DISCLOSURE / FILE
DIA's Negative Mass Propulsion Reference Document
DIA Defense Intelligence Reference Document DIA-08-1101-023, dated 3 January 2011, on negative-mass propulsion, produced under the AAWSA program for Senator Reid.
DISCLOSURE / FILE
DIA Defense Intelligence Reference Document DIA-08-1101-023, dated 3 January 2011, on negative-mass propulsion, produced under the AAWSA program for Senator Reid.
DIA AATIP Defense Intelligence Reference Documents (38 DIRDs), FOIA Release, DIRD_29-DIRD_Negative_mass_Propulsion.pdf
DIA Defense Intelligence Reference Document DIA-08-1101-023, dated 3 January 2011, on negative-mass propulsion, produced under the AAWSA program for Senator Reid.
This 43-page Defense Intelligence Reference Document, prepared by the Defense Intelligence Agency under the Advanced Aerospace Weapon System Applications (AAWSA) Program with an information cut-off date of 30 August 2010, surveys the physics of negative mass and its theoretical use for propulsion. The author argues that negative masses are 'all around us, albeit hidden behind positive masses,' and that freeing them would require either electromagnetic and gravitational fields beyond what is technically attainable, or mining naturally separated negative mass from gravitational potential wells. The document works through Bondi's mass-dipole metric, Hund's nonlinear Newtonian gravity, Schrödinger's Zitterbewegung as evidence of negative mass components in the Dirac equation, and the Planck Aether Hypothesis. It closes by proposing that a tunnel through the Moon, cut by 'a sequence of thermonuclear shape charges,' could reach a potential well shallow enough to access trapped negative mass and 'revolutionize interstellar space flight.'
It is easy to prove that there are negative masses all around us, albeit hidden behind positive masses.p.4
The first of these two possibilities is for all practical means excluded, because if possible at all, it would depend on electromagnetic or gravitational fields with strengths beyond what is technically attainable, or on extremely large particle energies likewise not attainable.p.4
Making a tunnel through the moon, provided there is a good supply of negative mass, could revolutionize interstellar space flight. A sequence of thermonuclear shape charges would be required to make such a tunnel technically feasible.p.4
It is this property of self-acceleration without expenditure of energy that has intrigued many researchers and raised the prospect of a propulsion system without limits.p.6
Over and over again we have found that what is possible, within the framework of the fundamental laws of physics, exists.p.6
The earth is therefore embedded in a sea of negative mass.p.11
We therefore see that there are huge amounts of negative masses bound to positive masses in Dirac spinors.p.17
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UNCLASSIFIED//FOR OFFICIAL USE ONLY Defense Intelligence Reference Document Defense Futures 03 January 2011 ICOD 30 August 2010 DIA-08-1101-023 Negative Mass Propulsion UNCLASSIFIED//FOR OFFICIAL USE ONLY
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Negative Mass Propulsion
The Defense Intelligence Reference Document provides non-substantive but
authoritative reference information related to intelligence topics or methodologies.
Prepared by:
(b)(3):10 USC 424
Defense Intelligence Agency
Author:
(b)(6)
Administrative Notes:
(U) COPYRIGHT WARNING: Further dissemination of the photographs in this publication is not
authorized.
This product is one in a series of advanced technology reports produced in FY 2010 under the Defense
Intelligence Agency, (b)(3):10 USC 424 Advanced Aerospace Weapon System Applications
(AAWSA) Program. Comments or questions pertaining to this document should be addressed to (b)(3):10 USC 424;(b)(6)
(b)(3):10 USC 424;(b)(6) AAWSA Program Manager, Defense Intelligence Agency, ATTN: (b)(3):10 USC 424;(b)(6)
Bldg 6000, Washington, DC 20340-5100.
ii
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UNCLASSIFIED//FOR OFFICIAL USE ONLY Defense Intelligence Reference Document Defense Futures 03 January 2011 ICOD 30 August 2010 DIA-08-1101-023 Negative Mass Propulsion UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY
Negative Mass Propulsion
The Defense Intelligence Reference Document provides non-substantive but
authoritative reference information related to intelligence topics or methodologies.
Prepared by:
(b)(3):10 USC 424
Defense Intelligence Agency
Author:
(b)(6)
Administrative Notes:
(U) COPYRIGHT WARNING: Further dissemination of the photographs in this publication is not
authorized.
This product is one in a series of advanced technology reports produced in FY 2010 under the Defense
Intelligence Agency, (b)(3):10 USC 424 Advanced Aerospace Weapon System Applications
(AAWSA) Program. Comments or questions pertaining to this document should be addressed to (b)(3):10 USC 424;(b)(6)
(b)(3):10 USC 424;(b)(6) AAWSA Program Manager, Defense Intelligence Agency, ATTN: (b)(3):10 USC 424;(b)(6)
Bldg 6000, Washington, DC 20340-5100.
ii
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY Contents 1. Introduction ................................................................................................ 2 2. The Theory by Bondi..................................................................................... 4 3. Hund's Nonlinear Newtonian Theory of Gravity .................................................. 6 4. The Theory of Bondi Revisited .....................................................................10 5. The Zitterbewegung Phenomenon as a Manifestation of Negative Masses ..............10 6. Planck Aether Hypothesis .............................................................................14 7. Dynamic Interpretation of Lorentz Invariance...................................................18 8. Negative Mass Interpretation of the Aharonov-Bohm Effect ...............................23 9. Negative Masses in Cosmology ......................................................................29 10. The Cusp/Core Problem in Galatic Halos ........................................................31 11. Searching for Negative Matter in the Gravitational Potential Well of the Moon......32 12. Making a Tunnel through the Moon ...............................................................33 13. Conclusion.................................................................................................38 Figures Figure 1. Forces..............................................................................................2 Figure 2. Translation of Mass Dipole..................................................................11 Figure 3. Circular Motion of a Pole-Dipole Particle................................................12 Tables Table 1. Interactions........................................................................................2 iii UNCLASSIFIED//FOR OFFICIAL USE ONLY
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Negative Mass Propulsion
Summary
It is easy to prove that there are negative masses all around us, albeit
hidden behind positive masses. But their use for propulsion by reducing the
inertia of matter, for example in the limit of macroscopic bodies with zero rest
mass, depends on a technical solution to free them from their imprisonment by
positive masses. It appears that there are basically two ways this might be
achieved: 1. By the application of strong electromagnetic or gravitational
fields or by high particle energies; 2. By searching for places in the universe
where nature has already done this separation, and from which the negative
masses can be mined.
The first of these two possibilities is for all practical means excluded,
because if possible at all, it would depend on electromagnetic or gravitational
fields with strengths beyond what is technically attainable, or on extremely
large particle energies likewise not attainable.
With regard to the 2nd possibility, it has been observed that non-
baryonic cold dark matter tends to accumulate near the center of galaxies, or
places in the universe which have a large gravitational potential well. Because
of the equivalence principle of general relativity, the attraction towards the
center of a gravitational potential well, produced by a positive mass, is for
negative masses the same as for positive masses. Large amounts of negative
masses might have over billions of years been trapped in these gravitational
potential wells.
Now it just happens that the center of the moon is a potential well, not
too deep that it cannot be reached by making a tunnel through the moon, not
possible for the deeper potential well of the earth, where the temperature and
pressure are too high. Making a tunnel through the moon, provided there is a
good supply of negative mass, could revolutionize interstellar space flight. A
sequence of thermonuclear shape charges would be required to make such a
tunnel technically feasible.
1
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1. Introduction
If we extend the law of gravity to negative masses, but hold onto the equivalence of
inertial and gravitational masses, we have to distinguish between the following four
cases, if a test particle is placed near a gravitational field producing mass (Table 1):
Table 1. Interactions
Case | Gravitational field producing mass | Mass of test particle | Motion of test particle
1 | + | + | attraction
2 | + | - | attraction
3 | - | + | repulsion
4 | - | - | repulsion
Under the principle of equivalence if a negative test mass particle would be placed in
the gravitational field of earth, it would not fall upwards, as happens in science-fiction
antigravity machines. A test particle, regardless of whether it has positive or negative
mass, would there always fall down. It would fall upwards only if placed in the field of a
large negative mass.
A somewhat different situation arises if both masses, the field producing mass and the
mass of the test particle, have the same absolute value but are permitted to have
different signs. There we have to distinguish between the cases shown in Figure 1.
Case
1 (+) → +(+) attraction
2 +(+) +(-) }
3 (-) → (+) → } selfacceleration
4 +(-) (-) → repulsion
Figure 1. Forces
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If both masses are positive, we have the usual Newtonian attraction. For negative
masses, the force has the same magnitude but is repulsive. A quite different situation
exists if one mass is positive and the other one is negative. With both forming a mass
dipole, the system becomes self-accelerating, because one mass is repelled and the
other one attracted. With the two masses having opposite sign, the total energy and
momentum of the combined system remains zero for all times, leaving intact the
conservation laws of energy and momentum. Under its self-acceleration, the mass
dipole would eventually reach the velocity of light. It is this property of self-acceleration
without expenditure of energy that has intrigued many researchers and raised the
prospect of a propulsion system without limits. We remark that even without an
appreciable gravitational interaction, a mass dipole with zero, or close to zero inertial
mass, could be accelerated to very high velocities with negligible jet power and energy.
No matter how strange the properties associated with negative masses appear to be,
there can be little doubt that they can be incorporated into Einstein's gravitational field
theory as long as they do not violate the principle of equivalence. In particular, the well
known Schwarzschild solution for a positive mass M
ds² = dr²/(1-2γM/c²r) + r²(dθ² + sin²θdφ²) - (1-2γM/c²r)c²dt² (1)
can be extended to a negative mass, simply by replacing M with -M:
ds² = dr²/(1+2γM/c²r) + r²(dθ² + sin²θdφ²) - (1+2γM/c²r)c²dt² , (2)
where γ is Newton's constant.
One therefore has to raise the question if nature has not made use of negative masses
somewhere. Over and over again we have found that what is possible, within the
framework of the fundamental laws of physics, exists. Only one important physical set
of laws, Einstein's special theory of relativity appears to forbid the existence of negative
masses. This is because in a relativistic quantum field theory the particle number is not
a conserved quantity, and the existence of negative masses would make all matter
unstable against decay into negative masses.
3
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY Apart from Einstein's purely kinematic interpretation of special relativity, being the expression of a Minkowskian space-time structure, there is an older alternative dynamic interpretation by Lorentz and Poincaré. In it space and time are absolute, but it can explain all relativistic effects as well. It assumes the existence of an aether, with all objects in absolute motion through the aether suffering a Lorentz contraction and time dilation. If this aether has a grainy structure, characterized by some smallest length (e.g., the Planck length ~ 10⁻³³ cm), then according to Heisenberg's uncertainty principle special relativity would ultimately break down at a high energy. If the length is very small, this energy can be so high as to be far beyond the capabilities of any existing particle accelerator or even beyond the high energy of cosmic ray particles, making both interpretations of special relativity experimentally indistinguishable at the energies presently available. 2. The Theory of Bondi The first attempt to introduce negative masses into general relativity to describe a mass dipole was made by H. Bondi [1]. For a uniformly accelerating mass dipole Bondi uses the axially symmetric metric by Weyl and Levi-Civita [2]: ds² = e²φdt²[e²σ(dr² + dz²) + r²dθ²] , (3) where φ = φ(r,z) and σ = σ(r,z) satisfy (∂²/∂r² + 1/r ∂/∂r + ∂²/∂z²)φ = 0 (4) ∂σ/∂r = r[(∂φ/∂r)² - (∂φ/∂z)²] (5) ∂σ/∂z = 2r(∂φ/∂r)(∂φ/∂z) . (6) 4 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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Inserting (3-6) into Einstein's nonlinear gravitational field equations, one obtains four
nonlinear partial differential equations (κ = 8πγ/c⁴) given by
-κρ = -κT₀⁰ = e^{2(φ-σ)}[-2∇²φ + ∇²σ - 1/r ∂σ/∂ρ₀ + (∂φ/∂ρ₀)² + (∂φ/∂z₀)²] (7)
-κp₁₁ = -κT₁¹ = κT₂² = κp₂₂ = e^{2(φ-σ)}[1/r ∂σ/∂ρ₀ - (∂φ/∂ρ₀)² + (∂φ/∂z₀)²] (8)
-κp₃₃ = -κT₃³ = e^{2(φ-σ)}[∇²σ - 1/r ∂σ/∂ρ₀ + (∂φ/∂ρ₀)² + (∂φ/∂z₀)²] (9)
-κT₁₂ = 2(∂φ/∂ρ₀)(∂φ/∂z₀) - r(∂σ/∂z₀). (10)
In solving these equations Bondi assumes that φ is small, which then also implies that
because of (5) and (6) σ is small by the second order, reducing the solution of the
problem to the linear Laplace equation of the scalar Newtonian potential in empty space.
Making this assumption, Bondi can reproduce the uniform acceleration of the mass
dipole, as it is expected from an elementary analysis. It is here that we must disagree
with Bondi¹, because it can be shown that the nonlinearity of the gravitational field
equation leads to a very different result. The nonlinearity also sheds light on why it is
so difficult to separate negative from positive masses, whereby negative masses are all
around us, but imprisoned by positive masses.
___________________
¹ The author had the pleasure to meet Prof. Bondi on a common flight from Graz,
Austria in 1993 (we both are members of an academy which had a meeting in that
year in Graz), and ask him how his solution can be correct since it does not include
the field of the positive gravitational field mass of a mass dipole. This problem will
be analyzed in the next section, and its solution has far reaching consequences.
5
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY 3. Hund's Nonlinear Newtonian Theory of Gravity As explained by Hund [3], already Newton's theory of gravity, in conjunction with the postulates of special relativity, leads to a nonlinear theory of gravity. With this model theory of gravity the nonlinearity of the gravitational field can be much better explained than with Einstein's theory. Hund begins with the force F acting on a mass m in a "merry-go-round": F = mg + mv/c × G . (11) If g is the gravitational acceleration by real masses with the density ρ, one has in Newton's theory div g = -4πγρ . (12) In the merry-go-round g has a vertical component from the earth's gravitational field, but also a radial component by the radial centrifugal acceleration which is not source- free. With |g| = ω²r, where ω = 2π/T, with T the time of revolution, and r the radial distance from the center of the merry-go-round, one has div g = 2ω² . (13) Comparing (13) with (12) one sees that the centrifugal force corresponds to a gravitational repulsion of a homogeneous mass density μ = -ω²/(2πγ) . (14) For a typical "merry-go-round" one has T = 10 sec, and hence, ω = 0.6s⁻¹. For this example one obtains from (14), μ = -10⁶ g/cm³, taken as an absolute value about equal to the mass density of a white dwarf. 6 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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The mass density (14) is not fictitious but represents physical reality. According to
Einstein's E = mc² one obtains for an electric field E the mass density
u = ε₀|E|²/(8πc²) > 0 . (15)
Replacing ε₀ with -(1/γ), one obtains the energy density of the gravitational field
u = -g²/(8πγ) < 0 . (16)
It possesses the (negative) mass density
μ = -g²/(8πγc²) . (17)
Besides the field g, we have on a "merry-go-round" the Coriolis force field
G = 2cω . (18)
By setting G = g and inserting G into (17) with g² = G², one obtains the mass density
(14). This means the centrifugal force is the gravitational force associated with the
mass density of the Coriolis field.
But where is this huge negative mass coming from? The obvious answer is by the very
large vacuum energy, making itself felt by going to an accelerated frame of reference.
In Mach's principle the motion of the distant galaxies as seen in an accelerated frame of
reference is responsible for the inertial forces. But this idea is wrong because if by some
miracle the distant galaxies were to be set into motion, it would take a long time before
their fields propagating with the velocity of light would reach the earth.
Adding the mass of the Coriolis field to the right side of the equation (12), we have by
putting it on the left hand side
div F - 1/(2c²) G² = -4πγρ . (19)
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In one further step one should have
div F - 1/(2c²) (F² + G²) = -4πγρ (20)
or if G = 0 and F = -∇φ, where φ is the Newtonian gravitational potential, one obtains
for Poisson's equation:
∇²φ = 4πγρ - 1/(2c²) (∇φ)² . (21)
According to (21) the positive mass as the source of the Newtonian potential is reduced
by the negative mass of its field. (Einstein's theory leads to almost the same, except
that there the negative gravitational mass density is twice as large.) One can then write
∇²φ + 1/(2c²) (∇φ)² = 4πγρ . (22)
The earth is therefore embedded in a sea of negative mass. Making the substitution
ψ = e^{φ/c²} (23)
transforms (22) into
∇²ψ = 4πγρ/c² ψ . (24)
If ρ is a delta function at r = 0, where
m = ∫4πr²δ(r)ρdr , (25)
then one obtains from (24):
d/dr(r² dψ/dr) = mγψ/c² (26)
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or φ = mγ/r , (27)
the same as in Newton's theory.
For a sphere of constant density ρ one obtains the solution
ψ = sinh kr/kr, k² = 4πγρ/c² . (28)
Hence,
φ = c² ln[sinh kr/kr]. (29)
For small values of r one has
φ = (2π/3)γρr² (30)
with the force per unit mass
|g| = -|∇φ| = (4π/3)γρr , (31)
as in Newton's theory.
In general one has
|g| = c²[1/r - √(4πγρ)/c · ctnh(√(4πγρ)/c · r)], (32)
which in the limit r → ∞ becomes a constant:
|g| = -√(4πγρ)c , (33)
which implies the self-shielding of a large mass by the negative mass of its own
gravitational field surrounding the mass. The shielding becomes important at the
distance
9
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY R = c/√(4πγρ) . (34) Inserting into (34) the average mass density of the universe, R becomes the radius of the universe. The negative gravitational mass of the universe there shields its positive mass. 4. The Theory of Bondi Revisited We are now in a position to revisit the theory by Bondi. In his treatment of the positive- negative-mass dipole two body problem, he did not take into account the gravitational field of this configuration. According to (27) the gravitational potential of a point mass remains the same as in Newton's theory. This means that the gravitational potential interaction energy for two positive equal masses, E_pot = -γ(m×m)/r = -γm²/r (35) is changed for a mass dipole into E_pot = -γ(m×(-m))/r = γm²/r . (36) In the theory of Bondi this positive field mass must be added to the positive mass, resulting in a mass pole-dipole, which is a mass pole with a superimposed mass dipole. As we will see in the next section this fact has very important consequences [4]. 5. The Zitterbewegung Phenomenon as a Manifestation of Negative Masses There is no fundamental physical principle standing in the way which forbids the existence of negative masses. If this is true, the question is: Where are these negative masses? The recently noticed large bubbles or voids observed in intergalactic space could possibly be explained by the repulsive force of negative masses assumed to occupy the voids, but alternative, less exotic, explanations have been offered as well. However, there is at least one fundamental phenomenon which strongly speaks for the 10 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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existence of negative masses. It is the spin of the fermions, like the spin of the electron.
Fermions are described by Dirac's relativistic wave equation. This equation has both
positive and negative energy components and because of the mass-energy relation, it
therefore must have negative mass components. According to Schrödinger [4], it is
these negative mass components which lead to the phenomenon of the spin. Since the
overall mass of the electron is positive, the occurrence of negative masses in the Dirac
equation must mean that the electron is a mass pole with a superimposed mass dipole
[5].
The spin is definitely not an intrinsic rotational motion of a finite size particle, as older
models had suggested it to be. The original model by Uhlenbeck and Goudsmit, for
example, cannot possibly be correct because it requires superluminal rotation velocities
for an electron with the classical radius r₀ = e²/mc² .
If we consider the linear motion of a mass dipole (Figure 2), we immediately see that
its translation generates angular momentum. Construction of a mass pole with a
superimposed mass dipole can simply be done by choosing the positive mass slightly
larger than the magnitude of its negative counterpart. For such a pole-dipole particle
the center of mass S is outside the line connecting both masses (Figure 3) with its
motion taking place on a circle of radius r_c.
[Figure showing: + → m⁺v (upward arrow)
→ v (center arrow)
-|m⁻|v ← (downward arrow with circle at right)]
Figure 2. Translation of mass dipole.
11
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY [Figure showing circular motion diagram with labels: v→c, m, m', S, r_c, r] Figure 3. Circular motion of a pole-dipole particle. Because it is self-accelerating the circular motion will eventually reach the velocity of light. The connection with Schrödinger's analysis is reached if one puts in Planck's constant and m as the electron mass r_c = h/(2mc) , (37) whereby the circular motion around S produces just the angular momentum h/2 as in Dirac's equation. The velocity-of-light result was first obtained by Breit [6], according to which Dirac's equation predicts a local electron velocity equal to the velocity of light. Even though its time-averaged velocity is always less than the velocity of light, this means that the electron, represented by the pole-dipole configuration, makes a circular luminal motion onto which a subluminal motion of the center of mass S is superimposed. The trajectory of the resulting motion is a screw-line, but it is the motion of S only which one identifies with the (time-averaged) electron velocity. The result derived from this simple pole- dipole model is in beautiful agreement with Schrödinger's analysis of the Dirac equation, 12 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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in which the luminal rotational motion emerges as a fluctuation of the electron
coordinate, called by Schrodinger Zitterbewegung (German for quivering motion).
We may analyze the simple pole-dipole model in more detail as follows. If the positive
and negative masses are m⁺, m⁻, with m⁺ - |m⁻| << m⁺, and with the mass pole given
by
m = m⁺ - |m⁻|, (38)
the mass dipole is
p = m⁺r ~ |m⁻|r (39)
where r << r_c is the distance of separation in between m⁺ and m⁻. The center of mass
is determined by
m⁺r_c = |m⁻|(r_c + r). (40)
The angular momentum then becomes (ω is the angular velocity of circular motion with
r_cω → c):
|J| = |[m⁺r_c² - |m⁻|(r_c + r)²]ω|
= m⁺r_crω
= m⁺rc (41)
Comparing (40) with (37) shows that
2m⁺rc = h (42)
and
m/m⁺ = r/r_c . (43)
13
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY Experimentally, the electron is indistinguishable from a point. This would make r = 0. In reality its size must be finite but in principle can be very small. This means that m⁺ (and |m⁻|) are likely to be very much larger than m. It has been conjectured by Hönl and Papapetrou [5] that the electron is a pole-dipole particle where the surplus positive energy comes from the positive gravitational interaction energy of a very large positive m⁺ mass with a likewise very large negative m⁻ = -|m⁺| mass. According to this hypothesis one would have for the electron rest mass energy mc² = γ|m'|²/r . (44) Combining (44) with (42) one can compute m'. The result is [5]: m' = ∛(mch/2γ) ~ 6 X 10⁻¹³ g , (45) larger by a factor 3.6×10¹¹ times the mass of the proton. We therefore see that there are huge amounts of negative masses bound to positive masses in Dirac spinors. It shows that it cannot be a simple matter to free the negative masses from the positive masses. And it explains why the masses of the elementary particles are so much smaller than the Planck mass, m_p ~ 10⁻⁵ g. 6. Planck Aether Hypothesis [7,8] We make here the proposition that the fundamental group is SU2, and that by Planck's conjecture the fundamental equations of physics contain as free parameters only the Planck length r_p, the Planck mass m_p and Planck time t_p (γ Newton's constant, h Planck's constant, c the velocity of light): 14 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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r_p = √(hγ/c³) ≈ 10⁻³³ cm
m_p = √(hc/γ) ≈ 10⁻⁵ g
t_p = √(hγ/c⁵) ≈ 10⁻⁴⁴ s .
The assumption that SU2 is the fundamental group means that nature works like a
computer with a binary number system. Since SU2 is isomorphic to SO₃, the rotation
group in R3, this explains why natural space is three-dimensional.
The Planck's aether conjecture is the assumption that the vacuum of space is densely
filled with an equal number of positive and negative Planck mass particles, with each
Planck length volume on the average occupied by one Planck mass, with the Planck
mass particles interacting with each other by the Planck force over a Planck length, and
with Planck mass particles of equal sign repelling and those of opposite sign attracting
each other. The particular choice made for the sign of the Planck force is the only one
that keeps the Planck aether stable. While Newton's action-reaction remains valid for
the interaction of equal Planck mass particles, it is violated for the interaction of a
positive with a negative Planck mass particle, even though globally the total linear
momentum of the Planck mass plasma is conserved, with the recoil absorbed by the
Planck aether as a whole.
It is the local violation of Newton's action-reaction which leads to quantum mechanics
at the most fundamental level, as can be seen as follows: Under the Planck
force F_p = m_pc²/r_p, the velocity fluctuation of a Planck mass particle interacting with a
Planck mass particle of opposite sign is Δv = (F_p/m_p)t_p = (c²/r_p)(r_p/c) = c, and hence
yields the momentum fluctuation Δp = m_pc. But since Δq = r_p and because m_pr_pc = h,
one obtains Heisenberg's uncertainty relation ΔpΔq = h for a Planck-mass particle.
Accordingly, the quantum fluctuations are explained by the interaction with hidden
negative masses, with energy borrowed from the sea of hidden negative masses.
15
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY According to Newtonian mechanics and Planck's conjecture, the interaction of a positive with a negative Planck-mass particle leads to a velocity fluctuation δv = a_pt_p = c with a displacement of the particle equal to δ = (1/2)a_pt_p² = r_p/2, where a_p = F_p/m_p. Therefore, a Planck-mass particle immersed in the Planck aether makes a stochastic quivering motion (Zitterbewegung) with the velocity v_D = -(r_pc/2)(∇n/n), (46) where n = 1/r_p³ is the average number density of positive or negative Planck mass particles. The kinetic energy of this diffusion process is given by (m_p/2)v_D² = (m_p/8)r_p²c²(∇n/n)² = (h²/8m_p)(∇n/n)². (47) Putting v = h/m_p ∇S , (48) where S is the Hamilton action function and v is the velocity of the Planck aether, the Lagrange density for the Planck aether is L = n[h∂S/∂t + h²/2m_p(∇S²) + U + h²/8m_p(∇n/n)²]. (49) Variation of (49) with regard to S according to ∂/∂t(∂L/∂(∂S/∂t)) + ∂/∂t(∂L/∂(∂S/∂r)) = 0 (50) leads to ∂n/∂t + h/m_p ∇(n∇S) = 0 (51) 16 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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or
∂n/∂t + ∇(nv) = 0 , (52)
which is the continuity equation of the Planck aether. Variation with regard to n
according to
(∂L/∂n) - ∂/∂r(∂L/∂(∂n/∂r)) = 0 (53)
leads to
h∂S/∂t + U + h²/2m_p(∇S²) + h²/4m_p[(1/2)(∇n/n)² - ∇²n/n] = 0 (54)
or
h∂S/∂t + U + h²/2m_p(∇S²) + h²/2m_p · ∇²√n/√n = 0 . (55)
With the Madelung transformation
ψ = √n e^{iS}, ψ* = √n e^{-iS} , (56)
one obtains from (51) and (55) the Schrödinger equation
ih∂ψ/∂t = -h²/2m_p ∇²ψ + Uψ . (57)
In the Planck aether hypothesis all particles, save and except the Planck mass particles,
are quasi-particles of the Planck aether, like the phonons, rotons, excitons, etc., of
condensed matter physics, and by the wave structure of the Planck aether are Lorentz
invariant. In forming quantized vortices, the Planck aether also has vortex waves,
simulating Maxwell's and Einstein's electromagnetic and gravitational waves. Dirac
spinors are made possible by the negative masses of the Planck aether.
17
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY Quantum mechanics predicts for each harmonic oscillator the zero-point energy (1/2)hω , which has to be multiplied with the volume element in frequency space 4πω²dω , to obtain the zero-point energy spectrum f(ω)dω = const × ω³dω. (58) Now (58) turns out to be just the only spectrum that is Lorentz invariant. But the spectrum (58) is also the only one which does not lead to a friction force on a charged particle moving through an electromagnetic spectrum with this frequency dependence. This means that special relativity is a consequence of quantum mechanics, leading to the zero point vacuum energy, and can be interpreted by saying that the zero-point vacuum energy "generates" the Minkowski space-time. 7. Dynamic Interpretation of Lorentz Invariance A cut-off at the Planck frequency generates a distinguished reference system in which the zero-point energy spectrum is isotropic and at rest. In this distinguished reference system, the scalar potential from which the forces are to be derived satisfies the inhomogeneous wave equation: -1/c² ∂²Φ/∂t² + ∇²Φ = -4πρ(r,t) , (59) where ρ(r,t) are the sources of this field. For a body in static equilibrium at rest in the distinguished reference system for which the sources are those of the body itself one has ∇²Φ = -4πρ(r). (60) If set into absolute motion with the velocity v along x-axis, the coordinates of the reference system at rest with the moving body are obtained by the Galilei transformation: 18 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY x' = x - vt, y' = y, z' = z, (61) t' = t transforming (59) into -1/c² ∂²Φ'/∂t'² + 2v/c² ∂²Φ'/∂x'∂t' + (1 - v²/c²) ∂²Φ'/∂x'² + ∂²Φ'/∂y'² + ∂²Φ'/∂z'² = -4πρ'(r',t'). (62) After the body has settled into a new equilibrium in which ∂/∂t' = 0 , one has instead of (60) (1 - v²/c²) ∂²Φ'/∂x'² + ∂²Φ'/∂y'² + ∂²Φ'/∂z'² = -4πρ(x', y, z). (63) Comparison of (63) with (60) shows that the left-hand side of (63) is the same if one sets Φ' = Φ and dx' = dx√(1-v²/c²) . This implies a uniform contraction of the body by the factor √(1-v²/c²) because the sources are contracted by the factor √(1-v²/c²) as well, whereby the right-hand side of (63) becomes equal to the right-hand side of (60). Since the zero-point energy is invariant under a Lorentz transformation, the quantum potential changes in the same way as Φ . The body therefore sustains its static equilibrium under a contraction by the factor √(1-v²/c²) if set into absolute motion, explaining the Lorentz contraction dynamically. The clock retardation effect can be derived from the contraction effect, and from there the Lorentz transformation. Following Builder [9] this original interpretation of Lorentz invariance by Lorentz and Poincaré has been worked out in every detail by Prokhovnik [10]. To derive the clock retardation effect from the contraction effect one considers a light clock, which is a rod with mirrors attached to its two ends in between which a light signal is sent forth and back. If the length of the rod is l, and if the rod rests in the distinguished reference system, the time needed for the light signal to be sent forth and back is 19 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY t₀ = 2l/c . (64) If prior to being set into motion the rod is inclined against the x-axis by the angle φ, it appears to be inclined against the x-axis by the different angle ψ after set into motion, with ψ expressed through φ by tan ψ = γ tan φ (65) γ = (1 - v² / c²)^(-1/2) . The absolute motion then contracts the rod from l to l': l' = l√(1-(v²/c²)cos²φ) = l / (γ√(1-(v²/c²)sin²ψ)) . (66) Relative to the moving rod the velocity of light is anisotropic, and for the to and fro directions given by c₋ = √(c² - v² sin² ψ) - v cos ψ c₋ = √(c² - v² sin² ψ) + v cos ψ (67) with the time t' for a to and fro signal given by t' = l'/c₊ + l'/c₋ = γt₀ . (68) Therefore, as seen from an observer at rest in the distinguished reference system the clock goes slower by the factor γ = 1/√(1-v²/c²) , independent of the inclination of the rod making up the clock. With solid bodies held together by electromagnetic forces, clocks made from solid matter should behave like light clocks. As it was claimed by Poincaré, it should for this reason be possible to obtain the Lorentz transformations solely from the contraction effect with a proper convention about the synchronization of clocks. According to Einstein, two clocks, A and B, are synchronized if 20 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY t_B = 1/2(t¹_A + t²_A), (69) where t¹_A is the time a light signal is emitted from A to B, reflected at B back to A, arriving at A at the time t²_A , and where it is assumed that the time t_B at which the reflection at B takes place is equal to the arithmetic average of t¹_A and t²_A . Only by making this assumption does the velocity of light turn out always to be isotropic and equal to c. From an absolute point of view, the following rather is true. If t_R is the absolute reflection time of the light signal at clock B, one has for the out and return journeys of the light signal from A to B and back to A, if measured by an observer in an absolute system at rest in the distinguished reference system: γ(t_R - t¹_A) = d/c. γ(t²_A - t_R) = d/c ' (70) where d is the distance between both clocks, and where c₊ and c₋ are given by (67). Adding the equations (70) one obtains c(t²_A - t¹_A) = 2γd√(1-(v²/c²)sin²ψ) . (71) If an observer at rest with the clock wants to measure the distance from A to B, he can measure the time it takes a light signal to go from A to B and back to A. If he assumes that the velocity of the light is constant and isotropic in all inertial reference systems, including the one he is in, moving together with A and B with the absolute velocity v, the distance is d' = (c/2)(t²_A - t¹_A). (72) And because of (71) d' = γd√(1-(v²/c²)sin²ψ) . (73) 21 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY Comparing this result with (66), one sees that he would obtain the same distance d', if he uses a contracted rod as a measuring stick, or Einsteins's constant light velocity postulate. The velocity of light between A and B by using a rod to measure the distance and the time it takes a light signal in going from A to B and back to A, of course, will turn out to be equal to c, because according to (72) 2d' / t²_A - t¹_A = c . (74) Rather than using a reflected light signal to measure the distance d', the observer at A may try to measure the one-way velocity of light by first synchronizing the clock B with A and then measure the time for a light signal to go from A to B. However, since this synchronization procedure also uses reflected light signals, the result is the same. For the velocity he finds d'/(t_B - t¹_A) = d'/(1/2(t¹_A + t²_A) - t¹_A) = 2d'/(t²_A - t¹_A) = c . (75) By subtracting the equations (70) one finds that t_R = t_B + (γ/c²)vd cos ψ , (76) which shows that from an absolute point of view the "true" reflection time t_R at clock B is only then equal to t_B , if v = 0. From an absolute point of view the propagation of light is isotropic only in the distinguished reference system, but anisotropic in a reference system in absolute motion against the distinguished reference system. This anisotropy remains hidden due to the impossibility to measure the one-way velocity of light. The impossibility is expressed in the Lorentz transformations themselves, containing the scalar c² rather than the vector c, through which an anisotropic light propagation would have to be expressed. 22 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY 8. Negative Mass Interpretation of the Aharonov-Bohm Effect [11] In Maxwell's equations the electric and magnetic fields can be expressed through a scalar potential Φ and a vector potential A: E = -1/c ∂A/∂t - gradΦ H = curlA (77) E and H remain unchanged under the gauge transformation of the potentials Φ' = Φ - 1/c ∂f/∂t A' = A + grad f (78) where f is called the gauge function. Imposing on Φ and A the Lorentz gauge condition, 1/c ∂Φ/∂t + divA = 0 , (79) the gauge function must satisfy the wave equation -1/c² ∂²f/∂t² + ∇²f = 0. (80) In an electromagnetic field the force on a charge e is F = e[E + 1/c v×H] = e[-1/c ∂A/∂t - gradΦ + 1/c v×curlA]. (81) By making a gauge transformation of the Hamilton operator in the Schrödinger wave equation, the wave function transforms as 23 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY ψ' = ψ exp[ie/hc f], (82) leaving invariant the probability density ψ*ψ . To give gauge invariance a hydrodynamic interpretation, we compare (81) with the force acting on a test body of mass m placed into the moving Planck aether. This force follows Euler's equation and is F = m dv/dt = m[∂v/∂t + grad(v²/c) - v×curlv]. (83) Complete analogy between (81) and (83) is established if one sets Φ = -m/2e v² A = -mc/e v (84) According to (78) and (82), Φ and A shift the phase of a Schrödinger wave by δφ = e/h ∫^t₂_t₁ Φdt δφ = -e/hc ∮ A · ds. (85) The corresponding expressions for a gravitational field can be directly obtained from the equivalence principle [3]. If ∂v/∂t is the acceleration and ω the angular velocity of the universe relative to a reference system assumed to be at rest, the inertial forces in this system are F = m[∂v/∂t + ω̇×r - ω×(ω×r) - r̈×2ω]. (86) For (86) we write 24 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY F = m[Ê + 1/c v×Ĥ], (87) where Ê = ∂v/∂t + ω̇×r - ω×(ω×r) Ĥ = -2cω. (88) With curl(ω̇×r) = 2ω̇ div(-ω×(ω×r)) = 2ω², (89) one has divĤ = 0 1/c ∂Ĥ/∂t + curlÊ = 0. (90) Ê and Ĥ can be derived from a scalar and vector potential Ê = -1/c ∂Â/∂t - gradΦ Ĥ = curlÂ. (91) Applied to a rotating reference system one has Φ̂ = -1/2 (ω×r)²  = -c(ω×r) (92) 25 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY or Φ̂ = -v²/2  = -cv (93) Apart from the factor m/e, this is same as (84). For weak gravitational fields produced by slowly moving matter, Einstein's linearized gravitational field equations permit the gauge condition (replacing the Lorentz gauge) 4/c ∂Φ/∂t + div = 0 ∂Â/∂t = 0. With the gauge transformation for Φ and A Φ' = Φ Â' =  + grad f (94) where f has to satisfy the potential equation ∇²f = 0. (95) For a stationary gravitational field the vector potential changes phase of the Schrödinger wave function according to ψ' = ψ exp[im/hc f], (96) leading to phase shift on a closed path δφ = -m/hc ∮  · ds 26 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY δφ = -m/ℏ ∮ v · ds (97) In the hydrodynamic interpretation suggested by the Planck aether hypothesis, the phase shifts caused by either the magnetic or the gravitational vector potential result from a circular flow of the Planck aether. The principle of equivalence can precisely relate this circular flow to the angular velocity of a rotating platform. According to (93), one has for the gravitational vector potential in a rotating frame of reference  = -ωcr with the phase shift given by (97). One can apply (97) to the Sagnac effect for photons of frequency v. By putting mc² = hv = 2πhv, with the result that δφ = -2πv/c² ∮ v · ds δφ = 2ωπr² 2πv/c² (97) the same as predicted by quantum mechanics. We now compute the phase shift (85) by a magnetic vector potential. To make a comparison with the gravitational vector potential in the Sagnac effect, we consider the magnetic field produced by an infinitely long cylindrical solenoid of radius R. Inside the solenoid the field is constant, vanishing outside. If the magnetic field inside the solenoid is H, the vector potential is A_φ = 1/2 Hr r < R A_φ = 1/2 HR²/r r > R. (98) According to (85) the vector potential on a closed path leads to the phase shift 27 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY δφ = -e/hc H πr² r < R δφ = -e/hc H πR² r > R. (99) As noted by Aharonov and Bohm [11], there is a phase shift for r > R, even though for r > R, H = 0 (because for r > R, curl A = 0). Expressing A by (84) through v, the hypothetical circular aether velocity, one has v_φ = -e/2mc Hr r < R v_φ = -e/2mc HR²/r r > R. (100) One sees that inside the coil the velocity profile is the same as in a rotating frame of reference, having outside the coil the form of potential vortex. If expressed in terms of the aether velocity the phase shift becomes δφ = -m/ℏ ∮ v · ds (101) the same as (97) for the vector potential created by a gravitational field, and hence the same as in the Sagnac experiment and neutron interference experiment. But for the magnetic vector potential the aether velocity can easily become much larger than in any rotating platform experiment. According to (102), the velocity reaches a maximum at r = R, where it is |v_max|/c = eHR/2mc². (102) For electrons this is |v_max|/c = 3×10⁻⁴ HR , where H is measured in Gauss. For H = 10⁴ G, this would mean that v_max ≥ c for R > 0.3 cm. If this would be same aether velocity felt on a rotating platform, it would lead to an enormous centrifugal and Coriolis field inside the coil, obviously not observed. 28 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY
The Planck aether model can give a simple explanation for this paradox. The Planck
aether consists of two superfluid components, one composed of positive Planck masses
and the other one of negative Planck masses. The two components can freely flow
through each other, making possible two configurations, one where both components
are corotating and one where they are counterrotating. The corotating configuration is
realized on a rotating platform, where it leads to the Sagnac and neutron interference
effects. This suggests that in the presence of the magnetic vector potential the two
superfluid components are counterrotating. Outside the coil, where curl A = 0, the
magnetic energy density vanishes, implying that the magnitudes of both velocities are
exactly the same. Inside the coil, where curl A ≠ 0, there must be a small imbalance in
the velocity of the positive over the negative Planck masses to result in a positive
energy density.
9. Negative Masses in Cosmology
In the Planck aether theory, the negative gravitational field energy surrounding a mass
is due to an excess of negative over positive mass, and the negative mass surrounding
a highly collapsed spherical body or black hole is of the same order of magnitude as the
positive mass accumulated inside the collapsed body. An assembly of interacting
positive and negative masses can, in general, not lead to a thermodynamic equilibrium.
If all the positive masses are separated from the negative ones, it is sufficient to
require that each mass species reaches thermodynamic equilibrium. It was shown by
Vysin [12] that an assembly of negative masses can acquire thermodynamic
equilibrium provided the temperature is negative. The kinetic energy of a negative
Planck mass is negative, and an assembly of negative Planck masses has, for this
reason, a negative temperature. It therefore can reach thermodynamic equilibrium.
We now make the following hypothesis:
If an assembly of positive and negative masses, with the total energy equal to
zero, is brought together, the temperature and hence entropy of the mixture will
go to zero.
29
UNCLASSIFIED//FOR OFFICIAL USE ONLYUNCLASSIFIED//FOR OFFICIAL USE ONLY This hypothesis is the only one consistent with Nernst's theorem. It, of course, implies that an assembly of positive and negative masses can perfectly mix because otherwise no equilibrium can be reached. To satisfy this hypothesis, we assume that the negative masses have negative entropy. Only the assumption that an assembly of negative masses has negative entropy permits an analytic continuation of the entropy from positive to negative temperatures. If the entropy for negative temperatures would be counted positive, the function dS/dT would be discontinuous at T = 0. For the entropy of a mixture of positive and negative masses to become zero requires an exact correlation in the disorder of the positive mass gas with the disorder of the negative mass gas. This is certainly true if the negative mass is equal to the negative mass of the gravitational field of the positive mass, because the Newtonian gravitational field of each particle, all the way down to the smallest dimension, is precisely correlated to the position of the particle. The entropy of the positive mass of matter and the entropy of the negative mass of its gravitational field (correlated to the entropy of the positive mass which is the source of this field) might therefore be called complementary (like a positive and negative photographic image). The expansion from a very small phase space volume would then be possible, because if the positive and negative masses are densely packed within the same volume, not only their energy, but also their entropy would cancel. The time needed to bring back the universe to its original low entropy state, is the Poincaré recurrence time. While under normal conditions this time is huge, it may in a dense mixture of positive and negative masses with a divergent acceleration become quite small. This might explain why the initial entropy of the universe is very small. The Planck aether hypothesis gives a plausible explanation for the observed vanishing of the sum of all charges, like the electric, color and weak charges. With the phenomenon of charge explained to result from the zero point fluctuations of Planck masses bound in vortex filaments and with an equal number of positive and negative Planck masses, the sum of all the charges must vanish. That this should be also be true for the gravitational charges finds its expression in the compensation of the positive energy in the universe by its negative gravitational energy. This compensation explains why the flatness parameter is Ω = 1. 30 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY Furthermore, with the sum of all gravitational charges equal to zero, the cosmological constant Λ playing the role of a kind of charge, demands that also Λ = 0. Finally, with the negative entropy of the negative Planck masses playing the role of a kind of photographic negative for the positive entropy of the positive Planck masses, the total entropy, made up from the sum of the positive and negative Planck masses, would also be equal to zero. In summary we may write ∑charges = 0 (103) with the cosmological consequence Ω - 1 = Λ = S = 0. (104) The horizon problem is here resolved by superluminal electromagnetic and gravitational shock waves during the high density phase of the cosmological evolution, not by an inflationary expansion of space. 10. The Cusp/Core Problem in Galactic Halos We have seen that there appears to be strong evidence for the existence of negative matter in the universe. And we have also given reasons that negative matter might be hidden behind positive matter, forming pole-dipole Dirac spinor configurations neutralizing the negative matter. This still leaves open the question as to whether in certain regions of space there might be a surplus of negative over positive matter, and if negative matter can be "mined" from such regions. It has been conjectured by Forward [13] that negative matter might be located in the intergalactic voids, explaining the "bubble" structure of the metagalaxy. The negative mass in the voids would produce gravitational potential hills repelling all matter, positive and negative, like the positive matter of the galaxies produces gravitational potential wells attracting all matter, positive and negative. But the accumulation of negative matter in the gravitational wells of positive matter reduces and flattens the 31 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY depth of the wells. This simple fact might explain the unsolved cusp/core problem of galactic halos [14]. But what happens in the center of galaxies must also happen to a lesser degree in the center of the sun, the planets, the earth and the moon. To "mine" negative matter, if it should exist there, excludes the sun, and also all planets, that have a hot molten core. This does not exclude the moon, however, having the deepest potential well near the earth, with only hot rocks in its center accessible by advanced nuclear mining technology, permitting in principle constructing a tunnel through the moon [15]. 11. Searching for Negative Matter in the Gravitational Potential Well of the Moon The radius of the moon is R = 1.74×10⁸cm and the gravitational acceleration at its surface is g₀ ≅ 1.62×10²cm/s². At a distance r < R from its center the gravitational acceleration is g(r) = g₀(r/R) (105) and the pressure balance equation is dp/dr = -ρg₀ r/R , (106) where p = p(r) is the pressure for r < R. The pressure at the center therefore is p_max = ρg₀/R ∫₀ᴿ rdr = 1/2 ρg₀R. (107) With ρ ≅ 3.33g/cm³ the average density of the moon, one finds that p_max ≅ 5×10¹⁰dyn/cm² ≅ 50,000atm. The temperature T can be estimated from the equation p_max = nkT, where k ≅ 1.38×10⁻¹⁶erg/K is the Boltzmann constant and n ≅ 10²³cm⁻³ the atomic number 32 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY density of the rocks. For p_max ≅ 5×10¹⁰dyn/cm² one finds T ≅ 4×10³K. Both the pressure and the temperature are technically manageable, the pressure with layers of shattered rocks around a tunnel passing through the center of the moon, and the temperature with some cooling. Seismic measurements suggest that the center of the moon is made up of hot rocks. If appreciable amounts of negative matter have accumulated over billions of years in the center of the moon, it is more likely that this matter is in the form of ultra-light matter, perhaps by an order of magnitude lighter than ordinary matter. There are indications that a Swedish research group has found evidence for the existence of an ultra-dense phase of deuterium, about more than 100,000 times more dense than water [16]. Suppose that in the center of the moon the accumulation of negative matter has led to a form of matter which is 100,000 times lighter than steel, but still has the strength of steel. This would not lead to a negative-positive mass self-chasing mass dipole as envisioned by Forward [13], but to something very important for space- flight, because it would dramatically reduce the energy requirements to accelerate a space craft made from such ultra-light material. The question as to whether there is such an unusual substance in the center of the moon can probably be answered by seismic wave tomography, obtained by nuclear explosions set off on the surface of the moon. 12. Making a Tunnel through the Moon [15] The cohesive energy of rocks is of the order ε_r ≈ 10¹⁰erg/cm³. Therefore, the explosive yield needed to shatter a spherical volume of radius r is E ≅ (4π/3)ε_r r³. (108) The energy released in a kiloton nuclear explosion is E ≈ 4×10¹⁹erg. With this energy, the radius of the crushed rocks would be r ≈ 10³cm (= 10 m), and with a 10 kiloton explosion it would be twice as large. 33 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY To make a tunnel, a cylindrical, rather than spherical, volume of crushed rocks is desired. For this reason a thermonuclear shape charge or an explosive lens is better suited to shatter the rocks. In the center of the moon the temperature is several thousand degrees centigrade. With the heat diffusion equation given by ∂T/∂t = χ∇²T , (109) where χ is the heat diffusion coefficient, the diffusion time for a layer of thickness x is τ = x²/χ. (110) For lunar rocks one has χ ≅ 4×10⁻³cm²/s. Taking the example x = 20m = 2×10³cm, one finds that τ = 10⁹s ≅ 30yr, and at the high rock temperatures the heat diffusion time would be uncomfortably long. The situation is drastically changed for a layer of crushed rocks, because there it is possible to remove heat by a coolant pumped through the porous medium of the crushed rocks. At the high temperatures of several thousand degrees centigrade, a liquid alkali metal, for example lithium, abundantly available on the moon, could be used as a coolant. The velocity the coolant diffuses into the crushed rocks is determined by Darcy's law v = -D grad p , (111) where p is the pressure, D = κ/ρg, with κ ~ 1 cm/s a typical value. If the pressure gradient is provided by the gravitational force one has grad p = ρg and hence |v| = κ ~ 1 cm/s. (112) 34 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY The time needed for the liquid metal to pass through a ~ 20 m thick layer is then ~ 2 × 10³ seconds ~ 1 hour. The specific heat per unit volume of the coolant is ρc_v ~ 3 × 10⁷ erg/cm³K, and for T = 3 × 10³ K one has ρc_vT ~ 10¹¹ erg/cm³. The heat per unit volume which has to be removed from the crushed rocks is of the order p, where p is the rock pressure. In the center of the moon where p = 5 × 10¹⁰ dyn/cm², this energy is 5 × 10¹⁰ erg/cm³. It thus follows that the volume of the liquid coolant must be about ½ of the rock volume to be cooled. For a rock volume of (20 cm)³ ~ 10⁴ m³, a coolant volume of about 5 × 10³ cm³ would be needed. The same coolant can be used many times over after the heat is removed from it, which could be done on the surface of the moon by radiation or perhaps better by heat exchangers transferring the heat to lunar shattered rocks surrounding the tunnel, the pressure acting on the tunnel wall would be large, in particular in the center of the moon. Because of friction between particles of the shattered rock, large shear stresses can be sustained changing the pressure distribution in the rock and reducing the pressure gradient and hence the pressure on the tunnel wall. A more detailed calculation [15] for the pressure distribution in the shattered rock tunnel wall gives p = (r/r₀)⁹, (113) where r₀ is the radius of the tunnel. Integrating eqn (108) one obtains for the pressure distribution in the moon p(r) = -ρg₀/R ∫ᴿ_r rdr = -ρg₀/R(R² - r²) (114) for which one can also write p(r) = p_max(1-(r/R)²). (115) 35 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY If r_s is the horizontal radius up to which the rocks at a certain depth have to be shattered (cylindrical case), one finds by equating p(r) in (115) and (117) r_s = r₀(p_max/p₀)^(1/9)(1-(r/R)²)^(1/9). (116) The total energy required to shatter the rocks to make a tunnel from the center at the moon where r = 0, to its surface where r = R, is then given by summing up over the slices with radius r_s and thickness dr E = πr₀²ε_r(p_max/p₀)^(2/9) ∫₀ᴿ(1-(r/R)²)^(2/9) dr , (117) where as in eqn (110) ε_r ≅ 10¹⁰erg/cm³ is the cohesive binding energy of the rocks. For (119) one can write E = πr₀²Rε_r(p_max/p₀)^(2/9) ∫₀¹(1-x²)^(7/9) dx. (118) With the help of Euler's betafunction one has ∫₀¹(1-(x)²)^(7/9) dx = (1/2)B(1/2, 11/9) ≅ √π/2. (119) Hence, E ≅ (π^(3/2)/2)r₀²Rε_r(p_max/p₀)^(2/9). (120) Inserting r₀ = 2×10³cm, R = 1.74×10⁸cm, ε_r ≅ 10¹⁰erg/cm³, (p_max/p₀) = 5×10⁴, one finds that E ≅ 2×10²⁴erg ≅ 5×10³ kiloton = 50 Megaton. It must be emphasized that this energy must be quite nonuniformly released along the tunnel shaft. Nuclear fusion explosions below a yield of 10 kiloton become uneconomical with only a fraction of the energy in the fissionable material (needed to 36 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY make a critical assembly) released. For a 10 kiloton fission explosion the shatter radius computed from (110) is ~ 20 m. With a tunnel radius r₀ ~ 10 m one would have (r_s/r₀) ~ 2 and from (118) that 1-(r/R) ≅ 2⁹(p₀/p_max). (121) Putting for the depth of the tunnel (if measured from the surface of the moon) δ = R - r, with 2δ/R = 2⁹(p₀/p_max) ≅ 10⁻² (122) or that δ ≅ 10 km. For a depth < 10 km the nuclear explosion with a yield < 10 kiloton would suffice, a yield which is uneconomical. It is for this reason suggested that one uses altogether thermonuclear explosive devices where the cost per yield is much lower. To penetrate and shatter the rocks more efficiently, jet-generating thermonuclear explosive lenses could be used. The thermonuclear detonation wave ignited at one point is there shaped into a jet-producing conical explosion by placing obstacles in the path of the wave. The ignition can be done by a fission explosive, but conceivably also by a powerful laser beam, with the laser beam projected down the tunnel shaft, triggering the thermonuclear explosive positioned at the lower end. With the above-given estimate of ~ 50 Megaton needed to dig the tunnel shaft, the number of thermonuclear explosive devices making use of the detonation wave lens technique could for this reason be quite reasonable, and certainly much less than the number of required fission explosives. After nuclear explosions have crushed the rocks and the heat is removed, the tunnel wall has to be made from some kind of ceramic material, since water with which to make concrete is only sparsely available on the moon. But for the wall to last, its temperature must be kept low. The low heat conductivity of rocks, requiring little cooling, is there of considerable help. For the crushed rocks the heat conduction coefficient should not be very different than for solid rocks. According to eqn (112) the 37 UNCLASSIFIED//FOR OFFICIAL USE ONLY
UNCLASSIFIED//FOR OFFICIAL USE ONLY heat diffusion time for a 20 m layer of rocks is ~ 30 years. This means that a small, continuous removal of the heat through the injection and circulation of a liquid metal into the crushed rocks should keep down the temperature of the tunnel wall and its environment. 13. Conclusion The purpose of this study is the question as to whether negative mass might exist, and if negative mass propulsion is possible at all. It is unlikely to be possible in the fashion speculated by Forward [13] (but also see Winterberg [17]), where a negative mass is chasing a positive mass without the expenditure of any energy. Rather, it might perhaps be possible through the existence of an ultra-light form of matter with the tensile strength of ordinary matter on a macroscopic scale where positive matter is bound to negative matter, as happens with Dirac spinor particles on a microscopic scale. This is the speculative existence of macroscopic bodies approaching zero rest mass, of importance for space flight because such matter would greatly reduce its energy requirements. 38 UNCLASSIFIED//FOR OFFICIAL USE ONLY
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