Talk:Bohr–Van Leeuwen theorem

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Restricted to thermal equilibrium?[edit]

Is this proof simply saying there is no magnetism in a classical system that is in thermal equilibrium? That sounds plausible. The usual example of a classically produced magnetic field, a solenoid (perhaps powered by any classic electrostatic generator), persists only while work is being done (i.e., while parts of the total system are very far from thermal equilibrium), and the field dissipates if the device is left to run down. However, the wording of the article (esp. introduction) seems to imply something stronger than this (i.e., that by thermodynamic argument, magnetism is not possible at all in the first place in any entirely classical system, so even a solenoid would not work except thanks to the quantum nature of the particles it is made of). Cesiumfrog (talk) 23:14, 13 October 2010 (UTC)[reply]

The theorem does say that there can't be any magnetism in solids - diamagnetism, paramagnetism or ferromagnetism - but I don't know if it rules out a magnetic moment in a solenoid. In view of this, I have made a more cautious statement in diamagnetism. --RockMagnetist (talk) 00:06, 14 October 2010 (UTC)[reply]
I don't see any basis for assuming thermodynamic equilibrium. For example, in what sense could electrons be in thermodynamic equilibrium with atomic nuclei? Maybe the existence of magnetic fields simply implies a deviation from a Maxwell-Boltzmann distribution? After all, one would think that the presence of a magnetic field would change the probability that some particle would have some momentum at some particular position. Such changes would be systematic in the sense that particular motions are to be preferred than others, particularly around the axis of the magnetic moment or around induced magnetic moments.siNkarma86—Expert Sectioneer of Wikipedia
86 = 19+9+14 + karma = 19+9+14 + talk 22:06, 12 April 2011 (UTC)[reply]
In a solid, electrons can be in thermal equilibrium in the sense that they occupy energy levels with the correct Boltzmann statistics. RockMagnetist (talk) 13:21, 13 April 2011 (UTC)[reply]

Guys, the answer to this is yes, simply in the formal proof itself. Cannot justify symmetric cancellation of momentum without thermal equilibrium averaging. It is much more amusing to note that no part of the argument depended upon it being a solid. Gases, liquids, and the article also references plasmas, are all subjected to it. It really is all classical systems in thermal equilibrium. And no net motion! 119.74.52.36 (talk) 07:20, 17 October 2019 (UTC)[reply]

Informal Proof - energy dependence[edit]

"Since the distribution of motions does not depend on the energy" Shouldn't this instead read: "Since the distribution of motions depends on the energy which is, in turn, independent of the field"? — Preceding unsigned comment added by 77.54.211.85 (talk) 18:18, 10 November 2011 (UTC)[reply]

Good catch. To avoid repeating earlier text, I shortened it to "does not depend on the magnetic field". RockMagnetist (talk) 18:25, 10 November 2011 (UTC)[reply]

Formal Proof[edit]

It is written "The above equation shows that the magnetic moment is a linear function of the position coordinates". I guess it is a typo? According to the rest of the proof, being linear function in coordinate is irrelevant, it is the fact of being a linear function of the "velocity" that matters. Could someone check? Gamebm (talk) 20:40, 19 September 2014 (UTC)[reply]

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