Can deuterons in palladium condense into a BEC?

In this earlier post, I introduced Yeong E. Kim’s Bose-Einstein Condensate (BEC) theory of cold fusion. According to this theory, when you pack lots of deuterons into palladium, they condense into a BEC, which makes nuclear fusion possible, and then the fusion energy is collectively absorbed by the BEC, thus explaining all the mysteries of cold fusion. In my earlier post I said that the two biggest problems with the theory are: (A) The deuterons do not actually condense into a BEC; and (B) Even if they did, it would not help explain cold fusion. I already blogged about (B) hereToday I will talk about (A). I have changed my mind: Although I suspect that deuterons would not condense into a BEC, I don’t know enough to say for sure!   😛

In this post I’ll mainly just summarize Kim’s argument. If anyone reading this is a BEC expert, please comment (or better yet write a guest post, just email me) with your opinion!

So… When room-temperature palladium is full of deuterons, will the deuterons condense into a Bose-Einstein condensate? Kim says yes. See especially this paper (Naturwissenschaften 2019). Should we believe him?

Let’s start with a bad argument against BEC formation:

Obviously, Bose-Einstein condensation can only occur very close to absolute zero, duh!

(See this page for an example of that argument—the sentence mentioning “millikelvins”.) Why is this a bad argument? Well, the original (and still most famous and common) example of BEC formation is in a dilute atomic gas, and these indeed require temperatures very close to absolute zero. But higher-density systems can condense at a higher temperature. For example, exciton-polaritons can form a Bose-Einstein condensate at room temperature. This is the basis for polariton lasers. Kim is arguing that the deuterons are at very high density, and therefore they can condense at a high temperature. That seems reasonable, or at least I don’t see why it can be dismissed out-of-hand.

Moving on, what is Kim’s argument that the BEC occurs at room temperature? I think I can summarize it in three steps. Again, I’m using this paper (Naturwissenschaften 2019).

Step 1: Assume that there is one deuteron per palladium unit cell. I don’t know whether or not this assumption is reasonable. If you have many hours to spare, Peter Hagelstein’s 2014 lectures have a detailed discussion of how many deuterons there are and how they’re configured. I don’t remember what he said. I think no one really knows for sure. Anyway, this assumption is how Kim derives a deuteron density of n \approx 4/(3.89\mbox{\normalfont\AA})^3 = 6.8\times 10^{22}\text{cm}^{-3} (p804). From this, Kim infers that there is an “average separation distance” between neighboring deuterons of about n^{-1/3} = 2.45\mbox{\normalfont\AA}.

Step 2: Assume that a BEC occurs when the thermal de Broglie wavelength equals the average separation distance between neighboring deuterons. This is a rule-of-thumb, one which (I think) is commonly invoked in discussions of BECs. I don’t know how accurate this rule-of-thumb is, or in what situations it does or doesn’t apply.

If we make this assumption, and plug in the 2.45Å figure from above, we conclude: A BEC will form if the typical deuteron thermal kinetic energy is less than 6.8meV. (Kim writes 6.3meV instead of 6.8meV, I’m not sure why.)

Unfortunately, according to the equipartition theorem (or equivalently, according to the Maxwell-Boltzmann distribution), the room-temperature root-mean-square kinetic energy of a deuteron is actually 39 meV, too big by a factor of 6. So I guess it won’t condense?

Well, maybe that the approximations we’ve already made—especially the rule-of-thumb in step 2—were too conservative by a factor of 6. I dunno, maybe that’s possible. But Kim makes a different argument instead:

Step 3: Argue that the equipartition theorem (or equivalently, the Maxwell-Boltzmann distribution) overestimates the deuteron kinetic energy. This is on p804-805 of the paper:

The [Maxwell-Boltzmann] distribution is originally derived for … [an] ideal gas. For strongly interacting and dense systems such as mobile deuterons in metal, the [Maxwell-Boltzmann] distribution is not applicable. There have been theoretical efforts for generalization of the [Maxwell-Boltzmann] distribution… described in the following two subsections…

Is this step a legitimate argument, or wishful thinking? I have my suspicions, but I’m not 100% sure.

Conclusion: Despite what I said in an earlier post, my current feeling is: I don’t know enough to say whether or not there is any chance that deuterons densely packed into palladium could possibly condense into a BEC.

I do not think this is a moot point. Even though I think that Kim’s explanation of cold fusion is wrong for other reasons (see this post), it is surely useful to know whether the deuterons might condense into a BEC. Can we dismiss that possibility out-of-hand, or not? Does anyone know?

2 thoughts on “Can deuterons in palladium condense into a BEC?

  1. Kevin M O'Malley

    Perhaps you should take it a step back and see if there is experimental or mathematical evidence that Deuterons will form into a Luttinger Liquid inside of condensed matter. Because there is evidence that a LL forms at much higher temperatures in one dimension (i.e. a line) than in a gaseous state with free movement. Since we know that Luttinger Liquids form at much higher temperature, postulating that the same kind of effects take place in the formation of a linear BEC inside condensed matter is not so much of a stretch. That is my V1DLLBEC hypothesis: Vibrating 1Dimensional Luttinger Liquid Bose Einstein Consensate.


    1. steve Post author

      A Luttinger Liquid is made of fermions, right? So deuterons cannot make Luttinger liquids, and nor can BECs. Or am I misunderstanding your point?



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