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Under what conditions does a system with many degrees of freedom satisfy the equipartition theorem?

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There are many comments on the conditions that must be satisfied in the wikipedia article: – Rafael Jan 24 '11 at 13:25
Thanks Rafael. The section on ET for systems described in terms of generalized coordinates makes all very clear. – Johannes Jan 25 '11 at 3:40
up vote 4 down vote accepted

the system must be in a thermal equilibrium - a condition that allows one to assign a uniform temperature to the whole system. Strongly interacting systems reach it more quickly, weakly interacting systems may need a longer time.

But once a system reaches the thermal equilibrium, it is guaranteed that each subsystem and/or each average degree of freedom of a certain kind carries the energy that is attributed to it by the equipartition theorem - e.g. $kT/2$ for a typical classical degree of freedom (translations or rotations). This is a statistical statement, so you need to look at many atoms or degrees of freedom of the same kind ($N$ of them) and take their average energy. When you do it right, the average energy will be given by the figure implied by the equipartition theorem plus minus an error - the relative error goes to zero as $1/\sqrt{N}$ for many atoms or degrees of freedom.

Of course, one must be careful what the actual prediction of the equipartition theorem for a "degree of freedom" is. The figure $kT/2$ only holds if the degree of freedom is de facto classical and more or less decoupled from others - at the given frequency. For very strongly coupled systems, the "number of degrees of freedom" is less clearly defined, and so is the energy predicted by the equipartition theorem for one of them.

Cheers LM

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Thanks for your answer Lubos. I was puzzled as (in another discussion here) the remark was made that equipartition holds only in very particular circumstances. The way I now understand it is that we need to distinguish two issues: the 'generic idea' of equipartition (which holds for virtually any equilibrium system) and equipartition with a precise energy of 1/2 kT assigned to each degree of freedom (which, strictly speaking, only holds for weakly interacting classical harmonic oscillators). – Johannes Jan 25 '11 at 3:35
Sure, @Johannes, one must carefully distinguish "any [but fixed] prediction for the distribution of energy" and "a particular value for a degree of freedom". The latter is a much stronger statement. – Luboš Motl Jan 25 '11 at 17:54
@Johannes: I just noticed this--- it's 1/2kT kinetic and 1/2kT potential for a Harmonic oscillation, and only 1/2kT kinetic for translation and rotational degrees of freedom, where there is no restoring potential. It interpolates between 1/2kT and kT for potentials that bend up more than quadratic. – Ron Maimon Feb 14 '12 at 13:09

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