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HOW
DO PHOTONS TRAVEL?
by Miles
Mathis milesmathis.com email:mm@milesmathis.com
 art
by David Kogan
Abstract: I develop a firm number
for the local wavelength of the photon, show how it is created by
the spin of the photon, and how this local spin is stretched by
the linear motion into the wavelength we see and measure. I prove
that the transform from the local wavelength to the measured
wavelength is c2,
showing why this term is found in the famous equation E =
mc2.
In
previous papers I have shown that photons have a number of
stacked spins, explaining the mysteries of superposition and so
on. But here I will look at the linear motion of photons. I will
study how the measured wavelength is caused by the wavelength of
the spins, and how different colors are created.
Before we get into that, I would
like to give credit to Descartes, who knew that color was caused
by photon spins in the early 1600's. I believe he made this
proposal in Dioptrique,
around 1638. This was before Newton's corpuscles, so the fact
that Descartes was able to intuit an answer so close to correct
at that time is extremely laudable. If I remember correctly,
Descartes got the energies reversed from red to blue, but
connecting color to spin was a masterstroke in itself. We will
see that color is due to the size of the photon, which is more a
matter of radius than spin velocity, but the two are inextricably
linked. It is spin collisions that cause the spin radius, so we
would obviously have no spin radius without spin. Since I
recently published strong evidence against Descartes' theory
of reflection by the back surface of raindrops to create
rainbows, I felt compelled to add this paragraph to this paper,
showing he was also often right against very long odds.
From
a previous paper, we know that the radius of the B-photon
is G times less than the radius of the proton. This gives us a
photon radius of 2.74 x 10-24m.
The z-spin is 8 times the radius, so we should find a basic
wavelength of 2.2 x 10-23m.
Obviously, we don’t find photons with a wavelength that small.
Why? Simply because the wavelength we measure has been stretched
out by the velocity of the photon. The photon would be measured
to have a wavelength of 2.2 x 10-23m
only if it were at rest.
You will say, “Even if we accept
that the photon is spinning, how can the z-spin be stretched? The
spin would give us a spin radius, which is just a length. A
length cannot be stretched by motion, unless you are proposing
some kind of relativity here.”
I am not proposing
relativity as the solution here. The answer to your question is
that a spin is not just a radius, and is not just a length. A
spin is a motion: a motion that takes time. Even if the photon
were spinning at velocity c, one rotation must take some real
time. We know that the linear velocity of light is not infinite,
so we must assume the speed of spin is also not infinite. If it
is not infinite, it must take time. If it takes time, then it
will be stretched by the linear motion. While the surface
of the photon is spinning, the photon as a whole is moving some
linear distance x.
So how much does the velocity
stretch out the wavelength? We can discover that most easily by
using this simple equation:
E = mc2 = hc/λ
λ = h/mc
Let us take an infrared photon, as
our first candidate. The mass equivalence of the infrared photon
is 2.77 x 10-37 kg, so we just solve:
λ = h/mc = 8 x 10-6m
If we compare that to the
wavelength at rest, we find the wavelength has been stretched out
by a factor of about 3.63 x 1017.
Since that is very nearly c2,
we assume that the transform is in fact c2, and that
the difference is a difference between the size of the B-photon
and of the infrared photon. Remember that we developed the
at-rest wavelength from the B-photon and the moving
wavelength from the infrared photon.
Our assumption is borne out by the
numbers, since if we divide 8x 10-6 m by c2,
we get 8.9 x 10-23 m, which is almost exactly 4 times
our B-photon wavelength. We may assume that the infrared
photon is about 4 times larger than our B-photon.
But why c2?
Let’s look at the mechanics, to find out. As usual, nothing
esoteric is going on here, so we can analyze just as if we were
analyzing pool balls. If the photon is spinning while it is going
c, and the radius is being stretched by a factor of c2,
then that must mean that the photon is spinning at a velocity of
1/c. It takes one full rotation to create a single wave, since
the rotation is physically creating the wave. And we want to find
one full rotation while the photon is moving 8 x 10-6m.
That will give us the wave we measure. At speed c, the photon
goes that far in 2.67 x 10-14
seconds. So the photon has spun once in that time. By my
new kinematic circular motion equations, the circumference is
4 times the diameter, so a point on the surface of the photon
travels 8.8 x 10-23
m. If we divide the circumference by the time, we get 1/c.
Which begs another question: Why
would the spin be the inverse of the linear velocity? Because
both are dependent upon the same fundamental factor: size. The
smaller the quantum is, the faster it goes. The photon goes c
precisely because it is so small. It can maximize its speed
because it can dodge most other quantum traffic. But this size
also determines its spin rate. Notice that we have found it to be
spinning extremely fast: 1 cycle every 2.67 x 10-14
seconds, which is equivalent to 3.7 x 1013 cycles each
second. That is extremely fast, from our point of view. But, as I
have just shown, from the photon’s point of view the surface is
moving incredibly slowly: 3 x 10-9 m/s. That is
because one cycle is such a tiny distance. With such a tiny
circumference, the photon can move with a tangential velocity of
1/c, and still achieve an astonishing local frequency.
That being true, it still doesn’t
provide a mechanical link between c and 1/c. I have shown that
the photon can move very slowly, in its own realm, and still
create the necessary wavelengths and frequencies we see, but
“very slowly” and 1/c are two different things. Why 1/c,
precisely? Because, as we know, velocity is a relative
measurement, and we are measuring the spin velocity relative
to the linear velocity. I will show this with a diagram:

Here
we have only the linear velocity and the tangential velocity. I
have drawn the linear velocity only a bit larger than the
tangential velocity, but let us say we are given a tangential
velocity of some magnitude x. We let that magnitude remain
constant and we begin increasing the linear velocity. So one line
gets very much longer than the other. Obviously, if we make one
line very much
longer, we no longer need be concerned about the real magnitude
of x. We only care about its magnitude relative to the linear
velocity. As the linear velocity approaches infinity, the
relative tangential velocity would approach zero. This is
sometimes called the math of very large numbers, and it allows us
to dispense with absolute values. Since velocity is not an
absolute value to begin with, this is very easy to do.
You may say that c is not really a
“very large number”, but relative to the radius of the
photon, it is. We have 31 orders of magnitude between the two,
which is a very large number by anyone’s reckoning.
So, as the linear velocity gets
larger, the tangential velocity gets smaller relative to it.
And the difference between the two is so large that the
relationship becomes nearly a perfect inverse relationship. In
other words, c2 is so large that the tangential
velocity can be estimated very successfully with just the
transform 1/c.
This also happens to be the reason
we can ditch the ½ in the equation E=mc2.
The two equations E=mc2
and E=½mv2
are supposed to be analogous, at least when the first one is
applied to a photon and the second one is applied to an electron,
say. Nearly all the energy of a photon is kinetic energy, so the
equation E=mc2
applied to a photon is an expression of kinetic energy, with m as
the mass equivalence of the photon. But the equation has no ½.
The reason it doesn’t is that because c2
is so large, ½ becomes negligible. One half of almost nothing is
still almost nothing, so we can ignore it as a factor. I
explain this in more detail in another paper.
To
clarify even further, I will show the math for this. All I have
to do is import my new equation which takes us from v to ω. I
developed this equation by correcting Newton's orbital equation
a=v2/r.
Current physics believes that the tangential velocity and the
orbital velocity are equal, but they are not. Newton assigns them
to different vectors or arcs, and then attempts to show they are
equal at the limit. I
have shown that his math fails. In my new equations, ω is
the angular velocity measured in meters/s, not radians/s, and we
can go from v to ω with a simple equation. I used this equation
in my paper on the Bohr
magneton to show the real radius of the elecron, and to
correct the Bohr radius.
ω = √[2r√ v2
+ r2) -
2r2]
If
we insert our numbers for the photon into this equation, using
v=c, we get
ω≈√(2rv) = 4 x 10-8m/s.
That is close to 1/c, so we see that it is not the
tangential velocity that is 1/c, it is the orbital velocity of
the photon that is 1/c. If we let the tangential velocity of the
spin of the photon equal c, then the orbital velocity is 1/c.
That is due to this equation, and the fact that the value of r is
so small.
And this has other major consequences. It means
that Einstein's famous equation must be clarified as well. It now
reads E=mc2.
But I have just shown that the two variable assignments of c in
that equation are not equivalent. So, although the equation
stands, we must subtly rewrite it. One of the c's is the linear
velocity of the particle and one of the c's is the tangential
velocity of the spin. This makes the expanded equation
E=mcLcT
Since both values of c are the same, the equation does
not change in number outcome, but we now have a much better
understanding of what the equation stands for, and what mechanics
it is representing. We are squaring c to represent both the
linear energy and the spin energy of the particle. The orbital
velocity of the particle is 1/c, and this allows us to see how
the wavelength is stretched by the linear motion. But when we
want to calculate the total energy of the particle, we have to
combine the linear velocity with the tangential
velocity. This is because the two velocities are straight-line
vectors and can be multiplied without a transform. As I have just
shown, when the tangential velocity is c, the orbital velocity is
about 1/c.
Some will say that I am assuming a
longitudinal wave for light, whereas Fresnel proved that light
has a transverse wave. If I am able to multiply my local spin
wavelength by c2
to get a visible wavelength, my local wave must be longitudinal.
But that is not correct. Since the wave of light belongs to each
photon, via spin, the wave is neither longitudinal nor
transverse. Longitudinal and transverse waves are defined as
field waves, and light is not a field wave. Light is a spin wave,
and the spin is neither transverse nor longitudinal. The local
wavelength is just a radius of spin. However, since I have shown
(in my paper on superposition) that any electromagnetic radiation
must have at least two stacked spins to show a physical wave,
this stacking can mimic either transverse or longitudinal waves,
depending on the experiment and the effect studied. Fresnel was
studying polarization, and although Young had already shown both
longitudinal characteristics and transverse characteristics, the
polarization experiments seemed to confirm only the transverse
part of this duality. And, indeed, polarization can be explained
with only the transverse characteristics of the stacked wave.
Other experiments and effects are better explained as the stacked
spins mimicking longitudinal waves. This is what is happening
with Tesla or plasma waves which are longitudinal. In plasmas,
the spins beneath the outer spin come into play, and the axial
spin of the moving electron is no longer hidden. The charge field
coheres or links these inner spins, creating uncommon effects. At
any rate, wave theory will not advance beyond its current wall
unless it comes to see that both transverse and longitudinal
waves are a misconception, built upon a mistaken field wave
theory that is an analogue of fluid or sound dynamics. Light
waves are not field waves, they are spin waves. Light is its own
field, since light is both
the linear motion and the spin motion of the photon.
So, I have explained the motion of
the photon in a simple manner, providing us with not only a spin
but a rate of spin. I have developed actual numbers, first
for the radius and mass
of the photon, and now for the rate of spin and the
tangential velocity and orbital velocity. I have shown how this
spin rate creates the visible or measurable wavelength of light.
The radius of the photon’s spin creates the actual wavelength,
and then this length is stretched out by the linear motion,
giving us the measured wavelength of the light.
But now we must move on to ask why
and how photons express different wavelengths. Electromagnetic
radiation, in the form of photons, comes in a wide range of
wavelengths, as we know. How is this achieved? It is achieved by
a wide variation of stacked spins. As
I began to show in a previous paper, it turns out that
photons can maintain a linear velocity very near c over a wide
range of sizes. The photon does not reach a size limit that
causes slowing until it approaches the spin radius just beneath
the electron. At that limit, the largest photons begin absorbing
the smallest photons, and the mass increase snowballs. This turns
the nearly massless photon into the small-mass electron.
The most common photons appear at
the size range of 18213
less than the proton mass and size. This is where we find the
infrared photons, as
I showed previously. But the small mass of the photon allows
it to stack spins over a wide range of radii. In this, it is
unlike the electron or proton. The proton cannot add extra spins
above the z-spin without creating instability. This
is why “mesons” over the baryon size are not stable. The
extra spins begin interfering with the energy of the inner spins.
But with the photon this appears not to be the case. Extra spin
levels do not cause appreciable slowing, nor do they cause
appreciable instability. We may theorize that smaller photons
would be more stable, but the difference in small photons and
large ones is not easily measured from our level.
What this means, specifically, is
that if we give the infrared photon a z-spin as its outer spin,
we can find a smaller photon whose outer spin is the y-spin. We
can also find a larger photon with another axial or x-spin on top
of the infrared’s z-spin. In this way, we find not only stacked
spins, we find stacked levels. In other words, we find spins of
a1, x1, y1, z1 and
a2, x2, y2, z2 and
a3, x3, y3, z3 and so
on. By this analysis, a2 has twice the spin radius of
z1. In fact, each spin has twice the radius of the
spin under it.
This means that photons do not come
in a continuous spectrum. No, they come in stepped levels, each
level double the one under it. For example, we found a wavelength
of 8 x 10-6 m for our infrared photon. You will
remember that number comes from (c2) 8.8 x 10-23
m. If we want the next photon larger than that, we double the
spin radius to 1.76 x 10-22 m and multiply by c2,
which gives us 1.6 x 10-5 m. There can’t be any
wavelength between 8 x 10-6 m and 1.6 x 10-5
m. If we measure light with an average wavelength in between
those numbers, we must have a mixture of photons. A single photon
cannot be found with a wavelength that is in between those
numbers. It must have one wavelength or the other.
This explains many optical effects,
including rainbows. Rainbows do not have a continuous spectrum,
and neither do prisms. We see bands of certain colors. Photons
only come in certain colors, and they do not come in the colors
in between. In between colors have to be created by mixing
photons.
It is known that light is
quantized, but no one before me has proposed an explanation of
this sort for it. Photons have always been considered point
particles, with no radius and no mass. For this reason, it was
thought that spin could not be a possible cause of quantization.
Therefore we have gotten many complex mathematical explanations
of quantization, with no mechanical explanation at all. I have
just shown you the simple mechanical cause of
quantization.
Addendum: many have asked why these charge
photons have not been discovered. My answer is that they have.
All the photons we already know about are part of the charge
field. The entire electromagnetic spectrum is the charge field.
We do not have to propose new photons, we can use the ones we
already have. I have given all photons mass and radius, so all
photons must cause mechanical forces by contact. This has long
been known (see the photo-electric effect) but not fully
interpreted.
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