Chapter 2. Properties of the Unifying Particle.  

This quest for a unifying theory started with a radically new approach to gravity, first in the Newtonian sense, and thereafter looking at relativistic phenomena. The key question here is whether a K particle with only positive momentum p_K   and positive energy E_K, and no attractive features, only repulsive impacts, can perform gravity. The answer is that not only can it do so, but it can explain all the basic forces of nature.

Properties of the K particle  

1. K carries quantized energy E_K and a momentum p_K.      
2. K propagates at the speed of light, and has some similarities to a boson.   
3. K only interacts with Elementary Particles (EPs).   
4. Ks are vastly abundant in the universe.     
5. Ks constitute a very uniform K-flux in empty space.   
6. Ks constitute the vacuum energy of quantum mechanics.     
7. Ks interact with elementary particles at an extremely high frequency.   
8. Ks are generally absorbed by EPs and emitted after a short retention time.   
9. EPs get their entire energy from Ks during the retention time of Ks in EPs.   
10. Regular Ks come with 2 different affinities or “sign”, K⁺ and K^-.      
11. The K neutrino = K⁰ represents the sum of Ks which have had their amplitude for EP interaction reduced by the equivalent of 1 net K interaction.     
12. Ks change only through interaction with EPs.      

Higgs particle attempts to explain some of the same functions as the K particle addresses, but Higgs particle is defined in a framework of gluons, which the K particle denies the ability to carry the strong force. Or, to be more specific, gluons can very well be suitable virtual particles for calculating the strong force, which is here explained by K vacuum. And they can be real intermediate particles of some kind, stripped of their attractive properties.

With K’s special definitions and its special way of interacting with elementary particles, differences are more obvious than similarities between Higgs particle and the K particle. K also seems to be a lot smaller than Higgs particle, although Higgs comes in many sizes. Higgs particle is not supposed to interact with every neutron more than 10²³ times per second, as the K must do.

To draw a parallel to how the K flux feeds for instance a neutron with energy, let us look at a segment of a river basin with a steady flow of water.

• The empty framework / structure of one neutron without Ks is analogue to the empty river basin.    
• The retention time of K is analogue to the time the water takes to flow from A to B.   
• The K-flux is analogue to the water flow.    
• The neutron’s energy is analogue to the amount of water contained between A and B.    
• The K particles are analogue to the H2O molecules.    

Fig.  3. A segment of a river basin between the lines A-A and B-B. The amount of water between the lines A and B will then be constant and the energy in the water will be constant, yet continuously changing. The energy in a neutron will have the same kind of constant, yet changing energy as the Ks flow in and out.

A deeper understanding of the nature of the laws of conservation of energy and momentum follows as a direct consequence of implementing forces by proxy. But for this to work out, the K particle must have a well defined set of features. And the same applies to the elementary particles (EPs) it interacts with, they must be re-examined regarding their basic properties.

It has been a long journey of checking properties of the K particle against known forces in different areas of physics. However, at this stage it could be helpful for the understanding of later chapters to present some properties before demonstrating why it must be exactly like this. Also a few corresponding properties in EPs (elementary particles = fermions and bosons) should be addressed at this stage.

As a starting point the K particle was given momentum and energy, and hence a mass according to:

E _K = m_Kc², or more correctly E_K = p_Kc,

since a particle travelling at the speed of light usually is regarded as having zero rest mass. What other properties were needed for K to be compatible with the concept of “opposite” gravitation? And all other phenomena in physics! The list of properties of the K particle shown above is a result from checking that it is compatible with every phenomenon it has been checked against. This list serves as a reference for later chapters. Se textbox: Properties of the K particle.

Why do we need a K neutrino? If some Ks were simply absorbed in matter to create the K pressure deficiency from the side of matter necessary for causing gravity, then all moving elementary particles, inclusive photons, would experience the K flux as resistance. The permanently absorbed K would go nowhere, and the world would eventually stop. Not only must the K transform to K neutrinos in order to generate gravity, but the K neutrino must be emitted in the same manner as regular Ks, with the same energy and momentum. This becomes evident in chapter 4-5, when we consider particles moving at constant speed through a dense flux of Ks.

An elementary particle (EP) gets all its energy from ever changing Ks which are temporarily retained in the EP. Let us take a look at the light particle - the photon. What is the connection between a photon’s energy and its number of K interactions? We assume that the average retention time is the same and constant for all EPs. Then we must have that the energy of a photon equals a constant times its frequency of K interactions, f_K 

E _photon = const · f_K 

This formula looks strikingly familiar to the general formula for the photon’s energy

E _photon = h · f

where h is Planck’s constant, and f is the photons general frequency. It would be strange if nature had 2 parallel equations for the photon energy, and it’s a fair assumption that the general frequency (f) of a photon equals its number of K interactions per second, f_K, and then we get that

const = h

However, when analysing K emission from photons, it seems that it would be more balanced if 2 Ks were released simultaneously in opposite directions, like at least 2 Ks should be emitted per round of emission. Then

f_K ≥ 2f     

and     

const ≤ h/2.

This issue will be treated in more details later, and for the time being we stick with our most likely outcome, that

f_K = 2f.  

The working mechanism for K interaction with a photon is based on the assumption that all energy in a photon is exchanged continuously and totally. For gravity to act on photons and fermions alike, the same volatility of matter must hold true for electrons, protons and neutrons as well. The way a high energy photon can split into an electron-positron pair also supports the notion of some sort of exchangeable energy.

We suppose that all elementary particles (EPs) like photons, electrons, protons, neutrons and the whole zoo of subatomic particles are just empty structures without energy of their own. These EPs thrive on the K flux which allows them to borrow Ks proportional to their energy. Hence the EP’s probability for K interaction must be proportional to the EP’s energy. If so, the frequency f_K of K interaction also for EPs with proper mass must therefore be given by

E = mc² = h · f = const · f_K 

where h is Planck’s constant.

If we stick to using only f_K =   2f, we get that the frequency of interaction for various common EPs can be expressed like:  

f _K(proton) = 2 · m_proton · c² / h = 4,5·10²³/s

f _K(neutron) = 2 · m_neutron · c² / h = 4,5·10²³/s

f _K(electron) = 2 · m_electron · c² / h = 2,5·10²⁰/s

These figures are shown to give a fair impression of the magnitude of K’s frequency of interaction with EPs.

A key feature for K interaction is the retention time of Ks in an EP. An EP must retain the Ks for a certain time in order to accumulate its energy. The way elementary particles may change from bosons to fermions indicate that all the energy in an elementary particle consists of Ks temporarily retained within the particle. Then we will show in a later chapter that the energy, E_K, of the K particle times its retention time, t_R, in EPs must equal Planck’s constant, h, divided by 2

t _R·E_K = h/2

This expression is based on f_K = 2f (Ks being emitted in pairs from a photon)  

By this we assume that no elementary particle has any fixed mass, only continuously changing energy consisting of retained Ks. Note that this is fundamentally different from saying that the photon or the Higgs particle give energy to EPs. In conventional terms one does not talk about energy which is exchanged 10²³ times per second, and that this energy exchange goes on also when nothing seems to happen to the EP.

The conservation laws describe how this energy may change. The retention time (tR) of Ks may be a natural constant valid for all Ks, or t_R may be the average retention time for Ks. Equally E_K may be a constant valid for all Ks, or an average value of K’s energy. Since we talk about quantum mechanics, it is almost imperative that Ks only exist with one or some very specific levels of energy. Then a K must have at least 3 different probabilities (amplitudes) for interacting with EPs, depending on whether it is a K⁺, K^- or a K⁰, and then of course the different types of EPs have different probability (amplitude) for K interaction.

To interact with Ks while following the laws of physics, the known elementary particles (EP) must have some special properties regarding how they interact with Ks, and how they transform energy. As we go on to analyse different known phenomena, the necessary properties of the EPs will be deducted.

Now we are getting closer to how Ks are emitted. Since we demand balance of momentum for elementary particle (EP) interaction, we see that an EP must increase or decrease its energy to conserve momentum. When a K is absorbed in an EP, it will be emitted after the retention time in a manner that conserves momentum, and in this process, energy may be added or taken away from the EP.

Consequence 1: The retention time of Ks in EPs constitute the key mechanism for the laws of conservation.  

Consequence2: Momentum is conserved at the elementary particle level. An EP must increase or decrease its energy in order to conserve momentum. When a K is absorbed in an EP, it will be emitted after the retention time in a manner that conserves momentum, and in this process, energy may be added or taken away from the EP.  

Consequence 3: At the elementary particle level energy is not conserved. Energy conservation must account also for systematic differences in the background K flux, which is then accounted for as potential energy.  

Consequence 4: Kinetic energy manifests itself in the emission angle in Ks from particles with proper mass. Energy must be added or taken away to change the emission angle.  

One should note that the Ks could just make an impact with an EP and deliver its momentum and energy like a quantum package. However, without a full absorption and a certain retention time with controlled angle of emission set by the speed of the EP, it would be hard to find a way that a fermion can move at all in the presence of a K-flux. Also, it would be hard to explain the add-on effect of energy at relativistic speed for fermions that way.

Furthermore, to explain the splitting of the EP into absorption centres that provide strong and weak nuclear forces, electric fields and electromagnetism seems impossible without a retention time for the Ks in the EP. To find out how the Ks are absorbed, retained and emitted in order to be compatible with the laws of energy and momentum, we need to consider bosons and fermions separately.

Do you have any comments? Click HERE

You are currently on page: Properties of the Unifying Particle

Previous page: Forces by Proxy                  Next page: What is Gravitation?