Understanding roles of energy and entropy in different polymer models Polymers can be modeled as ideal chains and bead-springs (among other various models). However, in an ideal chain, the interactions are entirely entropic, whereas in the bead-spring model, there are also energetic interactions. Does this lead to differences while calculating thermodynamic quantities like the partition function and free energy?
 A: The free energy of a polymer chain depends on the number of Kuhn monomers $N$ and on the end-to-end distance $\mathbf R$:
$$F(N,\mathbf R) = U(N,\mathbf R)-T S(N,\mathbf R)$$
An ideal chain is a chain which has no long-range correlation between its bonds. There are several ideal chain models: freely jointed, freely rotating, worm-like, bead-spring...
In the assumption that 
$$R \ll N b = \text{length of the stretched chain}$$
he distribution of the end-to-end vector of an ideal chain is Gaussian:
$$p(N,\mathbf R) = \left(\frac 3 {2 \pi N b^2}\right)^{\frac 3 2} \exp \left(-\frac{3 R^2}{2Nb^2}\right) \tag{1}$$
For a Gaussian chain (a chain with end-to-end vector distribution given by $(1)$), the entropy is
$$S(N,\mathbf R)=-\frac{3 k_B T R^2}{2N b^2}+S(N,0) \tag{2}$$
where $R\equiv|\mathbf R|$.(see Rubinstein, Colby, Polymer Physics, Chapter 2). For higher values of $R$, $(1)$ is in general not valid and the expression of the free energy becomes more complicated.
Eq. $(1)$ also follows from assuming that the distance between neighboring monomers $\Delta \mathbf r$ is a Gaussian:
$$p(\Delta \mathbf r) = \left(\frac 3 {2 \pi b^2}\right)^{\frac 3 2} \exp \left(-\frac{3 \Delta r^2}{2b^2}\right) \tag{3}$$
where
$$\langle \Delta r^2 \rangle = b^2$$
Equivalently, to obtain $(1)$ we can start by assuming that the system is described by the bead-spring Hamiltonian
$$H = E_{kin} + \frac{3 k_B T}{2 b^2} \sum_{n=1}^N |\mathbf r_n - \mathbf r_{n-1}|^2$$
(see also this document). In this sense, the bead-spring model is a mechanical realization of the Gaussian chain.
For an ideal chain, the energy does not depend on $\mathbf R$ since there is no long-range interaction:
$$U(N,\mathbf R)=U(N,0)$$
The exact form of $U(N,0)$ depends on the chosen model. For the bead-spring model, it can be calculated by taking the thermodynamic average of the Hamiltonian, which you can find for example in Tuckerman, Statistical mechanics - Theory and molecular simulation, Chapter 4.
However, this term is not important since the free energy is defined up to an additive constant. We can therefore set $U(N,0)=0$, so that the free energy will just be $F=-TS$. In this sense, the free energy of an ideal chain is purely entropic.
