
return to homepage return
to updates
A
COMPLETE REDERIVATION of E=MC^{2}
or
HOW CORRECTED TRANSFORMS in SPECIAL RELATIVITY affect MASS,
MOMENTUM and ENERGY EQUATIONS
Einstein
hiding behind his paper
by Miles
Mathis
click
here to go to a gloss of this paper
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 halfcentury, 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 m_{0}c^{2}
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  (v^{2}/c^{2})]
I will prove that by
correcting the math, the energy transform for Einstein's problem
is actually kappa κ
= 1 + [v'^{2}/(2c^{2}
+ cv'  v'^{2})]
or E_{T}
= m_{0}c^{2}[1
+ (v’/2c)]
[1 – (v’^{2}/c^{2})]
You can see that the difference is very small in most
situations*, and might pass for decades without final
experimental confirmation, especially in a milieu that considered
Relativity a settled question. Physical
Review Letters,
the primary publication of record in the US, doesn't even have a
category for Special Relativity. A scientist could not present a
finding if he had one.
*In
comparing kappa and
gamma, it is also
important to note that I prove below that Einstein's thought
problem is not directly analogous to the more common experimental
problem of a subatomic particle in an accelerator. I show that
the energy equations must vary from problem to problem, depending
upon the physical situation.
In
deriving this new transform I also discovered several other facts
of great interest. One of these is that E ≠ mc^{2}.
If we assume that the rest energy is given by the rest mass—as
in E_{r}
= m_{r}c^{2}—then
the moving energy cannot be given by the moving mass, in a
straight equation. A transform is required here as well, and it
is not gamma.
This is a consequence of Einstein's own variable assignments.
Einstein assumed, with no theoretical or mathematical backup,
that mc^{2}
must be the term that
is applied to the final energy E. It turns out that this is not
the case. Even more
astonishing is that using my new derivation, where all the
variables are rigorously assigned, I am able to prove that the
classical equation is precisely
equivalent to the
relativistic equation. In other words, K = κm_{r}c^{2}
 m_{r}c^{2}
= mv^{2}/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 doubleprimed system. At the end, when he
finds t = γt', he has mistaken a transform from the man to the
platform for a transform from the train to the platform. Einstein
completely ignores the direct transform from the platform to the
train. His given velocity v is the velocity of the train relative
to the platform, he tells us. But he does not say whether this is
the velocity as measured from the train or from the platform. The
two measurements must be different, but Einstein never includes
this difference in his calculations. t = γt' therefore applies
to a transform from the platform to the man, which is in fact a
transform of two degrees of relativity. He never provides
transforms for one degree of relativity. t = γt'
should read t = γt''
However, I show that gamma
is incorrect for two
degrees of relativity as well.
I was the first to
demonstrate firstdegree 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 seconddegree transforms,
although these do not enter into mass transform solutions.
My firstdegree
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 firstdegree transforms allowed me to
derive a firstdegree 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
One Einstein'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 farreaching 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 nonspecific 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
xaxis, and therefore to the observer. Let us call B the system
of the observer and A the system of the body. Using his
nomenclature, E_{0}
= 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 E_{0} is
at this point. But from the outcome of the equations, E_{0}
must be what he calls the initial
rest energy, as in E_{0}
= m_{0}c^{2}.
Since the body is at rest in A, E_{0}
is both the rest energy and the
total energy. E_{1}
is the energy in A after the
emission of the two planes of light.
H_{0}
is the initial energy of the body as
seen from B. That is, it is the initial rest energy plus the
kinetic energy. H_{1}
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.
E_{0}
= E_{1}
+ L/2 + L/2
This is the equation as calculated from A H_{0}
= H_{1}
+ 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
 (v^{2}/c^{2})]
Now,
Einstein says the initial kinetic energy of the body is
represented by the equation K_{0}
= H_{0}
 E_{0} And
the final kinetic energy is represented by K_{1}
= H_{1}
 E_{1} So
that the change in kinetic energy is
K_{0}
 K_{1}
= L{ 1
 1} = γL  L
√[1  (v^{2}/c^{2})]
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
H_{1} and
the term L to E_{1}.
He assumes that H_{1}
is mc^{2}
and E_{1}
is m_{0}c^{2}.
But look back up the series of steps: L ≠ E_{1}
H_{1}
≠ γL This is because K_{1}
≠ K_{0}
 K_{1}.
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)
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 willynilly, 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 m_{0}c^{2}
into the last equation above, as
Einstein did later and as history still does, this implies that L
= m_{0}c^{2}.
Not E_{0} but
L. In the beginning of the equations, E_{0}
is assumed to be the rest energy of
the particle. At the end, 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.
You
may say that the situation is different when Einstein expressly
assigns m_{0}c^{2}*
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 E_{0}
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 force E_{0}
= E_{1}
 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. H_{0}
= H_{1}
 bD
where b is
the transformation term.
But, the electron starts at rest relative to A and B, therefore
H_{0} =
E_{0} K_{1}
 K_{0}
= H_{1}
 E_{1}
 (H_{0}
 E_{0})
= H_{1} 
E_{1}
But K_{0} =
0 since the electron has no kinetic energy at rest in both
systems. So: K_{1}
= H_{1}
 E_{1}
= H_{0} +
bD
 (E_{0} +
D) K_{1} =
bD
 D
The 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 atrest system and is losing energy.
Here again, though, if
you insert m_{0}c^{2}
as it has been historically into the
last equation, you find that it is equal to D, not to E_{0}.
D is the energy gained from the field, by Einstein's own variable
assignments. E_{0}
≠ m_{0}c^{2}
*Einstein
assigns the variable in question (my D above) to mc^{2}
not m_{0}c^{2}.
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∫
β^{3}vdv. 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 longreaching 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
twoterm 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 xdirection. 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 massincreased. 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, K_{0}
 K_{1}
=
m_{0}c^{2}

m_{0}c^{2}}
√[1  (v^{2}/c^{2})] Einstein
says that the second term (the term on the right) applies to the
rest energy of the particle. There is no reason to do this. It is
an equation with two terms, but the terms are not divisible or
singly assignable without a very compelling reason and a full
explanation. I have shown that the two terms are simply the
outcome of a finessed equation. There is no necessary physical
reality to either term.
To clarify this, let's look again
at the lightplane problem. At the end Einstein finds that K_{0}
 K_{1}
= γ L  L
If L = m_{0}c^{2}
, then what is the value of E_{0}?
Let us see. E_{0}
= E_{1}
+ L
The initial rest energy = the final rest energy + the rest
energy? This only makes sense if the final term is understood to
be the change in rest energy. L is actually the mass equivalent
of the planes of light. Even if Einstein's final equation were
correct in form (it isn't), it would imply that m_{0}c^{2}
is the mass equivalent of the light,
not of the body. But this is not what m_{0}c^{2}
means in current energy equations.
Now let's look at the calculus derivation of E = mc^{2}
from a current textbook. It follows
Einstein pretty closely—meaning it makes all the mistakes he
makes, and then adds a few of its own. The problem for the
textbooks is that they try to clarify some of the things that
Einstein purposely kept in the shadows. They try to apply real
math to things that Einstein simply glossed over. Unfortunately,
they are no more thorough than he was.
The first thing they do is state that they take the Workenergy
Theorem as still valid in Relativity. This is false. It is valid,
but not with the same equations. Remember that Einstein throws
out the classical equation for kinetic energy. In fact, the
textbook finds, at the end of these very equations, that E ≠
mv^{2}/2
. But it assumes, for some reason, that the integral based on
this equation is valid! W = ∫ Fdx = ∫ dpdx/dt = ∫
vdp But this is absurd,
since according to Einstein, energy is not expressed in the same
terms in relativity as it is in classical mechanics. The integral
∫ Fdx works because F = ma and v = x/t , etc.
From this they get W = ∫ dpv  ∫ pdv but p ≠ mv in
relativity. You can't just juggle the same old variables for a
few steps and then suddenly introduce a mass substitution to make
it all right. But this is what is done. W = mv^{2}
 ∫
m_{0}v
dv
√[1  (v^{2}/c^{2})]
= mv^{2} +
mc^{2}[1
 (v^{2}/c^{2})]
 m_{0}c^{2}
and so on
Notice that if
we so much as lose the square root of the gamma
term, then the integration is
ruined. You have no third term from v = 0 as you do in the
current integration. So you have no m_{0}c^{2}
term. But it does not matter since
the integration was compromised long before that.
The textbook I have in hand says this (which is typical): "We
call mc^{2}
the total energy of the body, and we
see that the total energy equals the rest energy plus the kinetic
energy." mc^{2}
= m_{0}c^{2}
+ 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, m_{0}c^{2}
is the change in rest energy, not
the initial rest energy or the rest energy at zero. And notice
that mc^{2}
has been assigned to the total
energy with no mathematical or theoretical proof whatsoever. With
the givens we have in Einstein's thought problems, the real mass
of the body or electron is not calculable or assignable, beyond
the given E_{0}.
Part
Two A 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 lightplane problem. I will get rid of the
angles of emission, leaving the light to travel only along the
xaxis. 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 redshifted, 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 F_{0} and the L/2 receding G_{0}.
If F_{1} is the
energy of the light as measured by B, then αF_{1} =
F_{0} since F_{1} <
F_{0} F_{0} = G_{0}
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
is f' = αf Which
makes the transform for the energy of a light plane E' =
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 G_{1} must equal F_{1},
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: E_{0} = E_{1} + F_{0}
 G_{0} E_{0}  E_{1} = 2F_{0}
H_{0} = H_{1} + F_{1}  G_{1}
[We are dealing with energy as a vector,
remember!] H_{0}  H_{1} = 2F_{0}/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 redshift 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 observerwhich you see is the
same thing). A light plane traveling in the x direction is
neither blueshifted nor redshifted, 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 K_{0} = H_{0}  E_{0}
And the final kinetic energy is represented by K_{1} = H_{1}
 E_{1} So that
the change in kinetic energy is K_{1}  K_{0}
= H_{1}  E_{1}  (H_{0}  E_{0}) K_{1}
 K_{0} = 2F_{0}/α + 2F_{0}
Now, if we want to put L back in, and solve, we get L = 2F_{0} ΔK
= 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 xaxis, 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 m_{0}c^{2} is inserted into the equation, then
L = m_{0}c^{2}. 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 mc^{2}.
Currently, Einstein's theory assigns m_{0}c^{2}
to L and mc^{2} to γL (which is my L/α). He says that
E_{T} = γL. But this is false, according to his own
variable assignments. From the equations above and current
theory, we have E_{T} = H_{1} + K = H_{1}
+ γL  L H_{1} ≠ L Therefore, E_{T} ≠ γL
Above I showed that E_{T}
≠ H_{1}. Here I have shown that E_{T} ≠ γL.
The truth is that E_{T} is not singly assignable to any
of Einstein's energy variables; nor is it assignable to mc^{2}.
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
Three Thought 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 zeropoint 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
reabsorb 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 zeropoint 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 zeropoint to mirror E = energy of photon = 1
x 10^{19}J
(say) m' = mass of ball, measured by the ball v' = 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 zeropoint 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 zeropoint, then I am 150,000km from the
zeropoint. I went that far in 2.5s, therefore my average
velocity (relative to the system of the zeropoint) 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'_{av} m'
= E/cv'_{av}
= 1 x 10^{19}J/(3
x 10^{8}m/s)(6
x 10^{7}m/s)
= 5.55 x 10^{36}kg
Now
let us calculate from the zeropoint. The first photon arrives at
the zeropoint in 2 seconds, according to the clock at the
zeropoint. The observer at the zeropoint 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
tequation from my previous paper, to calculate the difference
between the period of the ball and the period of the zeropoint.
This would be the most direct way to calculate, since the only
data the zeropoint is receiving from the ball is ticks. [The
zeropoint is able to calculate velocity simply from receiving
ticks, since the zeropoint knows the local period of the ball.
When the ball was at rest at the zeropoint, 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'/c t' is a given as 1s. t is incoming data. Therefore
x' and v' and v may be calculated. v =
v'
1 + (v'/c) v_{av}
=
v'_{av}
1 + (2v'_{av}/c) v_{av}
=
6 x 10^{4}km/s
1 + (12 x 10^{4}/3
x 10^{5})
= 42,857km/s
(Again, the zeropoint could have arrived at
this number without knowing v'_{av}.
This is of some importance below.) m = E/cv_{av}
= 7.77 x 10^{36}kg
The
mass of the ball has appeared to increase, if measured from the
zeropoint, 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
zeropoint 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 zeropoint 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 zeropoint will calculate different rest masses,
unfortunately). m_{rB}
= rest mass of the ball, before emission m_{rAB}
= rest mass of the ball calculated by the ball, after
emission m_{rAZ}
= rest mass of the ball calculated by the zeropoint, after
emission m_{0B}
= mass equivalence of the photon, as measured by the ball m_{0Z}
= mass equivalence of the photon, as measured by the zeropoint
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
zeropoint. The ball is moving away from the photon when the
photon returns, so that its energy will be redshifted. E'<
E. E = m_{0Z}c^{2}
= the energy of the photon relative to the zeropoint E' =
m_{0B}c^{2}
= the energy of the photon relative to the ball m_{rAB}=
mr_{B}
 m_{0B}
We can solve since we also know that mv_{av}
= m'v_{av}'
= E/c m = m_{0Z}c/v_{av} m_{0Z}
= m(v_{av}/c) m_{0Z}
= (m  m')/2 = 1.11 x
10^{36}kg
which agrees with our given value for its energy. m_{0Z}
= αm_{0B}
m_{0B}
= 7.9 x 10^{37}kg
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 zeropoint 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 zeropoint must be calculated with E.
This is simply because we must imagine that the ball was not
moving at the instant of emission. The ball did not start moving
until the instant after T_{0}.
Emission took place at T_{0}
, therefore the light has its normal frequency relative to the
zeropoint. This seems
somewhat strange at first, since the ball is not at rest relative
to the zeropoint. How can we calculate a rest mass for it
relative to the zeropoint; 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 zeropoint. How can I find it, since I don't yet
have an equation for it? All I need is a mass after emission from
which to subtract my photon from. I have two masses, m and m',
but m' is the correct mass since it is connected to the correct
velocity. The variable v' was measured locally, meaning that the
t variable did not need to be transformed. That makes m' a
reliable mass. But it is not the rest mass itself. It is a moving
mass. To find the rest mass, we simply subtract the mass
equivalence of the photon from m'. m_{rAZ}
= m'  m_{0Z}
= 4.44 x 10^{36}kg
The rest mass before
emission is just the photon added back in: m_{rB}
= 5.55 x 10^{36}kg
= m' Whenever I speak of
rest mass from now on (concerning this problem with the ball) I
will be talking about m_{rAZ},
but I will simplify the notation, taking it back to m_{r}.
We should take note that all these masses were calculated
from an average velocity over the interval of acceleration up to
a final velocity. If we had used a final velocity, we would have
found a mass equivalence for the photon that was twice too
little. This final velocity is not used in the mass or momentum
transforms; but it will be used in the energy transforms, simply
so that I may be sure to derive equations that are analogous to
the ones that are currently used. The current energy equations
are used given a final velocity. In many experimental situations,
the scientist does not know or is not concerned with how the
particle reached velocity. His or her only data is a final
velocity.* Some might
complain that the ball must use m_{0B}
since that is the mass it would calculate from the frequency of
light it actually sees. But since we have as one of our givens
the fact that the ball knows it is moving and is already
calculating a velocity for itself relative to the background of
the zeropoint, 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 zeropoint 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 zeropoint would have found the rest mass of the
ball to be 5.55 x 10^{36}kg,
before the experiment. It then would have calculated the moving
mass to be 7.77 x 10^{36}kg,
from an experiment like this one—a mass that would appear to be
confirmed by any subsequent collision of the ball, since the
momentum equation used by the zeropoint would be assumed to be
correct. The momentum would in fact be correct, but neither the
velocity nor the mass would be.
Some may want to calculate a momentum using m_{r},
to find that the ball also miscalculated its velocity. p = m_{r}v_{r}.
But this cannot be done. A rest mass is at rest, by definition,
and can have neither velocity nor momentum nor kinetic energy.
The rest mass is defined as the mass at rest relative to the
normal frequency of light.
As you can see, the momentum
is the same measured from either the ball or the zeropoint,
which is just as it must be: mv_{av}
= m'v'_{av}.
It could hardly be otherwise, since the masses were calculated
from a momentum equation in the first place. All we have had to
do is keep our variables in order, so that we understand
precisely what we have been given and what we are seeking in each
event and with each solution.
Finally, let me address the
comment that E/c = m'v'_{av}
cannot be the correct equation describing the initial situation,
since the ball will not receive the photon back from the mirror
at energy E. It is true that when the first photon returns to the
ball its frequency will have changed, due to the movement of the
ball. Because E = hf, the ball will receive the photon at E', not
at E, and E'< E.
However, we do not use E' in this equation for this reason: we
are not concerned with the energy the photon has when it returns
to the ball, not from any vantage; we are concerned with the
energy the photon has when it leaves the ball. The equation E/c =
m'v'_{av}
describes an equality of numbers, when all the numbers are
relative to the same background. This background is the
background of the zeropoint, 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 zeropoint, as measured from the
ball. v is the velocity of the ball relative to the
zeropoint, as measured from the zeropoint.
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 preexisting 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.] mv_{av}
= m'v'_{av} mv'_{av} /[1 + (2v'_{av}/c)]
= m'v'_{av} m = m'[1 + (2v'_{av}/c)] = m'α
where m' is local mass and m is measured from a distance What
if we want to use v_{av} instead of v'_{av}? m
= m'/[1  (2v_{av}/c)] = m'α
However, these
equations tell only part of the story, as the above thought
problem made clear. The observer at the zeropoint would
calculate the ball to have a moving mass of m'/[1 
(2v_{av}/c)] but if the ball subsequently came to rest
relative to that observer and was weighed by him, it would weigh
m_{r} = m'  m_{0Z} m_{0Z} =
m(v_{av}/c) m_{r} = m'  m(v_{av}/c) 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)] m = m_{r}/[1 
(3v_{av}/c)]
I will call this transform beta. beta
= β = 1/[1  (3v_{av}/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
Five Mass 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 v_{1}'
and ends with a local velocity of v_{2}'. 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 zeropoint. 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 macrolevel, without a change in mass. However, we
will assume that some transfers are totally elastic (nothing
sticks or is emitted). At the macrolevel this will always be an
approximation (although often negligible); at the microlevel 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'v_{1}' and its final momentum by
m'v_{2}'. 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: p_{i}
= m_{i}v_{i} where the
i stands for initial p_{f} = m_{f}v_{f}
" final
Since there is only one energy output at collision, no matter
where it is measured from m_{i}v_{i} =
m'v_{1}' m_{f}v_{f} = m'v_{2}' v'
= v/(1 + v/c) v_{1}' = v_{i}/(1 + v_{i}/c) v_{2}'
= v_{f}/(1 + v_{f}/c) m' = m_{f}v_{f}/v_{2}' m_{i}v_{i}
= v_{1}'m_{f}v_{f} /v_{2}' m_{i}/m_{f}
= v_{f}/(1 + v_{i}/c)//v_{f}/(1 +
v_{f}/c)
= (1 + v_{f}/c)/(1 + v_{i}/c) m_{f}
= m_{i} (1 + v_{i}/c)/(1 + v_{f}/c) =
m_{i}(c + v_{i})/(c + v_{f})
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 v_{i} = c/4 and v_{f} = c/2
then gamma = 1.03 my term = 1.2 Somewhat greater
change in mass.
What if the initial velocity is zero? If
v_{i} = 0, then m_{f} = m_{i} /[1 +
(v_{f}/c)] = m_{i}/(c + v_{f})
Of
course, in the same way we can derive a transformation from local
velocities, if we want. v_{i} = v_{1}'/[1 
(v_{1}'/c)] v_{f} = v_{2}'/[1 
(v_{2}'/c)] m_{i}v_{i} = v_{1}'m_{f}v_{f}
/v_{2}' m_{f} = m_{i}(1
v_{2}'/c)
1  v_{1}'/c
You may be surprised to find that the
body can calculate its own mass increase due to velocity. But if
it can calculate its own velocity, it can calculate its own mass
increase. The body itself would of course interpret this not as a
real change in mass, but as a change in mass equivalence relative
to its background. The body, for itself, has not gained mass but
kinetic energy. The classical interpretation would be that this
is kinetic energy and nothing else. The modern interpretation is
that mass is a sort of energy, especially in a momentum equation,
so that they may be lumped together. I prefer to think of the
measurement of mass from the object itself as the moving mass.
The object must then do further calculations to obtain its own
rest mass.
The question is, can we also use these
equations to transform from a local mass at rest to a relative
mass at velocity? Let us set the initial velocity to zero, in
which case the initial mass in the relative system should equal
to the local mass or rest mass. m_{i} = m_{0}. We
know this not from the momentum equations, but by definition. In
which case m_{f} = m_{r}/(1 + v_{f}/c) This
is only if the object is moving toward the observer, since we
simplified an equation from that problem. The mass variables
would switch if the object were moving away: (Eq. 1) m_{r}
= m_{f}[1  (v_{f}/c)] However, we now have
two equations for the same situation, and they don't match. Even
if we switch directions, the equation we found above isn't
equivalent: (Eq. 2) m_{r} = m[1  (3v_{av}/c)]
How can we explain this?
It is because the experimental situations aren't the same. In the
first thought problem, the ball emits a photon in order to reach
velocity. In the second, it doesn't. Notice that the ball has
borrowed the energy of the photon in the first experiment. A
scientist wouldn't necessarily know this, if he came upon the
ball after emission, but it is an important fact of the
equations. In the second experiment we are just imagining that
the object goes from rest to a final velocity, and we calculate
the mass increase due to that velocity. But again it might be
asked, is this possible? Can an object gain or lose velocity
without borrowing the energy of another object, by collision,
emission, or other method? I don't think so. In any experimental
situation, we must assume that any object under
consideration—that is not at rest relative to our field—gained
its velocity by some means external to our initial measurement.
We may postulate emission, collision, or the influence of a
field, but we may not postulate a relative velocity that was
gained without energy transfer.
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
Six Einstein'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 zeropoint
was equal to the momentum measured from the object itself. p'
= p m'v_{av}' = mv_{av}
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'v_{av}' cannot equal mv_{av},
since Einstein has no v'. Instead, he finds that p = mv =
γm_{0}v 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 correctthat it is unaffected by relativity.
Einstein is transforming m_{0} (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 m_{0} is equivalent in math and theory to my
m_{r}. Therefore his equation for momentum is equivalent
to this p = xm_{r}v_{av} what does x equal,
using my transformation terms? p = mv_{av} x =
m/m_{r} = α p = αm_{r}v_{av}
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
Seven Energy 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
zeropoint to a final velocity of v' as measured from the ball or
v measured from the zeropoint. 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.
E_{0} = the initial energy of the ball (measured by the
ball) before emission of the photon.
E_{1} = the total energy of the ball measured by the ball
after the emission of the photon.
H_{0} = the initial total energy of the ball as seen from
the zeropoint. H_{1}
= the final total energy of the ball as seen from the
zeropoint. F_{0}
= the energy of the photon as measured by the ball
F_{1} = the energy of the photon as measured by the
zeropoint F_{1} = αF_{0}
since F_{1} > F_{0} E_{1} = E_{0}
 F_{0} H_{1} = H_{0}  F_{1}
E_{0} = H_{0} since
the ball is initially at rest in both systems, A and B H_{1}
 H_{0} = F_{1} = αF_{0} And the
final kinetic energy is represented by K = H_{1} 
E_{1} = H_{1}  (E_{0}
 F_{0}) = H_{1} 
(H_{0}  F_{0}) =
αF_{0} + F_{0} = (1
 α)F_{0} (The initial kinetic energy was zero.) K
= F_{0}{1 [1 + (v'/c)] } =
F_{0}(v'/c)
My kinetic energy is negative. It is
negative because the ball is moving away from the zeropoint. It
can do no work upon a body positioned at the zeropoint. To do +K
amount of work on the zeropoint, 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 F_{1} as the energy of my
photon. F_{1} = m_{0Z}c^{2} (though I
will drop the "z" after this).
We do not need Einstein's derivation of m_{0}c^{2}
here, nor the textbook's simplified calculus derivation. I have
shown that both are false. All we need is the equation we have
already used E/c = p_{L}
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 have E/c = m_{0}v E
= m_{0}c^{2}
where m_{0} is the mass equivalence of the light. My
photon has a mass equivalence of m_{0} in this particular
problem. Putting this into my equation above yields K =
F_{0}(v'/c) αF_{0} = F_{1} K =
m_{0}c^{2}(v'/αc)
But we want kinetic energy in terms of m' not m_{0}. m_{0}
= m'v'/c K = (m'v'/2c )c^{2}(v'/αc) K = (1/α)
m'v'^{2}/2
To put in some hypothethical numbers:
If v_{av}' =.2c, then v' =.4c, and K = .714 x 5.55 x
10^{36}kg x (1.2 x 10^{8}m/s)^{2}/2 =
2.85 x 10^{20}J
Now let me calculate equations
from the zeropoint K = (1  α)F_{0}
{For α we will use 1/[1  (v/c)] instead of 1 +
(v'/c)]} K = (v/c)m_{0}c^{2} m_{0}
= (m  m')/2 m_{r} = m'  m_{0} m_{0}
= [m  (m_{r} + m_{0})]/2 3m_{0} = m 
m_{r} K = (m  m_{r})(v/3c)c^{2}
3cK/v = mc^{2}  m_{r}c^{2} K ≠
mc^{2}  m_{r}c^{2}
Which means that if E_{T} = K + m_{r}c^{2}
E_{T} ≠ mc^{2} 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})] = 3.70 x 10^{19}J
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} 
{mr(v/c)c^{2}[(β/α)  1}
= m_{r}c^{2}  {mr(v/c)c^{2}[v/(2c 
3v) E_{T} = mrc^{2}{1  [v^{2}/(2c^{2}
 3cv)]} K = {m_{r}c^{2}{1  [v^{2}/(2c^{2}
 3cv)]}  m_{r}c^{2} I check this against my
previous numbers and find that indeed this also is 2.85 x
10^{20}J I will call this transformation term kappa_{2},
κ_{2}. κ_{2} = 1  [v^{2}/(2c^{2}
3cv)] Notice that it is not the equivalent of either gamma
or beta (although it is very close in output to gamma). @
v = .286c, κ_{2} = {1  [v^{2}/(2c^{2} 
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 kappa_{1}, which is κ_{1}
= 1 + [v^{2}/(2c^{2}  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 kappa_{1} to be 1.01 gamma (@
v = .143c) = 1.01
An 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 = mc^{2} 
m_{r}c^{2} (from above) mc^{2}  m[1 
(3v/2c)]c^{2} = 3cK/v multiply both sides by
v^{2}/c^{2} mv^{2}  m[1  (3v/2c)]v^{2}
= 3Kv/c (3v/2c)]mv^{2} = 3Kv/c K =  mv^{2}/2
= 2.85 x 10^{20}J
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 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 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 mv^{2}/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
m_{r}v^{2}/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 = mv^{2}/2
is not some leadup 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 zeropoint. m'v' is relative to the
zeropoint, therefore E/c must also be relative to the
zeropoint. E'/c is not relative to the zeropoint; 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, v_{av} 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 WorkEnergy 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 WorkEnergy 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 v_{av} in the
derivation of the energy equations. When I am transforming the
masses within the energy equations, I am using α with v_{av}.
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 v_{av} 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 v_{av}, 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 = 2F_{0} ΔK
= L[1  (1/α)] ΔK = L(v/c) Whereas my final equations
for the ball were: ΔK = (1  α)F_{0} ΔK =
F_{1}(v/c) So let us find the equations for
Einstein's problem ΔK = 2F_{0}(v/c) In his
problem, the body measured the normal frequency for the light, so
that F_{0} = m_{0}c^{2} ΔK =
2m_{0}c^{2}(v/c)
But we want kinetic energy in terms of m not m_{0}. 2m_{0}
= mv/2c (I used my equation m_{0}
= mv_{av}/c from above; but we must use the total mass of
the light planes, which is 2m_{0}.) ΔK =
2(mv/2c)c^{2}(v/c) ΔK = mv^{2}/2 E_{T}
= m_{r}c^{2} + 2(v/c)m_{0}c^{2} m_{r}
= m'  2m_{0} 2m_{0} = m_{r}β/α 
m_{r} E_{T} = m_{r}c^{2} +
{m_{r}(v/c)c^{2}[(β/α)  1}
E_{T}
= mrc^{2}{1 + [v^{2}/(2c^{2}– 3cv)]} K
= {m_{r}c^{2}{1 + [v^{2}/(2c^{2}–
3cv)]} – m_{r}c^{2} E_{T} = m_{r}c^{2}
+ 2(v/c)m_{0}c^{2} E_{T} = (1/β)mc^{2}
+ (v/c)[(m/α)  m/β)]c^{2} E_{T} = mc^{2}
[(1/β) + (v/αc)  (v/cβ)] E_{T} = mc^{2} [1
– (3v/2c) + (v^{2}/2c^{2})]
We have
proven that the classical equation also applies to Einstein’s
thought problem, and that E_{T} ≠ mc^{2} 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 WorkEnergy 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 noncritical 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 α_{2} = 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 α_{1}, and the new alpha
α_{2}. E_{1} = E_{0}  F_{0}
H_{1} = H_{0}  F_{1}
[Some will want to add F_{1} to H_{0} 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 = H_{1} – E_{1} F_{1}
= α_{1}F_{0} K = H_{0}  α_{1}F_{0}
(H_{0}  F_{0}) = (1  α_{1})F_{0}
= (v/c)F_{1} = (v/c)m_{0}c^{2} K
= mv^{2}/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 zeropoint 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 =
mv^{2}/2 E_{T} = mrc^{2} + (v/c)m_{0}c^{2}
m_{r} = m’  m_{0} m_{0} =
m_{r}β_{2}/α_{2}  m_{r} E_{T}
= m_{r}c^{2} + {m_{r}(v/c)c^{2}[(β_{2}/α_{2})
– 1} β_{2} = 1/[1 + (v/2c)] E_{T} =
m_{r}c^{2}{1 + [v^{2}/(2c^{2}+
cv)]} K = {m_{r}c^{2}{1 + [v^{2}/(2c^{2}+
cv)]} – m_{r}c^{2} E_{T} = m_{r}c^{2}
+ (v/c)m_{0}c^{2} E_{T} = (1/β_{2})mc^{2}
+ (v/c)[(m/α_{2})  m/β_{2})]c^{2} E_{T}
= mc^{2}[1 + (v/2c) + (v^{2}/2c^{2})]
~~~~~~~~~~
We
have now found three different transforms for three different
problems. The only thing that has remained constant is that K =
±mv^{2}/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 = mv^{2}/2 or K =
{m_{r}c^{2}{1 + [v^{2}/(2c^{2}–
3cv)]} – m_{r}c^{2}
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
Eight The 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 (m_{r}
=1.67 x 10^{27}kg).
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 = m_{r}/[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 108m_{r}.
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 (E_{1}
= E_{0}
 F_{0}).
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 superfield 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 superhigh 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
massequivalences. m_{0}
= mass equivalence of
the photon m_{ri}
= rest mass of proton
before absorption = 1.67 x 10^{27}kg m_{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_{0}c^{2} m
= 2m_{0}c/v m_{0}
= mv/2c 1/α =1 –
(v/c) v/c = 1 – (1/α) mv/c = m – m’ m_{0}
= (m – m’) /2 m_{rf}
= m_{ri}
+ m_{0} m’
= m_{ri}
+ 2m_{0} m’
= m_{rf}
 m_{0}
+ 2m_{0}
= m_{rf}
+ m_{0} m/α
= m_{rf}
+ m_{0}
= m_{rf}
+ mv/2c m_{rf}
= m[1 –
(3v/2c)] Still the term
beta.
But let us find m in terms of m_{ri}
and m_{ri}
in terms of m_{0}. m/α
= m_{ri}
+ mv/c m_{ri}
= m[1 – (2v/c)] m_{0}
= mv/2c m_{ri}
= 2m_{0}[(c/v)
– 2]
So we only need to return to Einstein’s equations
to make the proper corrections.
E_{0}
= the initial energy of
the proton before absorption of the photon (A as background).
E_{1}
= the total energy of
the proton after the absorption of the photon (A)
H_{0}
= the initial total
energy of the proton as seen from the zeropoint (B)
H_{1}
= the final total
energy of the proton as seen from the zeropoint (B)
F_{0}
= the energy of the
photon in A F_{1}
= the energy of the
photon in B F_{1}
= F_{0}α
since F_{1}
> F_{0} E_{1}
= E_{0}
+ F_{0}
H_{1}
= H_{0}
+ F_{1}
E_{0}
= H_{0}
since the proton is
initially at rest in both systems, A and B H_{1}
– H_{0}
= F_{1}
= αF_{0} And
the final kinetic energy is represented by K = H_{1}
– E_{1} =
H_{1}
– (E_{0}
+ F_{0})
= H_{1}
– (H_{0}
+ F_{0})
= αF_{0}
 F_{0}
= (α  1)F_{0}
= (v/c)F_{1} K
= (v/c)m_{0}c^{2} m_{0}
= (m – m’)/2 m_{ri}
= m’  2m_{0} m_{0}
= [m – (m_{ri}
+ 2m_{0})]/2 4m_{0}
= m – m_{ri} K
= (m – m_{ri})(v/4c)c^{2} 4cK/v
= mc^{2}
– m_{ri}c^{2}
K ≠ mc^{2}
– m_{ri}ic^{2}
4cK/v = mc^{2}
– m_{ri}c^{2}
mc^{2}
– m[1 – (2v/c)]c^{2}
= 4cK/v multiply
both sides by v^{2}/c^{2} mv^{2}
– m[1 – (2v/c)]v^{2}
= 4Kv/c (2v/c)]mv^{2}
= 4Kv/c K = mv^{2}/2
Which means that if E_{T}
= K + m_{rf}c^{2}
m_{rf}
= m_{ri}
+ m_{0} E_{T}
= K + m_{ri}c^{2}
+ m_{0}c^{2} E_{T}
= m_{ri}c^{2}[1
+ (v’/2c)]
[1 – (v’^{2}/c^{2})] E_{T}
= m_{ri}c^{2}{1
+ [(v^{2}
+ cv)/(2c^{2}–
4cv)]} E_{T}
= mc^{2}
[1 – (3v/2c) +
(v^{2}/2c^{2})] E_{T}
= mc^{2}
[1 +
(v’/2c)]
[1 + (2v’/c) + (v’^{2}/c^{2})] Notice
the last bolded equation above tells us why gamma
works so well in
accelerators despite being slightly incorrect and being derived
with so many mistakes.
In accelerators we are finding a
limit at 108. Therefore, we set my equation equal to 108 and see
what velocity the proton is really achieving. (v/c)m_{0}c^{2}
+ m_{0}c^{2}
+ m_{ri}c^{2}
= 108m_{ri}c^{2} (v/c)m_{0}c^{2}
+ m_{0}c^{2}
= 107m_{ri}c^{2}
This last step was allowed since m_{ri}
is the same in both
theories. [(v/c) + 1]m_{0}
= 107m_{ri} m_{ri}
= 2m_{0}[(c/v)
– 2] [(v/c) + 1]/[(c/v)  2] = 214 v = .4982558c v’
= .9930474c = 2.97708 x 10^{8}m/s
c
= 2.99792458 x 10^{8}m/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^{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 have F = 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 yields a = 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^{4}m/s
= 85
million years 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/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. This will
allow me to continue writing these "unpublishable"
things. Don't be confused by paying Melisa Smiththat is just
one of my many noms de plume. If you are a Paypal user,
there is no fee; so it might be worth your while to become one.
Otherwise they will rob us 33 cents for each transaction.
