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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?

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  • $\begingroup$ Why do you say that in the bead-spring model interactions are entirely energetic? $\endgroup$ – valerio Jan 9 '18 at 16:00
  • $\begingroup$ The way I've seen the bead-spring model used is that the Hamiltonian of a configuration is calculated, then properties like the probability of the configuration is given by $exp\{ \frac{-U}{kT} \}$, which seemed a purely energetic description $\endgroup$ – The Hagen Jan 9 '18 at 16:07
  • $\begingroup$ Are the entropic contributions to the free energy of a bead-spring chain the same as that for the ideal chain? $\endgroup$ – The Hagen Jan 9 '18 at 16:15
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    $\begingroup$ Yes, the entropic contributions have the same form. Also, strictly speaking the bead-spring chain is a kind of ideal chain. $\endgroup$ – valerio Jan 9 '18 at 16:41
  • $\begingroup$ Sorry, I replied too quickly and said something wrong: the bead-spring chain is a kind of ideal chain, so their free energy will be in both cases purely entropic. See my answer for more details. $\endgroup$ – valerio Jan 9 '18 at 18:29
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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.

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  • $\begingroup$ Thanks! I didn't realize bead-spring was a kind of ideal chain. I went back to my notes, and now have the understanding that a Gaussian chain assumes additionally that the length of each Kuhn segment is Gaussian-distributed, and the spring potential is a way of mechanically making this happen. However while there are no long-range interactions, there is a spring potential energy between neighboring beads, so there should be some potential energy of the entire polymer. $\endgroup$ – The Hagen Jan 9 '18 at 19:44
  • $\begingroup$ For example on slide 15 of this presentation theorie.physik.uni-muenchen.de/lsfrey/teaching/archiv/sose_06/…, the probability distribution of the length of one segment is calculated from this potential energy. Summing over all segments gives the same end-to-end probability distribution as the ideal chain, but in the former case the probability distribution came from the energy, whereas in the latter it came from an entropy calculation. $\endgroup$ – The Hagen Jan 9 '18 at 19:44
  • $\begingroup$ @TheHagen See updated answer. Hope it can clarify your doubts. $\endgroup$ – valerio Jan 9 '18 at 23:46

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