- The answer to the question reduces to the notion of electronegativity characterized by chemical potential. For example, oxygen is willing to accept electrons from less electronegative atoms and hydrogen, alkali atoms and carbon are ready to give them up in oxidation. In the process oxygen is reduced and one speaks of redox reactions. These reactions are basic processes of biochemistry.
In a typical situation, molecule A receives oxygen atoms from molecule B when it is oxidized. The process is more general: for instance, nitrogen, phosphorus or sulfur could replace oxygen. When an atom, say Carbon gives up oxygen atoms or C=O bonds transform C-OH bonds, reduction also occurs. When A is oxidized, B is reduced.
- When an electron moves from one electrode to another, the stored energy is determined by the difference of chemical potentials. Charge transfer occurs when the difference is non-vanishing. The difference in chemical potential between atom A and some reference atom determines the electronegativity. Depending on its sign, either A or B tends to give up electrons. At equilibrium (without load), the voltage between the electrodes at equilibrium equals the difference of chemical potentials. This voltage is created during charging and does not have to be the opposite of the charging voltage, which is decoupled from the system after the charging is complete.
Chemical potential is a thermodynamic parameter and can be considered as chemical energy that also includes electrostatic energy. It should be noted that we are talking about electrons now and electrons flow in the case of the Pollack battery when the battery is used because an ohmic electron current passes through the load.
Consider now the Pollack batteries.
- In the charging of a Pollack battery there is very little dissipation because the dark protons flow through the magnetic body and the chemical potential is reduced to the electrostatic energy determined by the voltage. For semiclassical Pollack batteries the dark protons transform to ordinary ones at the target electrode E2. For purely quantal Pollack batteries, the dark charges at MB do not transform to ordinary ones during the charging. It seems that different descriptions are needed at the level of the magnetic body and for ordinary matter, where the description would be purely chemical and involve chemical potentials.
- The difference of chemical potentials at the level of ordinary matter changing during charging should be equal to the changing voltage during charging. How does the initial charging voltage V0 relate to the voltage Vuse after charging?
One can start from a problem.
- Assuming a 100 g electrode and accepting the claims of Donut Lab, the energy gained by a dark proton in the Pollack effect is 1.44 eV, which corresponds rather accurately to the standard battery voltage of 1.5 V. 104 g gives a value of 1.5 V. Is it time to shout Eureka? Not quite yet.
- Marko informed that in some VTT based estimates the voltage was around 3.5 V. Is this voltage the charging voltage V0 or the post-charging voltage Vuse, whose magnitude need not be the same as that of the charging voltage. Which option is correct: V0=3.5 V and Vuse =1.5 V or vice versa?
Intriguingly, the pondering of this problem lead to a possible connection between pure quantum battery (see this) in which the capacitor behaves as if it had a net charge. The TGD explanation (see this) assumes that the capacitor indeed develops a net charge. The charge separation would be not only between the capacitor plates but also between the magnetic body (MB) and capacitor as a whole would be opposite. Biefeld-Brown effect would be also involved with the em drive (see this).
- The pure quantum version of the Pollack capacitor indeed assumes two kinds of charge separations so that the Biefeld-Brown effect is possible. The charge separation between E1 and magnetic body (MB) would induce charge separation between E1 and E2. Could the Pollack effect take place also at E2 and induce a charge separation between E2 and MB and make the Pollack battery charged and imply that the charges of E1 are not of the same magnitude so that the Biefeld-Brown effect is implied?
- But is the Pollack effect for E2 possible energetically? The TGD inspired quantum biology serves as a guideline here (see this and this). The proposal is that for DNA and cell membranes with a permanent negative state, the state formed by dark protons at the magnetic body forms what might be called dark nucleus having a binding energy much smaller than the ordinary nuclear binding energy.
Could the presence of dark protons at the magnetic body forming dark nuclei serve as a seed of phase transition for the Pollack effect also at E2 occurring spontaneously? The generation of dark nuclear binding energy would make the transition energetically possible.
This would increase the negative charge at electrode E2 but would not affect the negative charge at the electrode E1. Pure quantum batteries would allow more negative charge to E2 than the semiclassical Pollack battery for the dark protons drop to E2 during the charging. The Pollack effect at E2 would increase the effective charging voltage Veff during charging whereas Pollack effect at E1 would tend to reduce it.
- A reasonable expectation is that too much positive charge near E2 at MB makes the system unstable and some maximum dark proton charge can be loaded from E1 and E2 by the Pollack effect. There would also be an upper bound for the number of protons, which can be transferred from E2 to MB. The outcome would be a state in which MB and the battery would have opposite charges but this would not be the case for E1 and E2.
Veff= V0+VP, would be the sum of the charging voltage V0 and the contribution VP = -VP,E1 +VP,E2 due to the generation of negative charge by Pollack effect at both E1 and E2. This would give Veff= V0 -VP,E1 +VP,E2. The increase of Veff during charging would transfer more positive charge to E2.
- Intuitively it seems clear that saturation occurs and the decoupling of the original charging voltage V0 is possible. The user voltage would be Vuse= -VP,E1 +VP,E2 and should have a sign, which is opposite to that of V0. For VP,E2=0 the upper bound for | Vuse| would be | VP,E1| = | V0| corresponding to complete compensation of the positive charge at E1 by the generation of negative charge in Pollack effect.
- VP,E2 and VP,E1 have opposite effects on the magnitude of Vuse. The dielectric between capacitor plates is replaced now by a catalyst of the Pollack effect. A reasonable guess is that the Pollack effect at E2 has the same effect as the presence of dielectric.
For dielectrics, the voltage is given by V = CQ/εr and by εr>1 smaller than without dielectric. Same is expected now. This conforms with the estimate for the energy per dark proton if the final voltage is indeed about 1.5 V and smaller than the initial voltage. If so, Vuse=| -VP,E1 +VP,E2| would be about 1.5 V and by a factor 1.5/3.5= 3/7 smaller than the estimate 3.5 V by VTT using standard chemical picture if this estimate indeed corresponds to the charging voltage. Pollack effect at E2 would make possible a larger charge/voltage ratio just as the use of a dielectric.
See the article Are Pollack batteries possible? and the chapter with the same title.
For a summary of earlier postings see Latest progress in TGD.
For the lists of articles (most of them published in journals founded by Huping Hu) and books about TGD see this.
For a summary of earlier postings see Latest progress in TGD.
For the lists of articles (most of them published in journals founded by Huping Hu) and books about TGD see this.