UNITEL_7 - StealthSkater

archived as http://www.stealthskater.com/Documents/UNITEL_7.doc
more UNITEL is archived at http://www.stealthskater.com/UNITEL.htm
a Math Analysis of UNITEL's Aerodynamic Vehicle
(or a "Magnetic Field Final")
by Edward
Halerewicz, Jr.
The UNITEL Quantum Electromagnetic Laser Propulsion book is interesting. But it doesn't prove
the design would work, so I can't accept any of the claims as facts. I would say the ideas presented are
within the realm of possibility. But they are in no way proven. The laser propulsion system does seem
possible if -- and a BIG IF -- BECs can propagate QCD effects within a QED vacuum.
However, I had a problem with the chapter on "Macroscopic Quantum Tunneling" and the use of
hyperspace. According to string theory, only bosons should be able to propagate through extra
dimensions -- not fermions. This is called "brane jumping". Even in string theory, if certain fermions
can pass into certain branes, it wouldn't necessarily mean that all would. Also, hyperspace shouldn't
have a time frame which propagates linearly as seen from our dimension (at least according to existing
theory) but through complex space. However, if the ship can "phase" itself out of space-time, it would
appear as exotic matter and in that fashion could behave as a wormhole.
So it's interesting. It's based on real science and not fill-in-the-blank science like Star Trek. But
nothing really constable as proof. For that, they would have to get their 'Prototype 1-A' up-and-working.
If that does affect the ZPF, then it would prove the concept. I have yet to fully look at everything and
give a reply to Larry. But those are my thoughts at the moment.
StealthSkakter note: After a few months, Ed provided the following mathematical analysis
of UNITEL's quantum laser propulsion system. I did my best to convert his response to
a MS-Word document. If there are any glaring errors in symbols, exponents, etc., it's
probably due to my erroneous transcribing.
Originating e-mail (from Larry Maurer / UNITEL)
To: "Edward Halerewicz, Jr." <halgravity@yahoo.com>
From: "Larry Maurer" <lmaurer1101@hotmail.com>
Sent: Monday, May 26, 2003 3:50 PM
Subject: Re: Magnetic Field Final
According to Yang-Mills theory and the comparison to the moebius strip, one must go round the
integral path of the equator twice to return to the point of origin. Thus that fact implies Dirac's
quantiazation condition: 2q = integer. To remove the "twist" of phase in the nontrivial and trivial
connection, one has to go round the surface twice which equals 2 x 360 degree rotations that equal
720 degrees. You can find examples of the 720 deg. rotation to eliminate the kink or twist in the
well-know Aharanov-Bohm experiment.
Responding e-mail
From: "Edward Halerewicz, Jr." <halgravity@yahoo.com>
To: "Larry Maurer" <lmaurer1101@hotmail.com>
Subject: Re: Magnetic Field Final
Date: Wed, 28 May 2003 08:01:25 -0500
Greetings again,
I updated my spreadsheet file. Of interest, I found that when the electron cloud oscillates at 4π with
the laser, it is almost exactly phased shifted with the phase shift of the "green beam". Again, you are
free to edit and pass on the file to suit your needs. … … …
attachment :
Properties of UNITEL's Lens and Hull
Physical Constants in MKS System:
charge of an electron:
q = 1.602 x 10-19 C
speed of light in a vacuum:
c = 2.9979258 x 108
electrostatic charge constant:
4 0
permittivity of free of free space:
ε0 = 8.854 x 10-12
mass of electron:
me = 9.1093897 x 10-31 kg
(n m )
Planck's constant:
h = 6.67259 x 10-34 Js
Boltzmann's constant: kB = 1.380658 x 10-23
( KJ )
Basic Description
The UNITEL design requires that a magnetic charge to develop from phased-shifted adiabatic
electromagnetic forces. They line up at a distance from the ship which is approximate to the ship's
length r=5.5 ft through the parallel transport of exciton vectors. The phased-shifted EM beam then
causes the electromagnetic field to rotate at the speed-of-light, and hence the electron-hole system
created by the exciton particles which tunnel through the honeycomb RGB lens of the craft. It is
hypothesized by UNITEL that this rotation will force the excitons to contract down to the Planck length
lp=(Gh/2π c3) ~ 1.6 x 10-33cm (i.e., the theoretical quantum scale for gravitation), created by the extreme
Lorentz-Fitzsgerald length contraction induced on the excitons by the rotation of the laser's electric field
which has a propagation velocity of c. From this description, the magnetic field strength of the induced
singularity of the UNITEL craft is calculated. The strength is found to be 1.2 x 1040G.
determination of this value is described in the text below.
the Calculation
To begin, we must start with the electric field strength of the excitons at a distance of one Planck
length of separation. Thus the electric strength of the ships singularity is of order:
ES 
where ES = 5.624 x 1050
kg m
4       m  
(note that SI units
s A
kg m
uses Weber units).
s A
Next, one must consider the surface which the electric field covers. This would include the radius of
the UNITEL lens r=3 ft. The distance of the laser beam as it propagates from the ship must also be
included which is assumed to be roughly 5.5 ft. This then gives the basis for the Gauss Surface of the
UNITEL laser.
However, the lens-laser system would only appear to give a half-surface. Which is not acceptable.
The problem can be resolved if we view the ship as the half of the surface so that the ship and laser act
as a single Gauss Surface. This can be accomplished by charge quantization which can reproduce
Gauss' Law (Cohen M. Clifford Residues and Charge Quantization. gr-qc/0207107) by means of 2
spherical contour surfaces (which act as 2 spherically-charged monopoles, but with their vectors
reversed). This basically leaves one with an ellipse π a b, where 'a' is the length of the ship and 'b' is the
radius of the laser lens. Converting ab from ft to meters gives ab/m=5.029, so that area covered by the
Gauss Surface is AE = 4 π 5.025 . Gauss' Law therefore requires that:
ES AE = 3.554 x 1052
kg m
s A
Of course, as the laser propagates AE increases, thus decreasing the strength of the field as it
propagates outward.
(Gauss Surface schematic of the UNITEL Ship)
We now must consider the strength of the electrostatic force at the Planck length, which is
experienced by the length contracted excitons. The force is given in terms of:
Fe = k
q q
(1.6  10  30 m) 2
This is a huge force of Fe = 9.01 x 1031 N ! To find the magnetic strength, we must compare this
strength to charge and the velocity of the propagating field (which through phase shifting is assumed to
be c) so that:
= 1.876 x 1042 T
We can now check to see if this strength matches up with the electric field of the ship because
E=F/e. When we do so, we find that
Fe e = 3.315 x 1031 N
which is in agreement with our earlier value for E. However, this neglects the Gauss Surface of the
ship. When the Gauss Surface is input into the magnetic field, we have
ES A E q
= 1.186 x 1044 T
where ES AE q = 5.694 x 1033 N. If one multiplies this value by 10-4, the Gauss strength of the field is
ES A E q
1 G = 1.186 x 1040 4 2
s A
For comparison, the highest known magnetic field strength for a neutron star is 1015G . So the above
strength is roughly 10 times a trillion times a trillion the strength of a high energy neutron star!
Non-singular Field Strength
We now consider the field strength of the electric field which surrounds the ship (but which is not
Planck contracted):
Ef =
4 0 (1.676m ) 2
where Ef = 5.126 x 10-10
kg m
, which has a magnetic strength of
s A
Ef A Eq
q c
= 1.08 x 10-16 T .
The increase in magnetic energy of this value in comparison to the ship's singularity is small and can
be considered negligible. Nevertheless, the small field will produce forces on the electron cloud which
surrounds the hull of the ship. The electric frequency of field will produce a charge acceleration of
q E f cos(1) 4 s
qa =
where qa = 612.047
( ms ) . So that the ponderomotive force is
FP =
 q2
4m e (4) 2
(Ef)2 s .
Since Fp = -1.91 x 10-32 N is small, so is the acceleration
= -0.021
( m ).
This prevents the
magnetic singularity laser of the ship from creating "blow-up conditions" with the charged particles on
the hull of the craft which make up the electron cloud.
Phased-Shifted Lens
We now consider the effect of the phase-shifted RGB lens. First, let us consider the frequencies of
optical RGB light:
fR =
fG =
fB =
7  10 m
5  10 m
4  10  7 m
As the laser is activated with a microwave, one would expect a phase velocity of fR 0.075m = 3.212
for 'Red' light. We now wish to find the phase velocity of the electrons in the system to see if
they correspond with the microwave phases of the RGB laser. To find this result, we begin by finding
the de Broglie wavelength of the electrons. But first, we must know the momentum of the electrons.
UNITEL proposes acoustic control of electron waves through Semiconducting Diode Elements
(SDEs) so we can assume they move at the speed -of-sound. As the electrons act in a 4oKelvin
environment and the excitons simulate hydrogen (or a molecule of mass 3.58 x 10-27 kg), the speed-ofsound for the electron cloud is found to be:
k B ( 4K )
vS =
3.58  10  27 kg
where vS = 160.345
. This value allows us to calculate the de Broglie wavelength of the electrons:
λd =
me vs
where λd = 4.568 x 10-6 m which gives us the phase velocity of the electron by
m e 2c 2  d
evP = c
which is evP = 5.605 x 1014
1 .
. This is similar to the phase of the 'Green' phase of the microwave
beam of the lens when cycled through the Gauss Surface as seen by 4π fG 0.075m = 5.651 x 1014
From this, we see that when vS is reduced by 20 m/s we get the fB phase. And when vS is increased
by 20 m/s, we get fR (thus controlling the frequency of the SDEs would allow the hull to have the same
phase of the RGB lens). When we consider the magnetic strength of each lens, we find that
Ef A E
3 vS
= 6.734 x 10-11 T
So the total strength becomes of the non-singular portion of the RGB lens becomes
Ef A E
1 G = 2.02 x 10-14
The amount of voltage surrounding the hull of the ship can therefore be seen as
Ef A E
3 vS
(1.676 m) = 1.81 x 10-8 V .
As the "smart skin" can induce a high magnetic field of strength 14.7 T, the voltage in that case can
become vS (14.7 T) (1.676 m) = 3.95 x 103 V . When multiplied by q, one finds the energy in electron
volts which corresponds to 6 KeV which will ionize the hull of the craft.
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