Talk:Bohr–Van Leeuwen theorem
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Restricted to thermal equilibrium?
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)
- 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)
- 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)
- 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
- 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)
Informal Proof - energy dependence
"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)