Oscillatory nuclear fusion hypothesis

[Dmi3007]

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 ...

 

 

[exchemist]

1 hour ago, Dmi3007 said:

 

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 ...

 

 

This reads like a variant of the “cold fusion” idea - which went nowhere. The Coulomb barrier is too high and too thick for tunnelling of vibrational modes.

I presume no experiments have been done to test the idea.

 

Edited August 21, 2023 by exchemist

[Dmi3007]

1 hour ago, exchemist said:

This reads like a variant of the “cold fusion” idea - which went nowhere. The Coulomb barrier is too high and too thick for tunnelling of vibrational modes.

I presume no experiments have been done to test the idea.

 

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.

[exchemist]

24 minutes ago, Dmi3007 said:

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.

But you are not using muon catalysis, are you?

[Dmi3007]

15 minutes ago, exchemist said:

But you are not using muon catalysis, are you?

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?

[Steve81]

1 hour ago, Dmi3007 said:

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?

I’d need a proof of concept to believe any of this personally.

In conventional experimental fusion reactors, AFAIK they’re using lasers to fire up confined hydrogen isotopes to very high temperatures to achieve the conditions needed for fusion, and basically just waiting for the (incredibly small) nuclei to collide and fuse. They’ve actually achieved nuclear fusion in under these circumstances, indicating contrary to your statement, that temperature is a pretty big factor.

Everything you wrote reads like a bunch of pseudoscienctific gibberish to be quite honest.

Edited August 21, 2023 by Steve81

[Dmi3007]

1 hour ago, Steve81 said:

Everything you wrote reads like a bunch of pseudoscienctific gibberish to be quite honest.

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.

[exchemist]

1 hour ago, Dmi3007 said:

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?

I've only read it quickly but at first glance I think it is a mistake to the think the high permittivity of these SDM materials alters the permittivity of space at subatomic distances, when it is only a macroscopic, bulk property, arising from the polarisation of the  dipoles in the nanotubes, or whatever. Once you to get down to the atomic level, the applicable value of ε will still be that of free space, surely? 

 

 

[Dmi3007]

On 8/21/2023 at 3:54 AM, exchemist said:

I've only read it quickly but at first glance I think it is a mistake to the think the high permittivity of these SDM materials alters the permittivity of space at subatomic distances, when it is only a macroscopic, bulk property, arising from the polarisation of the  dipoles in the nanotubes, or whatever. Once you to get down to the atomic level, the applicable value of ε will still be that of free space, surely? 

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.

Edited August 23, 2023 by Phi for All
Fixed quote

[Steve81]

1 hour ago, Dmi3007 said:

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.

Without some kind of actual observation other than a personal pet theory, what am I supposed to make of your statement? When you observe cold fusion, let me know.

Edited August 21, 2023 by Steve81

[swansont]

9 hours ago, Dmi3007 said:

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

Earnshaw’s theorem tells us that you can’t confine a charge with static electric fields.

[exchemist]

6 hours ago, Dmi3007 said:

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.

 

I'm not a solid state physicist but I think you are making the mistake of thinking the +ve charges are brought closer together on the +ve plate of a charged capacitor. It is the electrons which are partially removed, surely, to create the net +ve charge? The nuclei do not move.

An SDM will simply allow more electrons to be removed, by the stabilisation of the +ve ions that arises from the alignment towards them of the -ve end of the dipoles in the dielectric. Even if you were able, somehow, make the ions flow freely through the +ve conductor material, I would expect this stabilisation will be maximised when the +ve ions are closely aligned with the dipole ends. This would be at normal interatomic separations.

 

 

Edited August 21, 2023 by exchemist

[Dmi3007]

15 hours ago, swansont said:

Earnshaw’s theorem tells us that you can’t confine a charge with static electric fields.

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)?

 

12 hours ago, exchemist said:

 

I'm not a solid state physicist but I think you are making the mistake of thinking the +ve charges are brought closer together on the +ve plate of a charged capacitor. It is the electrons which are partially removed, surely, to create the net +ve charge? The nuclei do not move.

An SDM will simply allow more electrons to be removed, by the stabilisation of the +ve ions that arises from the alignment towards them of the -ve end of the dipoles in the dielectric. Even if you were able, somehow, make the ions flow freely through the +ve conductor material, I would expect this stabilisation will be maximised when the +ve ions are closely aligned with the dipole ends. This would be at normal interatomic separations.

 

 

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).

[Dmi3007]

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

Edited August 22, 2023 by Dmi3007

[swansont]

6 hours ago, Dmi3007 said:

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)?

You said “the interacting nuclei are maximally immobilised and are held near the equilibrium position not by a magnetic field, but electrostatically” and I’m telling you you can’t immobilize a nucleus with an electrostatic field.

[Dmi3007]

25 minutes ago, swansont said:

You said “the interacting nuclei are maximally immobilised and are held near the equilibrium position not by a magnetic field, but electrostatically” and I’m telling you you can’t immobilize a nucleus with an electrostatic field.

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.

[swansont]

6 minutes ago, Dmi3007 said:

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.

You can’t trap electrons with an electrostatic field, either. Trapped does not mean they are at rest, but “maximally immobilised and are held near the equilibrium position” sounds an awful lot like trapped. If they are not, then why would the be immobilized?

 

[exchemist]

2 hours ago, swansont said:

You can’t trap electrons with an electrostatic field, either. Trapped does not mean they are at rest, but “maximally immobilised and are held near the equilibrium position” sounds an awful lot like trapped. If they are not, then why would the be immobilized?

 

To be fair, the nuclei of atoms bound by a chemical bond could be said to be trapped by the electrostatic field of the electron orbitals, couldn’t they?

13 hours ago, Dmi3007 said:

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).

Yes but you are making the mistake of thinking the permittivity, in the region between one deuteron and the next, will not be that of free space. As I said before, these SDMs exhibit a very high bulk permittivity because of their molecular structure (the way dipoles can align).  Such measurements take place perpendicular to the plates of the capacitor, or your needle tip and the plate. At the atomic level the permittivity in the plane of the plate, which is what determines the repulsion between adjacent deuterons, remains essentially that of free space, modified by whatever electron density remains between them. No element of the SDM dipoles penetrates that space and so the extra stability they confer on the ions is due to an attractive electrostatic force perpendicular to the plate, not parallel to it.

So your calculation, which assumes the bulk SDM value for the permittivity between deuterons, is using the wrong number and will give you the wrong answer.  

Edited August 22, 2023 by exchemist

[swansont]

5 hours ago, exchemist said:

To be fair, the nuclei of atoms bound by a chemical bond could be said to be trapped by the electrostatic field of the electron orbitals, couldn’t they?

No, you don’t have charges in a fixed position when you have orbitals. You can’t apply the theorem. The field is a time-average, similar to what you would have classically for charges in motion, which does allow for confinement. 

[Dmi3007]

10 hours ago, exchemist said:

No element of the SDM dipoles penetrates that space and so the extra stability they confer on the ions is due to an attractive electrostatic force perpendicular to the plate, not parallel to it.

  

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. 

5 hours ago, swansont said:

No, you don’t have charges in a fixed position when you have orbitals. You can’t apply the theorem. The field is a time-average, similar to what you would have classically for charges in motion, which does allow for confinement. 

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?

Edited August 23, 2023 by Dmi3007

[exchemist]

8 hours ago, swansont said:

No, you don’t have charges in a fixed position when you have orbitals. You can’t apply the theorem. The field is a time-average, similar to what you would have classically for charges in motion, which does allow for confinement. 

Agreed about Earnshaw’s theorem.  But the nuclei are “trapped” - though in an admittedly looser, time averaged sense. I’m not sure the OP meant “trapped” in the sense of classically stationary. Trapped can mean in motion but unable to escape, after all.

[Dmi3007]

13 hours ago, exchemist said:

To be fair, the nuclei of atoms bound by a chemical bond could be said to be trapped by the electrostatic field of the electron orbitals, couldn’t they?

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

Edited August 23, 2023 by Dmi3007

[exchemist]

2 hours ago, Dmi3007 said:

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

What hydrogen pressure gradient?

And no, it does not follow that that any of this can overcome the electrostatic repulsion between protons or deuterons. 

Edited August 23, 2023 by exchemist

[swansont]

8 hours ago, Dmi3007 said:

1. Eamshow theorem is not working here, this is not classical case.

What is inherently quantum-mechanical about holding a nucleus fixed with an electric field?

5 hours ago, exchemist said:

Agreed about Earnshaw’s theorem.  But the nuclei are “trapped” - though in an admittedly looser, time averaged sense. I’m not sure the OP meant “trapped” in the sense of classically stationary. Trapped can mean in motion but unable to escape, after all.

But that’s what you can’t do with static fields. There is an unstable equilibrium point, but as soon as the charge moves, it’s no longer confined.

[Dmi3007]

4 hours ago, exchemist said:

What hydrogen pressure gradient?

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.

4 hours ago, exchemist said:

And no, it does not follow that that any of this can overcome the electrostatic repulsion between protons or deuterons. 

What do you mean here?

1 hour ago, swansont said:

What is inherently quantum-mechanical about holding a nucleus fixed with an electric field?

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.

Edited August 23, 2023 by Dmi3007