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This question follows on previous work on connections between (other) areas of physics and thermodynamics as in here, here and even here.

P. Dirac (an electrical engineer initially) was one of the first to pose the possible existence of magnetic monopoles (original paper)(for example as solenoids). This would allow more symmetric forms of certain equations (physicist fetish?). Anyway, this (a-)symmetry is also the starting point for possible connections to thermodynamical concepts.

Is there any relation to the (non-) existence of magnetic monopoles and 2nd law?

Would the possible existence of magnetic monopoles violate the 2nd law, or does the 2nd law predict magnetic monopoles, or it is just irrelevant (sth many physicists would not assume lightly)?

Relevant physics areas include the intersection of materials magnetization and thermodynamics.

Relevant articles include this one on spin ice and this one on Brownian motion

NOTE: have not actually read the articles as i do not have access, however mentioned as part of existing literature on the question's subject.

Relevant papers and outlines: here, here, here and an arxiv paper by P. Davies on this issue (circa 2007)

UPDATE: An "unorthodox" critical view on mainstream mathematical physics (specifically Yang-Mills theory)

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  • $\begingroup$ If I believed in something like Electric-Magnetic duality, that seems to be a fairly fundamental support for the idea of magnetic monopoles (which might make one feel that technical trickery is missing something essential :-?) -- not to mention, that is the only reason we have till today, for why charge need be quantized. $\endgroup$ – Siva Nov 8 '14 at 18:10
  • $\begingroup$ @Siva, why is not current the dual to electric charge as it generates magnetic fields? Duality is here also. Irreversibility is here also $\endgroup$ – Nikos M. Nov 8 '14 at 18:46
  • $\begingroup$ What current configuration will give a magnetic field with the same isotropic $\frac{1}{r^2}$ profile that a point electric charge has? $\endgroup$ – Siva Nov 8 '14 at 19:04
  • $\begingroup$ @Siva, hmm, i guess we will have to see what symmetries and dualities are there and where and what not $\endgroup$ – Nikos M. Nov 8 '14 at 19:16
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No relation. The second law of thermodynamics can be proven on general statistical grounds. It is independent of what particles are in our universe or what equations describe their motion.

That is why physicists are very confident about the second law of thermodynamics, even though nobody knows for sure what particles are in our universe or what equations describe their exact behavior in all circumstances.

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  • $\begingroup$ Not quite correct, "proof" of second law on statistical grounds is only partial. Plus the lochsmidt paradozx still holds, how can a-symmetry stem from symmetric equations (or other symmetric considerations). Plus statistical arguments assume necessarily that at another level the law does not hold. Note that i am aware of "proofs" of 2nd law based on unitary dynamics et all. i would propose to check other approaches as well (e.g Prigogine, Hatsopoulos-Gyftopolos-Keenan quantum thermodynamics ) $\endgroup$ – Nikos M. Dec 10 '14 at 10:52
  • $\begingroup$ To sum up, the whole point is that 2nd law is not statistical, and a simple and fast argument to this is the fact that even statistical considerations necessarily (in most cases) assume (micro?)-violations of the 2nd law, which have never been observed. This in sequence leads also to the simplistic(?) interpretations of the 2nd law as disorder etc.. $\endgroup$ – Nikos M. Dec 10 '14 at 13:36
  • $\begingroup$ contd.., it may have statistical side-effects, but thermodynamic irreversibility is not statistical (as explained above) $\endgroup$ – Nikos M. Dec 10 '14 at 13:41
  • $\begingroup$ i can continue about what is (supposed to be) considered fundamental and what is phenomenological. The thesis implied by this answer is in the same line of thought where the 2nd law is phenomenological (unjustified i would add, see above), the thesis which is implied by my question (and partial answer and references) is that it is basic (or "fundamental" if you like) $\endgroup$ – Nikos M. Dec 10 '14 at 13:44

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