Plateau Moduli of Several Single-Chain
Slip-Link and Slip-Spring Models

In this work, we consider slip-link and slip-spring type models. In these models, the entanglement effect is mimicked by introducing transient objects which work like cross-links. These transient objects are called the slip-links or slip-springs. Various slip-link and slip-spring type models have been proposed for entangled polymer systems^{11, 12, 13, 14, 15, 16, 17, 18, 19}. Although these models can reproduce characteristic dynamics of entangled polymers, they are not equivalent. They are phenomenologically designed based on similar but different assumptions, and thus their statistical properties are different. As far as a model can reasonably reproduce some dynamic quantities such as the relaxation modulus, we can employ any model. Here we do not limit ourselves to a specific model and study some different models.

To make the analyses tractable, in this work we limit ourselves to the single-chain type models. In a single-chain slip-link model, we consider the dynamics of a single tagged chain and assume that the slip-links are attached to the tagged chain. A segment to which a slip-link is attached is spatially fixed and cannot move freely. (This corresponds to the affine network model in the rubber elasticity theory.) The polymer chain is allowed to slip (slide) through slip-links, and if a slip-link reaches to the chain end, it is destroyed. To compensate the destroyed slip-links, slip-links can be newly constructed and put on the chain. In the rubber elasticity theory, the fluctuation of the cross-links affects the modulus. Therefore, models with fluctuations of slip-linked points would be preferred. To incorporate the fluctuations of slip-linked points, we employ the slip-spring models in this work. In the case of a single-chain slip-spring model, slip-links are replaced by slip-springs. One end of a slip-spring is fixed in space (just like the case of a slip-link), and another end is attached to a segment. Two ends of a slip-spring is connected by a harmonic spring. (The positions of slip-springed points on the chain can fluctuate in space, and we expect that this may mimic the non-affine network models such as the phantom network model, in some aspects.) The slip-springs can be dynamically destroyed and constructed in the same way as slip-links. These slip-link and slip-spring models can reproduce various characteristic dynamical behaviors of entangled polymers. The rheological properties of a model depend both on static properties (such as the interaction potential model) and dynamic properties (such as the dynamic equations and transition models). As we show in the following subsection, however, the details of dynamics is not so important for the calculation of the plateau modulus. We need only the equilibrium probability distribution function to calculate the plateau modulus.

Without slip-links and slip-springs, a tagged chain is simply modeled as an ideal Gaussian chain. It would be fair to mention that the Gaussian chain statistics is not always reasonable. For example, if the chain is stiff, we should replace the Gaussian chain statistics by other statistics such as a freely jointed chain model and models with the bending rigidity. In this work we limit ourselves to the Gaussian chain statistics which is the simplest and suitable to theoretical analyses. For simplicity, in this work we assume that the total number of segments in a chain, $N$, is given as a multiple of $N_{0}$, as $N=N_{0}Z_{0}$ with $Z_{0}$ being a positive integer. (This assumption is not essential. The results are not changed even if $N$ is not a multiple of $N_{0}$. However, the expressions become a bit complicated in such a case.) $Z_{0}$ can be interpreted as the average number of slip-linked or slip-springed subchains in the tagged chain. Because we are interested in the plateau modulus, we assume that $Z_{0}$ is sufficiently large, $Z_{0}\gg 1$. (This assumption makes the statistical properties somewhat simple.)

We start from the single-chain slip-link models. The state of the system (the tagged chain) is expressed by the positions of chain ends and slip-linked points, the number of segments in slip-linked subchains, and the total number of subchains. We express the number of subchains as $Z$ ($Z=1,2,3,\dots$) and the $k$-th constrained point on the chain as $\bm{R}_{k}$ ($k=0,1,2,\dots,Z$). $k=0$ and $Z$ correspond to the chain ends and others correspond to slip-linked points. The number of segments between the $(k-1)$-th and the $k$-th points is expressed as $n_{k}$ ($k=1,2,\dots,Z$). An image of a tagged chain with slip-links is depicted in Figure 1(a). For convenience, in what follows, we use dimensionless units. We set $\sqrt{N_{0}b^{2}/3}$, $N_{0}$, and $k_{B}T$ as the units of length, segment number, and energy. (This is equivalent to simply set $N_{0}=1$, $b^{2}/3=1$, and $k_{B}T=1$.)

Even if we limit ourselves to the single-chain slip-link models, there are various similar but different models. For example, we can select the interaction potential model for slip-links, which directly affects the statistical properties²⁰. We start from the non-interacting slip-links. In this work, we call this model as an ideal slip-link model. The free energy of the ideal slip-link model is given as²¹

$$\mathcal{F}_{\text{IL}}(\{\bm{R}_{k}\},\{n_{k}\},Z)=\sum_{k=1}^{Z}\frac{\bm{Q}_{k}^{2}}{2n_{k}}+\sum_{k=1}^{Z}\frac{3}{2}\ln n_{k}.$$

(3)

Here, the subscript “IL” represents the ideal (“I”) slip-link (“L”) model. For convenience, we have introduced the bond vector $\bm{Q}_{k}\equiv\bm{R}_{k}-\bm{R}_{k-1}$ (see Figure 1(a)). The equilibrium probability distribution for the ideal slip-link model can be obtained by using the grand canonical type formulation for slip-links²¹. In the grand canonical type formulation, slip-links are treated as a sort of ideal gas particles in one dimension. The equilibrium probability distribution can be expressed as

$$P_{\text{eq},\text{IL}}(\{\bm{R}_{k}\},\{n_{k}\},Z)=\frac{\xi^{Z-1}e^{-\xi}}{\mathcal{V}}\delta\bigg{(}Z_{0}-\sum_{k=1}^{Z}n_{k}\bigg{)}\prod_{k=1}^{Z}\frac{1}{(2\pi n_{k})^{3/2}}\exp\bigg{[}-\frac{\bm{Q}_{k}^{2}}{2n_{k}}\bigg{]},$$

(4)

where $\xi$ is the effective fugacity (activity) for slip-links, and $\mathcal{V}$ is the volume of the system. $\xi$ is determined so that the equilibrium average number of subchains becomes $Z_{0}$.

We also consider the single-chain slip-link models with the effective interaction between slip-links²⁰. Although there are many possible interaction potential models between slip-links, in this work we limit ourselves to two special cases. One is the effective repulsion potential which cancels the logarithmic term in eq (3). We call this model as the single-chain repulsive slip-link (“RL”) model. The repulsive slip-link model can be related to the primitive chain network (PCN) model¹², which is one of the multi-chain slip-link models. Some properties of the repulsive slip-link model can be directly utilized to study the PCN model. Another is a very strong repulsive potential, with which slip-links form a sort of the Wigner crystal structure on the chain. We call this model as the single-chain equidistant slip-link (“EL”) model, because the number of segments between neighboring slip-links becomes constant and thus slip-links are placed equidistantly on the chain.

The free energy of the repulsive slip-link model is given as

$$\mathcal{F}_{\text{RL}}(\{\bm{R}_{k}\},\{n_{k}\},Z)=\mathcal{F}_{\text{IL}}(\{\bm{R}_{k}\},\{n_{k}\},Z)-\sum_{k=1}^{Z}\frac{3}{2}\ln n_{k}=\sum_{k=1}^{Z}\frac{\bm{Q}_{k}^{2}}{2n_{k}}.$$

(5)

Following the grand canonical type formulation, we have the formal expression for the equilibrium probability distribution of the repulsive slip-link model.

$$\begin{split}P_{\text{eq},\text{RL}}(\{\bm{R}_{k}\},\{n_{k}\},Z)=&\frac{\xi^{Z-1}}{\mathcal{V}E_{5/2,5/2}(\xi)}\delta\bigg{(}Z_{0}-\sum_{k=1}^{Z}n_{k}\bigg{)}\frac{1}{(2\pi)^{3Z/2}}\exp\bigg{[}-\sum_{k=1}^{Z}\frac{\bm{Q}_{k}^{2}}{2n_{k}}\bigg{]}\\ \end{split}$$

(6)

where $E_{5/2,5/2}(x)$ is the generalized Mittag-Leffler function²² (the explicit from of $E_{5/2,5/2}(x)$ is not required for the calculations below). As before, the effective fugacity is determined so that the average number of subchains becomes $Z_{0}$. Although eq (6) is not simple, if we focus on the statistics of just a single subchain in the system, the probability distribution can be largely simplified. (See Appendix A.) On the other hand, in the equidistant slip-link model, the equilibrium probability distribution function becomes quite simple:

$$P_{\text{eq},\text{EL}}(\{\bm{Q}_{k}\},\{n_{k}\},Z)=\frac{\delta_{Z,Z_{0}}}{\mathcal{V}}\bigg{[}\prod_{k=1}^{Z_{0}}\delta(n_{k}-1)\bigg{]}\frac{1}{(2\pi)^{3Z_{0}/2}}\exp\bigg{[}-\sum_{k=1}^{Z_{0}}\frac{\bm{Q}_{k}^{2}}{2}\bigg{]}.$$

(7)

Eq (7) means that the equidistantly slip-linked chain behaves as a simple bead-spring chain. ($\delta_{Z,Z_{0}}$ in eq (7) is the Kronecker delta which arises from the fact that $Z$ is constant and cannot fluctuate in the equidistant slip-link model.)

In the single-chain slip-spring models, slip-linked points are allowed to fluctuate in space around their average positions (the anchoring points). One end of a slip-spring is attached to a slip-linked point on the chain, and another end is anchored in space. We express the anchoring point of the $k$-th slip-spring as $\bm{A}_{k}$ $(k=0,1,\dots,Z)$. An image of a tagged chain with slip-springs is depicted in Figure 1(b). A slip-spring is modeled as an entropic spring which can be interpreted as a short polymer chain. We express the number of segments for this short polymer chain as $N_{s}$. For simplicity, we also attach slip-springs to chain ends. (This does not affect the plateau modulus for a sufficiently long chain with$Z_{0}\gg 1$.)

We consider three different interaction models between slip-springs, as the case of the slip-link models explained above. The free energy of the ideal slip-spring model is expressed as the sum of the free energy of the ideal slip-link model and the free energy of slip-springs:

$$\mathcal{F}_{\text{IS}}(\{\bm{R}_{k}\},\{n_{k}\},Z)=\mathcal{F}_{\text{IL}}(\{\bm{R}_{k}\},\{n_{k}\},Z)+\sum_{k=0}^{Z}\frac{(\bm{R}_{k}-\bm{A}_{k})^{2}}{2\phi}.$$

(8)

The subscript “IS” represents the ideal slip-spring (“S”) model. Here, $\phi$ is the parameter which represents the effective size of slip-springs. (In dimensional units, $\phi$ is expressed as $\phi=N_{s}/N_{0}$.) We call $\phi$ as the “slip-spring size parameter” in the followings. The free energies for the repulsive and equidistant slip-spring (“RS” and “ES”) models can be constructed in the same way.

The equilibrium probability distributions for the single-chain slip-spring models can be constructed by utilizing the equilibrium probability distributions for the single-chain slip-link models. The equilibrium probability distribution for the anchoring points is given as a Gaussian distribution:

$$P_{\text{eq},\text{S}}(\{\bm{A}_{k}\}|\{\bm{R}_{k}\},\{n_{k}\},Z)=\frac{1}{(2\pi\phi)^{3(Z+1)/2}}\exp\bigg{[}-\sum_{k=0}^{Z}\frac{(\bm{R}_{k}-\bm{A}_{k})^{2}}{2\phi}\bigg{]}.$$

(9)

Here, $P_{\text{eq}}(X|Y)$ represents the conditional probability distribution of $X$ under a given $Y$. The subscript “S” represents all the three slip-springs models (“IS”, “RS”, and “ES”). The equilibrium probability distribution of the slip-spring model is thus expressed as

$$P_{\text{eq},\text{S}}(\{\bm{R}_{k}\},\{\bm{A}_{k}\},\{n_{k}\},Z)=\frac{1}{(2\pi\phi)^{3(Z+1)/2}}\exp\bigg{[}-\sum_{k=0}^{Z}\frac{(\bm{R}_{k}-\bm{A}_{k})^{2}}{2\phi}\bigg{]}P_{\text{eq},\text{L}}(\{\bm{R}_{k}\},\{n_{k}\},Z).$$

(10)

The subscript “L” represents the slip-link models. (Eq (10) is common for all the three slip-spring models examined in this work.)

We can calculate the shear relaxation modulus by evaluating the relaxation process after the step deformation. Then the plateau modulus is determined by eq (1). Although this approach is intuitive, it requires us to evaluate the relaxation process directly. However, the relaxation process is generally model-dependent, and the analytic evaluation of the relaxation process is not easy. On the other hand, according to the linear response theory, the shear relaxation modulus of a given model can be evaluated systematically by evaluating the correlation function²³. The standard framework of the linear response theory is universal and we expect that we can proceed the calculation without knowing the details of the relaxation process. In this subsection, therefore, we derive the expression of the plateau moduli for the single-chain slip-link and slip-spring models from the view point of the linear response theory.

The linear response theory claims that the response function of a physical quantity can be expressed by using the correlation function of that physical quantity and another physical quantity which is conjugate to the applied external field. (In the case of the relaxation modulus, both of them are the shear stress.) For the single-chain slip-link models, the linear response theory gives

$$G(t)=\nu_{0}\langle\hat{\sigma}_{xy}(t)\hat{\sigma}_{xy}(0)\rangle_{\text{eq}}.$$

(11)

where $\nu_{0}\equiv\rho_{0}/Z_{0}$ is the average subchain number density and $\hat{\sigma}_{xy}(t)$ is the $xy$-component of the single-chain stress tensor at time $t$. $\langle\dots\rangle_{\text{eq}}$ means the statistical average in equilibrium. (The factor $\nu_{0}$ represents the fact there are many polymer chains in a bulk system. The single-chain model gives the modulus just for a single chain. However, a bulk system consists of many statistically independent chains. We should introduce the factor $\nu_{0}$ to reproduce the relaxation modulus correctly.) From the stress-optical rule, the single-chain stress tensor is simply given as follows:

$$\hat{\bm{\sigma}}(t)=\sum_{k=1}^{Z(t)}\frac{\bm{Q}_{k}(t)\bm{Q}_{k}(t)}{n_{k}(t)}.$$

(12)

Here, the stress-optical coefficient is taken to be unity. Now we consider to calculate the plateau moduli of the slip-link models by eq (11). Clearly it is not easy to calculate $G(t)$ exactly. (It requires the detailed information of the dynamics.) Nevertheless, we are able to obtain accurate approximate expressions under several conditions. From eq (1), it is sufficient for us to calculate the value of $G(t)$ only around $t=\tau_{e}$. Besides, at this time scale, $G(t)$ is expected to be almost constant. At the time scale of $t\approx\tau_{e}$, we can reasonably assume that the slip-links are not constructed nor destroyed. Thus here we assume that the positions of slip-links are not changed. However, the transport of segments between neighboring subchains occurs even at the relatively short time scale. Thus we expect that the segment numbers in subchains are locally equilibrated at this time scale. We consider that only the segment numbers are equilibrated and other variables are unchanged at $t\approx\tau_{e}$. Then we can set $\bm{R}_{k}(t)=\bm{R}_{k}(0)=\bm{R}_{k}$ and $Z(t)=Z(0)=Z$ in eq (11), and assume that $\{n_{k}(0)\}$ and $\{n_{k}(\tau_{e})\}$ are sampled independently from the local equilibrium distribution. Under these assumptions, the plateau modulus can be approximately expressed as

$$\begin{split}G_{N,\text{L}}^{(0)}&\approx\nu_{0}\sum_{Z=0}^{\infty}\int d\{\bm{R}_{k}\}\,\bigg{[}\int d\{n_{k}\}\sum_{k=1}^{Z}\frac{Q_{k,x}Q_{k,y}}{n_{k}}P_{\text{eq},\text{L}}(\{n_{k}\}|\{\bm{R}_{k}\},Z)\bigg{]}^{2}P_{\text{eq},\text{L}}(\{\bm{R}_{k}\},Z).\end{split}$$

(13)

For the single-chain slip-spring models, we should take into account the contribution of slip-springs (the virtual stress). The linear response formula becomes as follows^{24, 25}:

$$G(t)=\nu_{0}\langle\hat{\sigma}_{xy}(t)\hat{\sigma}_{xy}(0)\rangle_{\text{eq}}+\nu_{0}\langle\hat{\sigma}_{xy}(t)\hat{\sigma}_{xy}^{(v)}(0)\rangle_{\text{eq}}.$$

(14)

Here $\hat{\sigma}_{xy}^{(v)}$ is the virtual stress tensor by slip-springs:

$$\hat{\bm{\sigma}}^{(v)}(t)=\sum_{k=0}^{Z(t)}\frac{(\bm{R}_{k}(t)-\bm{A}_{k}(t))(\bm{R}_{k}(t)-\bm{A}_{k}(t))}{\phi}.$$

(15)

It would be fair to mention that the definition of the stress in the slip-spring models is not fully clear. From the linear response theory²⁵, $\hat{\sigma}_{xy}+\hat{\sigma}_{xy}^{(v)}$ is conjugate to the applied strain, and should be employed as the stress. However, $\hat{\sigma}_{xy}+\hat{\sigma}^{(v)}_{xy}$ is generally not consistent with the stress-optical rule. Thus, from the stress-optical rule, we should employ $\hat{\sigma}_{xy}$ as the stress tensor (as the case of the slip-link models). Here we assume that the stress-optical rule holds and employ $\hat{\sigma}_{xy}$ as the stress of the slip-spring models. This gives eq (14). At $t\approx\tau_{e}$, we expect that the positions $\{\bm{R}_{k}\}$ and segment numbers $\{n_{k}\}$ are locally equilibrated while other variables are unchanged. Then we have the following approximate expression for the plateau modulus:

$$\begin{split}G_{N,\text{S}}^{(0)}&\approx\nu_{0}\sum_{Z=0}^{\infty}\int d\{\bm{A}_{k}\}\,\bigg{[}\int d\{\bm{R}_{k}\}d\{n_{k}\}\,\sum_{k=1}^{Z}\frac{Q_{k,x}Q_{k,y}}{n_{k}}P_{\text{eq},\text{S}}(\{\bm{R}_{k}\},\{n_{k}\}|\{\bm{A}_{k}\},Z)\bigg{]}\\ &\qquad\times\bigg{[}\int d\{\bm{R}_{k}\}d\{n_{k}\}\,\bigg{[}\sum_{k=1}^{Z}\frac{Q_{k,x}Q_{k,y}}{n_{k}}+\sum_{k=0}^{Z}\frac{(R_{k,x}-A_{k,x})(R_{k,y}-A_{k,y})}{\phi}\bigg{]}\\ &\qquad\times P_{\text{eq},\text{S}}(\{\bm{R}_{k}\},\{n_{k}\}|\{\bm{A}_{k}\},Z)\bigg{]}P_{\text{eq},\text{S}}(\{\bm{A}_{k}\},Z).\end{split}$$

(16)

From the expressions shown above, we can calculate the plateau moduli of slip-link and slip-spring models only from the local equilibrium probability distributions.

To proceed the calculation and obtain the explicit expressions for the plateau moduli of the slip-link and slip-spring models, we employ the single-subchain approximation. We consider only a single tagged subchain in a chain. Such an approximation enables us to analytically calculate various statistical quantities including the plateau modulus.

For the single-chain slip-link models, the state of a single subchain can be fully expressed by the number of segments in the subchain and the bond vector of the subchain. We consider the $k$-th subchain as a tagged subchain and express $n=n_{k}$ and $\bm{Q}=\bm{Q}_{k}$ (as schematically shown in Figure 2(a)). The probability distribution function for a single subchain, $P_{\text{eq}}(\bm{Q},n)$, is already obtained for several slip-link models.²⁰ The segment number distribution function $P_{\text{eq}}(n)$ and the bond vector distribution function $P_{\text{eq}}(\bm{Q})$ are calculated straightforwardly from $P_{\text{eq}}(\bm{Q},n)$. We show the simple derivation which gives the same result as the previous work²⁰ in Appendix A. In the slip-link models, the segment number distribution function is given as follows:

$$P_{\text{eq},\text{L}}(n)=\begin{cases}e^{-n}&(\text{ideal}),\\ \displaystyle\frac{25}{6}\sqrt{\frac{10}{\pi}}n^{3/2}e^{-5n/2}&(\text{repulsive}),\\ \delta(n-1)&(\text{equidistant}).\end{cases}$$

(17)

The bond vector distribution under a given $n$ is simply given as the Gaussian distribution:

$$P_{\text{eq},\text{L}}(\bm{Q}|n)=\frac{1}{(2\pi n)^{3/2}}e^{-\bm{Q}^{2}/2n}.$$

(18)

The joint probability distribution function for the segment number and the bond vector is $P_{\text{eq},\text{L}}(\bm{Q},n)=P_{\text{eq},\text{L}}(\bm{Q}|n)P_{\text{eq},\text{L}}(n)$. The bond vector distribution function is calculated as $P_{\text{eq},\text{L}}(\bm{Q})=\int dn\,P_{\text{eq},\text{L}}(\bm{Q},n)$. The result is:

$$P_{\text{eq},\text{L}}(\bm{Q})=\begin{cases}\displaystyle\frac{1}{2\pi|\bm{Q}|}e^{-\sqrt{2}|\bm{Q}|}&(\text{ideal}),\\ \displaystyle\frac{25}{6\pi^{2}}|\bm{Q}|K_{1}(\sqrt{5}|\bm{Q}|)&(\text{repulsive}),\\ \displaystyle\frac{1}{(2\pi)^{3/2}}e^{-\bm{Q}^{2}/2}&(\text{equidistant}).\end{cases}$$

(19)

Here, $K_{1}(x)$ is the first order modified Bessel function of the second kind^{26, 27}. From the Bayes’ theorem, the conditional probability distribution function of $n$ under given $\bm{Q}$ is calculated from eqs (17) and (19), as $P_{\text{eq},\text{L}}(n|\bm{Q})=P_{\text{eq},\text{L}}(\bm{Q},n)/P_{\text{eq},\text{L}}(\bm{Q})$. Thus we have

$$P_{\text{eq},\text{L}}(n|\bm{Q})=\begin{cases}\displaystyle\frac{1}{\sqrt{2\pi}}\frac{|\bm{Q}|}{n^{3/2}}\exp\bigg{(}-\frac{\bm{Q}^{2}}{2n}-n+\sqrt{2}|\bm{Q}|\bigg{)}&(\text{ideal}),\\ \displaystyle\frac{\sqrt{5}}{2}\frac{1}{|\bm{Q}|K_{1}(\sqrt{5}|\bm{Q}|)}\exp\bigg{(}-\frac{\bm{Q}^{2}}{2n}-\frac{5n}{2}\bigg{)}&(\text{repulsive}),\\ \delta(n-1)&(\text{equidistant}).\end{cases}$$

(20)

Finally, by using the equilibrium probability distribution functions (19) and (20), the plateau modulus of the single-chain slip-link model becomes

$$\begin{split}G_{N,\text{L}}^{(0)}&\approx\nu_{0}\sum_{Z=1}^{\infty}Z\bigg{[}\int d\bm{Q}\,\bigg{[}\int dn\,\frac{1}{n}P_{\text{eq},\text{L}}(n|\bm{Q})\bigg{]}^{2}Q^{2}_{x}Q^{2}_{y}P_{\text{eq},\text{L}}(\bm{Q})\bigg{]}P_{\text{eq},\text{L}}(Z)\\ &=G_{0}\int d\bm{Q}\,\bigg{[}\int dn\,\frac{1}{n}P_{\text{eq},\text{L}}(n|\bm{Q})\bigg{]}^{2}Q_{x}^{2}Q_{y}^{2}P_{\text{eq},\text{L}}(\bm{Q}).\end{split}$$

(21)

Here we have defined the characteristic modulus $G_{0}$ as $G_{0}\equiv\nu_{0}Z_{0}=\rho_{0}$ (This characteristic modulus is common for all the single-chain slip-link and slip-spring models, as shown in Appendix B.) In dimensional units, $G_{0}$ is expressed as $G_{0}=\nu_{0}k_{B}TN/N_{0}=\rho_{0}k_{B}T/N_{0}$. Therefore, $G_{0}$ corresponds to the modulus of an ideal rubber with the average subchain segment number $N_{0}$. In eq (21), the integral over $n$ explicitly depends on the distribution function $P_{\text{eq},\text{L}}(n|\bm{Q})$. Therefore, models with different distribution functions have different plateau moduli.

Next we consider the single-chain slip-spring models. In the case of the slip-spring models, we need to specify the anchoring points in addition to the number of segments in the subchain and the bond vector. Under the single-subchain approximation, we cannot simply utilize the anchoring points and the slip-spring size parameters for the full chain model. The spatial fluctuation of a slip-linked point is affected by two subchains and one slip-spring connected to the point. If we simply cut the subchains connected to a tagged subchain, the fluctuation is affected only by one subchain and one slip-spring. This situation is clearly different from that of the original tagged subchain. To recover the correct fluctuation, we should employ the effective anchoring points and the effective slip-spring size parameters. We consider the $k$-th subchain as a tagged subchain, as before, and express the effective anchoring points for the $(k-1)$-th and the $k$-th anchoring points as $\bar{\bm{A}}_{k-1}$ and $\bar{\bm{A}}_{k}^{\dagger}$, respectively. Also, we express the effective slip-spring size parameters for the $(k-1)$-th and $k$-th slip-springs as $\bar{\phi}_{k-1}$ and $\bar{\phi}_{k}^{\dagger}$. The schematic image of the single-subchain approximation for the slip-spring model is shown in Figure 2(b). The effective anchoring points and the effective slip-spring size parameters are obtained by integrating-out the degrees of freedom of other subchains. Then we can express the equilibrium probability distribution of a single subchain as $P_{\text{eq},\text{S}}(\bm{R}_{k-1},\bm{R}_{k},n_{k},\{\bm{A}_{k}\})$. Unlike the case of the slip-link model, the bond vector can be equilibrated at the entanglement time scale in the slip-spring model. To calculate the plateau modulus, we consider that the number of segments and the bond vector obey the equilibrium distribution under a given configuration of the anchoring point, $P_{\text{eq},\text{S}}(\bm{R}_{k-1},\bm{R}_{k},n_{k}|\{\bm{A}_{k}\})$. We have the following expression for the plateau modulus:

$$\begin{split}G_{N,\text{S}}^{(0)}&\approx G_{0}\int d\{\bm{A}_{k}\}\,P_{\text{eq},\text{S}}(\{\bm{A}_{k}\})\bigg{[}\int dnd\bm{R}_{k-1}d\bm{R}_{k}\,\frac{Q_{k,x}Q_{k,y}}{n_{k}}P_{\text{eq},\text{S}}(\bm{R}_{k-1},\bm{R}_{k},n_{k}|\{\bm{A}_{k}\})\bigg{]}\\ &\qquad\times\bigg{[}\int dnd\bm{R}_{k-1}d\bm{R}_{k}\,\bigg{[}\frac{Q_{k,x}Q_{k,y}}{n_{k}}+\frac{2(R_{k-1,x}-\bar{A}_{k-1,x})(R_{k-1,y}-\bar{A}_{k-1,y})}{\bar{\phi}_{k-1}}\\ &\qquad+\frac{2(R_{k,x}-\bar{A}_{k,x}^{\dagger})(R_{k,y}-\bar{A}_{k,y}^{\dagger})}{\bar{\phi}_{k}^{\dagger}}\bigg{]}P_{\text{eq},\text{S}}(\bm{R}_{k-1},\bm{R}_{k},n_{k}|\{\bm{A}_{k}\})\bigg{]}.\end{split}$$

(22)

To calculate the plateau modulus by eq (22), we should evaluate the integrals over $n_{k}$, $\bm{R}_{k-1}$, and $\bm{R}_{k}$. Unfortunately, even under a single subchain approximation, such integrals cannot be easily evaluated in general. For the equidistant slip-spring model, however, the expression reduces to be simple. Since the number of segments is constant ($n_{k}=1$), eq (22) reduces to

$$\begin{split}G_{N,\text{ES}}^{(0)}&\approx G_{0}\int d\{\bm{A}_{k}\}\,P_{\text{eq},\text{ES}}(\{\bm{A}_{k}\})\bigg{[}\int d\bm{R}_{k-1}d\bm{R}_{k}\,Q_{k,x}Q_{k,y}P_{\text{eq},\text{ES}}(\bm{R}_{k-1},\bm{R}_{k}|\{\bm{A}_{k}\})\bigg{]}\\ &\qquad\times\bigg{[}\int d\bm{R}_{k-1}d\bm{R}_{k}\,\bigg{[}Q_{k,x}Q_{k,y}+\frac{(R_{k-1,x}-\bar{A}_{k-1,x})(R_{k-1,y}-\bar{A}_{k-1,y})}{\bar{\phi}_{k-1}}\\ &\qquad+\frac{(R_{k,x}-\bar{A}_{k,x}^{\dagger})(R_{k,y}-\bar{A}_{k,y}^{\dagger})}{\bar{\phi}_{k}^{\dagger}}\bigg{]}P_{\text{eq},\text{ES}}(\bm{R}_{k-1},\bm{R}_{k}|\{\bm{A}_{k}\})\bigg{]}.\end{split}$$

(23)

Here, we note that the equidistant slip-spring model looks similar to the rubber elasticity model by Rubinstein and Panyukov^{28, 29}. Eq (23) corresponds to the comb polymer model in the Rubinstein-Panyukov model. For the ideal and repulsive slip-spring models, we employ the decoupling approximation for $n_{k}$, and $\bm{R}_{k-1}$ and $\bm{R}_{k}$. Under the decoupling approximation, we calculate the contributions of the integral over $n$ and that over $\bm{R}_{k-1}$ and $\bm{R}_{k}$, separately. The former is the same as the case of the slip-link model, and the latter is the same as the equidistant slip-spring model. Therefore, we have the following approximate expression:

$$\frac{G_{N,\text{S}}^{(0)}}{G_{0}}\approx\frac{G_{N,\text{ES}}^{(0)}}{G_{0}}\frac{G_{N,\text{L}}^{(0)}}{G_{0}}.$$

(24)

To obtain the explicit expression of the plateau modulus for the equidistant slip-spring model, we need the equilibrium distribution function for the single subchain with the effective anchoring points and the effective slip-spring sizes. Although the exact expressions become quite complicated, we can reasonably approximate them by the following forms:

	$\displaystyle\bar{\phi}_{k-1}$	$\displaystyle\approx\bar{\phi}_{k}^{\dagger}\approx\bar{\phi}_{\infty}\equiv\frac{-1+\sqrt{1+4\phi}}{2},$		(25)
	$\displaystyle\bar{\bm{A}}_{k-1}$	$\displaystyle\approx\frac{\bar{\phi}_{\infty}}{\phi}\sum_{j=0}^{\infty}\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{j}\bm{A}_{k-1-j},$		(26)
	$\displaystyle\bar{\bm{A}}_{k}^{\dagger}$	$\displaystyle\approx\frac{\bar{\phi}_{\infty}}{\phi}\sum_{j=0}^{\infty}\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{j}\bm{A}_{k+j}.$		(27)

The detailed derivations of eqs (25)-(27) are shown in Appendix C. With the effective anchoring points and the effective slip-spring size parameters by eqs (25)-(27), the equilibrium distribution function for the single subchain becomes

$$\begin{split}P_{\text{eq},\text{ES}}(\bm{R}_{k-1},\bm{R}_{k}|\{\bm{A}_{k}\})&=\frac{(1+2\bar{\phi}_{\infty})^{3/2}}{(2\pi\bar{\phi}_{\infty})^{3}}\exp\bigg{[}-\frac{(\bm{R}_{k}-\bm{R}_{k-1})^{2}}{2}-\frac{(\bm{R}_{k-1}-\bar{\bm{A}}_{k-1})^{2}}{2\bar{\phi}_{\infty}}\\ &\qquad-\frac{(\bm{R}_{k}-\bar{\bm{A}}^{\dagger}_{k})^{2}}{2\bar{\phi}_{\infty}}+\frac{(\bar{\bm{A}}_{k}-\bar{\bm{A}}^{\dagger}_{k})^{2}}{1+2\bar{\phi}_{\infty}}\bigg{]}.\end{split}$$

(28)

We calculate the plateau moduli of the single-chain slip-link models. The simplest case is the equidistant slip-link model. In the case of the equidistant slip-link model, from eq (20), the integral in eq (21) is trivial:

$$\int dn\,\frac{1}{n}P_{\text{eq},\text{EL}}(n|\bm{Q})=1.$$

(29)

Then we have the following explicit form for the plateau modulus:

$$\frac{G_{N,\text{EL}}^{(0)}}{G_{0}}\approx\int d\bm{Q}\,Q_{x}^{2}Q_{y}^{2}P_{\text{eq},\text{EL}}(\bm{Q})=1.$$

(30)

The equidistant slip-link model has no fluctuation and thus this result is physically reasonable. Roughly speaking, this result corresponds to the pure reptation model (without the relaxation by the longitudinal motion).

For the ideal or repulsive slip-link models, explicit expressions for the integral in eq (21) become a bit complicated. For the ideal slip-link model, we have

$$\begin{split}\int dn\,\frac{1}{n}\,P_{\text{eq},\text{IL}}(n|\bm{Q})&=\int_{0}^{\infty}dn\,\frac{1}{n}\frac{1}{\sqrt{2\pi}}\frac{|\bm{Q}|}{n^{3/2}}\exp\bigg{(}-\frac{\bm{Q}^{2}}{2n}-n+\sqrt{2}|\bm{Q}|\bigg{)}\\ &=\frac{1}{\bm{Q}^{2}}+\frac{\sqrt{2}}{|\bm{Q}|},\end{split}$$

(31)

and thus the plateau modulus is approximately expressed as

$$\begin{split}\frac{G_{N,\text{IL}}^{(0)}}{G_{0}}&\approx\int d\bm{Q}\,\bigg{(}\frac{1}{\bm{Q}^{2}}+\frac{\sqrt{2}}{|\bm{Q}|}\bigg{)}^{2}Q_{x}^{2}Q_{y}^{2}P_{\text{eq},\text{IL}}(\bm{Q}).\end{split}$$

(32)

The integral over $\bm{Q}$ can be simplified by using the polar coordinates, ($Q_{x}=Q\cos\theta\cos\varphi$, $Q_{y}=Q\cos\theta\sin\varphi$, and $Q_{z}=Q\sin\theta$):

$$\begin{split}\frac{G_{N,\text{IL}}^{(0)}}{G_{0}}&\approx\int_{0}^{\infty}dQ\int_{0}^{2\pi}d\theta\int_{-\pi/2}^{\pi/2}d\varphi\,Q^{2}\cos\varphi\Big{[}(1+\sqrt{2}Q)^{2}\cos^{4}\varphi\sin^{2}\theta\cos^{2}\theta\Big{]}\frac{1}{2\pi Q}e^{-\sqrt{2}Q}\\ &=\frac{2}{15}\int_{0}^{\infty}dQ\,[Q+2\sqrt{2}Q^{2}+2Q^{3}]e^{-\sqrt{2}Q}=\frac{11}{15}.\end{split}$$

(33)

Then we have the following simple expression for the plateau modulus of the ideal slip-link model:

$$\frac{G_{N,\text{IL}}^{(0)}}{G_{0}}=\frac{11}{15}\approx 0.7333.$$

(34)

Unlike the case of the equidistant slip-link model, the plateau modulus of the ideal slip-link model is lower than the characteristic modulus $G_{0}$. This can be interpreted as the effect of the fluctuation of the segment number.

We consider the repulsive slip-link model. This model has probability distributions less simpler than other slip-link models. The integral in eq (21) becomes

$$\begin{split}\int dn\,\frac{1}{n}P_{\text{eq},\text{RL}}(n|\bm{Q})&=\frac{\sqrt{5}}{2}\frac{1}{|\bm{Q}|K_{1}(\sqrt{5}|\bm{Q}|)}\int_{0}^{\infty}dn\,\frac{1}{n}\exp\bigg{(}-\frac{\bm{Q}^{2}}{2n}-\frac{5n}{2}\bigg{)}\\ &=-\frac{2}{|\bm{Q}|K_{1}(\sqrt{5}|\bm{Q}|)}\frac{\partial}{\partial(\bm{Q}^{2})}\Big{[}|\bm{Q}|K_{1}(\sqrt{5}|\bm{Q}|)\Big{]}\\ &=\frac{\sqrt{5}}{|\bm{Q}|}\frac{K_{0}(\sqrt{5}|\bm{Q}|)}{K_{1}(\sqrt{5}|\bm{Q}|)}.\end{split}$$

(35)

As the case of the ideal slip-link model, the polar coordinates are convenient to calculate the plateau modulus. The plateau modulus becomes

$$\begin{split}\frac{G_{N,\text{RL}}^{(0)}}{G_{0}}&\approx\int d\bm{Q}\,\bigg{[}\frac{\sqrt{5}}{|\bm{Q}|}\frac{K_{0}(\sqrt{5}|\bm{Q}|)}{K_{1}(\sqrt{5}|\bm{Q}|)}\bigg{]}^{2}Q_{x}^{2}Q_{y}^{2}P_{\text{eq},\text{RL}}(\bm{Q})\\ &=\frac{1}{6\pi^{2}}\int_{0}^{\infty}d\zeta\int_{0}^{\infty}d\theta\int_{-\pi/2}^{\pi/2}d\varphi\,\frac{K_{0}^{2}({\zeta})}{K_{1}(\zeta)}\zeta^{5}\cos^{5}\varphi\cos^{2}\theta\sin^{2}\theta\\ &=\frac{2}{45\pi}\int_{0}^{\infty}d\zeta\,\zeta^{5}\frac{K_{0}^{2}({\zeta})}{K_{1}(\zeta)},\end{split}$$

(36)

where we have set $\zeta=\sqrt{5}|\bm{Q}|$. By numerically evaluating the integral over $\zeta$, we have the following expression for the plateau modulus for the repulsive slip-link model:

$$\frac{G_{N,\text{RL}}^{(0)}}{G_{0}}\approx 0.8214.$$

(37)

Thus we find that the plateau modulus of the repulsive slip-link model is smaller than that of the equidistant model yet is larger than that of the ideal slip-link model. The fluctuation of the segment number in the repulsive slip-link model is relatively suppressed (by the repulsive interaction), and thus the resulting plateau modulus becomes larger than that of the ideal slip-link model.

As we mentioned, we calculate the plateau modulus of the slip-spring models by utilizing the decoupling approximation (eq (24)). Therefore, what we need to calculate here is the plateau modulus of the equidistant slip-spring model by eq (23). The integrals over $\bm{R}_{k-1}$ and $\bm{R}_{k}$ in eq (23) become as follows:

$$\begin{split}&\int d\bm{R}_{k-1}d\bm{R}_{k}\,Q_{k,x}Q_{k,y}P_{\text{eq},\text{ES}}(\bm{R}_{k-1},\bm{R}_{k}|\{\bm{A}_{k}\})\\ &=\frac{(\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x})(\bar{A}_{k,y}^{\dagger}-\bar{A}_{k-1,y})}{(1+2\bar{\phi}_{\infty})^{2}},\end{split}$$

(38)

$$\begin{split}&\int d\bm{R}_{k-1}d\bm{R}_{k}\,\frac{(R_{k-1,x}-\bar{A}_{k-1,x})(R_{k-1,y}-\bar{A}_{k-1,y})}{\bar{\phi}_{\infty}}P_{\text{eq},\text{ES}}(\bm{R}_{k-1},\bm{R}_{k}|\{\bm{A}_{k}\})\\ &=\int d\bm{R}_{k-1}d\bm{R}_{k}\,\frac{(R_{k,x}-\bar{A}_{k,x}^{\dagger})(R_{k,y}-\bar{A}_{k,y}^{\dagger})}{\bar{\phi}_{\infty}}P_{\text{eq},\text{ES}}(\bm{R}_{k-1},\bm{R}_{k}|\{\bm{A}_{k}\})\\ &=\frac{\bar{\phi}_{\infty}(\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x})(\bar{A}_{k,y}^{\dagger}-\bar{A}_{k-1,y})}{(1+2\bar{\phi}_{\infty})^{2}}.\end{split}$$

(39)

Then eq (23) can be rewritten in a simple form as

$$\begin{split}\frac{G_{N,\text{ES}}^{(0)}}{G_{0}}&\approx\frac{1}{(1+2\bar{\phi}_{\infty})^{3}}\sum_{j,k=1}^{Z_{0}}\int d\{\bm{A}_{k}\}\,(\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x})^{2}(\bar{A}_{k,y}^{\dagger}-\bar{A}_{k-1,y})^{2}P_{\text{eq},\text{ES}}(\{\bm{A}_{k}\})\\ &=\frac{1}{(1+2\bar{\phi}_{\infty})^{3}}\langle(\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x})^{2}\rangle_{\text{eq}}^{2}.\end{split}$$

(40)

Thus we find that we can calculate the plateau modulus of the equidistant slip-spring model once the explicit expression for the variance for $\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x}$ is obtained.

The vector $\bar{\bm{A}}_{k}^{\dagger}-\bar{\bm{A}}_{k}$ can be rewritten in terms of the bond vector for the anchoring point, $\bm{U}_{k}\equiv\bm{A}_{k}-\bm{A}_{k-1}$, as

$$\begin{split}\bar{\bm{A}}_{k}^{\dagger}-\bar{\bm{A}}_{k-1}&=\frac{\bar{\phi}_{\infty}}{\phi}\sum_{j=0}^{\infty}\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{j}(\bm{A}_{k+j}-\bm{A}_{k-1-j})\\ &=\frac{\bar{\phi}_{\infty}}{\phi}\sum_{j=0}^{\infty}\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{j}(\bm{U}_{k+j}+\dots+\bm{U}_{k-j})\\ &=\sum_{j=-\infty}^{\infty}\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{|j|}\bm{U}_{k+j}.\end{split}$$

(41)

The bond vectors $\{\bm{U}_{k}\}$ obey the Gaussian distribution, and the mean and variance are given as

$$\langle\bm{U}_{k}\rangle_{\text{eq}}=0,\qquad\langle\bm{U}_{k}\bm{U}_{j}\rangle_{\text{eq}}=\begin{cases}(1+2\phi)\bm{1}&(k=j),\\ -\phi\bm{1}&(k=j\pm 1),\\ 0&(\text{otherwise}).\end{cases}$$

(42)

Here, $\bm{1}$ represents the unit tensor. The detailed calculations are shown in Appendix C. From eqs (41) and (42), the variance of $\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x}$ is calculated to be

$$\begin{split}\langle(\bar{A}_{k,x}^{\dagger}-\bar{A}_{k-1,x})^{2}\rangle_{\text{eq}}&=\sum_{j,l=-\infty}^{\infty}\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{|j|+|l|}\langle U_{k+j,x}U_{k+l,x}\rangle_{\text{eq}}\\ &=\sum_{j=-\infty}^{\infty}\left[(1+2\phi)\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{2|j|}-2\phi\left(1-\frac{\bar{\phi}_{\infty}}{\phi}\right)^{|j|+|j+1|}\right]\\ &=\frac{2(1+2\phi)}{1-(1-\bar{\phi}_{\infty}/\phi)^{2}}-(1+2\phi)-\frac{4\phi(1-\bar{\phi}_{\infty}/\phi)}{1-(1-\bar{\phi}_{\infty}/\phi)^{2}}\\ &=1+2\bar{\phi}_{\infty}.\end{split}$$

(43)