Question on the temperature dependence of the partition function Let's just say we're looking at the classical continuous canonical ensemble of a harmonic oscillator, where:
$$H = \frac{p^2}{2m} + \frac{1}{2}  m  \omega^2  x^2$$
and the partition function (omitting the integrals over phase space here) is defined as
$$
Z = e^{-\frac{H} {k_B  T}}
$$
and the average energy can be calculated as proportional to the derivative of ln[Z].
Equipartion theorem says that each independent coordinate must contribute R/2 to the systems energy, so in a 3D system, we should get 3R. My question is does equipartion break down if the frequency is temperature dependent?
Let's say omega = omega[T], then when you take the derivative of Z to calculate the average energy. If omega'[T] is not zero, then it will either add or detract from the average kinetic energy and therefore will disagree with equipartition. Is this correct?
 A: In the formulation of the question there are a couple of mistakes, or simply there is no definition of $R$. Statistical Mechanics says that for a classical system each independent quadratic coordinate of the phase space contributes to the energy with a factor $\frac{1}{2}k_BT$.
Now, what about statistical mechanics and equipartition if the Hamiltonian has some  parameter depending on $T$?
First of all, let's notice that this is something which may happen at least in two interesting situations:


*

*all mean field approximations introduce pseudo-Hamiltonians depending on one parameter which is obtained through some equation depending on $T$, thus making the pseudo-Hamiltonian a function of $T$;

*when a partial trace over some but not all the degrees of freedom is performed (for example when an effective interaction between solute particles is obtained by averaging over the solvent particles.


At the level of evaluating the partition function it is true that a possible dependence of the parameters of the Hamiltonian on $T$ does not change anything and, in the case of harmonic oscillators, calculations can be performed up the calculation of the partition function in analytic form. However, many other results and relations between thermodynamic functions may be heavily modified. In general some thermodynamic inconsistency appears, i.e. expressions which are equivalent in standard, non temperature-dependent Hamiltonians, become not equivalent and provide different results in the presence of $T$ dependence.
In general this is a serious problem. For example, in the usual Statistical Mechanics,  in the case of the canonical ensemble, it is trivial to show that 
$$
\frac{\partial{\log Z}}{\partial{\beta}}=-U,
$$
where $U$ is the internal energy and $\beta=\frac{1}{k_B T}$. If the Hamiltonian  $H$ depends on $\beta$ an additional term , equal to $-\left< \beta \frac{\partial{H} }{\partial{\beta}}\right> $ would be present in general ( $\left< \dots \right> $ indicates the canonical averge). This is equivalent to say that the relation between Helmholtz free energy and internal energy is not the same as provided by thermodynamics. Even if there are cases where this additional term vanishes (the situation 1. above), one has to take into account that also relations between second derivatives are heavily  modified.
In the canonical ensemble, in the case of a system of harmonic oscillators with $T$ dependent frequencies $\omega$, equipartition theorem still holds in the sense that the average energy per degree of freedom is still  $\frac{1}{2}k_BT$, but the relation between internal energy, Helmholtz free energy, $\log Z$ and specific heat is not the usual one.
A: The equipartion theorem states you get kb*T/2 for each quadratic degree of freedom, not just each dimension.  It is very specific and comes from doing the quadratic integrals explicitly, as these are Gaussian their result is well known.  So, to answer your question, if w(T) was included then the integral technique would not change but yes, you would get a different coefficient out front.  
