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262) H2O versus D2O


Ludwik Kowalski (10/5/05)
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043



As mentioned before, I belong to the International Society of Condense Matter Nuclear Science (ISCMNS). That society has a restricted discussion list, called CMNS, for its members. The purpose of this unit is to record what I learned from recent messages posted on this list. The topic is as old as cold fusion itself. Fleicshmann and Pons announcement was based on experiments with palladium in heavy water. Similar experiments with common water, if I recall correctly, produced negative results, as far as cold fusion was concerned. Later, however, cold fusion phenomena were observed by using other metals and common water. In what follows authors of messages are identified as researchers, R1, R2, R3, etc.

Voices from the CMNS list:

1) R1:
CMNC group - as you may know, I have been asked to discuss "CMN Effects in Light-Water and Hydrogen Systems" at the ICCF-12 short course. I would welcome any suggestions of recent work/results that you think should be included in this discussion. I anticipate positive results, but if you have arguments against this aspect of CMN, I would want to know those also.

2) R2:
To my knowledge, no real test has ever been performed with "pure" light water, which means that it is not possible to rule out the possibility that the residual (.02%) component of heavy water might be causing the Excess Heat. For this reason, I would suggest the term "normal water" be used in the future, not "light water."

3) R3 wrote:
I agree. The term light water should only be used for water with no D or T. To my knowledge no one has used light water. Also, most of the normal water- nickel, normal water- W, Mo, Zr systems I have used always appeared to work better with some D2O added.

4) R4:
I suggest when the argument that the deuterium in normal water is the source of heat in light water experiments, you should take into account that electrolysis concentrates light hydrogen in the cathode. If the electrolyte contains 0.02% D, the cathode will contain only about 0.001% D. This seems like too small a concentration to have much effect.

5) R3:
Yes the concentration will be small at the cathode. However, I still think the term light water should only refer to water devoid of any D or T. If not, we will have a problem later when someone does a true light water experiment.

6) R5:
My colleague and I, have effected some tests with the cell of Mizuno in which we have introduced some quantities of D2O. The amount of anomalies of energy did not change.

7) L.K. (addressing R2):
On the other hand, according to Fauvarque et al. "In any case, the device presented in this paper is a very simple device which can be used to rapidly verify hypotheses without sophisticated means. For example, we have verified that this phenomenon does not seem related with the heavy water which is found at one part in 7,000 in natural water. We increased the concentration of heavy water by a factor of 100 in our electrolyte, i.e. to 1 part in 70. We did not see any perceptible change in our results." This seems to be a good experimental proof that the role of deuterium is not dominant in Mizuno-type generation of excess heat.

8) R4:

I agree with D2 that we should make a distinction between normal water and pure light water. However, it has always concerned me that a little H2O in D2O will kill a heavy-water cell, but it is claimed that a little D2O in H2O will add to the excess energy. Why should it be so?

9) R2:
But if the effect is tied to crystal size (and considerable evidence exists that smaller, nanoscale structures are helpful) and/or magnetic effects, ideas associated with "how much" D in the cathode might not be relevant.

10) R3:
I have not made a study of it directly. However, it always seemed to me that the important part was the D2O purity on initial loading. Once it is loaded it was less sensitive to H2O. There have been a few times when I ran out of D2O and had to add H2O to an existing cell. They seemed to continue for some time even with 10 to 20 % H2O added if they were first loaded in good D2O and still running. - Now the counter to that is that if I had made a mistake in measurements, those errors just were continuing with the H2O addition. I know that X made several runs with H2O added to D2O runs and got reasonable results. In the Old CETI days, most of the "good" runs were those that where first run in D2O. If you added H2O they would continue about 2 or 3 days then "die out". R4- have you done some study at the replacement rate in a loaded cathode if the current is never removed after first loading with D?

11) R6:
If the claims are correct, the result is a function of the cathode material. With Pd, H2O kills the reaction. With Ni, ordinary water works, but increasing the deuterium content enhances the reaction. Apparently, with W glow discharge, the deuterium content makes no difference. This is a summary of what some people have reported. I do not think that these claims have been rigorously tested, except the first: H2O poisons the Pd heat producing reaction. What it does to the reactions that generate tritium and other nuclear effects I cannot say.

12) R4:
I think we need to consider two main points. If the nuclear reaction is simple fusion, the reaction rate will be proportional to the D concentration squared. If clusters are required, such as D4, then the rate will be proportional to the fourth power. This is presumably why the D concentration needs to be so high. Dilution with H would reduce the D concentration even though the total hydrogen concentration would remain high. The second point is that if, as we all believe, the effect occurs in the surface region, exchange with H will be very rapid because the electrolytic process consists of D dissolving in the Pd, diffusing to the nearest crack, and leaving the metal as D2. In other words, a steady flux of D(H) passes through the surface region. The only way D would not be rapidly replaced by H would be for the D to be trapped in a very stable lattice site, presumably by formation of a compound more stable than beta PdD. This is one of the reasons I have been suggesting that the NAE is PdD2 rather than PdD.

13) R7:
R1, you may want to talk about our work with CR39 detector plastic, showing alpha tracks after exposure to electrolysis using ordinary tap water and Li2SO4. The relevant papers can be found in the proceeds of ICCF-11.

14) R8:
If I were to take a long view of this, as Peter Gluck often does so well, I would say that the current body of knowledge and understanding of this field is only a mere fraction of what it will eventually be ... and that the semantics may not be so important.

15) R9:
Hi R4, would you please explain to the group (including me) how you arrived at the numbers above?

16) R2 (responding to R4):
This is one scenario. But there can be (and probably are) many scenario's. In band-like states, comparing concentrations of H vs D is like comparing apples and oranges because the statistics, possible modes of occupation, and forms of interaction are entirely different. In this kind of scenario, the concentration is of band state H's (protons) or band state D's (deuterons, d's). These concentrations can transient and/or (potentially) virtual (i.e., not long-lived). Within this scenario, many possibilities occur. Interfacial effects, however, will be important.

17) R4 (referring to R9’s question):
If you look at Fig. 1 in "The Effect of Hydriding on the Physical Structure of Palladium and on the Release of Contained Tritium" you will see how H is substituted for D in Pd at high concentrations of D. In this region the substitution factor is about 2 with a tendency to larger factors as the H/D ratio in the electrolyte increases. My suggestion of 20 for normal water is largely a guess which I suggest to get people thinking about this effect. Obviously, this measurement needs to be made more accurately if a model is based on the amount of D in Pd after electrolyzing in normal water.

18) R4 (addressinging R2):
It does not matter what changes the concentration of D, whether it be dilution by H or distribution between different states. It is the concentration of D in which ever state that allows the nuclear reaction to occur that matters. As long as several D nuclei must be in the same place at the same time for a reaction to occur, the concentration will influence this probability. The greater the number of D that are required too form the final product, the more the reaction rate is increased by increases by the concentration in the active state or environment, i.e. by concentration raised to increased power. Granted, the active state or environment may not have the composition obtained from measurement. If I understand your suggestion, you are proposing that the active state has a composition that is relatively independent of the measured composition and that no matter how much D is present in the electrolyte, the active state is always saturated. If this were so, heat production from the cold fusion should never be sensitive to the amount of H in the heavy water, which is clearly not the case

19) R2 (replying to R4):
Implicit in my comments are subtle points associated with band states and the fact that only a very small band state occupation is required and the fact that in a many situations, assuming such a picture is valid, when the band state occupation becomes to large, the reaction will turn off. Implicit in my comments is also the notion that in a true, many-body situation, many charges are involved, and this can drastically alter the situation in such a way that basic notions about being "localized" at the same place can be replaced by a counter-intuitive notion: being at "same" place simultaneously at many different locations. Below, I am providing specific answers to your questions.

It does not matter what changes the concentration of D, whether it be dilution by H or distribution between different states. It is the concentration of D in which ever state that allows the nuclear reaction to occur that matters.

Not true if some of the deuterons are in in band states and some are not.

As long as several D nuclei must be in the same place at the same time for a reaction to occur, the concentration will influence this probability.

Not true when the deuterons interact with many charges coherently; in this case the d's need not "be" in the same place in order to react. Waves are not localized.

The greater the number of D that are required too form the final product, the more the reaction rate is increased by increases by the concentration in the active state or environment, i.e. by concentration raised to increased power.

Not true when only a small amount of d are required and when the reaction turns off if too many are involved, which is the case when "ion band states" (or even ion band state fluctuations) are involved.

Granted, the active state or environment may not have the composition obtained from measurement.

There need not be a single "active state." In fact, Quantum Mechanics requires that "all" accessible states be included.

If I understand your suggestion, you are proposing that the active state has a composition that is relatively independent of the measured composition and that no matter how much D is present in the electrolyte, the active state is always saturated.

This is partly true. The solid can (and probably does) filter out potentially reactive d's

If this were so, heat production from the cold fusion should never be sensitive to the amount of H in the heavy water, which is clearly not the case

This pre-supposes saturation is required. The key point is that the electronic structure dictates what goes on. The electronic structure dictates the chemical potential, and this determines if the d's (and/or protons) can occupy band states. The point is that the solid (which can have characteristic dimensions as small as 10's of nanometers) plays a key role.

20) R4 (replying to R2):

a) I wrote:
It does not matter what changes the concentration of D, whether it be dilution by H or distribution between different states. It is the concentration of D in which ever state that allows the nuclear reaction to occur that matters.
You replied:
Not true if some of the deuterons are in in band states and some are not.

I see that we are talking past each other. I will try to make my point clearer. You point out that some of the deuterons are in band states and some are not. Those that are not in an suitable band state can not enter into a nuclear reaction, hence are irrelevant to the discussion. This is point I was making. We must only concentrate on those D that are in a state that allows the nuclear reaction to occur. This means that the total measured D concentration is not the main variable we must discuss.

b) I wrote:
As long as several D nuclei must be in the same place at the same time for a reaction to occur, the concentration will influence this probability.
You replied:
Not true when the deuterons interact with many charges coherently; in this case the d's need not "be" in the same place in order to react. Waves are not localized.

Granted, the waves are not localized so that my description does not apply. Nevertheless, a probability exists that describes the interaction of waves. Would not this probability be sensitive to the number of waves present in the active state?

c) I wrote:
The greater the number of D that are required too form the final product, the more the reaction rate is increased by increases by the concentration in the active state or environment, i.e. by concentration raised to increased power.
You replied:
Not true when only a small amount of d are required and when the reaction turns off if too many are involved, which is the case when "ion band states" (or even ion band state fluctuations) are involved.

I do not understand this argument. You seem to be adding additional variables that allow anything to be explained, i.e. the model can never be tested because its failure to predict behavior can be explained by too much D or too little, neither of which can be measured.

d) I wrote:
Granted, the active state or environment may not have the composition obtained from measurement.
You replied:
There need not be a single "active state." In fact, Quantum Mechanics requires that "all" accessible states be included.

Are all accessible states active in initiating nuclear reactions? My point is that only those states that are active are important.

e) I wrote:
If I understand your suggestion, you are proposing that the active state has a composition that is relatively independent of the measured composition and that no matter how much D is present in the electrolyte, the active state is always saturated.
You replied:
This is partly true. The solid can (and probably does) filter out potentially reactive d's

What do you mean by "filter out"? All d are identical until one converts to a wave by being in a band state. Presumably a mechanism exists that converts one d to a wave while another d that is dissolved in Pd does not convert. What determines how many d form waves?

f) I wrote:
If this were so, heat production from the cold fusion should never be sensitive to the amount of H in the heavy water, which is clearly not the case.
You replied:
This pre-supposes saturation is required.

No, this only presupposes that the number of d waves available for interaction determine the reaction rate. If the number of d waves is reduced by the presence of H, the nuclear reaction rate will be reduced.

g) You wrote:
The key point is that the electronic structure dictates what goes on. The electronic structure dictates the chemical potential, and this determines if the d's (and/or protons) can occupy band states. The point is that the solid (which can have characteristic dimensions as small as 10's of nanometers) plays a key role.

This is all true, but does not address the issue.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

21) LK (myself):
OK, I will stop recording this thread. It turned out to be a good sample of interactions between CMNS researchers. It does not seem to differ from interactions between scientists in other areas. So why do people continue to think that the field is pseudoscientific? Why are reports from investigations in that field rejected by mainstream journals? Why do their research proposals not supported by the DOE or NSF grants? Who benefits from this highly abnormal situation?

By the way, as I type this I recall a recent conversation with a mathematician in Austin. He said that the situation cold fusion finds itself in is not unique. Something similar is taking place in astronomy. A group of highly qualified people, rejecting the “big bang” idea and proposing another interpretation of experimental data, is practically excommunicated from the community of astronomers. Their publications, he said, are rejected by mainstream journals and they find it difficult to get access to large telescopes. That is interesting. My impression was that only the CMNS field is treated in that way.

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