferable. My corrections will actually have a greater impact on the math and theory of General Relativity. The basic theory of Special Relativity—that is, relativity caused by velocity—has not been greatly altered by my corrections; only the math is changed, and, for the most part, subtly. Nor is my
resuscitation of mv2/2 to be understood as a return to
Newton. Newton’s equation is not a transform and does not allow us
to go from measurement by a object to measurement by an observer.
The variables in Newton’s equations are all as measured by the
observer. They are therefore not relativistic and are useless as
transforms. My only point in re-proving the old equation is show the
true mathematical link from classical kinetic energy to
relativistic. A simple correction to Einstein’s math shows why
Newton’s equation worked and continues to work in the proper
situations. Part One We will start with Einstein’s second major 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. 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.
a = γ[1 +
(v/c)cosφ] 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 is K0 - K1 =
L{
1
- 1} =
γL -
L That is the whole paper. It takes up less than three pages in Annalen der Physik. The cardinal error is in the final two steps. 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. 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
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) If you insert m0c2 into Einstein’s last equation (as Einstein did later and as history still does) this implies that L = m0c2. Not E0 but L. In the beginning of the problem, E0 is assumed to be the rest energy of the particle: 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. 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. Part Two 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. It is true that this new thought problem implies an acceleration over an interval, but we can get around that by using an average velocity over that interval. Let us first relist our variables, using Einstein’s as a guide. E0 = initial energy of the
body (measured by the body) before emission of the photon.
F1
=
αF0
since F1 > F0 H1 – H0 = -F1 = -αF0 And the final kinetic energy is represented by K = H1 –
E1 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. Furthermore, I may assume that F1 = m0c2. If we assume that light can have a mass equivalence, then from the equation for the momentum of light we have E/c = m0v where m0 is the mass equivalence of the light. [Notice that I am accepting Einstein’s postulates concerning mass and energy equivalence; I am simply not accepting his math. To achieve E = m0c2 we do not need any of Einstein’s math; we only need the momentum equation for light and the postulate that light has a mass equivalence—both of which Einstein explicitly accepted.] In order to proceed, we need an equation for α. My new thought problem shows us that we need two measurements of mass and two measurements of velocity. That means we need two more variables than Einstein worked with. We need m, m’, v, and v’. This is the only possible way we can do logical mass and velocity transforms. The equations for v and v’ I must import from my first paper on Special Relativity. I will include here a gloss of the math I used to achieve them. In these equations, primed variables are as observed by A (the object itself); unprimed variables are as observed by B (the observer). A simple correction to Einstein’s parameters gives us xt =
x’t’ The reason v ≠ x/t is that these unprimed variables must be defined as variables as observed by B, not in B. For instance, x is how x’ looks to the observer. It is not how the observer sees his own distances. This is a huge difference, and it is the main problem of Einstein’s and Lorentz’s original variable assignments. The variable x is a distance in A observed by B. It is not a distance in B observed by B. You must understand this fully before you can proceed with any of my corrections. t = t' + (x'/c) mvav = m’vav’ = E/c Now we solve in terms of rest mass. To do this we need an equation for rest mass in this particular problem. I say this particular problem because it turns out that there are several variations of mass increase and decrease, and the equations vary a bit depending on whether the bodies are emitting or absorbing and whether they are moving toward or away from an observer. Notice that in Einstein’s thought problem, the body would be measuring the normal frequency of light and the observer would be measuring a redshift. His observer is moving away from the point of emission. But in my thought problem, the observer is at the point of emission already. It is the body that measures the redshift. The rest mass is the moving mass minus the mass equivalence gained by moving. The mass equivalence gained by moving can only be the mass equivalence of the ejected photon. What the body lost in the photon it gained in movement. But we have two moving masses, m and m’. We also have two possible mass equivalences for the photon, since A and B will get different numbers for its energy (F0 and F1). We must choose m’ since it is connected to v’, which is the correct velocity (I have shown that v and m, although calculable, are not correct since they have been skewed by c). Likewise, we must choose F1 since it is connected to the normal frequency of light. It gives us the correct mass equivalence of the light. I have defined the normal frequency of light as the frequency measured from any point at rest relative to the point of emission. mr = m’ -
m0 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.
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 in a
classical equation 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 were
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. -3cK/v =
mc2 –
mrc2 Which means that if the total energy, ET = K
+ mrc2 ET =
mrc2 -
(v/c)m0c2 The transformation term here is 1 – [v2/(2c2– 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 + [v2/(2c2– 3cv)], which is also not gamma. A body moving toward an observer would have the term 1 + [v2/(2c2 + 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 = ±mv2/2. Amazingly, this was the one constant, no
matter what variations of energy transformation I was dealing
with. Part Three
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. m0 = mass
equivalence of the photon m = moving mass of proton,
measured by an observer mv/2 = E/c [remember that we must use the average velocity] E =
m0c2 Still the
term beta. But let us find m in terms of mri and
mri in terms of m0. 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). H1 – H0 = F1 = αF0 And the final kinetic energy is represented by K = H1 –
E1 multiply both sides by v2/c2 Which means that if ET = K + mrfc2
(v/c)m0c2 +
m0c2 + mric2 =
108ric2 This last step was allowed since mri is the same in both theories. [(v/c) + 1]m0
= 107mri According 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. 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 6.06 x 10-13m/s2. Here is a gloss of that math. Given two equal spheres of radius r touching at a point, we have F = Gmm/(2r)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/Δt2 = Gm/4r2 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 =
4Δr3/GΔt2 For the proton, this yields a = 6.06 x 10-13m/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. Fortunately, both the velocities above fit into a possible range for the estimate of the age of the universe and the proton. This would seem to corroborate my math. Furthermore, my velocity is a much better fit. Using this acceleration due to mass and gamma, we get an age of the proton of 1.9 x 1016 seconds or about 628 million years. 1.2 x
104m/s
= 628 million years My corrected numbers give an age of the proton of about 110 billion years. 2.1 x
106m/s
= 110 billion years
Neither confirms current models, which put the age of the universe at 12
to 20 billion years. However, current theory based on gamma
is clearly wrong, since the proton cannot be as young as 600 million
years. That would make protons 8x younger than the earth. My
prediction, though quite high, is only about 5x above current
estimates. These estimates have continued to rise throughout the
last hundred years. It is probable that current estimates are too
low. My finding here, if true, would suggest that estimates may
still be off by a factor of five or more. Conclusionp
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