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Re: [Phys-l] equipartition riddle



Regarding:

Here's a scenario ... with a riddle in part (d).

a) Suppose we have a harmonic oscillator i.e. a particle in a
parabolic potential well. It is in thermal equilibrium at
temperature T. We are interested in the classical limit, i.e.
where kT is high compared to ℏω. In this case the average
energy is kT i.e. 1/2 kT per degree of freedom, where the PE
is one degree of freedom and the KE is another.

Actually, to be precise, the Equipartition Theorem (for the classical limit for thermally equilibrated degrees of freedom) claims the average energy is (1/2) kT per *quadratic* degree of freedom. And even in that case, each quadratic degree of freedom in question is required to have a positive coefficient on the quadratic term in the Hamiltonian for the quadratic polynomial in which that quadratic degree of freedom appears as an argument. The thermal energy in question is measured from the classical ground state energy for that degree of freedom, which is at the minimum of the quadratic polynomial in question (which only *has* a minimum, rather than a maximum, when the quadratic coefficient is positive). The Equipartition Theorem does have a number of stipulations for its validity which, if not honored, will typically compromise the validity of its conclusion.

b) Same as above, except that the particle is in a square well
potential. In this case the average energy is 1/2 kT, not kT.
Apparently in this case the KE counts as a degree of freedom
but the PE does not.

The KE does count as a quadratic degree of freedom, assuming kT << mc^2 so that the KE function actually *is* a quadratic function. Otherwise, the KE function for a particle in a relativistically hot environment results in an average thermal energy which involves a modified Bessel function (as I recall) whose order depends on the dimension of space in which the particle moves. But if the temperature becomes relativistically hot then particle-antiparticle pair creation and destruction processes can become relevant, and a single particle system can turn into a Grand Canonical Ensemble of multiple particles and antiparticles whose average number is regulated by the relevant chemical poential. In such a situation a fixed particle number Canonical Ensemble becomes untenable.

The PE for the square well is always zero for all realizable states of the particle. Its thermal average value is zero (relative to the bottom of the well, that is). Such a degree of freedom with a restricted domain is not a quadratic degree of freedom.

c) By way of formalism, note that we can consider the two
previous cases together, as follows: We say the potential goes
like x^N, where in case (a) we have N = 2, and in case (b) we
have N = ∞.

I think you might want it to go like K*|x/a|^N where x = 0 corresponds to the center of the well and 2*a is the width of the well in the N -> ∞ limit. The constant K is the positive PE the particle has at x = +/- a for any finite N-value.

For the N = 2 SHO PE function the 2*K/a^2 corresponds to the spring constant's value. When N=2 the PE is a quadratic degree of freedom giving (1/2) kT for its average thermal energy. But for any other N-value the degree of freedom is not quadratic, and its average energy is (1/N) kT instead. In the square well N -> ∞ limit its average energy goes to zero.

d) So the question arises, what about the intermediate case?
What happens in the case where the potential goes like x^10,
where 10 is bigger than 2, but a lot less than infinity?

In that case (in the classical and Newtonian limit) the PE contributes (1/10) kT to the energy and the KE contributes (1/2) kT to the energy (for the single dimension of motion defined by the excursion of the variable x). The total average thermal energy in this N = 10 case is (3/5) kT.

How do we interpolate between two degrees of freedom and one
degree of freedom?

It's not a matter of interpolating between integer numbers of degrees of freedom. It's a matter of deciding which degrees of freedom can be reasonably approximated as being of a *quadratic* form, and which can not. Those that cannot be considered as quadratic will not be contributing a (1/2) kT to the total thermal energy. They will be contributing something else, the precise value of which depends on the details of the struction of the functions defining the contributions of those degrees of freedom to the system's microscopic Hamiltonian. If a degree of freedom contributes a power law function to the Hamiltonian for the absolute value of that degree of freedom, then the contribution to the average total thermal energy will be (1/N) kT when N is the exponent of the power law in question. If the functional form of the contribution to the Hamiltonian is something else, then its contribution to the total average thermal energy will be something else [perhaps involving a modified Bessel Function as for a contribution to the Hamiltonian that goes like E = sqrt((m*c^2)^2 + (p*c)^2)].

BTW, for a generic dynamical degree of freedom that has a ground state configuration, we typically expect that the minimal energy configuration, if it is a nonsingular point, to be, typically, a quadratic minimum assuming the degree of freedom is a normal sufficiently smooth and differentiable parameter. This means that for states whose energy value is close to this minimal ground state value the functional form for the contribution of the degree of freedom will be approximately a quadratic function that curves upward. This means that if kT is sufficiently close to the minimum value the degree of freedom will act like a quadratic one, and count as one, for the purposes of the Equipartition Theorem. But we need to have a degree of freedom that is massive enough so that there is a temperature range such that the temperature is high enough so quantum mechanical complications at low temperatures will not spoil the Equipartition result, but yet have a temperature not so very high that an-harmonic non-quadratic terms in the energy function (such as relativistic corrections or anharmonic vibrational corrections near the melting point of a normal solid) will also not spoil it.

For valence electrons in solid crystals at room temperature the temperature is still much too low to be able to treat them classically, and that is the main reason why the classical Drude model for conductors gives such bad results for typical solids. Such electrons need to be properly treated as quantum entities obeying Fermi-Dirac statistics, not independent classical particles. For electrons in a white dwarf near the Chandrasekhar limit the Fermi energy is high enough that relativistic considerations become relevant. In this case both quantum *and* relativisitic non-quadratic corrections are relevant, and the Equipartition Theorem breaks down two ways--spoiled by both quantum effects and relativistic ones.

For massless photons each one of them contributes a KE to the Hamiltonian of the form E = |p|*c in a 3-dimensional space, and each one by itself, would contribute 3 kT to the total energy. But massless photons don't conserve particle number in their environment, have a zero value for their chemical potential, have their quantum mechanical Bose-Einstein statistics enumerated in a Grand Canonical Ensemble rather than each particle acting like an independently conserved classical particle in a canonical ensemble. This is why the Planck blackbody energy in a given volume of space has a total energy E = 2.701178033*N*k*T rather than simply E = 3*N*k*T (where N is the number of photons in the volume), which would be the case for simple independent classical massless particles.

Note: This riddle bugged me for many many years. I learned the
answer a few days ago.

===========

Give-away: Sturge Appendix G.

David Bowman