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A
Final Simplification of the Problem of Special
Relativity
by
Miles Mathis
Part
One The Current Derivation
In other
papers I have extensively critiqued the mathematical proofs of
Special Relativity by Einstein, Lorentz and Minkowski. In this
paper I will present the shortest, most concise explanation of
the problem and its solution.
Only one illustration is necessary, and I will again use the
illustration of the spaceship flying by an earthbound
observer—the one that I borrowed from a textbook and modified
for use in my longer paper.
In this illustration a man walks from the near end of the
spaceship to the far end. In the coordinate system of the
spaceship he has walked directly away from the observer on earth,
in a straight line. This straight line is x'. But from the point
of view of the observer in the second coordinate system he has
walked in a slant, x. This slant and the first straight line are
two sides of a triangle. The third side is the distance the
spaceship went from the beginning of the walk to the end. This
distance may be represented by vt.
Given x' and v, we seek x.
In order to solve, we must
assume several other things. These are the first assumptions of
Einstein and Lorentz and Minkowski: x = ct x' = ct'
These equations
are supposed to describe the behavior of light in the two
coordinate systems. The only other assumption you need to solve
is that we have created a right triangle here, allowing us to us
the Pythagorean theorem
x = √x'^{2}
+(vt)^{2}]
c = √x'^{2}
+ v ^{2 }t^{2 }]/t
c² =
(x'^{2} + v^{2 }t^{2 })/t^{2}
c² =
x'^{2}/t^{2} + v^{2}
c^{2}
 v^{2} = x'^{2}/t^{2} = x'^{2 }c^{2
}/x^{2}
(c^{2}
 v^{2 })/c^{2} = x'^{2}/x^{2}
x/x' = γ
= 1/√(1 – v^{2 }/c^{2 })
And
there you have it, γ, gamma, the famous transformation term. It
is also the transform for the t variable. Equally simple math
gives us
c^{2}
= x'^{2}/t ^{2} + v^{2}
c^{2}
= c^{2 }t'^{2}/t^{2} + v^{2}
t/t' =
1/√(1 – v^{2 }/c^{2 })
That is
all there is to Special Relativity. This series of equations is
equivalent to the math and assumptions of Michelson, Lorentz,
Einstein, Minkowski, and everyone else in the 20th century. This
simple algebra underlies the tensor calculus and has never been
corrected to this day. Every explanation of Special Relativity
you will find, no matter how complex, can be boiled down to this.
Part Two The Critique
Unfortunately
it is wrong in several places. The first place that it is wrong
is in the light equations: x = ct and x' = ct' cannot both be
true, because together they imply that x and t change in direct
proportion, where in fact they change in inverse proportion.
Einstein even admits this. In the book Relativity, he says
(Ch.XII, p. 37) "As judged from K, the clock is moving with
the velocity v; as judged from this reference body, the time
which elapses between two strokes of the clock is not one second
but [γ] seconds, i.e. a somewhat larger time. As a
consequence, the clock goes more slowly than when at
rest." Time
dilation and length contraction are now cliches. Everybody knows
that time slows down and lengths get shorter. But time slowing
down is a lengthening of time. When time slows down the period
increases, so that the length of time between two ticks is
longer. This sets up an inverse relation between x and t, and
makes one of the two light equations false. It turns out that the
false one is the first one, x = ct. This is simply because the x
in this equation is not a length measured in its own coordinate
system. x is the way a length in another coordinate system looks
to an observer. x is defined as the way x' looks to the observer.
Therefore x = ct is not a parallel construction to x' = ct'. The
two equations are not analogous. In fact, if we are given that x'
= ct' (and we must be, that is one of the postulates) then x =
ct'^{2}/t. This assures that xt = x't'. Using the current
equations, xx'= tt', which cannot be.
The second mistake is in assigning the variable v in the term vt.
What velocity is this? You will say it is the velocity of the
spaceship, in this example. But is it the velocity measured by
the spaceship or by the observer? The two will measure different
velocities, but we are not told which it is. The variable is
undefined. Likewise the t variable in vt. Because v and t are
unprimed we assume they are measured by the observer, but if the
observer already has a velocity in hand, why do we need a
velocity transform later on? The velocity transform of Special
Relativity claims to transform a v' into a v. But if we already
have a v then what do we seek? You will say that v is the
velocity of the spaceship and that the velocity transform allows
us to calculate the velocity of the man walking. But shouldn't
the spaceship require transforms too, not just the man? The
spaceship is moving, therefore it is not in our coordinate
system, therefore by the laws of Special Relativity it must
require transforms. But Special Relativity never supplies these
transforms. And
this takes us to the third mistake. Gamma and the velocity
transform are both generated from a conceptualization that yields
two degrees of Relativity. You can see that the spaceship should
require a transform itself. Then the man inside the spaceship
should require a second one. The spaceship moves relative to the
observer; the man moves relative to the spaceship. Relativity
provides the twodegree transform but ignores the onedegree
transform. In addition, Special Relativity provides the wrong
transform for two degrees, as I have shown. Its math fails
because its postulate equations are incorrect, its variables are
undefined, and it has only two coordinate systems when it needs
three. The final
mistake is giving the spaceship in the drawing a trajectory at
right angles to the observer on earth. If you will notice, the
man cannot be made to walk directly away from the observer, even
in his own field. His orientation relative to the observer on
earth changes as time passes, so that the line x is not an
accurate representation of his movement relative to the earth.
The trajectory of the man would actually describe a curve in this
problem. The line of equal distance from a point is a circle;
therefore to remain at a constant distance from the observer on
earth, the spaceship would have to orbit the observer, not go in
a straight line tangent to his line of sight. This skews the
whole problem. The right triangle does not have an angle of 90
degrees, for one thing, so that the Pythagorean theorem may not
be used in the way it has. Furthermore, correct math will show
that a spaceship passing on a tangent like this will require a
variable transformation equation, one that changes every moment
depending on its angle to the line of sight of the observer. A
man moving on the spaceship will require even more complex
transforms. The simple constant transforms of Special Relativity
can only apply to movement directly away from an observer, and
even there they require the corrections below.
Part Three The Correction
It turns out
that the time as measured by an observer of a moving body is
simply the time of the moving body plus the time it takes for
light to go from the moving body to the observing body. t =
t' + t_{c} t_{c} = x'/c
x' is used in this equation, because although it would seem to be
only a measurement of distance by the moving body, it happens to
be equivalent to the measurement of the background by the
observer. In this way it becomes the background of all three: the
moving body, the light, and the observer. This is the secret
information that has always existed behind the problem, although
it has never come to light until now. You will say that the
measurement of the background by the observer is x, but that
analysis is another that has been historically incorrect. In the
transformation equations of Special Relativity, x stands for—and
always has stood for—the way that the observer measures the
moving body. It does not stand for the way the observer measures
its own lengths and distances. x is the distance that the
observer calculates the body to have gone, using its data. This
is not the same conceptually as a length or distance within the
system of the observer.
The fact is that in SR, x' must be one of the givens of the
thought problem. Einstein gives it to us by giving us v, although
this has never before been clear. But if you give someone a
velocity, you have given them some x over some t. This is
conceptually the x that Einstein has given us: x' (although he
never puts it in those terms). Of course, x' is also the variable
used for length within the system of the body. This is also one
of the givens or postulates of the problem, and therefore does
not need to be proved.
x' = distance in moving body's system, when measured by the
moving body = distance in observer's system when observer is
measuring itself.
x = distance in moving body's system when it is measured by the
observer. This is the body as seen by the observer. x' thus
belongs to what I call a local system. Even relativists believe
in the local system. It is not a return to Galileo. Feynman, for
instance, calls it the proper system.* In the local system,
magnitudes are always measured from a negligible distance, so
that the speed of light does not enter the equation or the
measurement. In addition, light always travels in the local
system. This is simply because every observer observes light in
his own system. It is impossible to see light in another system.
Our data arrives on electromagnetic waves, which waves must be in
our system when we receive them, by a tautology. We measure
everything relative to our own background, and we are stopped
relative to that background. That is what makes it our
background, of course. A background that is moving is a
contradiction. Therefore, since light travels c relative to any
unmoving background, light travels c in every local system. This
means, of course, that there is a universal local system, defined
by the speed of light. This universal local system, which is
simply the equivalent of the classical universal system of
Galileo and Newton, pertains whenever we are making measurements
in our own system, by a method that does not require a transform.
If you are measuring your own velocity relative to a given
background, for instance, you are in the universal system. Only
when you are measuring the velocity of a distant object are you
no longer in the universal system. In this case you require a
transform. The transform may be derived like this:
t = t' +
(x'/c)
xt = x't'
v = x'/t
= x/t'
x' = v't'
= vt
t = t' +
(v't'/c)
= t'[1 +
(v'/c)]
t = t'/[1
– (v/c)]
v =
x'/[t'(1 + v'/c)]
v =v'/[1
+ (v'/c)] = cv'/(c + v')
v' = v/[1
 (v/c)] = cv/(c – v)
x = x' [1
– (v/c)]
x' = x[1
+ (v' /c)]
The
common term (which I have dubbed alpha) in all these
transforms is α = alpha = 1/[1  (v/c)] = 1 +
(v'/c) t = αt' x' = αx v' = αv
Alpha
has long been used in optics to transform frequency from one
system to another. Feynman uses the equation in his Lectures
on Gravitiation in a proof of gamma. f' = f[1 +
(v'/c)] That is, he uses the correct term to prove the
incorrect term. No one before me has seen that the accepted
transform for frequency must be equivalent to the transformation
term for onedegree Special Relativity.
Part Four An
Approaching Body & a Trajectory at an Angle
All these
transforms apply only when the moving body is moving directly
away from the observer. You can see that the observer measures
the period of the clock of the moving body to be greater than the
period measured by the body itself: t > t'. [These time
variables stand for periods, not instants, as even Einstein
admitted (see The Meaning of Relativity, chapter 2, eq.
22a)]. In other words, there has been a redshift. If the clocks
are thought of as waves—and clocks certainly may be called
waves—then the period of the clock has been stretched by its
movement away from the observer. It has been redshifted.
Extending this reasoning, a clock approaching an observer must be
blueshifted. If t_{d} = period of departing
object and t_{a} = period of approaching blinker t_{d}
= 1/t_{a} t = t'  x'/c v = x'/t v = x'/(t' 
x'/c) = v'/(1  v'/c) v' = v/(1 + v/c)
In this way we see that a body approaching an observer will be
timecontracted rather than time dilated. In other words, the
observer will measure a smaller time than the body will measure
for itself: t' > t. He will also measure a larger x: x >
x'. This contradicts the current interpretation of SR.
For simple experimental proof of this assertion, consider binary
pulsars. Pulsars are clocks in the sky, which clocks create
arriving waves of data here on earth. When the pulsar is moving
toward us in its orbit, its period increases. The data is
blueshifted. This directly contradicts the standard
interpretation of SR, which says that all relative motion causes
time dilation. Time dilation is a larger period, but the motion
of the binary pulsar creates a smaller period, which is time
contraction. There is no way for the standard model to answer
this, since if they answer that time dilation is a smaller
period, then you point to the same pulsar moving away in its
orbit. We have two opposite shifts here, no matter how you define
them. You can't logically define both of two oppositie shifts as
redshifts.
In addition, bodies moving at an angle to an
observer will require complex transforms—ones that do not match
either of the simple ones we have found. Trajectory must always
be considered in SR. Trajectories at an angle will yield variable
transforms, since the time and distance differences are not
constant. These variable transforms will be dependent upon the
speed, angle, and trajectory (approaching or receding) of the
moving body.
Part Five The SecondDegree
Transform
To find
seconddegree transforms, like Einstein's velocity transform, we
must expand our problem to three coordinate systems and five sets
of variables: A = man's system B = train's system C =
system of the embankment and observer
We need five
velocity variables to solve: 1) The man's velocity relative to
the train, measured by the man, v'''. 2) The man's velocity
as seen from the train, v''. 3) The train's velocity relative
to the embankment, measured by the train, v''''. 4) The
train's velocity as seen from the embankment, v'. 5) Only
then can you ask about the man's velocity as seen from the
embankment, v.
Let v of A rel B (man to train) = v"
let v of B rel C (train to embankment) = v' what is v of
A rel C (man to embankment)? = v Given v' and v'', seek
v. v''' = the velocity of A measured by A.
= v''/[1  (v''/c)] And we can calculate the velocity of B
measured by B in the same way. v'''' = B measured by B,
= v'/[1  (v'/c)] The velocity of A relative to C, if ABC is a
straight line, would be v = v''''
+ v'''
1 + [(v'''' + v''')/c] = [v'//1
 (v'/c)] + [v''//1  (v''/c)]
1 + {[v'//1  (v'/c)] + [v''//1  (v''/c)]}/c v = v'
+ v"  (2v'v"/c)
1  (v'v"/c^{2})
Equally simple math (not shown) allows us to find a
twodegree time transform, one that is not gamma: t/t''
= c^{2}
 v''v'
= 1
 v''v'/c^{2}
(c 
v'')(c  v')
(1  v''/c)(1  v'/c)
These transforms apply only to
objects moving away from an observer in a straight line. Remember
that we are dealing with observation by the use of light rays. In
the observation of A from C, the light rays will travel directly
from A to C. They will not necessarily pass through B. B has its
own light rays from A that it is dealing with. But we should only
be concerned with the light rays coming to us. That is, visual
observations are made directly, and indirect evidence is
dangerous in relativity. We must deal only with our own light
rays, the ones entering directly into our eyes. The relativity
equations apply only to these rays.
This is not so clear when you are dealing with relative
velocities all in the same line. In this case, the light rays do
pass through B. But this will not always be the case, obviously.
In seconddegree transforms, the trajectories of both objects
must be taken into account.
*He says
(Feyman Lectures on Gravitation,
p.94), "How much is the time difference at various points in
space? To calculate it we compare the time rates with an absolute
time separation, defined in terms of the proper times ds."
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