Abstract
In a previous paper (ref 3, AP4.7), a self-consistent, cyclical theory for the universe called Model 1 was developed, that is complete, self-constructing and regenerating. The unique features of this model are:
- There are two spaces in the universe, particle space and 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 our space through black holes where they are converted into super particles that operate with unified force. They then pass into the high-energy vacuum space where they become dark matter operating behind the potential barrier.
- Dark matter particles interact with each other and form a slowly moving halo centered on a galaxy. The halos of dark matter are connected to each other by corridors of dark matter, which form a cosmic web. This web facilitates and guides the development of new galaxies.
- There, behind a potential barrier, the dark matter particles gain energy, build up in number and eventually exceed the ability of the barrier to contain them. They then explode back into particle space as a big bang. This process repeats to make an endless series of new universes.
- After the big bang exhausts itself, super particles continue to tunnel through the barrier into particle space. The super particles are unstable and break down into particles (protons) with extreme kinetic energy. In doing so, they give up potential energy into particle space. The potential energy gradually builds up to become the dark energy that we observe as the cause of our accelerating, expanding universe. The extreme energy protons are observed as cosmic rays (UHECRs) with energy between the energy of force unification and the Planck energy, which is beyond the GZK cutoff.
Several observables were noted in the paper that support the model (see ref 1, AP4.7D), but nothing would be more compelling as support, than to actually extract a super particle from its barrier shell and observe its properties. A paper proposing this extraction was written (ref 2, AP4.7L). This extraction paper has shown us that the properties of the super particle make it useful to us in areas other than the ones that nature found. Here we investigate the possibility that super particles might be used as a starship drive and energy source.
The Problem
In ref 4, AP4.2, a space energy supply system based on a momentum beam and energy from decaying plutonium is described that has application for a starship. It has the advantage that the starship does not have to carry its fuel to use this system. It also has the disadvantage that it requires the construction and maintenance of the momentum beam by people in the solar system to operate successfully. In addition, the maximum speed achievable is severely limited. Model 1, however, hints at the possibility of an energy supply system based on dark matter that does not require that the starship to carry its fuel, nor does it require the momentum beam and plutonium of the above-mentioned system.
Model 1describes a separate vacuum space filled with high vacuum energy as well as high-energy super particles. Furthermore, this space can export its energy into particle space under some circumstances–by tunneling, if the particle energy is low, and spilling if the super particle kinetic energy exceeds the shell potential energy. Thus if we can work out a convenient procedure to release the super particles from behind the barrier where they could be captured, this procedure would provide an exceptionally convenient source of drive particles and energy for a starship. These super particles will be difficult to release, however. They are held inside an extremely high potential barrier, so the only way it can come out is by:
- Tunneling the barrier, an extremely low rate process.
- Overflowing the barrier, a process that requires the super particle to achieve extremely high kinetic energy (~1019 GeV)
In this paper we will explore the possibility of building a stellar drive and energy supply system by adding enough energy to super particles to allow them to overflow the barrier and thus provide both thrust and energy for a starship. Here, we build on ideas described in ref 2, AP4.7L.
The Solution
In reference 2, AP4.7L, we described a possible method for extracting super particles from behind their barriers for study. Here, we will describe a system that may be useful for extracting super particles from behind their barriers for production of thrust and operational electrical energy for a starship.
In reference 2, AP4.7L, we noted that the rate of passage of a super particle through its barrier shell to particle space is:
R = T h/2p ko nsp / mr particles/sec
Where:
k1= (8p2m(V0-E)/h2) 1/2
ko = (8p2m(E)/h2) 1/2
And:
T = 1/(1+V02sin2(k1w)/4E(E-V0), if E>V
T = 1/(1+Vo2 sinh2(k1w)/4E(V0-E), if E<V
Clearly the transmission T is very low unless E is close to or greater than V0
Now in order to get the super particles out, we want to give the super particles enough energy so that they will flow out from behind the barrier at a rate higher than the natural tunneling rate so we can use them for energy and thrust. We don’t have a particle or gamma source with enough energy to knock the super particle out with one ultra high-energy particle or photon, so we must take advantage of resonance within the barrier shell to pump the energy up one low energy quantum at a time to obtain a kinetic energy near the barrier potential energy. This pump must operate fast enough to overcome energy leakage so buildup can occur, however. Thus there are certain conditions that must be satisfied (Wylie, 75).
- We must use electromagnetic energy, which will penetrate the barrier shell because of the zero rest mass of photons, and because it can act on super particle ions to increase their energy.
- We must pump at the resonant frequency to insure the rapid buildup of energy of the particles, and take advantage of resonance magnification.
- We must pump fast enough to overwhelm the damping for low energy super particles due to the potential energy in vacuum space.
With these conditions, we can pump with gamma photons at less than V0 energy
Experiment to Extract Super Particles from Behind their Barriers
Now we must decide if we can generate gamma photons with the proper characteristics and at the correct rate. Ref 10 shows that generating gammas for Case 1 appears feasible using an undulator or wiggler type synchrotron radiation generators. Gammas for Case 2 appear much harder, however, because of the extreme gamma energy. It does not appear feasible to obtain enough photon energy to boost the super particle over the barrier potential with a single energetic photon. Thus we will build an apparatus in which a wiggler synchrotron generator fires gammas down the length of a rectangular volume with a face area of W x H, and a length of L. The gammas fired are polarized with the polarization vector along the W dimension. Thus the excited super particles will exit preferentially along the W dimension. The apparatus will be oriented so that the super particles (dark matter) pass preferentially along the L dimension.
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 no passing through the experimental apparatus in a time td that has energy greater than Em. This is the number of particles that pass through the reactor tube that can obtain enough energy to break through the barrier shell.
no = F(Em)ρd vd WH = F(Em) 102 super particles / sec
= 1 super particle / sec at two standard deviations from the mean
Where:
F(Em) = fraction of the distribution with energy > Em
~ 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.
ρd = dark matter density in galaxies
~ 10-7 super particles / cc
vd = average dark matter velocity in galaxies
~ 109 cm/sec
W = 1 cm
H = 1cm
This rate of super particle passage through the barrier is determined by ρd and vd which are beyond experimental control, and W and H which are controllable. We have chosen to operate at 1 super particle / sec.
Now we must calculate nγa.
2). The Number of Gammas nγa per Super Particle Needed to Reach Vob
In order to push the super particle through the barrier potential, we need to increase the kinetic energy by an amount
∆E = Vob (1- Em/Vob)
= 0.1 x 1019 GeV = 0.1 x 1028 ev
Where:
Em = average starting energy of the super particle in GeV
~ 9×1018 with Em at two standard deviations above the super particle mean.
Vob = 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
nγa = ∆E / Er = Vob (1- Em/Vob) / Er
~ 0.1 x 1028 / 106
~ 1021 photons
Where:
Er = energy of the resonant gamma photons
= 106 ev
Note that if Em = 9×1017, then nγa ~ 1022 photons instead of 1021photons, and F(Em) = 0.16 instead of 0.025 so nγ is not very sensitive to Em.
So a super particle needs to absorb an average of 1021 photons in a reactor tube of length L according to the equations:
A = Ro(1- R/Ro) = (Ro-R)
nγ = nγa no A= nγa no [Ro– Ro exp(-μL L)]
Where:
nγ = gamma absorption rate
A = number of photons absorbed
Ro = gamma generation rate
nγa = number of absorbed gammas needed to push a super particle out
no = number of super particles passing through the reactor per second = 1/sec
μL = ρd σsp / cm
ρd = dark matter density in galaxies ~ 10-7 super particles / cc
σsp = 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 σsp(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) = πΓγ Γsp / k2 [(Eγ – E)2 + ¼ Γ2]
Where:
Esp = variable operation energy with values less than but close to Vob
Γτ = ђ
τ = mean lifetime
Γγ = energy width of the resonance level for the γ
~ 1026 ev = 1014 erg
Γsp = energy width of the resonance level for the bare super particle
~ T Esp = T 1028
T = 1/(1+Vob2 sinh2(k1w)/4E(Vob-E)
~1/ [1 + (Vob/4E)(2w2m/ ђ2)]
w = barrier width = 10-31 cm
k1 = (2m(Vov-E)/ђ2) ½ = momentum of the super particle in vacuum space
= (2m Vov(1-E/ Vov)/ђ2) ½ =2×1031(1-E/Vov) ½ 1/cm
Vob ~1028 ev
Vov ~1026 ev
Γ = Γγ + Γsp + Σ Γn = sum of Γ’s for all possible modes of decay
kγ = wave number of the incoming gamma
= Eγ /ђ c= 106 10-12 / 10-27 1010 = 1011 1/cm (for Eγ ~106 ev)
At resonance (Eγ = Eo), the cross section is:
σ(γ,sp) ~ Γγ/ kγ2([1 + (Vob2/4 Esp)(2w2m/ ђ2)/ Esp])
~ (Vob/ Esp) cm2
~ 1 cm2
Thus the gamma generation rate needed for a length of the reactor tube L is
nγ = nγa no[1 – exp(-μL L)]
= 1021 1/sec
Where:
[1-exp -μL L] = [1-exp(-ρd σspL)] = [1- 1/e)] ~1
ρd = dark matter density in galaxies ~ 10-7 super particles / cc
no = super particle interaction rate = 1/sec
L = 102 cm
nγa = 1021
Note that the cross section is the size of 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. Relatively few are not absorbed
3). Generation of the Gammas
So in order to generate no super particles in each second, the power required under the most efficient conditions is as follows.
Pi = nγ Er /e = nγa Er no [1- exp(-μL L)] / e
= 1021 x 106 x 10-12/ 10-1 = 1016 erg/sec
= 109 watts = 106 KW
Where:
nγa = 1021
no = 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:
Po = noF(Em) 1028 erg/sec
= 1 x 0.025 x 1028 erg/sec
= 1019 watts = 1016 KW
Where:
no = 1 super particle / second
F(Em) = 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 efficiencies can be made large enough. The black hole generated the energy taken out. However, the efficiencies 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 [tγ(n)] is:
tγ(n) ~ ђ/nEγ = 10-27 / n10-6 = 10-21 / n sec
Where:
ђ = 10-27 erg sec
Eγ = gamma energy = hυ
= 106 ev = 106 x 10-12 = 10-6 erg
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 (vsp) after absorbing n gammas is:
vsp(n)= (mspvsp – npγ) / ms
Where:
msp = shielded super particle mass
vsp = shielded super particle velocity
The distance traveled after absorbing n gammas is:
L = Σ tγ(n) vsp(n) Δn = Σђ/nEγ [(mspvsp – npγ) / msp] Δn
~ Σ (vspm / nγ)tγ Δn
~ ∫(vspm / nγ)tγ dn
~ vspm tγ ln n
Where:
vspm = maximum shielded super particle velocity
For n = 1021,
L = vspm tγ ln nγ = 10910-21 50 = 5×10-11 cm
Where:
L = tube length, choose ~100 cm for convenience.
vi = 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:
ђνγ / c = 10-61021 / 1010 = 105 gm cm/sec
The momentum of the shielded super particle at 109 cm/sec is:
psp < msp vsp = 1019 109 / 1023= 105 gm cm/sec
Where:
vsp < 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.
Use of Super Particles for an Example Starship Drive.
We have seen that theoretically, super particles can be extracted from their barrier shells, and used to obtain energy and thrust in particle space. Now let us do an example starship reaction motor. We take 1of the above experimental reaction tubes for the motor, so W and H each become 1 cm, and L is 102 cm. Then the drive parameters become:
Po = noF(Em) Esp
= 1 x 0.025 x 1028 erg/sec
= 1019 watts = 1016 KW
Where:
no = 1 super particle / second
F(Em) = 0.025 for two standard deviations above the mean.
Esp = 1019 GeV = 1028 erg = energy of the exiting super particle
The super particles that exit the reaction tube exit at right angles to dimension L because the gamma photons used to excite the super particles are polarized in that direction and so preferentially add energy to the super particles in that direction. So we get two beams of super particles from the LH faces of the reaction tube moving in opposite directions in order to balance momentum. We set up a thick blanket of material on one side of the reaction tube to absorb the super particles (now high energy protons) from one beam. This blanket then becomes heated. The heat is used to generate electricity as follows.
Pe = Po ee
= 1015 KW
Where:
ee = 10-1 = conversion efficiency from heat to electricity
Of this electricity, Pi is used to generate the gammas needed to excite the shielded super particles, where:
Pi = 109 watts = 106 KW
The rest can be used for household energy for the starship.
The super particles from the other beam are vented to space, and they become the exhaust products of the super particle reaction motor that produces thrust for the starship. The thrust produced is:
F = starship force or thrust = noћko
= 1 x 10-27 x 1034
= 107 dynes
= 10 KGF
Where:
no = the number of super particles/sec generated = 1 sp/sec
ko = 1034 1/cm for an energy mean of 1019 GeV
So for a 105 KG starship, we need 104 of these 1x1x102 cm reactor tubes to give a 1-G acceleration.
Now we need to see what happens to the top speed of the starship. The momentum of the starship must equal the momentum of the exhaust products, so
Ms Us / (1- Us2/vo2)½ = noћkoΔt
Where:
Ms = mass of the starship
Us = velocity of the starship
vo = velocity of light
c = velocity of light at low energy (see ref 5 AP4.7M)
Δt = thrust time for the starship drive
With a 1-G thruster, a 108 gm starship could obtain a speed 0.99 of the speed of light at low energy in the following amount of time.
1081010 / (1- 0.99 1010/1010) = 107 104 Δt
Δt = 105 sec = 10-2 yr
For the starship to go beyond the low energy speed of light, it must achieve an energy nearing the Planck energy. This will take an impossibly long acceleration time. This subject is addressed in reference 4, AP4.7T.
Summery and Conclusions
We have described a system that may be useful for extracting super particles from behind their barriers for production of thrust and operational electrical energy for a starship. This system has the advantage that it does not require the construction and maintenance of the momentum beam by people in the solar system to operate successfully (see ref 4, AP4.2). In addition, the final speed it can achieve is limited only by the acceleration time that can be expended, although to exceed c would require an unacceptable acceleration time. (see ref 4, AP4.7T).
References
- L. H. Wald, “AP4.7D HOW TO PROVE A THEORYS CORRECTNESS” www.Aquater2050.com/2015/12/
- L. H. Wald, “AP4.7L EXTRACTING SUPER PARTICLES FROM THE BARRIER SHELL” www.Aquater2050.com/2016/01/
- L. H. Wald, “AP4.7 DARK MATTER AND ENERGY-FUNDAMENTAL PROBLEMS IN ASTROPHYSICS” www.Aquater2050.com/2015/11/
- L. H. Wald, “AP4.7T ADDING ENERGY TO PARTICLES NEAR THE PLANCK ENERGY” www.Aquater2050.com/2016/8/
- L. H. Wald, “AP4.7M VARIABLE LIGHT SPEED IN MODEL 1” www.Aquater2050.com/2015/12/