Photon - K Interaction

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 |p_K| 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 45^o from ahead do not interact more often with a photon than Ks coming in 45^o 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 m_Kc² = p_Kc, 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 p_K 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:

1. 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.

2. 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:

1. 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.

2. A photon has spin 1, which is the sum of an electron spin (½) and a positron spin(½).

3. Let the straightened mini-electron emit K^- and the straightened mini-positron emit K⁺.

4. 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.

5. Hence we have a model for the photon consisting of a straightened mini-electron and a straightened mini-positron.

6. 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.

7. 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.

8. 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.

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.

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

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.

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.

A strongly redshifted galaxy. Credit: NASA, ESA, P. van Dokkum (Yale University), M. Franx (Leiden University, The Netherlands), and G. Illingworth (University of California, Santa Cruz, and Lick Observatory)