High Current Densities; Environmental Conditions for “Cold Fusion” (Gerhard Hunf)

The following post has been submitted by Gerhard Hunf

High current densities; Environmental conditions for “cold fusion”

The effect of high current densities (J = I / F) on an atom is an e-capture, which leads to the transmutation of the atom. Examples of this are transmutations in conductor explosions and flashes. The cause of the high current density here are extreme currents when compensating extreme potential differences.

However, high current densities can also be achieved by small currents with small potential differences if the current-carrying area is extremely small. A prerequisite for this is, however, a carrier material which can channel this current flow without damage. These are superconductors with Sc> 10exp6A/cm2 and Tc> 293 K.
For practical application, a standard conductor (Cu, Ni, Pd ..) – diameter in the millimeter range – is coated with a superconductor in the nanometer range.

If a current of 1 A flows in the circuit of such a coated conductor, the current branches on the basis of Kirchhoff’s law:I1/I2=R2/R1

Assume (R1=10exp-2Ωmm2/m; R2=10exp-20Ωmm2/m), than you get:

10exp-18 A to the standard line and 0.99999 A on the superconductor line..

This means, the current flows almost completely through the nm cross-section of the superconductor.

To determine the area of the nm cross-section, calculate the area of the circular ring that the superconductor forms around the normal axis:

FKreisring = π(R2 – r2)
R2-r2 = 0.000000000001 m2 = 10exp-11 m2

The current density in the superconducting layer is thus:

J=1A/10exp-11 m2=10exp11 A/m2

The coating of the substrates with the necessary superconductors can performed as described in DE102008047334B4.

15 Replies to “High Current Densities; Environmental Conditions for “Cold Fusion” (Gerhard Hunf)”

  1. Please explain the statement:”The effect of high current densities (J = I / F) on an atom is an e-capture, which leads to the transmutation of the atom.”

    1. A good question, because a simple electronic capture is simply ionization, and doesn’t last very long at all in a metal lattice (or any conductive matrix, but metals are certainly the extreme).

      1. By “electron capture” he means a reaction which is also known as “inverse beta minus decay”:

        p + e- -> n [+ ve]

        (The neutron could subsequently be captured by an adjacent nucleus, causing transmutation via ‘normal’ beta decay in case that the resulting nucleus is unstable.)

        This looks very similar to the theoretical concepts of Widom-Larsen and Robert Godes (see Gordon’s comment above).

        However, electron capture by a single proton is usually considered highly improbable, since it would require plenty of energy (at least 782 kEV). Perhaps an extremely high current density, as proposed by Gerhard, could help to make the electrons ‘heavy’ enough. Somebody should do the math, though, and the presence of neutrons would have to be confirmed experimentally. Maybe the following paper, which is about light hydrogen that was exposed to an electric arc, could be relevant in this context: https://arxiv.org/pdf/physics/0608229v1.pdf

          1. : )
            I suppose they really are nearly the same thing, Li + p just requires one random collision instead of two for Li7 to be forced to decay.

        1. As shown in the post above the current flow at the contact point (conductor/ superconductor) is almost completely changed from the conductor to the superconductor. For this, the velocity of the electrons must change in the ratio of the cross sections mm2 / nm2 (conductor / superconductor). The branching point (conductor / superconductor) thus acts as a particle
          accelerator for electrons. The electron becomes heavier.

          The “e-capture reaction” can be formulated as follows
          (p + + e) + E1 = n + v.

          E1 can be calculated from the mass difference Δm = m (neutron) – m (p + + e). The mass difference is 1.39 x 10exp-30 kg. Substituting this into E mc2 gives E1 = 1.249x10exp-13J or 0.78MeV.

          This work W = U I t. must be supplied by the system. In order to
          check whether a current density (J) of 10exp11 A / m2 is sufficient, we use for the superconducting layer

          W = U I t = U x (J x F) x s / v => J = W x v / U x F x s

          W = 1.249×10-exp-13J; v = 10exp2 m /sec; U = 12V (battery);
          F = 10exp-12 m2; s = 10exp-10 m. ! Attention: 1Volt is also 1J/C, 1Ampere is also 1C/sec!

          From this follows current density (J) is about 10exp11 A / m2

          This means current density for the “e-capture” is the same achieved in the superconducting layer.

  2. In the presence of high temperature and current density gradients – there should be thermo-magnetic conversion as seen with the nernst effect.

  3. “However, high current densities can also be achieved by small currents
    with small potential differences if the current-carrying area is
    extremely small.”?
    So still very high current density?

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