Chapter 5. The
Photon.
The photon is an elementary particle with zero proper mass, meaning that it cannot be at rest
and still exist. It always moves at a constant speed c, which is the speed of light. Depending on its
frequency, it is known as radio wave, micro wave, infrared radiation, light wave, X-ray, gamma ray,
etc.
Photons have been studied in detail, and any theory of physics of the kind
presented here must be consistent with a long list of well documented results regarding the properties of the
photon. Quantum electrodynamics is the formalism that seems to work almost perfect for describing photon
interaction, even though it presents no real explanation for what goes on.
Basically it is a technique for calculating the probability for all possible ways
a photon may move from A to B, including photon-electron interactions. In this formalism probabilities are
calculated from Amplitudes (A), where the square of the Amplitude is the probability for the event to take
place.
P(event) = A2
It will take quite some work to develop a formal theory, before we can show how
the K particle matches the results of quantum electrodynamics. To start with, this presentation will focus on
giving an understanding of how the photon is built up, and how it interacts with Ks.
First of all, our mechanism for K interaction must allow the photon to travel at
the speed of light, interacting with Ks at an amazingly high rate without loosing any energy, as long as the
K flux is homogeneous.
This is not as straight forward as it may seem, since we suppose the K flux to be
homogeneous, and Ks move with speed c, and the K flux has an average 0 velocity relative to the universe,
while the photon travels at speed c. Then an absorbed K must change direction of its momentum with a factor
|pK| during its retention time when it travels along with the photon. At emission, this change in
momentum must be nullified by the emission pattern.
Photons must be like lines (or wavy lines) in the sense that they have no
geometrical dimension seen from the front or rear. A photon has no forward facing target for K interaction,
it is seen by Ks only from the side.
By expressing it this way, the intention is that the reader shall envision a
school of Ks inside a photon, probably with some Ks in parallel, but no part of the photon is rounded up to
form any target in the forward direction. In a homogeneous K flux, the average K flux has zero velocity
relative to the universe. Irrespective of the direction of the K, if a K happens to pass the trajectory of
the photon at any point, it will need the same time from it enters this point of crossing, till it has passed
the point. Therefore Ks 45o from ahead do not interact more often with a photon than Ks
coming in 45o from the rear of the photons direction.
A K hitting the photon’s line of motion at a given point will use the same time
passing this point, and hence the K will stay put for an interaction with the photon for the same amount of
time independent of its direction. But the number of Ks which hit the photon’s line of motion varies with the
K’s angle of inclination, because the number of Ks from a given space angle will vary.
The photon may, or may not, have the same affinity for absorption of Ks from all
directions, but it is quite likely that it has higher amplitude for K interaction from the side than from the
front or rear. If we should take a guess here, the affinity for interaction may increase with the sinus of
K’s angle of inclination. This would be compatible with the way amplitudes work.
We must also show the mechanism which causes the photon to experience red shift
or blue shift as it travels away from or towards gravitational matter respectively. We shall demonstrate how
well our model fits with the present notion of the photon as a wave package. In a later chapter we shall also
argue why our model fits well with the cosmic lens effect on light passing close by galaxies, and finally we
will also look briefly at quantum electro dynamics.
Let us start with the basic equation for the energy of a
photon,
E = hf = mc2.
For a photon to react proportionally to gravity regarding its energy / mass, its target size for
K interaction (~ A2) must increase and decrease proportionally to its total energy /
mass.
Consequence 9:
The probability (~ A2) for a photon – K interaction increases and
decreases proportionally to its energy E = hf = mc2.

Fig. 8. K interaction with a photon. The figure shows 10
arbitrary Ks in three different phases of interaction with a photon; before absorption, during retention, and
after emission. At absorption the Ks have the direction of their momentum changed, and they move parallel to
the photon during the retention time. At emission their velocity along the line of the photon becomes zero,
and they are emitted perpendicular to the direction of the photon.
To visualize how this works, let us look at a photon in a zone with homogeneous
K-flux. The photon will absorb Ks from random directions, and the absorbed Ks will be part of the photon for
the retention time. Thus the K is accelerated to the speed of c in the
direction of the photon.
But since the photon moves at speed c and the K does so as well, we realize that
if the target of the photon or the K were like ball targets, the photon would get a lot of hits from straight
ahead, but no hits straight from the rear. If K can only be emitted with an energy
mKc2 = pKc, this would constitute a resistance which would drain the
photon of energy, which we know is not the case. So with respect to the interaction with Ks, the photon must
take on the form of a particle with some length, and only the side of the particle (or wave) make up the
target for K interaction - only then will the hits of Ks from behind balance the hits from ahead in numbers,
since only the transverse component of a K’s speed will count for the time the K is available for interaction
when it crosses the path of a photon.
The direction of the K will not matter, if it crosses the path of the photon it
will be available for a hit for the same length of time independent of its direction, because the transversal
velocity of a K from the front and rear will be the same if they come in at the same angle, only in opposite
directions. See Fig. 8.
Consequence 10:
To enable a photon to travel at the speed of light without loosing energy, the
photon can not have any forward facing amplitude (√probability) for K interaction, and the amplitude must be
symmetrical around 90 degrees to the photon’s direction, possibly like a sinus function of the angle of
inclination.
Note that even though a photon has no target facing forwards, it may still have a
sideways dimension. Ks move with the photon as a school, and the photon may consist of two or more structures
of Ks in parallel. We suppose the K with momentum pK is absorbed, then the direction of the
momentum is turned to be parallel to the direction of the photon, and then K is emitted after a certain
retention time.
This process will require that all Ks are emitted with velocity –c relative to
the photon to balance the momentum, since they are initially taken from an average zero velocity in the
direction of the photon to +c. So a photon must emit all Ks with a backward velocity of c relative to the
photon, which is then a 0 velocity component backwards relative to the universal K-flux. The momentum of the
emitted Ks will be sideways, and it may be balanced out statistically over a large number of interactions,
hence the sum of the sideways momentum is zero for the incoming K flux as well as the outgoing K flux over
time. But there are good reasons to assume that Ks are emitted in pairs from photons.
Since both the Ks and the photons move at velocity c, the emitted Ks will move
perpendicular to the photon’s direction, and if emitted in pairs, the two Ks move in opposite directions. But
still, emitted Ks moving away perpendicular to the line of motion of the photon will form a wave at 45
degrees angle backwards, since the photon moves as well. See Fig. 9 under emission.
Consequence 11:
A photon absorbs Ks at random with 0 average velocity in the direction of the
photon, accelerates the Ks to c in the direction of the photon, and after the retention time it emits Ks with
a velocity c straight backwards relative to the photon, but with 0 backwards velocity relative to the general
reference frame. Hence the Ks are emitted straight sideways in a way which balances out the momentums of
incoming and outgoing Ks.
Characteristics of photons
Our model postulates that the photon has the following
characteristics:
-
A photon only interacts from the side, meaning its entire probability
(~Amplitude2) for K interaction faces sideways.
-
Regardless of incoming direction, all Ks use the same time to pass the
line of motion of the photon.
-
At K absorption by a photon: If the K-flux is homogenous, then the
average K velocity parallel to the photon is 0.
-
During K retention in a photon: Ks travel with the photon and have
velocity c parallel to the photon.
-
At K emission from a photon: To balance energy with new Ks being
accelerated from 0 to c in the direction of the photon, all Ks must be emitted backwards with
velocity –c relative to the photon, meaning Ks are emitted at 0 backward velocity relative to the
universal K-flux (only velocity relative to the universal K-flux is shown
here).
-
Ks are emitted in a rotating manner, and we assume that they will be
emitted sideways in pairs of Ks with opposite directions but with the same
spin.
-
Emitted Ks move with a velocity of c.
-
To enable a photon to travel at the speed of light without loosing
energy, the photon cannot have any forward facing amplitude (probability) for K
interaction.
But the photon can also be envisioned differently:

Fig. 9. This illustration shows a
photon as a school of Ks (or a package of Ks). At the time of absorption, all Ks assume the
direction of the propagating photon. All the short lines in all the figures symbolise retained Ks, but during
retention we show the few Ks we follow by adding arrows.
One could argue that emission could take place with arbitrary velocity components forwards and
backwards in addition to a sideways component, much like the spread of the incoming Ks. A phenomenon like
Laser strongly suggests that the K emission is steered, since it is hard to imagine a steered absorption of a
random K flux. The Laser-effect seems to represent steered emission of Ks from many photons moving in the
same phase. If there is not a steered emission, it is hard to see how the Laser phenomenon can be
understood. Within our model, LASER can be explained in 2
ways:
-
A synchronised rotational K emission from a beam of photons creates the Laser
effect. Then the wave nature of photons is represented by a rotational emission of Ks, and the
wavelength is how far the photon travels until the emission has made a round of 360
degrees.
-
If we suppose that Ks are emitted in pairs, it is the
timing of emission which is synchronised in Laser photons, hence all photons will drop their K pairs at
the same time, and thereby create strong K flux pulses. Such pulses seem to depend on the rotational
phase as well. The simultaneous timing of the K emission is the key to the Laser effect in this
alternative.
In alternative 1 of the photon, it seems like
a photon has a standard length, and a standard amount of Ks to rid of at each rotation. When the energy
increases, it absorbs more Ks per time unit, thus the photon must rotate faster to rid of more
Ks.
Alternative 1 says that:
f = the number of photon rotations per
second.
λ = the distance travelled between each
rotation.
Alternative 2 says that photon frequency and
wavelength is:
f = the number of K-pairs emitted per
second.
λ = the distance travelled between each K-pair
emission.
Alternative 3 is that both alternatives above
are true:
f = the number of photon rotations, and the
number of K-pairs emitted per second.
λ = the distance travelled between each
rotation and each K-pair emission.
The energy equation E = hf states that there
is 100% proportionality between the frequency f and the Energy, and the strict proportionality of the equation
gives rather strong indications that there is not a bunch of Ks per emission, rather 1 distinct pair of Ks or
just 1 K per emission. Why should nature choose to emit 4 or perhaps 1040 Ks at a time? And always that same
amount? Until we come up with a plausible reason for any other figures, we’ll stick with 2 Ks emitted per
frequency, with 1K as the fairly probable alternative.
Let us look a bit closer at the 3 alternatives
above. Are f and λ average values, or absolute quantities for every instance of the photon’s existence? Take the
example with a super thin glass plate. Suppose this type of glass has an average reflection of 4% for a specific
type of light, meaning the glass will reflect 0 – 8% depending on how its thickness matches the wavelength of
the light. Make a plate with thickness exactly 5 λ, and the QED theory predict that we have 0 reflection and
100% transmission. If this holds true, f and λ are absolute quantities without fluctuations. But we will still
have to determine just what is embedded in the terms frequency and wavelength; we still have 3 alternatives, and
in the 3rd alternative above, there may be a fixed rotation with exactly 1 K-pair emission per round, or there
may be a fixed rotation with an average of 1 K-pair emissions per round. Again the LASER indicates that emission
is quantized and is related to a fixed wavelength and a steady emission pulse pattern. But the best reason for
assuming that there should be a certain photon rotation, is the way QED uses rotating arrows to explain electromagnetic interaction. Let
us see how this can make our photon model for K pair emission fit with the concept of amplitudes of the existing
QED formalism.
A suggested model for a photon based on the
assumption that Ks are emitted in simultaneous pairs could be:
-
A high energy photon may
split into a positronelectron pair. If each of the two photon parts consists of a straightened
positron-electron pair, they could emit one K each.
-
A photon has spin 1,
which is the sum of an electron spin (½) and a positron spin(½).
-
Let the straightened mini-electron emit
K- and the straightened mini-positron emit
K+.
-
Suppose that the reason 1 photon interferes with
itself, is that it splits in its positron and electron part when it moves through 2 slots, and to
remain 1 photon, it must unite in the same phase of rotation.
-
Hence we have a model for the photon consisting of a
straightened mini-electron and a straightened mini-positron.
-
K pair emission would require that the 2 parts of the
photon rotate and are in position to emit outwards in opposite directions once per
rotation.
-
Because of the 2 parts of the photon, they rotate 360
degrees between each emission. With one part only they could in principle emit any time they
like.
-
If a photon splits up through a slot, the two parts
must reunite in a way which secures that they emit their Ks in opposite directions, hence they must
unite in whole numbers of rotational differences, which means the parts can be shifted a whole number
of wavelengths.
Suppose the wave functionality of photons is linked to a fixed rotating pattern of K emission.
Perhaps also other photon properties are embedded in the rotation. The photon’s rotational amplitude can be
represented by a unit arrow turning around the photon’s direction of motion. We suppose that the surroundings
which the photon interact with, are represented by preferred directions of interaction. To make it very simple,
let us say the photon travels in direction z and it has a specific interaction in direction x. Then the unit
circle will follow the equation
x2 + y2 =
1
The component of the photon’s unit amplitude
in the direction of this specific interaction is then x, which is the amplitude for this interaction. To find
how much of the photons wave is in this direction, we use the unit equation. Since this equation is normalised
to 1, it can also represent probability. Because we see that the amplitude for the photon’s rotation and the
amplitude of QED is the same, it is quite likely that photons have this kind of rotation.
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Gravitational lens.
In a gravitational field there will be a
deficit in the K-flux from one side of the photon because a massive body transforms some of the regular K-flux
to K0. Therefore there will be a net surplus K momentum from the opposite side, pushing the photon
towards the massive body. Again the retention time between absorption and emission is essential. The Ks must
travel along with the photon long enough to set the course of the photon before being emitted. Most likely the
photon is hit by Ks at the nose at absorption, and it emits by the tail, which will give the maximum bending of
the light. If the photon is hit at the middle, it will still deviate sideways, but always continue in a
trajectory parallel to its original trajectory. The gravitational push on light can be observed in gravitational
lenses in the universe.
Consequence 12:
A photon interacts with Ks coming
in with a surplus K-flux from one side, and the net sideways momentum of the Ks alters the path of the photon
all the time. Since the Ks are emitted in a neutral direction relative to the course at the time of emission,
the incoming net momentum of all Ks will decide the bending of the course of the photon.
Red
Shift.
Close to a heavy mass, like a star, there will
be a surplus flux of Ks towards the star. Instead of interacting with the same number of Ks coming from the
front and from the rear, a photon moving away from the star will have more hits from the front than from the
rear. The problem is that it cannot emit backwards with higher velocity than c relative to its own velocity, so
it will constantly loose energy by emitting more Ks than it absorbs in order to balance incoming and outgoing
momentum. Its target size will shrink in proportion to lost energy, and there will be a red
shift.

A gravitational lens system photographed by the Hubble
Space Telescope. Credit: NASA, ESA, R. Gavazzi and T. Treu (University of California, Santa Barbara ),
and the SLACS team
Consider a photon moving away from a black
hole. Since the target size of the photon will shrink, the photon will at some point have lost half its energy,
then at some later point it has lost half of that energy again, which is ¼ of the original energy, etc. It will
never loose all its energy, and therefore it cannot be confined by any black hole if it is heading straight
out.
Consequence 13:
A photon moving away from a
massive body will have a net flux of Ks against it, so it will loose energy by emitting more Ks than it absorbs
in order to balance incoming and outgoing momentum, and the photon’s amplitude for K interaction will diminish
continuously in proportion to the lost energy, and there will be a red shift as it moves away from the massive
body.
The laws of conservation of energy and
momentum for the elementary particle must take into account K-flux variations, usually treated as gravitational
potential in this context.
Then what with a photon moving towards a
massive body? It will experience the push from the K surplus, therefore increasing its energy and its amplitude
for K interaction, and thereby experience an even bigger push. Therefore this theory predicts that the push
towards a massive body is not equal to the push from the variable net K-flux, but all energy added to the
elementary particle due to the extra push will be added continuously, as extra frequency (Amplitude2)
of K interaction. When the growing amplitude is taken into account, the photon will experience an extra push
compared to what will be predicted by Newtonian mechanics. As a general principle, if one photon is emitted at
the surface of a planet and travels outwards with a red shift, when it meets another photon going the opposite
way having a blue shift, and if the photons have exactly the same frequency when they meet, then the two photons
must also be equal in other respects, like energy, momentum and amplitude for K interaction.

Lasers are used as a means of improving the quality of
ground-based astronomical observations. What is the physics behind lasers? Credit: Gerhard Hudepohl
(ESO)
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