I think such a gravitation may even create artificial black holes. Here are my calculations. Did I make a mistake? Just for the case someone is interested in that topic and likes to think about it. This gravitation press is my favourite mindgame at the moment.
To remind you: I questioned if it could be possible to compress a massive iron globe to a massive stellar object by using the Teller-Ulam principle. One could shoot thousands of fusion detonators from all directions to form a sphere of plasma around a metal globe that would implode. With a staggered approach it would be possible to a) control and damp the pressure shock slope to prevent simple thermal explosion of the globe b) only to use complicated bombs for the first wave, the second wave and so on could be simple metal cylinders, that are triggered from the previous ones c) a certain control of the concentration of the pressure in the center of the object
Our sun is not heavy enough to become a neutron star or a black hole at the end of it’s lifetime after it has consumed all it’s fusion fuel and collapses. We don’t have this power, but if some extraterrestrials had a technology to compress it additionally from outside with the appropriate pressure it would of course become a black hole. This can be derived directly from Einsteins law of gravitation and it’s energy-momentum tensor. It shows us that any form of energy becomes part of the gravitational field – not only rest mass energy.
Ricci-Tensor and it’s non-relativistic Limit
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For the non-relativistic limit the energy-momentum tensor is normally calculated in the following way. We start with the definition of the Ricci-Tensor
R_kn = D_n C_jkj – D_j C_jkn + C_pkj C_jpn – C_pkn C_jpj (1)
where C_mnp is the Christoffel Symbol
C_mnp = 0.5 g_mk (Dg_kn/Dx_p + Dg_pk/Dx_n + Dg_np/Dx_k) (2)
D is the partial derivative symbol, x_n is the four dimensional coordinate. In the non-relativistic limit the approximate metric is given by
ds^2 ~= (1 + 2 Phi/c^2) dx_0^2 + g_ab dx_a dx_b (3)
with the gravitational potential Phi, the time coordinate ‘_0′ and the space coordinates ‘_a’, ‘_b’. We see
g_00 ~= (1 + 2 Phi/c^2) (4)
Thus the only non-trivial equation for the Ricci-Tensor is
R_00 = D_0 C_j0j – D_j C_j00 + C_p0j C_jp0 – C_p00 C_jpj ~= – D_a C_a00 (5)
because the first term vanishes due to the static nature of the metric and the last two terms are approx. zero. For a static gravitational field the Cristoffel symbol can be calculated as
C_n00 = -0.5 g_nk Dg_00/Dx_k (6)
and with Dg00/Dx_0 = 0 it is
C_a00 = -0.5 g_ab Dg_00/Dx_b (7)
n,k are the four-dimensional coordinates and a,b the three dimensional coordinates.
(4) in (7) leads to
C_a00 = – 1/c^2 D_a Phi (8)
(8) in (5) let us obtain
R_00 ~= 1/c^2 D_a D_a Phi = 1/c^2 H^2 Phi (9)
where H = D_n is the Hamilton operator in three-dimensional Euclidean metric space.
Energy-Momentum Tensor
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The general and total energy-momentum tensor T_kn in general relativity for a perfect fluid is
T_kn = (p + rho c^2) U_k U_n – g_kn p (10)
where k and n are the indices, p is the pressure rho is the mass density, U_n is the four-velocity vector and g_nm is the metric of the space-time. The non-relativistic case mostly with negligible gravitation pressure p << rho c^2, g_00 ~= 1 becomes
T_00 ~= rho c^2, T ~= rho c^2 (11)
In most practical calculations this is sufficient and the contribution from the pressure can be neglected. But not in our case, where we want to squeeze an object into infinity with ablation pressure
T_00 ~= rho c^2, T ~= p g_kn U_k U_n – g_kn g_kn p = rho c^2 – 3p (12)
Einsteins General Law of Gravitation
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The gravitational law can be written with contracted energy-momentum tensor
R_kn = – (8 pi G / c^4) (T_kn – 0.5 g_kn T) (13)
or with a contracted Ricci-Tensor Rkn, the Ricci-Scalar R
G_kn = R_kn – 0.5 g_kn R = - (8 pi G / c^4) T_kn (14)
Here G = 6.67e-11 Nm2/kg2 is the universal gravitation constant. G_kn is called the Einstein-Tensor.
Newtons Special Case Law of Gravitation
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From (13), (11), (9), and g_00 ~= 1, with k=n=0 we obtain
T_00 – 0.5 g_00 T = rho c^2 – 0.5 rho c^2 = 0.5 rho c^2 (15)
R_00 = 1/c^2 H^2 Phi = – (8 pi G / c^4) 0.5 rho c^2 (16)
H^2 Phi = – 4 pi G rho (17)
The solution of this differential equation is
Phi(r) = – G V-Int( rho dV/r) (18)
Proof by inserting:
Phi(r) = – G V-Int( rho dV/r) = – G Int( rho A(r) dr /r)
D/Dr(Phi) = – G rho A(r) /r = – G rho 4 pi r^2 /r = -G rho 4 pi r
D/Dr(D/Dr(Phi)) = – 4 pi G rho
Integrating (18) with uniform mass distribution over the volume results in
Phi = – GM/r (19)
which is Newton’s law of gravitation. With
F = m g = – m Grad(Phi) (20)
we get
F = m g = G m M / r^2 (21)
Special Case Law of Gravitation for a Gravitation Press
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From (13), (12), (9), and g_00 ~= 1, with k=n=0 we obtain
T_00 – 0.5 g_00 T = rho c^2 – 0.5 rho c^2 + 1.5 p (22)
R_00 = 1/c^2 H^2 Phi = – (8 pi G / c^4) (0.5 rho c^2 + 1.5 p) (23)
H^2 Phi = – 4 pi G rho (1 + 3 p / (rho c^2)) (24)
The solution of this differential equation is
Phi(r) = – G V-Int( rho dV/r) – 3 p G/c^2 V-Int(dV/r) (25)
We can proof this again by inserting and see that it is a solution of the differential equation.
Integrating (25) with constant density within the volume results in
Phi = – GM/r – 3 p G V/(r c^2) (26)
With
F = m g = – m Grad(Phi) (27)
we get
F = m g = G m M / r^2 + 3 p G m V/(r^2 c^2) (28)
Event Horizon
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To form a miniature black hole, the surrounding matter must be compressed to such high densities that it’s circumference is smaller than the event horizon, also known as Schwarzschild radius. The Schwarzschild radius is the point where the escape velocity equals the speed of light. The escape velocity for a body is in the Newtonian field
E_kin + E_pot = 0 (29)
0.5 m v_esc^2 – GMm/r = 0
v_esc = sqrt(2GM/r) (30)
The escape velocity in the Gravitation Press or Shark field is
E_kin + E_pot = 0
0.5 m v_esc^2 – GMm/r – 3 p G V m/(r c^2) = 0 (31)
v_esc = sqrt(2GM/r + 6pGV/(r c^2)) (32)
G = 6.67e-11 Nm2/kg2 is the universal gravitation constant. M is the mass of a spherical body and r is the radius from the center of the body, p is the pressure, V is the volume of the body rho is the density of the body. To find the Schwarzschild radius one has simply to replace v_esc with the speed of light:
r_sch = 2GM/c^2 (33)
r_sch = 2GM/c^2 + 6pGV/c^4 (34)
(33) is well known (34) is a variant that is appropriate for our problem.
Lowest Energy Limit
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The volume of an iron globe of 250 m radius would be
V = r^3 * pi * 4/3 = 65e6 m3 (35)
With the density of iron 7.8 g/cm3 it is 5e11 kg. Using Einstein’s rest mass energy equivalence formula
E=mc^2 (36)
the mass or 5e11 kg of the 250 m diameter iron globe is equivalet to an energy of 4.5e28 J. I assume that to squeeze the mass of a body to an infinitely small point at least the same energy the mass is equivalent to (36) is needed. If we provide fusion energy to squeeze a 250 m radius iron globe we need at least
m = E/0.003c^2 = 4.5e28 J /0.003c^2 = 1.2e14 kg (37)
fusion fuel, i.e. deuterium, for such a prozess. This is because 0.003 of the fraction of mass is converted to energy at nuclear fusion. Or we can also say: for any kg black hole we need at least 1/0.003 kg = 333 kg deuterium fuel to squeeze it to infinity.
For a 250m radius iron globe this would be a correspondend liquid deuterium sphere with deuterium density of 70 kg/m3 of 8300 m radius and 1.2e14 kg mass. But as I said, after the first wave of detonations any other material that is able to produce fusion energy can be uses. Because only a maximum of 0.3 % of the plasma is converted into energy, there is 0.997 * 1.2e14 kg of additional plasma mass.
Event horizon without pressure
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The iron globe weighs 5e11 kg. This gives with G = 6.67e-11 Nm2/kg2 and
r_sch = 2GM/c^2 = 7.5e-16 m
or 0.75 fm (femtometer) for the Schwarzschild radius. 1.5 fm diameter is the size of a single proton. For example using the mass of the Sun as 2e30 kilograms, gives theoretically 2964 meters for the Schwarzschild radius of a solar mass black hole. But our sun can never become a black hole, it is too small for that.
The density for transforming a 250 m radius iron globe of 65e6 m3 into a black hole of the diameter of 1 neutron, 1.5 fm, with 1.8e-45 m3 volume is 5e11 kg / 1.8e-45 m3 = 2.8e56 kg/m3 = 2.8e53 g/cm3. For the sun with 2e30 kg and 2964 m Schwarzschild radius it would be 1.8e19 kg/m3 = 1.8e16 g/cm3.
The plasma creates a sphere around the globe and the plasma is faster (has more kinetic energy) than the vapourizing plasma from the surface of the sphere, so I assume that half of the plasma of 1.2e14 kg is becoming part of the mass of the massive object, that is 6e13 kg. The new event horizon is then
r_sch = 2GM/c^2 = 9e-14 m
It is at least bigger than an atom now. The density for transforming 6.05e13 kg into a black hole of the diameter of 9e-14 m with 3e-39 m3 volume is 6.05e13 kg / 3e-39 m3 = 2e52 kg/m3 = 2e49 g/cm3.
Event horizon with pressure
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After all detonators are fired the object has a mass of 6.05e13 kg, it has started to implode but due to the additional mass from all detonations I assume that the Volume is still the same at this time. It should be possible to calculate a fast detonation rate that let’s the radius and volume be constant for a short time. The maximum pressure that fusion detonations can provide can be calculated as follows. The ablation pressure on a Teller-Ulam tamper in a hohlraum is typically
P = m_evap_rate * v_ex
and because both, the mass evaporation rate and the plasma expansion velocity, are only functions of temperature, it is possible to reduce the ablation pressure formula for a tamper in a hohlraum to
P = 0.3 E[eV]^3.5 [bar]
E ist the energy in eV. For fusion reactions we have typical energy outputs of 4.8MeV. That means with fusion it would be possible to achieve this energy in a hohlraum, if the hohlraum is very small. An alternative to a small hohlraum is to increase the number of energy sources. With 4.8MeV we get
P = 0.3 * 4.8e6^3.5 = 7.3e22 bar
or 7.3e27 Pa. We can use this value for calculating the Schwarzschild radius
r_sch = 2GM/c^2 + 6pGV/c^4 = 9e-14 m + 2.3e-8 m (!) = 2.3e-8 m
This means the radius of the event horizon is 260,000 times bigger than without that tremendous pressure from outside. The density for transforming 6.05e13 kg into a black hole of the diameter of 2.3e-8 m with 5.1e-23 m3 volume is then ‘only’ 6.05e13 kg / 5.1e-23 m3 = 1.2e36 kg/m3 = 1.2e33 g/cm3.
First Phase – White Dwarf
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The atom mass of iron is 55.8 u * 1.66e-24 g/u = 9.26e-23 g. The density for a typical white dwarf is 1e9 kg/m3 = 1e6 g/cm3. This means for iron 1.1e28 atoms would be squeezed into one cm3. Approximately half of the atom’s fermions are protons. The electron density would be therefor n = 3e29 electrons/cm3. The Fermi pressure in bar inside the white dwarf would become theoretically
P_fermi = 2.34e-33 * n^(5/3) = 3.2e16 bar (38)
The Fermi energy in eV inside the dwarf would become
E_fermi = 3.65e-15 n^(2/3) = 164 keV (39)
And the Fermi temperature within the artificial dwarf star
T_fermi = E_fermi/k = E_fermi/8.62e-5 = 2e9 K (40)
or 2 billion degrees Celsius/Kelvin/Fahrenheit with the Boltzman constant k = 8.62e-5 eV/K.
Due to the mass of the globe of 6.05e13 kg the volume of the white dwarf star that is created from that mass will be 60,500 m3 and the radius 29 m. The geometrical amplification factor is (250/29)^3 = 640. If the pressure on the surface of the globe was 7.3e22 bar it is 4.7e25 bar on the surface of the white dwarf. That is more than enough to let it implode further.
Second Phase – Neutron Star
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The atom mass of iron is 55.8 u * 1.66e-24 g/u = 9.26e-23 g. The density for a typical neutron star is 1e18 kg/m3 = 1e15 g/cm3. This means for iron, after all protons have been transformed to neutrons, 1.1e37 neutrons would be squeezed into one cm3. The neutron density would be therefor n = 1.1e37 neutrons/cm3. The Fermi pressure in bar inside the white dwarf would become theoretically
P_fermi = 2.34e-33 * n^(5/3) = 1.3e29 bar
The Fermi energy in eV inside the dwarf would become
E_fermi = 3.65e-15 n^(2/3) = 1.8e10 eV
And the Fermi temperature within the artificial dwarf star
T_fermi = E_fermi/k = E_fermi/8.62e-5 = 2e14 K
or 200 trillion degrees Celsius/Kelvin/Fahrenheit with the Boltzman constant k = 8.62e-5 eV/K.
Due to the mass of the globe of 6.05e13 kg the volume of the neutron star that is created from that mass will be 6.05e-5 m3 and the radius 29 mm. The geometrical amplification factor is (250/0.029)^3 = 6.4e11. If the pressure on the surface of the globe was 7.3e22 bar it is 4.7e34 bar on the surface of the white dwarf. That is more than enough to let it implode further.
Third Phase – Black Hole
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For the density of the black hole of 1.2e36 kg/m3 = 1.2e33 g/cm3 the relativistic energy density is 1.1e53 J/m3 and therefor the pressure 1.1e53 Nm/m2 = 1.1e48 bar.
Which geometrical amplification factor do we need to reach the pressure that is needed for the formation of the black hole?
1.1e48 bar/ 7.3e22 bar = 1.5e25
That means the radius of the region that has a sufficient density for a formation to a black hole is
r = 250 m / (1.5e25)^1/3 = 1e-6 m
That is a 44 times bigger radius than the Schwarzschild radius of 2.3e-8 m. This means the formation of the black hole would work. One could reduce the theoretical maximum pressure on the surface of the globe of 7.3e22 bar to a lower technically more feasible level [7].
Hawking radiation
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A formula for the evaporation time exists, it is the time until all mass of a black hole has evaporated due to this effect:
t_ev = 5120 * pi * G^2 * M^3 / h * c = 8.4e-17 * M^3 [s/kg3]
where h is the Planck constant. A black hole with the mass of the sun, 2e30 kg, with an event horizon of 2964 m would therefor evaporate 6.7e74 sec or in 2e67 years, much longer than the universe existst. For a black hole of a 500 m diameter iron globe with the additional bomb mass of together 6.05e13 kg is is ‘only’ 2e25 seconds or still much longer than the universe exists.
Let’s assume our machine may build black holes that exist at least as long as the universe, it is with the standard theory 13.73e9 years or 4.3e17 s, then the minimum mass of our black holes the Shark should build are
M = (1.2e16 * t_ev)^1/3 = (1.2e16 * 4.3e17)^1/3 = 1.7e11 kg
If we allow only 100 years – that should be enough for an economic power plant – or 31.5e6 s then the minimum mass of the black hole is 72e6 kg or an iron globe of 4.6 m diameter and 432 metric tons weight on which a mass of 143,136 tons had been fired and half of it became additional mass. Because the flying platforms I mentioned, that are propelled by pure fusion pulse detonations, will have a payload of 200,000 metric tons such a machine will be of no problem at all after the advent of civilian pure fusion detonations. Even if the material that is needed for the bomb cylinders, space stations and space harbour, is several times heavyer.
Schwarzschild Space-Time Distortion
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The Schwarzschild space-time metric for a centrally spherical symmetric massive object is given by
ds^2 = (1 – r_sch/r) c^2 dt^2 - (1 – r_sch/r)^-1 dr^2 – r^2 (dtheta^2 + sin^2theta dphi^2)
If we compare it with the flat pseudo-Euclidean space-time metric
ds^2 = c^2 dt^2 – dR^2 – R^2 (dtheta^2 + sin^2theta dphi^2)
we see that in the flat space-time metric the radial coordinate is the direct measure of the radial distance from the origin of the coordinates. This is not the case for the Scharzschild metric for a black hole. If we compare the second term of the metrics
dR^2 = (1 – r_sch/r)^-1 dr^2
and integrate dR we get
R_12 = Int_r1r2( (1 – r_sch/r)^-1/2 ) dr) = [ sqrt( r (r-r_schw) ) + r_sch ln( sqrt(r) + sqrt(r-r_sch) ) ]_r1r2 > R2 – R1
The last inequality is crucial. It means that only for a small r_sch/r or a big r the distance between points in a flat space and distorted space are equal. If r approaches r_sch the distance in the distorted space becomes bigger. That means when looking to the spot of the black hole from outside, the width and height of the space are the same as in flat space, but the length becomes allways longer when approaching the event horizon. It also means the volume of a sphere that is approaching the event horizon or Schwartzschild radius becomes bigger than in the 3-dimensional flat space.
Would that mean that the volume of the small neutron star increases and the pressure to let it implode becomes lesser with 1/V as (36) shows us:
p = (c^2 / 3V) (c^2r_sch/2G – M)
In principle yes, but the volume increase will be very small because the neutron star is still very big in relation to it’s Schwarzschild radius that the effect is negligible.
Open Questions
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a) not sure with calculating the Fermi pressures, especially in the neutron star
b) not sure calculating the minimum matter densities for the formation of a black hole
c) not sure if I'm right that half of the added mass from the detonators would add to the mass of the object
d) will the white dwarf or neutron star allow a pressure gradient in the inside or will it more or less of homogenous pressure due to quantum effects