Computing heat capacities in Bohmian Mechanics user7348 Junior Member Posts: 12 Threads: 1 Joined: May 2016 Reputation: 0 05-26-2016, 09:47 PM I created a similar thread on physics.stackexchange, see here: http://physics.stackexchange.com/questio...364#256364 I'm asking for a heat capacity calculation in Bohmian Mechanics. Please do not respond with commentary about how the theories of BM and QM are equivalent. Please do not respond with commentary about how it should be possible to calculate it. The challenge here is to compute the heat capacity of aluminum using Bohmian Mechanics. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-27-2016, 02:02 AM (This post was last modified: 05-27-2016, 07:55 AM by Schmelzer.) Hello user7348, I have to thank you for willingness to talk! LM ([...]) banned me from his blog so you unfortunate guys have to get by without my input. My main question is: how much will you pay for this calculation? Since I have no particular interest in it otherwise. A thousand dollars should be enough; half up front. But, on a more serious note. I don't know how to do the calculation. Schmelzer does, but may not have the time to deal with it. So let me make a few comments. Most important, I don't know that it actually is possible to get the right answer. Long ago I studied "Undivided Universe" and decided dBB was valid as far as that presentation went. It did NOT go into a lot of details, like heat capacity, entropy, and more. Bohm clearly admitted that. Since then there's an ongoing debate whether dBB can deliver the goods in these detailed areas. My opinion is, it probably can, but that hasn't been demonstrated - not very clearly anyway. Considering the resources available to work on it - just a few over-worked workers - that's understandable. I also think that when dBB really gets into these details a few more "assumptions" will be required. Seems there are details which, examined closely, aren't going to work. That doesn't bother me, depending on what assumptions may be necessary. They may be few and natural, or quite a few and ad-hoc. Obviously, if the latter, dBB should be dropped. So at this time, to me, dBB's viability remains to be proven. Here's an example. In the 2-slit experiment, look at the Bohmian trajectory for an electron as it travels to the detector. It jerks around almost chaotically. Why doesn't it emit radiation during these extreme accelerations? Classically, it should; obviously, it doesn't. Isn't it clear some extra assumption is needed to explain this? Or am I missing something? I figure most observables actually apply to the wave function (pilot wave) not the particle. The electron "beable" is a different sort of beast, whose only function, really, is to be detected someday. It never emits radiation, nor does it absorb it. Whenever we speak of that happening, we really mean the pilot wave. The situation is identical to QFT. In QFT, there is no particle as such, rather it's represented as an excitation in the electron field. All observables such as position and entropy are expressed in terms of the field. I suppose dBB is similar. The pilot wave (which corresponds to QFT's field) is the entity which counts. The particle just doesn't participate very much, except when detecting position. Its role is simply to be guided by the pilot wave as appropriate. I intend to make some comments relevant to your question soon! But it occurs to me that like LM, you may realize you're dealing with someone sensible, who may be able to debate rings around you if he feels like it. LM's technique is to run and hide; a sensible person's is to welcome intelligent discourse, especially when it disagrees with his view. So let me stop here and find out which approach you favor. Any comments - any response at all - welcome. user7348 Junior Member Posts: 12 Threads: 1 Joined: May 2016 Reputation: 0 05-27-2016, 04:06 AM Well, your response included things like "it can be done, but I won't show it here". That's exactly what my op said not to do. I've reported your post for this reason. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-27-2016, 04:11 AM (This post was last modified: 05-27-2016, 04:35 AM by secur.) secur previous: LM's technique is to run and hide; a sensible person's is to welcome intelligent discourse, especially when it disagrees with his view. So let me stop here and find out which approach you favor. Now I know. user7348: That's exactly what my op said not to do. I've reported your post for this reason. You, on the other hand, answered my question unequivocally, so I'm not reporting your post :-) user7348 Junior Member Posts: 12 Threads: 1 Joined: May 2016 Reputation: 0 05-27-2016, 04:14 AM This is not supposed to be for sociological discussions. I just want to see this calculation that has been routinely done in QM. If nobody can compute the heat capacity for aluminum in dBB, then fine. If someone can calculate it, fine, but I need to see the calculation. Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-27-2016, 07:43 AM (This post was last modified: 05-27-2016, 07:54 AM by Schmelzer.) Sorry, but I see no possibility to answer your question as requested without being misleading.   The situation is close to a request "please tell me what is 2+2 but don't say it is 4".  The answer is the same as in quantum theory, for the reason that the two theories are equivalent.  I could copy-paste some derivation, without saying this, but this would violate another ethical rule, you know, against plagiarism.  I could invent a new derivation, and, then, claim out of nothing that it is somehow a dBB-based derivation.  Which would be misleading, because the new derivation could be easily applied by proponents of the Copenhagen interpretation too.  So, sorry for not completely following your request.  The answer is: See if there is an equivalence theorem between some dBB variant which is appropriate for the given situation and standard quantum theory.  Once there is one, one can use standard techniques from quantum theory.  So, if you have some computation using standard QM techniques, fine, take it, copy-paste it and use it.   What could be wrong with this answer?  First, there could be some error in the equivalence proof.  This would mean that one has to find an error in a published and often cited paper like Bohm 1952.  Then, maybe we have a situation where we have simply no dBB theory to prove an equivalence.  This may be considered a problem for relativistic fermion fields, but even here we have proposals, see this thread. But if the equivalence has been proven, then this equivalence automatically includes thermodynamics too. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-27-2016, 04:48 PM (This post was last modified: 05-27-2016, 06:04 PM by secur.) Heat capacity depends on available degrees of freedom (among other things). Since the pilot wave is identical to the traditional wave function, it will have, at least, the same dof's, viz., the orthogonal eigenfunctions. But dBB includes another entity, the beable particle. On the face of it this provides at least 3 more dof's. Furthermore the pilot wave is considered "real" instead of ... whatever it's considered in the traditional interpretation. Interpreting that wave as "real" potentially introduces infinite extra dof's. I suppose that when analyzing an atom's heat capacity similar dof's are introduced? So this introduces some doubt about dBB's calculation of heat capacity. Looking at Bohm's 1952 paper, this concern is not directly addressed. He proves equivalence for standard measurements, but doesn't consider heat capacity or dof's. Actually I have only his first paper Part I, but am pretty sure it's not addressed in Part II either. Nor is it in "Undivided Universe". It's easy to resolve this issue by making an additional, reasonable assumption: those extra dof's don't count. They're not capable of absorbing energy thus raising heat capacity. I can think of a couple ways to formulate that. Schmelzer, probably this is a non-issue, but can you explain why? Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-27-2016, 07:47 PM In quantum equilibrium, the probability distribution of the particles is completely defined by the wave function. In this sense, these additional degrees of freedom are irrelevant for anything what matters in physics. Thermodynamics is only about rules of approximation if one is no longer able to distinguish microscopic variables but has to handle only averages. If an equivalence is proven for all physical variables, one does not have to proof a separate theorem for averages. We do not need any additional assumptions here. Once we have quantum equilibrium, the wave function defines the complete state completely. So all the functions which matter physically can be computed from the wave function. And forget about counting the degrees of freedom. This is meaningful and correct if we have the usual classical formulas for energy of type $$H = \frac{1}{2m}p^2 + V(q)$$ for these classical degrees of freedom. The formula for energy in dBB theory in quantum equilibrium is, of course, the same as that of quantum theory. Which is what matters. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-27-2016, 08:13 PM Ok, makes sense. And quantum equilibrium, of course, is well-accepted, first demonstrated by Bohm in 1953 or thereabouts. I'm still puzzled on a couple points. In the Physics Stack Exchange thread referenced by user7348, LM gives the normal von Neumann entropy S. Whereas you call it H (of course the letter used is not important) but your formula is quite different. It's an integral, and also doesn't include the Trace function. Robert Mastragostino points this out, saying your formula is "wrong" and so it seems to me also. If so, of course, you can just correct it to the "right" formula and there's no problem. Or is there more to it? Finally, you may not be able to respond to the following: you're not responsible for other's misunderstandings. But, do you have any idea why LM does seem to insist on "counting dof's" in this situation? Your approach seems fine, why don't they accept it? If you have no idea why, please ignore. Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-27-2016, 09:15 PM Ok, the von Neumann entropy $$-Tr \rho \ln \rho$$ is anyway zero for a pure state. But the formula I have given there is what you get if you if you make a measurement in such a pure state, this would reduce the pure state into a density matrix \[ \rho \to \sum \rho_a |a>