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Creating artificial ultradense objects by ablation implosion?

  1. Sep 28, 2012 #1
    An interesting mindgame, I hope.

    I'm wondering if it could be possible to build a "machine" for creating even higher densities than that of a smaller star.

    For example one could take a 491 m radius massive iron globe. It doesn't matter how to build it. You could surround it with a sphere of hydrogen detonators that would vapourice the outer surface as fast that you get a typical 1e15 Pa of ablation pressure.

    The spherical shock wave of the imploding surface would travel with the speed of sound of iron (4910 m/s) in 0.1 s to the center of the globe where it would amplify geometrically with the factor (491/0.1)^2 = 24.1e6 in a 10cm radius sphere. Nuclear fusion could not occur, because the whole globe is of iron.

    The shock wave would be reflected in the center and travel back to the surface, where it is partly reflected due to phase transition. Normally the outer layer of the globe would now explode, but..

    After ca. 0.2 s a new spherical wave of staggered nuclear fusion explosions would be exactly triggered (considering the loss of material due to ablation) that the new shock wave superimposes the old. This new ablation induced pressure impulse should also hinder the outer material of the iron sphere to explode. It should be theoretical possible to increase the shock wave pressure from blast to blast.

    Let's say 50% of the shock wave would be reflected at the surface. Then it would be still possible to increase the pressure (neglecting the flattening of the shock wave peak with time) by a factor of 1.5^100 = 4e17 with 100 spherical blasts from outside.

    This means after 100 blasts or 20 seconds one would have theoretically a pressure in the 20 cm diameter central region of the iron globe of 1e15 * 24.1e6 * 4e17 = 9.64e39 Pa. That seems to be enough for a neutron star.

    What do You think? Where is the error in reasening? Is it theoretically possible to build an artificial 20cm diameter neutron star in this way?
  2. jcsd
  3. Sep 28, 2012 #2
    Erratum: geometrically factor for a sphere should be (491/0.1)^3 = 1.2e11

    Addendum: It's possible to increase the induced ablation pressure to certain value by increasing the heat input per cm^2 and therfor the vapourized mass (but not by the exhaust velocity). But after a while it runs into a classical impulse problem, I think, where the plasma layer from outside is the first opponent and the wave front from inside is the second opponent that is additionally also vapouriced by it's own pressure, I guess. Will a certain amount of the wave travel back to the center or will it loose all it's energy at the surface?
  4. Sep 28, 2012 #3


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    This is not true. Fusion of iron could occur in your sphere, but it would use up energy instead of give it off.
  5. Sep 28, 2012 #4
    How about this: The crucial phase transision zone may be WITHIN the globe. From the first implosion on, a certain radius of the center will become plasma. At the transision zone between this inner plasma and the outer solid iron, the reflection of the shock waves occurs, that determine the triggering rate of the spherical explosions from outside to let the amplitude grow from pulse to pulse.

    The inner plasma sphere also grows from pulse to pulse. When the inner plasma sphere reaches the limiting surface of the steel globe (that was ablating from pulse to pulse and shrinking), no more impulse feeding to increase the core pressure is possible. How high can the core pressure become?
  6. Sep 28, 2012 #5
    I think even if you can momentarily compress material to those densities, it will immediately expand back.
  7. Sep 28, 2012 #6
    That's ok. As long as the pressure is able to grow further.
  8. Sep 28, 2012 #7
    You are right. I think that is the reflecting shock wave from the core, I mentioned, that would normally blast away the surface of the globe.

    That was exactly what I first wanted to prevent. But now I think it's impossible. At least after a while.

    Now, the theoretical question would be: is it possible to induce new implosion shocks, even so. The goal would be now, to amplify the innerst oscillating shock wave, that is running between the core and the growing plasma/solid transition zone. If a part of the outer surface is blasted away, doesn't matter, as long it is still possible to induce mechanical impulse. But how?
  9. Sep 28, 2012 #8
    Uh, I just read that weapon designers do the thing I'm talking about on a smaller scale: they cascade chemical explosives spherically to increase the density of fissile material up to 4 times for reducing the critical mass.

    So, it MUST be possible to do the compression with eigenfrequency of a spherical wave at least for a certain time with that mindgame-device before it blasts in all directions. I only have to find out for how long and if this time is sufficient for temporally creating a white dwarf, a neutron star or other superdense celestial objects.

    Does anyone wan't to help me?

    Let's begin:
    - is the iron globe stiff or is it rather elastic and oscillates?
    - does is vapourize during the shock wave runs through it?
    - or does it only vapourize locally and immediately hardening or liquifying again?
    - how much of the shock wave will be dissipated into heat?
    - how much of the shock wave will be reflected from the surface?
    - how much of the surface will be blasted away from the shock wave alone?
    - what happens in the core?

    Do you know other questions of interest?
  10. Sep 28, 2012 #9


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    I think your line of reasoning has several points which do not work:
    And infinity in the center? While there is some amplification, your sphere and the shock waves are not perfect. I would be surprised to see an amplification of a factor of 25 million.

    As already posted by Drakkith, fusion would occur and reduce both energy density and pressure. You could use lead instead (uranium is a bit tricky with fission).

    Engineering question: 0.2s ago, you disintegrated everything around the sphere with several million nuclear explosions. How exactly do you get new explosives into that ultra-hot, expanding plasma of [strike]iron[/strike] lead?

    Neglecting the engineering issue: No. If every stage gives 50% of the old value plus some fixed new value, it tends towards an equilibrium at twice the added value. In addition, I don't see where this reflection should occur in your big ball of hot plasma.
  11. Sep 28, 2012 #10
    Thank You for Your interest. Your question is simple to answer. The detonators fly fast enough that 0.2s is enough distant not to get blast in parts. The hot plasma of the last explosions is not a problem for them, if they are not to long inside the plasma. In a fraction of a second only a thin layer of their steel casings will vapourize.
  12. Sep 28, 2012 #11
    1. ;)

    2. Why are You sure the whole iron globe would become a plasma?
  13. Sep 28, 2012 #12


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    I'll just say this. While it may be theoretically possible to compress a sphere to the density of neutron star matter, the engineering challenges make it entirely unfeasible currently. (As well as economic, social, and political issues) How would you even hold the sphere up, as that would require supports that would throw off your shockwaves.Furthermore, what's the point? You've detonated a bunch of nukes to momentarily create a small volume of super compressed material that then immediately loses that compression. Keep in mind that we can't even achieve the compression of a tritium-deuterium pellet without extremely precise lasers and a huge facility. I find it extraordinarily unlikely that your idea would work as you think it would.

    I'm all for helping people learn science, but this is simply a ridiculous scenario to try to learn from.
  14. Sep 28, 2012 #13


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    Rather than mess around with globes of iron and nuclear explosions would a particle accelerator not work? Not necessarily one we have now but theoretically.
  15. Sep 28, 2012 #14
    1. I thought it could be a good starting point for playing theoretically with the phenomenons inside a star without the need of gravitation!! Why must a mindgame be realistic or technically feasible?

    2. Or think about it how to produce possibly _great_ quantities of exotic matter..

    3. By the way: what would happen to the mini-neutron star, after the pressure around it vanishes again. Would it still exist or immediately explode?
  16. Sep 28, 2012 #15


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    Why not use a particle accelerator? I genuinely want to know if I'm missing something.
    It would explode. Neutron stars are held together by their immense gravity. Without that the matter would violently expand.
  17. Sep 28, 2012 #16


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    Because you are coming up with ideas that simply don't make sense, that's why. If you want to ask what would happen if we could compress a large iron sphere, irregardless of how we do it, then that's different. But you are suggesting methods of creating this compression which are not only unfeasible, but unrealistic to the point of bordering on absurd. May I suggest you split your question up into multiple pieces? Something like the following.

    1. How could we compress a sphere to extreme densities?
    2. What would happen to the material inside it if we could compress it to neutron star density?
    3. What challenges do we have to overcome to do these things?
    4. Is there any possible benefit we could get out of this?

    Those kinds of questions are MUCH easier to discuss and I guarantee you will learn more. Generally it will be of much more benefit to everyone if you ask about something instead of trying to create a possible scenario to get your idea to work.
  18. Sep 28, 2012 #17
    Ok. 1. With the Teller-Ulam principle the only known method to compress greater masses than particles to conditions like in the core of a star.
    2. I really don't have the faintest idea.
    3. and 4. don't matter for a mindgame - you are writing like an engineer not a physicist ;)

    And I think You have forgotten the most importan question:
    5. How far could we go with the method I mentioned: white dwarfs or even neutron stars

    In my mindgame it would be simply possible to pulse impulse for impulse from outside and to use the natural frequency to let the amplitude grow. It doesn't matter how this would be done in detail. But there is a possible way to do it - to produce large quantities of ultra dense matter. That's what I think is so interesting on that principle.


    By the way: why would the neutrons of a neutron star recombine partly to protons and explode, as You said, if the high pressure from outside would vanish again? I really don't know, that's why I ask.
  19. Sep 28, 2012 #18


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    The decay neutron -> proton gives 800keV of energy plus 2 additional particles. Without a gravitational field, they simply decay. In addition, I think some energy levels could be unbound - they are fermions, and with so many fermions some of them have to occupy quite high-energetic states.
  20. Sep 28, 2012 #19


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    Neither white dwarf nor neutron star densities, if achieved, could be maintained once the pressure was relieved. For white dwarf densities, material resists further compression due to electron degeneracy pressure. For neutron star densities, material resists further compression due to neutron degeneracy pressure. Only the immense gravity of these stars prevents them from immediately decompressing
    Last edited: Sep 28, 2012
  21. Oct 8, 2012 #20
    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
    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
    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
    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
    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
    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)


    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
    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
    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
    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
    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
    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
    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
    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
    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
    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
    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
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