Dmi3007

No definetly like you tell. I think, the temperature or pressure may be reason for deuterons to go away from the needle. But if the needle have a positive charge the way out is the needle tip. Electric forces (attraction to the negative charges inside the tip will hold the deuterons and SDM will make deuterons density critical for fusion. Why insane? Did you measure the energy of the pressure or temperature? Lets now use this energy not for all deuterons in the needle, but only for deuterons near the tip. This is very small quantity (much-much less than 1 Gramm). And this is very huge energy for this quantity to make it closer for fusion distances.

I think I understand the probable mistake Pd could start to be semiconductor al the exceeding hydrogen concentration inside. The critical larval is D/Pd = 0,5

Why? I described deuteron behaviour in the attached video. If we have some exceed of their concentration, they are oscillating near the equilibrium position in the first approaching. But of course the real behaviour is more complicated as they have wave properties too. But I think that the experiment with negative charged needle inside SDM and the supplying of deuterons to the tip through the PEM could be more lucky and simple.

No, with all due respect to you, if we are solving a task with quants, this is a quantum mechanic task, and Earnshaw's theorem does not work there.

Are we talking about quants? How can this theorem be applied here?

But also If a syringe with deuterium is attached to the blunt end of a positively charged needle and a certain excess concentration of deuterons is created in the needle. then near the tip of the needle a certain amount of them will appear, held by the forces of electric attraction. The situation may be reversed. On the one hand, the negatively charged needle is stucked into the SDM; on the other side of the SDM is a proton exchange membrane that supplies deuterons from pressurized deuterium. A dense cloud of deuterons is formed on the membrane opposite the needle, the distances between which can be sufficient for nuclear fusion.

Imagine that there are hydrogen purification technologies: a pressurized gas is directed into a palladium membrane and purified hydrogen is produced on the other side of the membrane. That is gradient of pressure. But I think, it is not necessary. If we solve deuterium in palladium needle and than create the lack of electrons there, place negative charged cathode near the tip of the needle, deuterons inside the needle will go to the tip as they can move. What the process is inside fuel cells? Proton from palladium catalyst goes inside membrane to cathode. Same process could be made in this case. Ceramic PEM needle can allow only deuterons to go to cathode and create deuterons high density cloud near the tip. What do you mean here? In the case under consideration, deuterons are held by electrons. The interaction of nuclei and electrons is a quantum mechanical process. This interaction is stable contrary to Eamshaw's theorem, since it does not work there. Electrons can be stably smeared around nuclei contrary to this theorem. Theorem will not work for quants.

Yes it is so. I know that s-level of hydrogen sole-electron is matched to p-level of palladium electron it may course high mobility of hydrogen nuclei inside the palladium. And the hydrogen pressure gradient toward to the tip of the needle as well as positive charge on the needle can create the situation when deuterons will leave the tip for about 1 angstrem but can not take electron because of positive charged needle does not want to give it ) So high density deuterons cloud can be created near the tip of the needle and the root-mean-square distance between deuterons may be sufficient for fusion synthesis

I understand it well and moreover it is very good for the experiment. If SDM will be between the deuterons the Coulomb barriers start to be very-very wide between them. I do not need SDM between deuterons at all. I need SDM to prevent deuterons emission from the tip of the needle only (to minimise analog of Shottky effect). Usual capacitor also does not have dielectric between the electrons on the cathode and this situation does not hesitate to create larger electron density using dielectric gasket. Look, to come back to experiment condition: 1. Eamshow theorem is not working here, this is not classical case. 2. Positive or negative charges will always create bigger density in the tip of the charged needle and if the needle is stuck in dielectric this density will be more. Will you argue?

OK, but my words "maximally immobilised and are held near the equilibrium position" does not mean that the charged particle is fixed (the particle is at rest) at one point (as Earnshaw’s theorem tells). Also I think this theorem is not for this case as nuclei and electron experiences quantum (not classical electrostatic) effects.

The deuterium pressure gradient from the bottom of the needle to the tip will create an excess of deuterons at the tip too

Please explain where do you see the contradiction of this case with this theorem? Do you think that an increased charge density cannot be created at the tip of the needle (especially if the needle is stuck into a dielectric)? Of course I understand, that regular capacitor anode just has lack of electrons. Palladium atoms are covalently bonded, while deuterons within palladium may be less bound. I think deuterons mobility depends on concentration of deuterium inside Pd. and temperature. At 1 atm we can get about D/Pd = 0,7. Also fuel cell technology allows tome proton through the proton-exchanging membrane (PEM). This is prove the possibility of the moving of hydrogen nuclei without electrons in electrostatic field and I think, some kind of PEM could also be used in this experiments (before I made electrolyse of deuterium water, using Chinese PEM).

The USA government patent on SDM describes that it could be used like a gasket in capacitors to increase their charge. It does mean that electron quantity is increasing and they start to be closer to each other without any kind of emission from. And also we should remember, that during the solution in palladium hydrogen/deuterium will come inside like proton/deuteron. So if we use deuterons instead of electrons we should have the same effect, nuclei will start to be much closer to each other thanks to SDM. The distance between deuterons will be about 10^(-13) m = wide of Coulomb barrier. Oscilation of deuteron will rapidly increase the quantity of attempts to overcome the barrier. One second is much more than enough to do it.

It looks like your personal opinion, but not like a scientific refutation. Once again, for nuclear fusion, the distance of approach of the nuclei and the time of keeping at this distance are important. Lasers, temperature, etc. are means of approach, not reaction conditions.

No, and what? The necessary and sufficient condition for any nuclear fusion is not temperature and not muons, but the approach of nuclei to a distance sufficient for fusion and retention there for some time. With the above calculations, I showed that such an approach is possible using super dielectric materials. If I made a mistake somewhere, show me where?

My own laboratory (from Alibaba) 😀 does not allow to make quality experiment. But I will be very grateful, If you can refute the calculations given here. By itself, cold nuclear fusion now exists within the framework of muon catalysis. Calling something a pathological science (like it was done 40 years ago) is no longer scientific. Scientific language is - proven or not proven.

In contrast to the approach of nuclei under thermal conditions, the interacting nuclei are maximally immobilised and are held near the equilibrium position not by a magnetic field, but electrostatically, while thin Coulomb barriers are overcome during quantum vibrations of the nuclei. The approach of nuclei to the minimum distances to (10^(-13) m) becomes possible due to the using of recent discoveries in the field of composite dielectric materials (SDM, super dielectric materials). Description of the proposed processes The essence of proposed method is the creating of conditions for maximum approaching of nuclei and hold it approached as long as possible. The method can be implemented on the basis of a device resembling a capacitor and having a conductor of hydrogen isotopes (CHD), in which hydrogen is partially or completely ionised (1); a voltage generator (VG) (2) connected to the specified conductor, and a heat removal system (HRS) (3), which can simultaneously convert thermal energy into its other forms. For the implementation of the method, a CHD with hydrogen isotopes dissolved in it is required (when dissolved, hydrogen molecules first decay into atoms, then the atoms donate electrons to the CHD and diffuse in it already as nuclei) [4]. Further, a positive potential is created on the CHD, which, however, is not sufficient for ionisation/breakdown in the space surrounding CHD. Composite dielectrics with a high relative permittivity and electric strength (https://patents.google.com/patent/US9530574B1/en) can be placed near the surface of the CHD to create an electrostatic field sufficient for the method. In this situation, the following processes will take place. An electrostatic field is created near the CHD surface, and some of the nuclei of hydrogen isotopes will be placed near the CHD surface, being attracted to the negative charges induced by them in the CHD. Isotope nuclei find themselves in potential wells created by the Coulomb forces of neighbouring nuclei, where they oscillate. As long as the electrostatic field on the surface and near the CHD is not strong enough, and the nuclei tempera- ture is low enough, zero oscillations will occur near the bottom of the potential well. If the temperature increases, then the nuclei in the well will take a higher energy level and the oscillatory process will change with the appearance of the second mode. If we further increase the strength of the electrostatic field on the surface of the CHD and select the temperature of the nuclei, then it is expected that meetings of neighbouring nuclei will occur due to their wave properties. The distances between neighbouring nuclei will decrease so much that tunnelling effects become possible. These processes can be enhanced if deuterium predominates among the isotopes of hydrogen dissolved in CHD. Deuterium nuclei are bosons; as the temperature on the CHD surface decreases, their velocities decrease. If deuterium is dissolved in the CHD in a concentration much higher than other isotopes, and also with the exclusive presence of deuterium in the CHD, the motion of non-boson nuclei near the surface has little effect on the speed of deuteron motion. And a lot of deuterons appear with significant de Broglie wavelengths and low particle velocities. 4. Assumed conditions for observing the process of electrostatic wave fusion. To save voltage on the CHD the shape of the CHD can have a sharply convex tip. A high positive electrostatic field will be created near the tip at a relatively low positive potential on it. The reactions described above are expected near the tip. The tip can be coated with a composite dielectric with a high relative permittivity and dielectric strength. It should be taken into account that Schottky barriers [3] appear at the boundaries of the CHD - dielectric. However, if a dielectric is used, Schottky barrier may have a width and height sufficient to prevent field ion emission of nuclei in a high external electrostatic field. To minimise the overheating process, it is necessary to use HRS. Approximate calculation of processes. Let's choose the values of the electrical strength of the dielectric material and its relative permittivity, which are close to the maximum possible for dielectric material at the present time: E = 10^7 V/m, ε = 3*10^10 It should be mentioned that in present time the materials with the relative permittivity ε can reach values more than 10^11 have been synthesised (https://apps.dtic.mil/sti/pdfs/ADA632380.pdf) . These materials are called Super Dielectric Materials (SDM). In fact, they are composites. High dielectric permittivity is achieved there due to combinations of properties of porous dielectrics and ionic solutions. Porous dielectrics can be in the form of nanotubes containing an electrolyte. When an external electrostatic field is created, the ions in the electrolyte diverge in different directions, but cannot leave the nanotubes. Accordingly, an electric field, directed oppositely to the external one, arises inside such these tubes and due to the relatively large distance between electrolyte ions, this field, and, accordingly, the relative permittivity, are high. The use of such dielectric coatings will reduce the distance between nuclei on the surface of the needle by thousands of times, while preventing auto-ion emission from the tip of the needle. Let the CHD be implemented in the form of a palladium needle with a tip radius r =10^(-8) m. Then the strength of the electrostatic field near the positively charged tip of the needle can be described with sufficient accuracy by the formula: Е= kQ/εr^2 From where we determine the magnitude of the charge at the tip of the needle: Q = εЕr^2/k And the number of nuclei (let it be deuterons) can be calculated as the total charge divided by the elementary charge e: N = Q/e = εЕr^2/ek = 2*10^10 Let's estimate the distance between deuterons at the tip of the needle. To do this, we calculate the area of the needle tip and di- vide by the number of deuterons - we get the area occupied by each deuteron. The square root of this elementary area will give the approximate distance between deuterons. If the tip is hemispherical, the tip square could be calculated as: S = 2πr^2 Hence the elementary area occupied by one deuteron on the tip: s = S/N = 2πek/εЕ Then the distance between deuterons is: d = s^(1/2) = (2πek/εЕ)^(1/2) = 1,7*10^(-13) m Three times more than this distance between the nuclei can cause nuclear fusion with muon catalysis. The approach of nuclei is carried out by an increase in their potential energy in an electrostatic field, and not by an increasing of their kinetic energy. The approach of nuclei to comparable distances in the process of thermonuclear fusion would mean heating them up to 8,000 eV (about 93 million kelvins). In combination with an unlimited time for keeping nuclei at close distances, the described method seems worthy of attention for practical application in order to obtain the implementation of nuclear fusion. Will be glad to read any regarded comments/critics/amendments ...