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How New Transforms in Special Relativity affect
Mass, Momentum and Energy Equations
In this paper I will derive new transformation equations
for mass, momentum and energy. I will show that Einstein, despite using a
thought problem that was useful and mostly correct in variable
assignments, made several crucial errors that compromised his final
equations. The thought problem I am mainly concerned with here is in his
short paper of 1905, Does the Inertia of a Body Depend upon its Energy
Content? Fully half of my paper is devoted to analyzing, critiquing
and expanding this thought problem and its math. The rest of the paper is
devoted to a variant thought problem I devised to clarify Einstein’s
variable assignments and conceptual assumptions. This thought problem
yields new equations that answer many of the embedded mysteries of
relativity and mass transformation. Einstein’s paper Does the Inertia of a Body Depend upon
its Energy Content? has long been a source of confusion. It’s brevity
and opacity have made its underlying concepts quite difficult to unravel.
As with the time and length transforms of Special Relativity, the mass
transforms that this paper yielded have never been corrected. They have
been confirmed to the satisfaction of most experimental scientists and
therefore the math to derive them has become a moot point. It was long ago
swallowed up by much more complex math, including hyperbolic fields,
imaginary numbers, Hilbert spaces, Hamiltonians, Lagrangians, and the
tensor calculus. Although thousands of papers have been written on the
mass transforms, no one has so far offered a crystal clear explanation of
Einstein’s algebraic variables and equations. In the past half-century, no
famous physicists or mathematicians have even attempted to do so. Some
have glossed the derivation as presented by Einstein, but none who
accepted his final equations have provided a superior groundwork for them.
Now, a century later, only those
who do not accept the final equations spend time on the mass transforms.
And they do not attempt to clarify Einstein’s mistakes. Rather, they
present a variant math that makes more sense to them. Some of these
variant maths have a certain validity, but I believe that none will be
looked at seriously until Einstein’s math is proven to be false. That is
what my paper does. A falsification of Einstein’s algebra will be a
falsification of all the higher maths that rest upon it.
Einstein’s paper is a compound—and sometimes a compensation—of
several basic algrebraic errors. Although in the body of the paper I will
prove these errors exhaustively, here I will just gloss them. Firstly, he
incorrectly applies his time transform gamma to the planes of
light. Secondly, he misapplies the term m0c2 at the
end of the derivation, giving it to the body rather than to the planes of
light. This is difficult to understand, since the final equation contains
the variable L, which he has explicitly given to the light. Despite these
two errors Einstein arrives at a transform that is very nearly correct.
That transform is again gamma. Einstein then solves down from this energy transform to find
a mass transform, which is likewise gamma. But in this case he is
wholly mistaken: his misassignment of variables has cost him needed
clarity, and gamma is not even an approximation of the correct mass
transform. This mistake has rarely been seen, since in experimental
situations mass is always calculated from energy equations. In working
with subatomic particles, for instance, the naked mass transform equation
is never used. Values are arrived at from energy equations. As I said,
Einstein's energy equation is almost correct. The term for gamma is κ = 1 + [v'2/(2c2 + cv' -
v'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. If we assume that
the rest energy is given by the rest mass—as in Er =
mrc2—then the moving energy cannot be given by the
moving mass, in a straight equation. A transform is required here as well.
This is a consequence of Einstein's own variable assignments. Einstein
assumed, with no theoretical or mathematical backup, that mc2
must be the term that is applied to E. It turns out that this is not the
case. Even more astonishing is
that using my new derivation, where all the variables are rigorously
assigned, I am able to prove that the classical equation is precisely
equivalent to the relativistic equation. In other words, K =
κmrc2 - mrc2 =
mv2/2 Simply by
correcting the math of Einstein's own thought problem, I arrive at a new
energy transform κ that is nearly equivalent to γ. This new transform
allows me to derive the classical equation directly, by a straight
substitution. In doing so, I prove that the classical equation is not an
approximation at low speeds, as has always been assumed. It is an exact
equation. The binomial expansion of the differential in gamma is a
manufactured proof, since gamma itself does not exist as a correct
transform in any part of Special Relativity. Before I get to Einstein's
thought problem, I must first gloss the findings of my earlier paper,
since they are crucial to understanding this paper. In that paper I showed
that Einstein misunderstood his initial coordinate system and variable
assignments, so that his transforms end up being unassignable. He applies
gamma to his time and distance transforms, in this way: t = γt'
and x = γx'.
Unfortunately, these transforms
are not correct. In his various thought problems—the most famous of which
is the man on the train—Einstein has three coordinate systems. He has the
man's system, the train's system, and the system of the platform, for
example. But he tries to solve from only two systems. In his equations, he
has only a primed system and an unprimed system, but no double-primed
system. At the end, when he finds t = γt', he has mistaken a transform
from the man to the platform for a transform from the train to the
platform. Einstein completely ignores the direct transform from the
platform to the train. His given velocity v is the velocity of the train
relative to the platform, he tells us. But he does not say whether this is
the velocity as measured from the train or from the platform. The two
measurements must be different, but Einstein never includes this
difference in his calculations. t = γt' therefore applies to a transform
from the platform to the man, which is in fact a transform of two degrees
of relativity. He never provides transforms for one degree of relativity.
t = γt' should read However, I show that
gamma is incorrect for two degrees of relativity as well. t = αt' and
x = x'/α
I was also the first to offer corrected second-degree transforms, although
these do not enter into mass transform solutions.
My first-degree transforms are in
inverse proportion between x and t, whereas Einstein's were in direct
proportion. His mistake came about by borrowing the light equations of
Lorentz, x = ct and x' = ct', which I have shown are incorrect.
Finally, my discovery of
first-degree transforms allowed me to derive a first-degree transform for
velocity, which Einstein never derived. His transform for velocity is for
two degrees of relativity, v to v'', as he admitted, and as has never been
questioned. Until my paper, there had been no v', nor any idea that it was
necessary to the solution. 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 Now let us proceed to the mass and energy transforms. The
best place to start is with Einstein's second paper of 1905, Does the
Inertia of a Body Depend upon its Energy Content? In this paper he has
a body at rest emit two planes of light in opposite directions. The two
planes of light have equal energies; therefore the body remains at rest
after the emission. He then asks how the energy of this body before and
after the emission would look to an observer moving directly away from the
body at velocity v. To be precise, he never specifies that the observer is
moving away from the body (in the positive x direction, with the body at
the origin) but it is implied by analogy to his previous paper. I will
say, in passing, that his failure to specify a direction in this paper has
had far-reaching consequences, since it has been assumed (without much
argument one way or the other) that the direction is not important. That
is, all the transforms of Special Relativity are now assumed to be
non-specific regarding direction. This is too bad, since I have shown (and
will show again, below) that Relativity must be specific regarding
direction. Einstein lets the two
planes of light emit from the body at angles to the x-axis, and therefore
to the observer. Let us call B the system of the observer and A the system
of the body. Using his
nomenclature, E0 = the
initial energy in A. This is not kinetic energy, since he states that the
body is not moving in A. It is unclear what E0 is at this
point. But from the outcome of the equations, E0 must be what
he calls the initial rest energy, as in E0 =
m0c2. Since the body is at rest in A, E0
is both the rest energy and the total energy.
E1 is the energy in A
after the emission of the two planes of light. H0 is the
initial energy of the body as seen from B. That is, it is the initial rest
energy plus the kinetic energy. K0
= H0 - E0 And the final kinetic energy is
represented by K1 = H1 - E1 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. Einstein assigns
the term γL to H1 and the term L to E1. He assumes
that H1 is mc2 and E1 is
m0c2. But look back up the series of steps: L ≠
E1 An even greater error
is made in assigning values to the light angle transforms a and
b. Notice that the magnitudes of a
and b are not equal. The observer in B would therefore expect Einstein's
body to change course, since one of the planes of light would have more
energy than the other, measured from B. Einstein ignores this. The body
must not change velocity, because then the change in kinetic energy would
be due to that velocity change and not to a change in mass—which is of
course what he is trying to prove. By a mathematical trick Einstein gets
the two planes of light to add to unity in both systems, but in B the two
light planes do not have equal energies.
Another crucial error in this
thought problem is that Einstein applies his transform γ only to the
planes of light, L/2 and L/2. He does not transform the mass, velocity, or
energy of the body directly. Those transforms are implications of the
thought experiment, but they are calculated indirectly, as results of
these very energy equations. In truth, the masses are applied to the
energies somewhat willy-nilly, and a rigorous explanation has never yet
been provided. The problem can be
solved down from the energy equations, of course, but it is a curious
method, especially as it stood (and still stands, until the publication of
this paper) as the first and only method. To solve from the energy
equations one must be extremely careful to keep all the hidden variables
in order. Einstein does not do this, as I show in the paragraphs that
follow. But the greater problem is that solving by this method keeps those
variables in the dark. In solving a problem for the first time, a
scientist or mathematician should put all the variables in plain view,
showing how they are transformed directly. He should not derive them
indirectly by a compact but impenetrable method. This problem is the
perfect example of that. Einstein has not been corrected for a century due
to the obtuseness of his proofs. In my opinion, it would have been more
helpful to do transforms on the basic variables, those being mass and
velocity, and then to build energy equation from those. As it is the
conceptual basis for relativistic mass, momentum and kinetic energy has
been keep under a cloud from the
beginning. As a first example of
this cloud, notice that if you insert m0c2 into the
last equation above, as Einstein did later and as history still does, this
implies that L = m0c2. Not E0 but L. In
the beginning of the equations, E0 is assumed to be the rest
energy of the particle. At the end, Einstein and history have assigned
m0c2 to E0. But according to these
equations, L = m0c2. That is,
m0c2 is not the rest energy before or after the
emission of the light, it is the change in rest energy. It is the energy
equivalence of the planes of light. No, there is not. But
the situation is directly analogous, otherwise how could it yield the
exact same equation? In it, the electron starts at rest with a given
energy. Let us call it E0 again, as above. If we apply all the
electric force at the first instant, to complete the analogy to the light
planes being emitted, then we can follow the problem in the same way,
without calculus.** The body reaches v instantaneously, and we want to
know how much energy it has gained from the force. Einstein has his
electron accelerate slowly, but that is only to avoid giving off
radiation. That is, it is an experimental concern, not a theoretical
concern. D = the energy gained from the electrical
force Einstein says the field imparts
a velocity to the electron. So the electron is now the moving body. Let us
assign it to B, the observer being at rest in A. It is the electron that
is moving, not us. It would be even more precise to say that the electron
is B. It is not moving in B; it is the system B itself. H0 =
H1 - bD where b is the transformation
term. But, the electron starts at
rest relative to A and B, therefore H0 =
E0 K1 - K0 = H1 -
E1 - (H0 - E0) = H1 -
E1 But
K0 = 0 since the electron has no kinetic energy at rest in both
systems. So: K1 = H1 - E1
= H0 + bD -
(E0 + D) Here again, though, if you insert
m0c2 as it has been historically into the last
equation, you find that it is equal to D, not to E0. D is the
energy gained from the field, by Einstein's own variable assignments.
E0 ≠ m0c2 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.
Einstein's equations do not
distinguish between movement toward and movement away. He says that moving
things act the same, no matter the direction. Therefore the energy of both
planes of light should increase from the point of view of B. If the body
had sent out electrons instead of planes of light, Einstein would have
found both the electron receding and the electron approaching the observer
in B to be slowed and mass-increased. But this is false, as I will show.
Finally, Einstein finesses the
equations by assigning (in the last equation of either problem) the first
term to the kinetic energy and the second term to the rest energy. Like
this, K0 - K1 =
m0c2 -
m0c2} 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.
If L =
m0c2 , then what is the value of E0? Let
us see. The initial rest energy = the
final rest energy + the rest energy? This only makes sense if the final
term is understood to be the change in rest energy. L is actually the mass
equivalent of the planes of light. Even if Einstein's final equation were
correct in form (it isn't), it would imply that m0c2
is the mass equivalent of the light, not of the body. But this is not what
m0c2 means in current energy equations. The first thing they
do is state that they take the Work-energy Theorem as still valid in
Relativity. This is false. It is valid, but not with the same equations.
Remember that Einstein throws out the classical equation for kinetic
energy. In fact, the textbook finds, at the end of these very equations,
that E ≠ mv2/2 . But it assumes, for some reason, that the
integral based on this equation is valid! 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 = mv2 - ∫
m0v
dv The textbook I have in hand
says this (which is typical): "We call mc2 the total energy of
the body, and we see that the total energy equals the rest energy plus the
kinetic energy." mc2 = m0c2 +
K But this assigning of physical
realities to the terms in the equations is completely ad hoc,
whether it is done by Einstein or the current textbooks. As I have shown
above, the equation works in the same way whether the particle starts from
rest or not. In these equations, m0c2 is the change
in rest energy, not the initial rest energy or the rest energy at zero.
And notice that mc2 has been assigned to the total energy with
no mathematical or theoretical proof whatsoever. With the givens we have
in Einstein's thought problems, the real mass of the body or electron is
not calculable or assignable, beyond the given E0.
Part Two The first thing to do, before I derive new equations for
mass increase and energy, is to correct the thought problem I have just
critiqued. If I have asserted that Einstein has made mistakes, I should
rerun the problem in the right way.
Let us return to the light-plane
problem. I will get rid of the angles of emission, leaving the light to
travel only along the x-axis. One plane of light travels directly toward
the observer in B, and one plane directly away. Since with light the
energy is dependent on the frequency, not the speed, we need linear
transforms for frequency, not velocity. The light moving in the +direction
of v will be red-shifted, since although it is moving toward the observer
in B, the observer is moving away from it. As regards the other plane of
light, the case is a bit more subtle. That light is not moving toward the
observer at all. It is wrong to say that an object moving away from an
observer has a kinetic energy, since that object cannot possibly do any
work for the observer. To be even more precise, light moving away from an
observer cannot be known to exist at all. However, we can measure the
energy of the incoming light, and we can see—or we are given—that the
emitting body has not changed speed or direction. Therefore, the receding
light must have an equal but opposite energy to the incoming light. This
is only an inference though, and may not be measured or seen directly. Let
us see if we can express this in equations. I am assuming the given
velocity is B as measured from A. Einstein's nomenclature is (purposefully)
confusing so I am going to call the L/2 incoming F0 and the L/2
receding G0. If
F1 is the energy of the light as measured by B,
then αF1 = F0 since
F1 < F0 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 G1 must equal F1, otherwise the observer
in B would see the body change velocity or direction after emission. We
are told that it does not change velocity. It stays at rest in A, and
keeps velocity v in B. We could express the direction of the planes of
light as angles of 0 and 180, to mirror Einstein, but notice that it is
completely unnecessary in this sort of problem. We are only interested in
vectors, not in angles. Both planes of light end up being subtracted from
the mass of the body. Einstein's use of 1 + cos and 1 - cos, etc. was just
false bombast. This is the way the equations should go: E0 =
E1 + F0 - G0 You may say,
shouldn't the light plane traveling in the -x direction have a blue shift,
and a transform that is the inverse of the red-shift transform? No. Light
is blue shifted if it is traveling toward the observer and the observer is
traveling toward it (or if the point of emission is traveling toward the
observer-which you see is the same thing). A light plane traveling in the
-x direction is neither blue-shifted nor red-shifted, nor subject to any
possible transformation. It is invisible and undetectable, except by
inference. Now, Einstein says the
initial kinetic energy of the body is represented by the equation
K0 = H0 -
E0 And the final
kinetic energy is represented by K1 = H1 -
E1 So that the change
in kinetic energy is K1 - K0 = H1 -
E1 - (H0 - E0) Now, if we want to
put L back in, and solve, we get L = 2F0 Einstein
had to finesse his equations to get a positive number at the end. I have
shown how to analyze the vectors correctly. Furthermore, if
m0c2 is inserted into the equation, then L =
m0c2. In this problem, according to Einstein's own
assignments, this is the mass equivalence of the emitted planes of light,
not the rest mass of the object. Finally, my corrections make it clear that L/α cannot be assigned to
mc2. Currently, Einstein's theory assigns
m0c2 to L and mc2 to γL (which is my
L/α). He says that ET = γL. But this is false, according to his
own variable assignments. From the equations above and current theory, we
have ET = H1 + K = H1 + γL -
L Above I showed that ET
≠ H1. Here I have shown that ET ≠ γL. The truth is
that ET is not singly assignable to any of Einstein's energy
variables; nor is it assignable to mc2. Since I have thrown out
Einstein's method for deriving mass increase equations, I must now derive
them on my own, using my own thought problems. Part Three Let us say that we have a tiny ball containing a device
that emits light. It is able to emit light one photon at a time, with a
known energy. At a distance of 300,000km from this ball is a mirror that
reflects directly back to the ball. This distance has been measured
locally (by walking it, say). It is a given, not a measurement by the ball
after emission. Also, the zero-point of the experiment is marked on the
ground with a white line, so that an observer may be placed there.
We run the experiment twice. The
first time the device in the ball emits a photon toward the mirror at T' =
0s, and then receives the same photon upon its return from the mirror (it
does not re-absorb the photon, it simply measures it with an instrument).
T' is the time on the ball's clock. At the beginning of the experiment,
just before the emission of the photons, T' = T. That is, the clocks of
the zero point and the ball are synchronized.
The second time, the device emits
a photon at T' = 0s, another at T' = 1s, and then another at T' = 2s. The
observer at the zero-point intercepts the second and third photons from
the ball in order to calculate where the ball is after T' = T = 0. This
observer also intercepts the first photon returning from the mirror.
By the conservation of momentum,
the ball must recoil in the opposite direction from the emission of the
photon. When the photon returns, the distance the ball has traveled may be
measured, and the inertial mass of the ball may be determined.
L = distance from
zero-point to mirror Let us
calculate from the ball, first of all. In this case, the ball is the
measurer, and the system of the ball is therefore the S' system—the primed
system (I make the local system the primed system simply to be consistent
with my other paper). What does the ball see?
The simplest thing to do is to
let the photon return all the way to the ball. We could let the photon
return to the zero-point and then let a signal be triggered, but that
seems redundant, since the signal would have to be a light
signal. Let the ball be very tiny,
to be sure it travels a nice long distance. But do not assume it reaches
velocity instantaneously (this will be important later). When the photon
arrives back at the ball, the ball looks at its clock and discovers that
2.5 seconds have elapsed. The ball thinks, "This is very easy. The light
took one second to get over to the mirror and one second to get back, and
half a second to reach me. If the mirror is 300,000km from the zero-point,
then I am 150,000km from the zero-point. I went that far in 2.5s,
therefore my average velocity (relative to the system of the zero-point)
is: v'av = 150,000km/2.5s =
60,000km/s By the conservation of
momentum, the momentum of the light must be equal to the momentum of the
ball: E/c = m'v'av t =
t' + x'/c v =
v' mrB = rest mass of the ball, before
emission The photon will have two mass
equivalents, since the photon will have a different energy relative to the
ball than it will have relative to the zero-point. The ball is moving away
from the photon when the photon returns, so that its energy will be
redshifted. E'< E. E = m0Zc2 = the energy of
the photon relative to the zero-point We
can solve since we also know that mvav = m'vav' =
E/c which agrees
with our given value for its energy. m0Z = αm0B
Now all we need
is the rest mass. Some will think that is just the mass measured by the
ball, since that is the only mass that is truly at rest with regard to its
background. But the ball, using the equation above, is calculating with
redshifted light. This means that its value for the mass equivalence of
the photon is incorrect. In this way my thought problem is not like that
of Einstein. In the light planes problem, the body is at rest and the
observer is moving away. Therefore the body measures the normal frequency
and the observer sees a redshift. But in my thought problem, the observer
at the zero-point sees the normal frequency and the ball sees the
redshift. Upon emission the ball
lost a certain amount of energy. This amount of energy is expressed by E,
not by E'. Therefore, the rest mass relative to the zero-point must be
calculated with E. This is simply because we must imagine that the ball
was not moving at the instant of emission. The ball did not start moving
until the instant after T0. Emission took place at
T0 , therefore the light has its normal frequency relative to
the zero-point. This seems
somewhat strange at first, since the ball is not at rest relative to the
zero-point. How can we calculate a rest mass for it relative to the
zero-point; and what is more, why would we want to? We want to in order to
get the correct energy equations. If we work with the wrong rest mass, we
will get the wrong equations. We must use the rest mass that is at rest
relative to the light. That is the only true rest mass, in any problem
whatsoever. You will say, "doesn't Relativity imply that all bodies are at
rest relative to light, since light travels c relative to all bodies?"
Relativity does say that light travels c relative to all bodies, and it is
correct to do so; however, it is quite obvious that a body that is
measuring red or blueshifted light is not at rest relative to that light.
The true rest mass of any body will be calculated from unshifted
light-that is to say, light with a normal frequency (see a full definition
of "normal frequency" below). And
so, in this particular problem, I must seek the rest mass relative to the
zero-point. How can I find it, since I don't yet have an equation for it?
All I need is a mass after emission from which to subtract my photon from.
I have two masses, m and m', but m' is the correct mass since it is
connected to the correct velocity. The variable v' was measured locally,
meaning that the t variable did not need to be transformed. That makes m'
a reliable mass. But it is not the rest mass itself. It is a moving mass.
To find the rest mass, we simply subtract the mass equivalence of the
photon from m'. mrAZ = m' - m0Z = 4.44 x
10-36kg The rest mass
before emission is just the photon added back in: mrB = 5.55 x
10-36kg = m' Whenever I
speak of rest mass from now onconcerning this problem with the ball-I will
be talking about mrAZ, but I will simplify the notation, taking
it back to mr. Some
might complain that the ball must use m0B since that is the
mass it would calculate from the frequency of light it actually sees. But
since we have as one of our givens that the ball knows it is moving¡ªand
is already calculating a velocity for itself relative to the background of
the zero-point, it is not difficult to require that the ball notice that
the normal frequency of light is f rather than f'. Using other methods than this
experiment (such as a gravitational method), the zero-point would have
found the rest mass of the ball to be 5.55 x 10-36kg, before
the experiment. It then would have calculated the moving mass to be 7.77 x
10-36kg, from an experiment like this one—a mass that would
appear to be confirmed by any subsequent collision of the ball, since the
momentum equation used by the zero-point would be assumed to be correct.
The momentum would in fact be correct, but neither the velocity nor the
mass would be. Some may want to
calculate a momentum using mr, to find that the ball also miscalculated
its velocity. p = mrvr. But this cannot be done. A
rest mass is at rest, by definition, and can have neither velocity nor
momentum nor kinetic energy. The rest mass is defined as the mass at rest
relative to the normal frequency of light.
However, we do not use E' in this equation for this reason: we are not
concerned with the energy the photon has when it returns to the ball, not
from any vantage; we are concerned with the energy the photon has when it
leaves the ball. The equation E/c = m'v'av describes an
equality of numbers, when all the numbers are relative to the same
background. This background is the background of the zero-point, or the
background of the ball before it gained a velocity. You may say, no, the
variables as measured against that background are unprimed variables, by
definition. The primed variables I have said are measured from the ball.
However, if you think this, you are not being rigorous enough in your
variable assignments. Just as in my first paper, the variable assignments
here are very subtle, and we now must write them out in full, to avoid
confusion. v' is the velocity of the ball relative to the zero-point,
as measured from the ball.
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.
Part Four These then are the new mass transform equations, for one
degree of relativity, if the object is moving away from the measurer.
[alpha must be modified if the object is moving toward the
measurer-see below for modification process; or see paper on velocity
transforms for full proof of modification.] mvav =
m'v'av where m'
is local mass and m is measured from a distance What if we want to use
vav instead of v'av? m = m'/[1 -
(2vav/c)] = m'α However, these equations tell only part
of the story, as the above thought problem made clear. The observer at the
zero-point would calculate the ball to have a moving mass of mr
= m' - (m0Z) I will call this transform
beta. 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 Now let us find equations for a velocity change that is
not from zero. Let us imagine an even simpler situation. Let us say that a
ball of local mass m' starts out with a local velocity of v1'
and ends with a local velocity of v2'. Will its mass appear to
increase from a distance? Let us assume (at first) that its local mass
will not change, since no particle is being emitted in order to accomplish
a higher velocity, as with the photon emission above. First we must
specify a direction. Let us say it is moving directly toward an observer
or a zero-point. In this case we will not have to make the velocity or the
momentum negative. For notice that once we start talking about momentum
and kinetic energy, we must think in terms of vectors. Objects moving away
will have negative momenta and negative kinetic
energies. Now let us take a closer
look at these givens. Are they possible? Is it possible for a ball to
change velocity without changing its total energy? Of course not. But can
it change total energy without changing its local mass? That is a subtler
question. As we saw above, the ball gained a velocity by emitting a
photon. Its rest mass therefore changed. In many other situations,
especially in particle physics, the local or rest mass of the body in
question will be affected by a field or by bombardment, since photons or
positrons or neutrinos or other small particles will be emitted or
absorbed. It may be that no transfer of energy is possible, even on the
macro-level, without a change in mass. However, we will assume that some
transfers are totally elastic (nothing sticks or is emitted). At the
macro-level this will always be an approximation (although often
negligible); at the micro-level it will likely always be a falsification.
But for this part of the problem, we will assume that the ball changes
velocity without changing its rest mass or local mass.
The initial momentum of the ball
as measured by the ball is given by the equation m'v1' and its
final momentum by
m'v2'. But in an
experiment where energy or momentum is the yield, then the mass will be
calculated down from the momentum equation. In this case, the velocity
will be measured from a distance, obviously. Scientists do not measure the
local velocity of quanta, or anything else. So these scientists will be
using these equations for the initial momentum and the final
momentum: pi = mivi
where the i stands for initial Since there is only one
energy output at collision, no matter where it is measured
from mivi =
m'v1' And there is your mass
transform. It has two v's, unlike Einstein's equation; and this is very
convenient, since it allows us to calculate from initial to
final. Now let's see if my term
causes more change than Einstein or less. If vi = c/4 and
vf = c/2 then Somewhat greater change in mass. If vi = 0, then mf =
mi /[1 + (vf/c)] = mi/(c +
vf) vi =
v1'/[1 - (v1'/c)] mf = mr/(1 +
vf/c) This is only if the object is moving toward the
observer, since we simplified an equation from that problem. The mass
variables would switch if the object were moving away: (Eq. 1)
mr = mf[1 - (vf/c)] However, we now
have two equations for the same situation, and they don't match. Even if
we switch directions, the equation we found above isn't
equivalent: (Eq. 2) mr = m[1 - (3vav/c)]
How can we explain this? It is
because the experimental situations aren't the same. In the first thought
problem, the ball emits a photon in order to reach velocity. In the
second, it doesn't. Notice that the ball has borrowed the energy of the
photon in the first experiment. A scientist wouldn't necessarily know
this, if he came upon the ball after emission, but it is an important fact
of the equations. In the second experiment we are just imagining that the
object goes from rest to a final velocity, and we calculate the mass
increase due to that velocity. But again it might be asked, is this
possible? Can an object gain or lose velocity without borrowing the energy
of another object, by collision, emission, or other method? I don't think
so. In any experimental situation, we must assume that any object under
consideration—that is not at rest relative to our field—gained its
velocity by some means external to our initial measurement. We may
postulate emission, collision, or the influence of a field, but we may not
postulate a relative velocity that was gained without energy transfer.
Therefore, I hypothesize that eq. 2 is always the correct one.
This equation implies some rather
shocking things, of course. The most important being that there is a limit
at .5c for v. The mass m becomes infinite when v = .5c. But this is the
same limit for v that I found with my velocity transforms. [There is not
really a limit for v' in my velocity transforms, but when v' = c, v = .5c.
When v' is infinite, v = c.] So at least my equations are consistent.
Whether I can reconcile this with the findings of particle physics is yet
to be seen (below). & One thing that
makes it slightly easier to accept 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. Part Six I said that according to my equations, momentum does not
need to be transformed. In order to find our initial transforms for mass,
we had to assume that the momentum of our object from the zero-point was
equal to the momentum measured from the object itself. p' =
p 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. m'vav' cannot equal mvav, since
Einstein has no 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 correct-that it is unaffected by relativity.
Einstein is transforming m0 (which is in the coordinate system
that is going v) to the coordinate system of the observer (which is the
unprimed system here). The unprimed system is the system of the scientist
measuring the particle whizzing by. But Einstein does not transform the
velocity. He finds a velocity transform in Special Relativity, but he does
not use it in the momentum equation. Why? One must suppose it is because
the velocity transform he finds there is for two degrees of relativity,
and he does not think it applies in this situation. I have shown in my
previous paper that it does apply. The given velocity v is affected by
relativity and must be transformed. It is affected by the speed of light.
Why would the speed of light affect mass but not velocity, requiring a
mass transform but no velocity transform?
Einstein's m0 is
equivalent in math and theory to my mr. Therefore his equation
for momentum is equivalent to this p =
xmrvav 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 Let us now return to my correction of Einstein's energy
equations and see if we can apply them to my problem with the ball and the
photon. First, notice that
Einstein's thought problem is analogous to mine except for one thing. Upon
emission of the planes of light, his body does not change position in
system A or velocity in B. My ball, however, does change velocity.
It goes from rest at the zero-point to a final velocity of v' as measured
from the ball or v measured from the zero-point. Einstein's two planes of
light cancel out. My one photon has no twin in the opposite direction,
therefore the ball is given a push and it achieves a velocity. In this way
my ball is more like Einstein's slowly accelerated electron. So we only
need to return to Einstein's equations to make the proper
corrections. E0 = the
initial energy of the ball (measured by the ball) before emission of the
photon. And the final kinetic energy is represented by K =
H1 - E1 (The
initial kinetic energy was zero.) Now that I
have brought Einstein's problem into line with my own thought problem, I
may use F1 as the energy of my photon. F1 =
m0Zc2 (though I will drop the "z" after this).
We do not need Einstein's
derivation of m0c2 here, nor the textbook's
simplified calculus derivation. I have shown that both are false. All we
need is the equation we have already used E/c =
pL which says that the
momentum of a photon is expressed by E/c. This equation comes from
previous theory and has nothing to do with relativity. If we assume that
light can have a mass equivalence, then we have E/c =
m0v where
m0 is the mass equivalence of the light. My photon has a mass
equivalence of m0 in this particular problem. Putting this into
my equation above yields K = -F0(v'/c) But we want kinetic energy in
terms of m' not m0.
K = (1 - α)F0 {For α we will use 1/[1 - (v/c)] instead of 1 + (v'/c)]}
K = -(v/c)m0c2 Which means that
if ET = K + mrc2 ET =
mrc2 -
(v/c)m0c2 I
will call this transformation term the inverse of kappa, κ. Notice that it is not the
equivalent of either gamma or beta (although it is very
close in output to gamma). @ v = .286c, κ = {1 -
[v2/(2c2 - 3cv)]} = .929 If we had been in an experimental situation where the kinetic
energy had been positive, then we would have found the inverse of this
number using kappa, which is κ = 1 +
[v2/(2c2 - 3cv)] 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 to
be 1.01 -3cK/v =
mc2 - mrc2 (from above) K ≈
mrv2/2 Which is absurd. What should have been intended is to show that K ≈
mv2/2 at slow speeds This latter equation is the classical expression of kinetic energy. As I
have shown, expressing kinetic energy in terms of a rest mass isn't even
sensible, once it is understood what the different terms mean. The
relativistic equation would have to resolve to either mv2/2 or
m'v'2/2 at slow speeds, even if gamma and Einstein's
theory were correct. Having it resolve to mrv2/2 is
just further proof that no one knew what was going on with the math and
the variable assignments. In the momentum equation m'v' =
E/c, I say that the ball does not use E'. And this is true. In this
equation, the equality applies to two numbers that are both generated by
the same field, that field being the field of the zero-point. m'v' is
relative to the zero-point, therefore E/c must also be relative to the
zero-point. E'/c is not relative to the zero-point; it is relative to the
coordinate system of the ball.
But in Einstein's thought problem, we are not creating a momentum
equality, or conserving momentum. We are transforming from one system to
another, A to B. We are transforming the energy of the light from E to E'
at the same time that we are transforming masses and total energies. We
must therefore include in the derivation E', which is the energy of the
light measured from the ball. E' is a necessary variable in his problem.
In my initial thought problem it is not. ~~~~~~~~~~ The 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. L = 2F0 Whereas my final equations for the ball
were: ΔK = (1 - α)F0 So let us
find the equations for Einstein's problem ΔK = 2F0(v/c) In
his problem, the body measured the normal frequency for the light, so that
F0 = m0c2 But we want kinetic energy in terms of m not
m0. 2m0 = mv/2c (I used
my equation m0 = mvav/c from above; but we must use
the total mass of the light planes, which is 2m0.) But we are finally in a position to show that Einstein chose
his thought problem carefully. He wanted avoid an acceleration and the use
of average velocity that my problem entailed. So he chose a problem with
no velocity change at all. Kinetic energy changes only because the mass
has changed. But this has the effect of oversimplifying the problem of
energy transformation due to Relativity. Notice that it is difficult to
understand where to apply the Work-Energy Theorem in Einstein’s problem,
since it is unclear where there is any force. You can’t have a force
without an acceleration, and there is no acceleration here. What has
happened is that the two forces from the two light planes have cancelled
eachother out. You have forces, but they have added to zero. Some may say
that is the beauty of the problem. It sidesteps all non-critical issues.
But by sidestepping them it has cloaked them, historically. Einstein’s
problem was too subtle by half. It was so subtle that it confused Einstein
himself. What is more, by generalizing his findings from this one very
unique thought problem (which was not at all general—it was not a standard
problem of mass increase or energy transformation) he hid all the
variations of mass and energy in Relativity. My thought problem is both
more standard and more complex, so that it shows all the issues involved
in solving problems of this nature. E1 = E0 -
F0 K = H1 –
E1 What happened?
We still got a negative kinetic energy. Everything worked out just like
our first problem. But we know that the body must have a positive kinetic
energy relative to the observer, since it is now moving toward it (it was
at rest to start, of course). The reason we got a negative value at the
end is that the equations don’t know the difference between one white line
in our system and the other. The equations only know the difference
between one system and another. This series of equations gives us values
relative to the zero-point of the experiment, which has not changed. We
moved our observer, but we did not change the point of emission. The
equation does not recognize that our observer has moved, since we did not
add any pertinent information to the equation. The movement of the
observer took place only in our heads, not in our math. The point of
emission is behind the ball, therefore the kinetic energy, as a vector, is
still negative. In order to get the right direction, we must make the
change by hand. Anyone who has done a lot of vector equations knows that
this kind of thing is a common feature of directional problems. The
equations don’t always yield the desired information, since it is often
impossible to include the pertinent postulates into the math. That is why
it is so critical to be able to visualize vector problems and other
geometric problems. Juggling equations is not sufficient. No problem in
history has made this clearer than Special Relativity. K =
mv2/2 ~~~~~~~~~~ We have now found three different transforms for three
different problems. The only thing that has remained constant is that K =
±mv2/2 which can be
considered ironic in that this was the one equation that was thought to be
an approximation; also the one equation that was thought to have been
superceded. One very important thing is different from classical theory,
though, and that is that I have shown that v has a limit at .5c. Using
gamma, physicists now think that v (the velocity as measured from a
distance) can approach very close to c—since this is the value for v that
gamma gives them. I have shown that v’—which is the true velocity
of the body—may approach c; but v may not. This is not a great difference
in theory—in that currently the variable v is thought to be the real
velocity of the object. But it does take some getting used to,
experimentally. An experimental physicist must now use either one of the
equations K = mv2/2 or 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.
by Miles
Mathis
γ =
1 .
√1 - (v2/c2)]
I will prove that by correcting
the math, the energy transform for Einstein's problem is actually
kappa
In deriving this new transform I also discovered several
other facts of great interest. One of these is that E ≠
mc2.
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.
t = γt''
I
was the first to demonstrate first-degree relativity, as well as the first
to offer transforms for it.
Einstein's Solutions
H1 is the final total energy of the body from B, being the
final rest energy plus the final kinetic energy.
L/2 = the energy of each plane of light, as
measured from A.
E0 = E1 + L/2 + L/2
This is the equation as calculated from
A
H0 = H1 + aL/2 +
bL/2 This is the equation from B, where a is
the negative angle transform and b is the positive angle transform
a =
γ[1 + (v/c)cosφ]
b = γ[1 - (v/c)cosφ]
where γ = gamma =1/√[1
- (v2/c2)]
Now, Einstein says the initial
kinetic energy of the body is represented by the equation
So that
the change in kinetic energy is
K0 - K1 =
L{ 1
- 1} =
γL -
L
√[1 - (v2/c2)]
That is the whole
paper. It takes up less than three pages in Annalen der Physik. It
will take me somewhat longer to show all the mistakes in
it.
H1 ≠ γL
This is because K1 ≠
K0 - K1.
Once you have digested the enormity
of that, notice that in the final step Einstein has subtracted the final
kinetic energy from the initial. This is backwards. It is standard
practice to subtract the initial energy from the final to find a change in
energy. Corrected, the equation should read, K1 - K0
= L(1 - γ)
You may say that the situation
is different when Einstein expressly assigns m0c2*
to the rest energy. In that problem ("Dynamics of the Slowly Accelerated
Electron," last part of section 10 of On the Electrodynamics of Moving
Bodies, 1905) he applies a force from an electrostatic field, taking
the electron from rest to v. There is no L
involved.
E0 = E1 -
D
K1 = 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
at-rest system and is losing energy.
**Einstein actually uses
calculus, and provides us with a single equation: K = ∫ εXdx = m∫
β3vdv. This kind of math is not helpful in creating a new
theory, since precisely none of the concepts are enumerated.
Also notice that, just as in the proofs of
Special Relativity, Einstein has failed to assign v to either system A or
B. This must affect his calculations. Nor does he consider that kinetic
energy can be calculated from either system, A or B. If A can calculate a
velocity relative to B, then A can also calculate a kinetic energy. He
does not specify where K is measured from. The form of the equations
implies that K is measured from B, but this is not a necessity. The fact
that Einstein does not carry into this problem a v', as I do, has had
long-reaching consequences.
Because he does not have enough variables or coordinate systems,
Einstein has once again been forced to finesse his math. He has done so in
several places, in fact. In the first thought problem, the initial trick
is letting L/2 stand for half the emitted light. Splitting his variable so
that it yields a two-term equation is done only to ensure that it cancels
properly. The second trick is using a transformation term that has a 1+
and a 1- in the numerator that also cancels out. This is not just luck.
Nor is it necessity. As I will show, it is much more convenient to choose
the send the planes of light straight ahead and straight back, since then
they are all in the same line as the given v. Everything is then in the
x-direction. Why does Einstein choose an angle? He chooses it because it
is the best way to finesse this equation. If he lets the planes of light
be emitted in a line, he gets into all kinds of trouble. His split
equations won't cancel out in that case, according to his own faulty
theory, since Einstein's transforms are the same regardless of direction.
In using the angles, as he has, he ensures they cancel, but only at the
cost of theoretical consistency. Mathematically they cancel. Conceptually
they do not, as I have shown. The observer in B should see the body change
direction, and Einstein cannot explain why this does not happen.
√[1 -
(v2/c2)]
To clarify this, let's look again at the light-plane problem. At
the end Einstein finds that K0 - K1 = γ L -
L
E0 = E1 + L
Now
let's look at the calculus derivation of E = mc2 from a current
textbook. It follows Einstein pretty closely—meaning it makes all the
mistakes he makes, and then adds a few of its own. The problem for the
textbooks is that they try to clarify some of the things that Einstein
purposely kept in the shadows. They try to apply real math to things that
Einstein simply glossed over. Unfortunately, they are no more thorough
than he was.
W = ∫ Fdx = ∫ dpdx/dt = ∫
vdp
√[1 - (v2/c2)]
=
mv2 + mc2[1 - (v2/c2)] -
m0c2
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 m0c2 term. But it does
not matter since the integration was compromised long before
that.
A Correction for Einstein's Thought
Problem
F0 =
-G0
E0 -
E1 = 2F0
H0 = H1 +
F1 - G1 [We are dealing
with energy as a vector, remember!]
H0 - H1 =
2F0/α
K1 -
K0 = -2F0/α +
2F0
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.
You can see that I
have done a lot of housecleaning. The way I dealt with the planes of light
was quite different than Einstein. Notice, for one thing, that I would
never let the planes of light be emitted in any other way than the way I
did. Why? Because any other planes of light, emitted at any angle to the
x-axis, will be undetectable from B. Einstein assumes that B can perform
transformation equations on light that never even comes to B. Light
emitted at any angle will never reach B, and is therefore not a source of
possible calculation. In that case all energy changes will be inferences;
none will be measurements. Which would make Einstein's thought problem a
fantasy from beginning to end, rather than a meaningful potential
experiment.
H1 ≠ L Therefore, ET ≠ γL
Thought Experiment 3
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?
E = energy of photon = 1 x 10-19J
(say)
m' = mass of ball, measured by the ball
v' = velocity of ball,
measured from the ball
m' = E/cv'av = 1 x
10-19J/(3 x 108m/s)(6 x
107m/s)
= 5.55 x
10-36kg
Now let us calculate from the zero-point. The
first photon arrives at the zero-point in 2 seconds, according to the
clock at the zero-point. The observer at the zero-point then must measure
the distance the ball has appeared to travel. The observer does this by
receiving the other photons from the ball. We could use the t-equation
from my previous paper, to calculate the difference between the period of
the ball and the period of the zero-point. This would be the most direct
way to calculate, since the only data the zero-point is receiving from the
ball is ticks. [The zero-point is able to calculate velocity simply from
receiving ticks, since the zero-point knows the local period of the ball.
When the ball was at rest at the zero-point, at the beginning of the
experiment, it's period was 1s.] However, since we have already calculated
the velocity of the ball according to the ball, I am going to skip this
step and use my velocity transformation equation instead. If the ball
calculates its own velocity to be 60,000km/s, then the observer at the
zero point will calculate (by receiving ticks) the velocity to be:
t' is a given as 1s. t is incoming data. Therefore x' and v'
and v may be calculated.
1 + (v'/c)
vav =
v'av
1 +
(2vav/c)
vav =
6 x 104km/s
1 + (12 x 104/3 x
105)
= 42,857km/s
(Again,
the zero-point could have arrived at this number without knowing
v'av. This is of some importance below.)
m =
E/cvav
= 7.77 x
10-36kg
The mass of the ball has appeared to increase,
if measured from the zero-point, as compared to measurement from the ball
itself. This much is consistent with the findings of Einstein: mass
appears to increase as time dilates. But the transformation term is
obviously different. I have used a variation of my velocity term
alpha rather than gamma.
Now, one may ask, which mass
is correct? The mass measured from the zero-point or the mass measured
from the ball? Either mass conserves momentum, as long as we keep it in
its own equation. But you can see that the mass as calculated by the ball
itself must be the correct moving mass, since it is connected to the
correct velocity. The zero-point calculates a larger mass only because it
has used an incorrect velocity. Its visual data has been skewed by time
dilation, making the velocity wrong and then the mass.
Next, one
may ask, what was the rest mass in this problem? Well, there must be three
calculable rest masses: the rest mass before the emission and two rest
masses after (the ball and the zero-point will calculate different rest
masses, unfortunately).
mrAB = rest mass of the ball calculated by the
ball, after emission
mrAZ = rest mass of the ball calculated
by the zero-point, after emission
mOB = mass equivalence of
the photon, as measured by the ball
m0Z = mass equivalence
of the photon, as measured by the zero-point
E' = m0Bc2
= the energy of the photon relative to the ball
mrAB=
mrB - m0B
m = m0Zc/vav
m0Z =
m(vav/c)
m0Z = (m -
m')/2
= 1.11 x
10-36kg
m0B = 7.9 x
10-37kg
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.
Both the
observer at the zero-point and the ball itself are calculating a moving
mass when they use a momentum equation, since the momentum equation
includes a velocity. The variable m' could hardly be understood as a rest
mass, since it was calculated from an equation that describes movement.
As you can see, the
momentum is the same measured from either the ball or the zero-point,
which is just as it must be: mvav = m'v'av. It could
hardly be otherwise, since the masses were calculated from a momentum
equation in the first place. All we have had to do is keep our variables
in order, so that we understand precisely what we have been given and what
we are seeking in each event and with each solution.
Finally, let
me address the comment that E/c = m'v'av cannot be the correct
equation describing the initial situation, since the ball will not receive
the photon back from the mirror at energy E. It is true that when the
first photon returns to the ball its frequency will have changed, due to
the movement of the ball. Because E = hf, the ball will receive the photon
at E', not at E, and E' less than E.
v is the velocity of the ball relative to
the zero-point, as measured from the zero-point.
There is no
velocity of the ball measured by the ball, relative to the ball. In the
same way, E' is not just the energy as measured by the ball, it is the
energy of the returning photon relative to the ball. It is not the energy
we want for any of our equations.
Before we continue, I wish to make one final comment regarding
this problem. We have just seen that light may have a different frequency
depending upon who measures it. Of course this is not news: we have known
of redshifts for decades. But our experiment above has shown us that
frequency may be privileged just as I have privileged certain measurements
of velocity and mass. What is the privileged measurement of light? The
measured frequency of light is normal, and therefore privileged, when the
system that measures the ray or photon is at rest relative to the point of
emission. That is fairly straightforward, I think, besides appealing to
common sense. This effectively privileges the point of emission of light
regarding measurement of the light's frequency. Notice it is just the
opposite of the privileging of time, velocity and mass to the local
system. Local time cannot be wrong. But the measurement of the frequency
of light can be wrong, from what we have heretofore called a local system.
The ball was the local system above, but it would have measured f', which
is not the normal frequency.
If you say, we can't privilege
certain fields like that-how can we know if we are moving relative to the
light source? Well, I say, we can't always know. But it is possible to
know in certain situations, from spectra shifts. The fact that it is
possible to know means that there is a pre-existing fact. Light does have
a normal frequency. For instance, we know, due to stability, that the sun
is not moving relative to us. It is neither approaching the earth, nor
fleeing it. Therefore measurements of the frequencies of sunlight from the
earth are privileged. Notice, however, that measurements of sunlight from
the sun are not privileged, since the sun is moving through space. You
will say, it doesn't matter, since the sunlight is moving away from the
sun, and is therefore undetectable from the sun. But sunlight reflected
back to the sun could be measured from the sun. [See my paper on the
mirror experiment to replace Michelson/Morley].
New Mass Transforms
mv'av /[1 + (2v'av/c)] =
m'v'av
m = m'[1 + (2v'av/c)] = m'α
m'/[1 -
(2vav/c)], but if the ball subsequently came to rest relative
to that observer and was weighed by him, it would weigh
m0Z = m(vav/c)
mr = m' - m(vav/c)
m' = m[1 -
(2vav/c)]
mr = m[1 - (2vav/c)] -
m(vav/c)
=
m[1 - (2vav/c) - (vav/c)]
mr = m[1 -
(3vav/c)]
m = mr/[1 -
(3vav/c)]
beta = α = 1/[1 - (3vav/c)]
This
is a very important equation, since it mirrors many experimental
situations. Already you can see that there are many equations involved
with mass increase, and the correct one must be chosen for the situation.
Just as with velocity, we must take into account the direction of relative
motion. In addition, we must take into account which mass we are seeking,
which mass or momentum we are given, and precisely what we are
transforming to and from.
In the thought problem we have just
solved, the mass changed twice, for two reasons: firstly, it changed
because the ball emitted a photon. This changed the mass even from the
point of view of the ball, of course. So this is not a consideration of
Relativity. Secondly, it changed from the point of view of the observer,
since a velocity was involved. This second change required a mass
transform due to Relativity.
Mass Transforms from one velocity
to another
pf =
mfvf "
final
mfvf = m'v2'
v' =
v/(1 + v/c)
v1' = vi/(1 +
vi/c)
v2' = vf/(1 +
vf/c)
m' =
mfvf/v2'
mivi =
v1'mfvf
/v2'
mi/mf = vf/(1 +
vi/c)//vf/(1 +
vf/c)
= (1 + vf/c)/(1 +
vi/c)
mf = mi (1 + vi/c)/(1
+ vf/c) = mi(c + vi)/(c +
vf)
If the final velocity is greater than the initial
velocity, the final mass must be less than the initial mass. For an
approaching object, there is an apparent mass decrease. Obviously this is
just to keep the momentum the same. If you are measuring its velocity and
getting a number that is too high (compared to the real value) then you
must measure the mass to be too low, so that when it hits you, the real
momentum and your calculated momentum are the same thing. If the object
were moving away, then you would once again calculate a mass
increase.
gamma = 1.03
my term =
1.2
What if the initial
velocity is zero?
Of course, in the same way we can derive a
transformation from local velocities, if we want.
vf =
v2'/[1 - (v2'/c)]
mivi =
v1'mfvf /v2'
mf
= mi(1
-v2'/c)
1 - v1'/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. mi =
m0. We know this not from the momentum equations, but by
definition. In which case
Einstein's Momentum Transformation
Equation
m'vav' = mvav
p = mv = γm0v
what does x equal, using my transformation
terms?
p = mvav
x = m/mr = α
p =
αmrvav
Energy Transformation
Equations
E1 = the total
energy of the ball measured by the ball after the emission of the photon.
H0 = the initial
total energy of the ball as seen from the zero-point.
H1 = the final total
energy of the ball as seen from the
zero-point.
F0 = the
energy of the photon as measured by the ball
F1 = the energy of the
photon as measured by the zero-point
F1 =
αF0 since F1 >
F0
E1 = E0 - F0
H1 = H0 - F1
E0 =
H0 since the ball is initially at rest
in both systems, A and B
H1 - H0 = -F1
= -αF0
= H1
- (E0 - F0)
=
H1 - (H0 -
F0)
= -αF0 +
F0
= (1 - α)F0
K = F0{1- [1 + (v'/c)]
}
= -F0(v'/c)
My kinetic
energy is negative. It is negative because the ball is moving away from
the zero-point. It can do no work upon a body positioned at the
zero-point. To do +K amount of work on the zero-point, a force would have
to be applied to the ball creating energy in the amount of 2K. In other
words, a force sufficient to turn the ball around and give it v' in the
opposite direction.
E =
m0c2
αF0 =
F1
K = -m0c2(v'/αc)
m0 = m'v'/c
K = -(m'v'/2c
)c2(v'/αc)
K = -(1/α) m'v'2/2
If
vav' =.2c, then v' =.4c, and K = .714 x 5.55 x
10)-36kg x (1.2 x 10)8m/s)2/2 = -2.85 x
10-20J
Now let me calculate equations from the zero-point
m0 = (m - m')/2
mr = m' - m0
m0 = [m - (mr + m0)]/2
3m0 = m - mr
K = -(m - mr)(v/3c)c2
-3cK/v = mc2 - mrc2
-K ≠ mc2 - mrc2
ET ≠
mc2
ET = mrc2 -
(v/c)m0c2
= mc2/β -
(v/2c)c2 [m - (m/α)]
= mc2/β -
(v/2c)[mc2 - (mc2/α)]
= mc2[(1/β) -
(v/2c) + (v/2αc)]
ET = mc2[1 - (3v/2c) -
(v2/2c2)]
= 3.70 x 10-19J
Now
let us find ET in terms of mr, so that we can
compare the transform to gamma.
mr = m' -
m0
m0 = mrβ/α -
mr
ET = mrc2 -
{mr(v/c)c2[(β/α) - 1}
= mrc2 - {mr(v/c)c2[v/(2c -
3v)
ET = mrc2{1 - [v2/(2c2
- 3cv)]}
K = {mrc2{1 -
[v2/(2c2 - 3cv)]} - mrc2
-2.85 x 10-20J
1/κ
= 1 - [v2/(2c2- 3cv)]
= 1.07
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:
mc2
- m[1 - (3v/2c)]c2 = -3cK/v
multiply both sides by
v2/c2
mv2 - m[1 -
(3v/2c)]v2 = -3Kv/c
(3v/2c)]mv2 = -3Kv/c
K =
- mv2/2
= -2.85 x
10-20J
Absolutely incredible! Once Einstein's variable
assignments are corrected it turns out that the classical equation is
precisely correct. Einstein and current wisdom both treat the classical
equation as an approximation at slow speeds relative to c. As supposed
proof of this, they expand the square root in gamma using the
binomial expansion, the first uncancelled term being
v2/2c2. But this is once again a fortuitous
collision of luck and bad math. I have shown that gamma is an
incorrect transformation term, so that expanding the square root of the
term is pointless. If there is no gamma, there can be no expansion of the
square root and no proof of the approximation of mv2/2.
Besides, this expansion proposes to find that
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?
The second rough spot
concerns the variable v'. In my mass transform equations, vav
was the average velocity. But in using Einstein's thought problem, I show
that v must be the final velocity of the body. Some may ask why I did not
simply let v be the final velocity in my own thought problem. It was
foresight that made me do it (and the Work-Energy Theorem). Remember that
the ball, when it is calculating its own velocity, is in possession of
only two pieces of data. It has an elapsed time and a distance. It also
knows it started from rest. It therefore must assume an acceleration over
one part of the distance or all of it. According to the Work-Energy
Theorem, the ball may not assume that a final velocity was achieved
instantaneously or over no distance. There is a kinetic energy because
there is work (and vice versa) and there is work because there is time and
distance involved. A force cannot be exerted over zero time or distance.
Since I knew that kinetic energy would be both the end product and the
driving "force" of my thought problem, I was astute enough to let v be
what it must be under the situation—an average velocity. Of course this
gives us just one more thing to be very careful about. Each problem has
its own specific variable assignments, which have to be written out in
full and kept track of. Even α does not always contain the same variables.
In my mass transforms, the velocity variables in α are average velocities.
In my time, distance and velocity transforms it does not matter, since no
acceleration is involved. In the light frequency or light energy
transforms, the velocity variables are commonly assumed to be final
velocities.
The last rough spot concerns the use of both v and
vav in the derivation of the energy equations. When I am
transforming the masses within the energy equations, I am using α with
vav. But the final equation is expressed in terms of v. Isn't
this an illegal mathematical substitution? No. It is perfectly legal to
use v and vav in the same equations, as long as you keep track
of them. As you can see, α has the same value whether you use v or
vav, as long as you use the correct form of alpha each
time. Therefore canceling alphas from one equation to the next is
not a problem.
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:
ΔK
= L[1 - (1/α)]
ΔK = L(v/c)
ΔK = -F1(v/c)
ΔK =
2m0c2(v/c)
ΔK =
2(mv/2c)c2(v/c)
ΔK = mv2/2
ET =
mrc2 +
2(v/c)m0c2
mr = m' -
2m0
2m0 = mrβ/α -
mr
ET = mrc2 +
{mr(v/c)c2[(β/α) - 1}
ET =
mrc2{1 + [v2/(2c2– 3cv)]}
K =
{mrc2{1 + [v2/(2c2– 3cv)]} –
mrc2
ET = mrc2
+ 2(v/c)m0c2
ET = (1/β)mc2
+ (v/c)[(m/α) - m/β)]c2
ET = mc2
[(1/β) + (v/αc) - (v/cβ)]
ET = mc2 [1 – (3v/2c) +
(v2/2c2)]
We have proven that the classical
equation also applies to Einstein’s thought problem, and that
ET ≠ mc2 there either.
Now, what if we have a thought
problem where the velocity is toward an observer? The kinetic energy will
be positive, but the masses will show a decrease. Let us return to our
ball and our photon. We will imagine two white lines drawn on the ground
now, instead of one. The ball starts at the first white line, as before,
but this time it is propelled toward the second white line, where we put
an observer. Everything else is the same as in the first experiment. The
momentum equalities will be the same, except that alpha for both
the velocities and the masses will be inverted. Alpha in this
problem will take the value α2 = 1 – (v’/c) = 1/[1 +
(v/c)]
Alpha in the energy
transform of the photon must stay in the original form, however, since the
ball will still be measuring a redshift. The observer will be measuring a
normal frequency, just as in the first problem. We must be very careful
here, since we have two values for alpha. Not just two equal
constructions, as in the first problem; now we actually have different
values. We will call the original alpha α1, and the new
alpha α2.
H1 = H0 - F1
[Some will want to add F1 to
H0 here, since we have changed directions. But if we subtract
the photon’s energy in one system, we must do so in the other system as
well. The body cannot lose the mass equivalence of the photon in one
system and gain it in the other.]
F1 = α1F0
K =
H0 - α1F0 -(H0 -
F0)
= (1 - α1)F0
=
-(v/c)F1 = -(v/c)m0c2
K =
-mv2/2
ET = mrc2 +
(v/c)m0c2
mr = m’ - m0
m0 = mrβ2/α2 -
mr
ET = mrc2 +
{mr(v/c)c2[(β2/α2) –
1}
β2 = 1/[1 + (v/2c)]
ET =
mrc2{1 + [v2/(2c2+ cv)]}
K
= {mrc2{1 + [v2/(2c2+ cv)]} –
mrc2
ET = mrc2 +
(v/c)m0c2
ET =
(1/β2)mc2 + (v/c)[(m/α2) -
m/β2)]c2
ET = mc2[1 +
(v/2c) + (v2/2c2)]
K =
{mrc2{1 + [v2/(2c2– 3cv)]} –
mrc2