Abstract

There are currently three important connected major unanswered questions in physics and astrophysics.

(1)  How can the theories of symmetry and the Higgs field be used to calculate the masses of the fundamental particles?

(2)  How can dark matter be explained and described?

(3)  How can dark energy be explained and described?

A self-consistent theory (called Model 1 in this paper) has been developed that appears to answer questions 2 and 3 quantitatively. In order to derive and justify Model 1, however, it became necessary to calculate the mass-energy of the proton and other fundamental particles that make it up, which answers question 1. This procedure then gave a path for calculating the mass of the Super Particle, which is the primary particle of Model 1. The super particle was found to be an SU(6) or Grand Unified Particle, which unifies the electromagnetic, weak and the strong forces. The super particle was found to be unstable, however, but a stable particle can be formed by wrapping the super particle in an SU(10) Totally Unified potential energy barrier which unifies the electromagnetic, weak, strong and gravitational forces. In investigating the properties of this shielded super particle, it became obvious that it had the properties of dark matter, and when it breaks down, it generates dark energy. In a sequence of related papers (Wald, Model 1-A, Model 1-B; Wald, Model 1-C; Wald, Model 1-D; Wald, Model 1-E; and Wald, Model 1-F), Model 1 is detailed and expanded. In order to demonstrate the correctness of model 1, it may be necessary to extract the Super Particle from its shell and observe it, however. This paper will show that it may be possible to do just that.

 

The Problem

The unique features of Model 1 are:

• There are two spaces in the universe, low-energy particle space and high-energy quantum vacuum space. A potential barrier separates them. One space contains visible matter, and the other space contains dark matter.

• There is a cycling of mass-energy between these spaces through the black holes that connect them. Particles pass from particle space to vacuum space through black holes where they are converted into super particles (energy ~10¹⁷GeV).  They are then wrapped with a potential energy barrier shield (~10¹⁹GeV) to stabilize them. The barrier shield forms the boundary of high-energy vacuum space. These stabilized, shielded super particles are then able to escape from the black hole into particle space. The shielded super particles have a low interaction cross-section with ordinary particles except through gravity, and so are observed as dark matter.

• Dark matter particles interact with each other and form a slowly building bubble centered on a galaxy that stabilizes its outer edges. Corridors of dark matter are also generated which form a cosmic web between the galaxies. These corridors guide the development of new galaxies.

• The super particles can tunnel through the barrier into particle space. Upon reaching particle space, the super particles become unstable and break down into particles (cosmic ray protons) with ultra high kinetic energy (UHECR’s). In doing so, they give up potential energy from their barrier shields into particle space which becomes the dark energy that we observe as the cause of our accelerating, expanding universe.

If the super particle could be extracted from its shell and observed, the resulting high energy particle particle would provide proof of the existence of super particle. In order to determine how this extraction might be accomplished, consider the Schrödinger equation in the following form: 

            [-(h²/2m d²/dr² + V(r)]U(r) = EU(r)    

            Where

            E = p²/2m = the energy of the particle with mass m and momentum p.

            V(r) = V_o[Q(r) – Q(r-w)] = the barrier potential.

            Q(x) = the Heavyside step function of width a starting at x =0

            w = the width of the potential zone.

The solution can be used to determine how to force the transmission through the barrier T (see Wald, Model 1-E), which is as follows:

            If E> V_o,

            T = 1/(1+V₀²sin²(k₁w)/4E(E-V₀)           

            If E< V_o,

            T = 1/(1+V_o² sinh²(k₁w)/4E(V_o-E),

            Where (if E< V_o)

            k₁= (2m(V₀-E)/ђ²) ^½ = momentum of the super particle in vacuum space

            k_o = (2m(E)/ ђ²) ^½ = momentum of the super particle in particle space 

The best fit for the vacuum space sphere and the barrier shell of the rough data available is (see Wald, Model 1-E for details):           

            r_ov = 10^-20 cm which defines the inner boundary of the vacuum shell

            r_ob = 10^-11 cm which defines the outer boundary of the vacuum shell

            a = 10^-31 cm which defines the thickness of the barrier shell.    

Clearly the transmission T of a particle through the barrier is very low unless E is close to or greater than V₀. When a particle passes (tunnels) through the barrier, it is available for observation, and shows itself as an ultra high energy proton. 

Thus, the high-energy space can export part of its energy into particle space under some circumstances. This exporting of energy can happen by tunneling, if the particle energy is low, and spilling over the barrier if the super particle kinetic energy exceeds the shell potential energy. If the super particle enters particle space, it would appear as an Ultra High Energy Cosmic Ray (UHECR) (see Wald, Model 1-E). Thus if we can determine a convenient procedure to release the super particle from behind the barrier on demand, where it would be detected as an ultra high energy proton, its extreme energy would indicate its origin, and this procedure would provide exceptional proof for Model 1.

We note that the barrier shell has no electric charge, and no magnetic moment (see Wald, Model 1-A), so it will not interact with electromagnetic photons and thus impede the passage of photons through the shell to the super particle where interaction can take place. Further, a portion of the super protons inside the barrier will be ionized, so a photon can interact with the super proton. Thus, for example, if the photon energy is high enough, photons (gammas) can reach ionized super protons inside the barrier in vacuum space, and interact with them to give them enough energy to force them through the barrier. We don’t have a gamma generator that can generate a gamma with enough energy to do this by the interaction of a single gamma, however. Since a low energy photon in particle space can reach through the barrier to a super particle in vacuum space, we ask if it is possible to store the energy from many low energy photons in a resonance in a shielded super particle until the total energy is high enough to force the super particle through the barrier. If so, it might be possible to use lower-energy gammas to excite the super particle inside its shell enough to drive it through the barrier shell so we can observe it. 

Here, we will pursue this concept a bit further to establish its potential. 

Classical Force Equation Results

First we consider the classical analog of a gamma exciting a super particle inside a potential barrier shell. This is like an ionized ball of mass m bouncing back and forth in a rubber spherical shell under the influence of an oscillating electric field. We start by constructing a classical force equation for an oscillating mass m. This force equation has the form (Wylie, 82):

            1/2m d²r/dt² + C_o dr/dt + kr = F_ocos ωt

            Where:

            k = return force of the rubber shell

            C_o = damping force

            F_ocos ωt = electric forcing function

Now we define a natural or resonant frequency as:

            ω_r /2p = 1/2p [k / m]^½ cycles / sec

Also we define critical damping as:

            C_oc = 2(k m)^½

Then we define the magnification ratio as:

M = the factor by which a deflection of r_ov by a constant force must be multiplied in order to give the amplitude of the vibrations that result when the same force acts dynamically with frequency ω.

We note that the ratio of the successive maximum displacements remain constant throughout the entire motion of an underdamped system. Also, more energy per cycle is taken out of the energy source (F_ocos ω) with each cycle 

According to Wylie (Wylie, 82):

            M = 1 / [(1 – ω²_/ ω_r²) ² + (2 ω/ω_rC/C_oc )²] ^½ 

Furthermore, note that the damping ratio C/C_oc can approach (but never reach) zero, so the magnification ratio can approach (but never reach) infinity.  

We note that the envelope of the oscillations rises exponentially under the influence of the forcing function to a maximum value r_e until the mass breaks free if the system is under damped as follows. 

            r / r_e = 1 – Exp [-t(C_o /2m)]

            Where:

            C_oc< 2(k m)^½  

Also the envelope of the oscillations falls exponentially if the system is over damped as follows: 

            r / r_e = Exp [-t(C_o /2m)]

            Where:

            C_oc > 2(k m)^½  

Note that the system has a long life (it lasts for many cycles) if it is under damped, and the energy in the system builds with each cycle. Also, if the system is under damped, the driver will eventually increase the energy of the mass enough to force it through the rubber shell. If it is critically damped, the driver increases the energy of the mass enough to force it through the rubber shell in one cycle. 

Now we pause to answer a fundamental energy conservation question. The equation we used above is a force equation. We must ask where the energy that we gain at resonance comes from when we obtain a magnification. If we multiply the original force equation by an increment of radius, it becomes an energy equation. By using the resulting energy equation the physical process becomes clearer. The mass oscillates by trading potential with kinetic energy. When the mass position is at its maximum radius, the potential energy is at its maximum, and the kinetic energy is at its minimum. When the mass is moving through the mean radius, the kinetic energy is at its maximum, and the potential energy is at its minimum. At resonance, the electromagnetic drive pushes the mass on each cycle at just the right time to reinforce the motion, and thus pushes the mass position to a greater radius. Thus the potential energy is magnified. The magnification we obtain is the magnification of the envelope of the oscillations of the mass within the rubber shell. If the damping is enough to absorb the energy of each push (critical damping), the energy is used up in damping, the mass returns to its original position, and the magnification is one. If the damping is less than the amount needed to absorb each push, the radius will increase each cycle, and magnification will be greater than one. The smaller the damping, the longer the lifetime and the larger the magnification will be. Thus as the number of cycles increases, the radius will continue to increase until the return force can no longer contain the mass, and the mass will break free through the rubber shell. As time goes on, an equilibrium is established in an ensemble of these systems in which free masses are generated at a rate determined by the rate of energy entry into the system through the forcing function. 

This is only a classical example. It is real, however if the elements of the apparatus are macroscopic. Let us see if this example can be transformed into quantum nuclear theory. 

Compound Nucleus Theory

Consider a nuclear analog of a gamma stimulating a super particle inside a barrier shell followed by emission of a super particle. This analog is called the compound nucleus hypothesis. In the compound nucleus hypothesis (Halliday, 327), a particle enters a nucleus, excites it and becomes subject to strong internal forces. Its direction is changed and its excitation energy dissipated, so its identity is lost. A compound nucleus is thus formed, and its dominant feature is its long life. Sooner or later, a nucleon comes close to the nuclear surface with enough energy and it escapes. If this process takes too long, a photon is emitted to relieve the excess energy. This hypothesis can be expressed as follows. 

            σ(a,b) = σ(c)P_b(ε) 

            Where:

            σ(a,b) is the cross section for the complete reaction X (a,b)Y

            σ(c) = cross section for the formation of compound nucleus with excitation energy           

                       ε by absorbing the particle a

            P_b(ε) = the normalized probability that the nucleus will decay by emission of b

Here it is assumed that P_b(ε) is independent of the mode of formation of the compound nucleus.

For the case in which all spins are assumed zero, and the energy of a is low, σ(a,b) is given by the expression (Halliday, 344): 

            σ(a,b) = πΓ_a Γ_b / k² [(E_a – E_o)² + ¼ Γ²] 

            Where:

            Γτ = ђ

            τ = mean lifetime of the system

            Γ_a  = energy width of the resonance level for a

            Γ_b = energy width of the resonance level for b

            Γ = Γ_a + Γ_b + Σ Γ_n = sum of Γ’s for all possible modes of decay

            k = wave number of emitted particle b

This hypothesis is commonly used for a neutron or proton absorption and gamma and/or particle emission. However, it appears to be useful for photon absorption and particle (electron) emission. In this case, we call it the photoelectric effect. Note, that a photon cannot transfer its energy entirely to a free particle (Halliday, 367). It is not possible to balance both momentum and energy for such a reaction. The emitted particle must be initially bound to some structure that remains after the particle is emitted. For example, the particle can be an electron initially bound to an atom, and the electron is kicked in one direction and the atom in the other to balance the momentum, after initial excitation by a photon. It does appear useful for our shielded super particle problem, however, when we compare it with the classical results.

In this example, however, only one particle or photon is used for excitation. For the case of interest to us here, the absorption of many low energy photons are required for a super particle to gain enough energy to be pushed out of its shell. The classical case shows how the buildup can happen. Small increments of energy are captured in the resonance generated by the exchange of kinetic and potential energy. Each increment of energy (gamma photon) is added at the proper time, and at the proper frequency, so the resonance builds. Eventually enough energy is absorbed so that the kinetic energy exceeds the ability of the shell to return it. The bouncing mass then breaks free of the shell and the momentum of the mass is balanced by pushing the containing shell in the opposite direction. Thus in order for a super particle to break free by use of low energy photons, there must be a possibility for a low energy resonance, and a sequence of coherent energetic pulses that can be absorbed at the proper time. There must also be a massive structure containing the super particle to absorb the momentum of the escaping super particle. Most important, the lifetime of the excited state must be long enough to build up the necessary energy, i.e.-there must be no other modes of decay that will bleed energy before the system reaches its critical energy. 

Now we will see if Model 1 gives the super particle the necessary requirements to force a super particle out of its barrier shell. 

Super Particles Passing Through the Reactor Tube

Here we need to derive an expression for the number of super particles passing through the walls of an experimental reactor tube. We recall that the shielded super particles are in a thermal (Gaussian) distribution, so they are distributed between a mean energy somewhat greater than 10¹⁷ GeV, and a high-energy tail cutoff slightly less than 10¹⁹ GeV. We want to know the number of super particles n_o passing through the reactor tube interior in a time t_d that has energy greater than E_m. This is the number of particles that pass through the reactor tube that can gain enough energy from photons to break through the barrier shell.         

            n_o = F(E_m)ρ_d v_d WH = F(E_m) 10² super particles / sec

                 ~ 1 super particle / sec at two standard deviations from the mean

            Where:

            F(E_m) = fraction of the super particle distribution with energy > E_m

                      ~ 0.16 at one standard deviation from the mean, and

                      ~ 0.025 at two standard deviations from the mean.

                      ~ 0.0015 at three standard deviations from the mean.

            ρ_d = dark matter density in galaxies

                ~ 10^-7 super particles / cc

            v_d = average dark matter velocity in galaxies

                ~ 10⁹ cm/sec

            W = 1 cm

            H = 1cm          

If enough gammas are present in the reactor at a resonance frequency to increase the super particle energy to the barrier potential energy, this is the rate of super particle passage through the barrier and it is determined by ρ_d and v_d which are beyond experimental control, and W and H which are controllable. Note that ρ_d and v_d are only roughly known, and F(E_m) is even less well known (see ref 4, AP4.7I), so the accuracy of estimates from this equation is limited, but n_o can be bounded with the numbers shown. We have chosen to operate at 1 super particle / sec.

Gamma Absorption Probability by the Super Particle

Now for each shielded super particle that absorbs enough gammas to break through the barrier, an easily measured UHECR proton appears. This event is what we are searching for. However, this will not happen until the proper number of gammas is absorbed, so we must calculate the number of gammas absorbed by each super particle in the reactor tube. We start by letting R be the number of gammas in a collimated beam that falls on the front face of the reactor tube. Then we must determine the rate at which photons disappear from the beam of gammas  (-dR) by absorption by the shielded super particles (Halliday, 165). A group of shielded super particles moves slowly compared to a gamma photon, so we can think of a beam of gammas moving through and being absorbed by a slab of shielded super particles. So:

            -dR = μ_L R dL

Integration gives 

            R = Ro exp(-μ_L L) 

Then the total number of photons that are absorbed by particles (A) is:

            A = Ro(1-R/Ro)

            Where:

            μ_L = ρ_d σ_sp / cm

                ρ_d = particles / cc in the slab

            σ_sp = interaction cross section of the super particle (cm²)

The interaction cross section is expected to be much larger than the size of the shielded super particle (10^-22 cm²) because it is a resonance phenomenon. Thus with a 100 cm tube, a large fraction of the incoming super particles is expected to be absorbed. Next we hope to determine the expression that will allow for calculating the interaction resonance cross section.

The Solution

The Resonance

Recall from classical theory that we defined a natural or resonant frequency as:

            ω_r /2p = 1/2p [k / m]^½ cycles / sec 

We can use quantum equivalents of these constants to get a shielded super particle resonant frequency of:

            ν_r = [V_ov / r_ov² m]^½

                = 10²¹ 1/sec

Photons with this frequency have an energy of 

            E_r = hν_r

                = 10^-6 erg

                = 10⁶ ev

We would expect this frequency from a super particle mass bouncing inside a barrier shell of the size of the shielded super particle shell (r_ob = 10^-11 cm).

This resonance is the energy storage device we need to store low energy photon energy until it can force the super particle out through the barrier shell. 

Gamma Excitation and Super Particle Emission

We can now develop the equations for absorption of gammas by shielded super particles, and emission of bare super particles. The energy of the incoming photons is captured sequentially in the resonance of the super particle bouncing back and forth inside the barrier shell, and the energy builds up. When the energy is high enough to equal the potential of the barrier shell, the super particle escapes. The momentum of the recoiling barrier shell can balance the momentum of the expelled super particle. Thus the expulsion can take place, and the shielded super particle can break down. Note that if the gamma photon energy is high enough, the resonance takes place inside the barrier walls. This reaction is equivalent to critical damping. Then only one photon is required and the reaction is photoelectric type reaction in which a super particle is emitted. We don’t have a gamma generator that produces gammas with enough energy for a single photon reaction, however. Thus we hope to use many low energy resonant gammas to store up a kinetic energy equal to the barrier potential energy. The sequence of events is as follows.

1). Super Particles Passing Through the Reaction Tube

Recall that the super particles are in a thermal (Gaussian) distribution, so they are distributed between a mean energy somewhat greater than 10¹⁷ GeV, and a high-energy tail cutoff slightly less than 10¹⁹ GeV. We want to know the number of super particles n_o(E_m) passing through the reactor tube window (WH), and remaining in the tube length (L) for a time t_d long enough to collect n_γa gamma photons. The energy E_m  is the minimum energy that allows the super particle to reach the barrier breakthrough energy V_ob (10¹⁹ GeV) after absorbing an average of n_γa gamma photons. This number is:          

            n_o(E_m) = F(E_m)ρ_d v_d WH = F(E_m) 10² super particles/sec

                          ~ 1 super particle/sec (at two standard deviations from the mean)

            Where:

            F(E_m) = fraction of the distribution with energy > E_m

                      ~ 0.16 at one standard deviation from the mean, and

                      ~ 0.025 at two standard deviations from the mean.

                      ~ 0.0015 at three standard deviations from the mean.

            ρ_d = dark matter density in galaxies

                ~ 10^-7 super particles / cc

            v_d = average dark matter velocity in galaxies

                ~ 10⁹ cm/sec

            W = 1 cm

            H = 1 cm        

Now we must calculate n_γa.

2). The Number of Gammas n_γa per Super Particle Needed to Reach V_ob 

In order to push the super particle through the barrier potential, we need to increase the kinetic energy by an amount 

            ∆E = V_ob (1- E_m/V_ob)

                   = 0.1 x 10¹⁹ GeV = 0.1 x 10²⁸ ev

            Where:

            E_m = average starting energy of the super particle in GeV

                   ~ 9×10¹⁸ with E_m at two standard deviations above the super particle mean.

            V_ob = barrier potential = 10¹⁹ GeV = 10²⁸ ev

Then, for each super particle passing through the barrier, the number of photons coming in at the resonance frequency with energy 10⁶ ev must be absorbed

            n_γa = ∆E / E_r = V_ob (1- E_m/V_ob) / E_r

                    ~ 0.1 x 10²⁸ / 10⁶  

                    ~ 10²¹ photons

            Where:

            E_r = energy of the resonant gamma photons

                = 10⁶ ev         

Note that if E_m = 9×10¹⁷, then n_γa ~ 10²² photons instead of 10²¹photons, and F(E_m) = 0.16 instead of 0.025 so n_γ is not very sensitive to E_m.   

So a super particle needs to absorb an average of 10²¹ photons in a reactor tube of length L according to the equation:

            A = R_o(1- R/R_o) = (R_o-R)

            n_γ = n_γa n_o A= n_γa n_o [R_o– R_o exp(-μ_L L)]

            Where:

            n_γ = gamma absorption rate

            A = number of photons absorbed

            R_o = gamma generation rate

            n_γa = number of absorbed gammas needed to push a super particle out

            n_o = number of super particles passing through the reactor per second = 1/sec

            μ_L = ρ_d σ_sp / cm

                ρ_d = dark matter density in galaxies ~ 10^-7 super particles / cc

            σ_sp = resonance cross section (>> 10^-22 cm² the size of a shielded super particle)

            L = length of the reactor tube = 10² cm

Now, we need to calculate the resonance cross section σ_sp(a,b).

Resonance Cross Section

Here, gamma photons are the particles (a) that excite the shielded super particle (c), and the emitted particle is a super particle (b). Two paths appear possible, n low energy gammas build up the energy of the super particle to breakout energy, or one high-energy gamma gives up enough energy to achieve breakout. We will calculate the many gamma case. Remember that the many gammas must be coherent, and maybe entangled in order to maintain resonance, although this latter requirement is not sure.

So particle a is chosen to be a gamma photon. Particle c is the excited super particle. Particle b is the bare super particle that is passed through the barrier. For the case in which all spins are assumed zero, and the energy of a is low, σ(γ,sp) is given by the expression: 

            σ(γ,sp) = πΓ_γ Γ_sp / k² [(E_γ – E)² + ¼ Γ²] 

            Where:

            E_sp = variable operation energy with values less than but close to V_ob

            Γτ = ђ

            τ = mean lifetime

            Γ_γ  = energy width of the resonance level for the γ

                 ~ 10²⁶ ev = 10¹⁴ erg

            Γ_sp = energy width of the resonance level for the bare super particle

                  ~ T E_sp = T 10²⁸

            T = 1/(1+V_ob² sinh²(k₁w)/4E(V_ob-E)

                ~1/ [1 + (V_ob/4E)(2w²m/ ђ²)]

            w = barrier width = 10^-31 cm

            k₁ = (2m(V_ov-E)/ђ²) ^½ = momentum of the super particle in vacuum space

                      = (2m V_ov(1-E/ V_ov)/ђ²) ^½ =2×10³¹(1-E/V_ov) ^½ 1/cm

            V_ob ~10²⁸ ev

            V_ov ~10²⁶ ev

            Γ = Γ_γ + Γ_sp + Σ Γ_n = sum of Γ’s for all possible modes of decay

            k_γ = wave number of the incoming gamma 

                 = E_γ /ђ c= 10⁶ 10^-12 / 10^-27 10¹⁰ = 10¹¹ 1/cm (for E_γ ~10⁶ ev)

At resonance (E_γ = E_o), the cross section is:

            σ(γ,sp) ~ Γ_γ/ k_γ²([1 + (V_ob²/4 E_sp)(2w²m/ ђ²)/ E_sp])

                          ~ (V_ob/ E_sp) cm²

                          ~ 1 cm²

Thus the gamma generation rate needed for a length of the reactor tube L i 

            n_γ = n_γa n_o[1 – exp(-μ_L L)] 

                       = 10²¹ 1/sec         

            Where:

            [1-exp -μ_L L] = [1-exp(-ρ_d σ_spL)] = [1- 1/e)] ~1

            ρ_d = dark matter density in galaxies ~ 10^-7 super particles / cc

            n_o = super particle interaction rate = 1/sec

            L = 10² cm

                    n_γa = 10²¹

Note that the cross section is the size of the reactor tube, so the length required is only 1 cm, rather than 10²cm. What this result means is that absorption rate is controlled by the dark matter density. Each time a super particle meets a resonance gamma photon in a meter long reactor, it will have a very high probability of absorption because the resonance cross-section is so high. Thus we only need to generate enough photons to match those needed for the shielded super particles passing through the reactor tube. Few are not absorbed.

3). Generation of the Gammas

So in order to generate n_o super particles in each second, the power required under the most efficient conditions is as follows 

            P_i = n_γ E_r /e = n_γa E_r n_o [1- exp(-μ_L L)] / e

                = 10²¹ x 10⁶ x 10^-12/ 10^-1  = 10¹⁶  erg/sec

                = 10⁹ watts = 10⁶ K 

            Where:

            n_γa = 10²¹

            n_o = 1 super particle / second

            e = efficiency of conversion of electrical to gamma energy

               ~ 10^-1          

Note that the power being dumped out into particle space by the super particles is:

            P_o = n_o F(E_m) 10²⁸ erg/sec

                 = 1 x 0.025 x 10²⁸ erg/sec

                 = 10¹⁹ watts = 10¹⁶ KW

            Where:

            n_o = 1 super particle / second

            F(E_m) = 0.025 for two standard deviations above the mean.

Thus it is theoretically possible to get more energy out of this apparatus than is put into it, if the gamma generation and absorption efficiences can be made large enough. The black hole generated the energy taken out. However, the efficiences must be investigated, because their values are important.   

4). The Dwell Time Problem

There may be a problem in obtaining the dwell time needed to absorb 10²¹ gammas. The dark matter is believed to rotate very slowly around the galactic center because it travels out from the central black hole rather than falling in from the outside under the influence of gravity as visible matter does. Now the visible matter is known to rotate around the galactic center at a speed of ~100 km/s at our radial distance from the center (Peebles, 47) (earth rotation speed around sun is ~ 30 km/sec, so galactic speed dominates). Thus there is a relative interaction speed between photons and particles of ~100 km/s (10⁷ cm/sec) In addition, the super particles passing out of the black hole have significant velocity heading toward the edge of the galaxy. They start just inside the event horizon at near light speed, but lose a significant portion of that speed in passing through the event horizon and escaping the black hole. We estimate they retain less than a tenth of that speed (<10⁹ cm/sec). Thus a shielded super particle passing along the length of the reactor tube L might pass out of the reactor before it has had enough time to absorb 10²¹ gammas. Remembering that the gammas will reduce the super particle velocity as they are absorbed, we will now calculate the time a shielded super particle will remain in the reactor.

According to the uncertainty principle, the time necessary to absorb the nth photon [t_γ(n)] is:

            t_γ(n) ~ ђ/nE_γ = 10^-27 / n10^-6 = 10^-21 / n sec

            Where:

            ђ = 10^-27 erg sec

            E_γ = gamma energy = hυ

                  = 10⁶ ev = 10⁶ x 10^-12 = 10^-6 ev 

Now 10²¹ gammas must be absorbed. If they are absorbed one at a time and the shielded super particle velocity remains constant, the shielded super particle must remain in the beam for ~1 sec. However, the shielded super particle velocity does not remain constant. It is reduced by the impinging gammas. While the shielded super particles are moving down the length of the reactor tube, the gammas are moving up, and when the gammas are absorbed, they give their momentum to the shielded super particle and slow it down. A momentum balance shows that the shielded super particle velocity (v_sp) after absorbing n gammas is:

            v_sp(n)= (m_spv_sp – np_γ) / m_sp           

            Where:

            m_sp = shielded super particle mass

            v_sp = shielded super particle velocity

The distance traveled after absorbing n gammas is:

            L = Σ t_γ(n) v_sp(n) Δn = Σђ/nE_γ [(m_spv_sp – np_γ) / m_sp] Δn

               ~ Σ (v_spm / n_γ)t_γ Δn

               ~ ∫(v_spm / n_γ)t_γ dn

               ~ v_spm t_γ ln n

            Where:

            v_spm = maximum shielded super particle velocity

            For n = 10²¹,

            L = v_spm t_γ ln n_γ = 10⁹10^-21 50 = 5×10^-11 cm 

            Where:

            L = tube length, choose ~100 cm for convenience.

            v_i = interaction speed < 10⁹ cm/sec

It is important to note that the total momentum given to the shielded super particle by the 10²¹ gammas is: 

            ђν_γ / c = 10^-610²¹ / 10¹⁰ = 10⁵ gm cm/sec 

The momentum of the shielded super particle at 10⁹ cm/sec is:

            p_sp < m_sp v_sp = 10¹⁹ 10⁹ / 10²³= 10⁵ gm cm/sec

            Where:

            v_sp < 10⁹ cm/sec

So by the time all of the photons have been absorbed, the shielded super particle velocity has been reduced to zero. Thus the shielded super particle velocity does not change direction, and the integral does not have to be segmented. 

Summary and Conclusions

In this paper, we have investigated the possibility of extracting super particles from the barrier shell using low energy gamma photons in a resonance excitation to boost the super particle energy up to the barrier potential and thus force it over the barrier into particle space. We found that it might be possible to use the resonance of the bouncing of the super particle back and forth inside the barrier shell to store up gamma energy until it is high enough in the super particle to break through the barrier shell. This resonance is at a frequency of ν_r = 10²¹ 1/sec which is in the gamma range. This frequency gives an energy of E_r = 10⁶ ev. We also found that the gamma energy put into shielded super particles in vacuum space is less than the energy that comes out in particle space in the form of high-energy protons (cosmic rays). Thus it may potentially be a means of obtaining energy from dark matter. We caution, however, that these results are possible only if there is no alternative decay mode that would bleed energy from the resonance faster than it is being added by the gamma source. Currently, no such decay mode is known.

  

References

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4.  L. H. Wald, Model 1-D “The Recycling Universe” www.Aquater2050.com/2017/02/
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