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New  Energy & Mass
Transforms  in
Special  Relativity:

a  compressed  argument

by  Miles Mathis

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.
       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. 

Let me start by reminding the reader that by making a couple of small corrections in Einstein’s derivation, I was able to derive the classical energy equation from Einstein’s own equations.  That is, I showed that
K  = κm
rc2mrc2   = 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 mv
2/2  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.  I only mean to 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. And, perhaps most importantly, this new derivation shows that c is not a limit where mass goes to infinity. Since Einstein's equation is not an infinite series expansion of Newton's, c is no longer a limit in the math. In other words, my proof below shows that gamma is incorrect, which means that there is no power series or infinite terms, which means that c is not a mass limit. Einstein's equation and Newton's equation are mathematically equivalent, and one does not "approximate" the other. This solves the so-called mass gap in Yang-Mills, among many other things.

Part One
Locating the Errors

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. 
      E1 = the energy in A after the emission of the two planes of light.
      H0 = the initial energy of the body as seen from B.  That is, it is the initial rest energy plus the kinetic energy.
      H1 = 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 A
H0 = H1 + aL/2  + bL/2      This is the equation from B, where a is the negative angle transform and b is the positive angle transform
a = 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 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
                       √[1 - (v2/c2)]
       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 line with the next to the last line, 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  
1  ≠ γL 
This is because K
1  ≠ 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 =  m0c2Not  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
the Corrections

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. 
      E1 = total energy of the body measured by the body after the emission of the photon.
      H0 = initial total energy of the body as seen from the observer. 
      H1 = final total energy of the body as seen from the observer.
      F0  = energy of the photon as measured by the body 
      F1 = energy of the photon as measured by the observer
α = light frequency transform
F1 = αF0            since F1 > F0
E1 = E0 - F0 
H1 = H0 - F1  
E0 = H0     since the body is initially at rest in both systems, A and B
H1 – H0  = -F1  = -αF0
And the final kinetic energy is represented by
K  = H1 – E1
     = H1 – (E0 - F0)
     = H1 – (H0 - F0)
     =  -αF0 + F0
K   =  (1 – α)F0
      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
E =  m0c2
        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.  I must import the equations for v and v’ 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’
        v =  x'/t  =  x/t'  ≠ x/t
        v’ = x’/t’
The reason  v ≠ x/t   is that these unprimed variables must be defined as variables as observed by B, not in B.   To say that another way, 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)
  = t' + (v't'/c) = t' (1 + v'/c)  = t'/( 1 - v/c)
v = x'/[t'(1 + v'/c)]
v =      v'   
      1  + (v'/c)    
v' =     v          =  α v
       1 - (v/c)     
α = alpha  = 1/[1- (v/c)] = 1 + (v'/c)  =  c/(c - v) = (c + v')/c

Alpha is the current term for light frequency transform, as I said above.  I have shown that it is also the correct term for one-degree velocity transform, replacing the term gamma in the first part of Special Relativity. 
Now we can continue with our energy transforms.  Next we need mass transform equations.  These are as simple to derive as the velocity transforms.  When the body emits the photon it must recoil, achieving a velocity.  This velocity can be found by using the conservation of momentum—the momentum of the light must be equal to the momentum of the ball:
mvav = m’vav’ = E/c = m0c
m = m0c/vav
m0 = m(vav /c)
v’av  =  αvav   = mvav/m’
m = αm’
1/α =1 – (2vav/c)
We assume a constant acceleration over the interval in question, so that the average velocity must be half the final velocity, so
vav/c = [1 – (1/α)]/2
mvav/c = [m – (m/α)]/2
m0 = (m – m’)/2
And from above we had these equations:
K  =  (1 - α)F0
F0 = m0c2
K  =  -(v/c)m0c2
         So now we have almost enough to solve. But remember that m0 is not the rest mass of our body in question; it is the mass equivalence of the emitted photon. We need to solve in terms of the body's mass, not the photon's mass. To do this, we need another mass variable beyond the ones above. 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. You will have to consult my longer paper to see why this is so. But you may 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. 
         By definition–both mine and Einstein's–the rest mass of the body is the moving mass minus the mass equivalence gained by moving.  In this problem, 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
m0 = [m – (mr + m0)]/2
3m0 = m – mr
K  =  -(m – mr)(v/3c)c2
-3cK/v   = mc2   –   mrc2   
           express mr in terms of m
mc2   –   m[1 – (3v/2c)]c2  = -3cK/v   
           multiply both sides by v2/c2
mv2   –   m[1 – (3v/2c)]v2  = -3Kv/c 
(3v/2c)]mv2  = -3Kv/c 
K = - mv2/2       
           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 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 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 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.

Now let us derive the new energy transforms.  Above we found that
-3cK/v   = mc2   –   mrc2   
-K  ≠  mc2    –   mrc2
       Which means that if the total energy,
ET  =  K +  mrc2
ET  ≠  mc2
m’ = m[1 - (2vav/c)]
mr  =  m[1 - (2vav/c)] –  m(vav/c) 
       =  m[1 - (2vav/c) – (vav/c)]
mr  =  m[1 - (3vav/c)]
beta = β = 1/[1 - (3vav/c)]
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)]

Now let us find ET in terms of mr, so that we can compare the transform to gamma.
ET  =   mrc2   - (v/c)m0c2
mr = m’ - m0
m0  =  mrβ/α - mr
ET   =   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/(2c23cv)]} – mrc2    
         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
The Accelerator
Why 108?

Let us now apply our equations to a real experimental situation—the accelerator.  To do this we must reverse the situation of our thought experiment above—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 (m
r =1.67 x 10-27kg).   We know from experiment 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 once again differentiate between our different masses and mass-equivalences. 
0 = mass equivalence of the photon
ri = rest mass of proton before absorption = 1.67 x 10-27kg
rf = rest mass of proton after absorption, measured from B
m = moving mass of proton, measured by an observer
m’ = moving mass of proton, measured by the proton, relative to the observer

By 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 = m
m = 2m
0 = mv/2c
1/α =1 – (v/c)
v/c = 1 – (1/α)
mv/c = m –  m’
0 = (m – m’)/2
rf = mri + m0
m’ = m
ri + 2m0
m’ = m
rf  - m0 + 2m0 = mrf  + m0
m/α = m
rf  + m0 =  mrf  +  mv/2c
rf  =  m[1 – (3v/2c)]
         Still the term beta.  But let us find m in terms of m
ri and mri in terms of m0.
m/α = m
ri  + mv/c
ri  =  m[1 – (2v/c)]
0 = mv/2c
ri  = 2m0[(c/v) – 2]

So we only need to return to Einstein’s equations to make the proper corrections.
0 = the initial energy of the proton before absorption of the photon (A as background). 
1 = the total energy of the proton after the absorption of the photon (A)
0 = the initial total energy of the proton as seen from the zero-point (B)
1 = the final total energy of the proton as seen from the zero-point (B)
0  = the energy of the photon in A 
1 = the energy of the photon in B
1 = F0α       since F1 > F0
1 = E0 + F0 
1 = H0 + F1  
0 = H0     since the proton is initially at rest in both systems, A and B
1H0  = F1  = αF0
And the final kinetic energy is represented by
K  = H
     = H
1(E0 + F0)
     = H
1(H0 + F0)
     =  αF
0 - F0
     =  (α - 1)F
     =  (v/c)F
K = (v/c)m
0 = (m – m’)/2
ri = m’ - 2m0
0 = [m – (mri + 2m0)]/2
0 = m – mri
K  =  (m – m
4cK/v   = mc2   –   m
K  ≠  mc
2    –   mric2
4cK/v   = mc
2    –   mric2    
2   –   m[1 – (2v/c)]c2  = 4cK/v   
           multiply both sides by v
2   –   m[1 – (2v/c)]v2  = 4Kv/c 
2  = 4Kv/c 
K =  mv
       Which means that if
T  =  K +  mrfc2
rf = mri + m0
T  =  K +  mric2 + m0c2
ET = mric2[1 + (v’/2c)]
                [1 – (v’
T = mric2{1 + [(v2 + cv)/(2c24cv)]}
T = mc2 [1 – (3v/2c) + (v2/2c2)]
T = mc2 [1 + (v’/2c)]
              [1 + (2v’/c) + (v’
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. 
0c2    +  m0c2 + mric2 =  108ric2
0c2    +   m0c2  =  107mric2
         This last step was allowed since m
ri  is the same in both theories.    
[(v/c) + 1]m
0   =  107mri
ri  = 2m0[(c/v) – 2]
[(v/c) + 1]/[(c/v) - 2] = 214
v  = .4982558c  
c = 2.99792458 x 10
v’ = .9930474c = 2.97708 x 108m/s
        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.

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
4m/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 have
F = Gmm/(2r)
ma = 2Gmm/(2r)
a = 2Gm/4r
a/2 = 2Δr/2Δt
         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. 
2 = 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 + Δr
m = Δr
a = 2Δr/Δt
a = 2mG/Δr
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
      If the proton has a radius of 10
-13m, this yields
a = 8.88 x 10
         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
          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.]


My mathematical connection of this paper with my other papers on mass and gravity 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.  Mass is the acceleration of a volume.  Mass therefore has a calculable velocity over any interval. 
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 explaining why gravity doesn’t seem to exist at the atomic level.  It doesn't seem to exist simply because we have assigned the motion to another "field" or cause. At the atomic level we have decided to call this force "charge", but it is the same force as gravity, it is just hiding under another name.  But at both the quantum level and the macrolevel, gravity is not a force at all.    According to the new theory, you can assign mass, gravity, charge, strong force, and inertia all to the same basic motion.   All these concepts are not separate ideas, they are different expressions of the same thing.  And they all resolve to length over time. These last claims were not proved in my long paper, much less here, but they are proved in links to other papers, and my mass increase paper is one of the central pillars in that proof.

*Feynman Lectures on Gravitation, eq. 7.2.1

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