Monomer motion

We shall now demonstrate that according to our model the mean quadratic displacement  of a typical monomer behaves like in Fig. (8.3). This behaviour has been qualitatively verified by computer simulations for times up to somewhere in the third regime. Of course the final regime should be simple diffusive motion. The important prediction is the dependence of the diffusion constant on N.

  

Figure 8.3: Double-logarithmic plot of the mean square displacement, , in case of the reptation model (solid line) and the Rouse model (dashed).

In Fig. (8.3) $\tau _{R}$ is the Rouse time  which is equal to $\tau_1$ in chapter 6. The meaning of $\tau _{e}$ and $\tau _{d}$ will become clear in the remaining part of this section.

We shall now treat the different regimes in Fig. (8.3) one after another.
i) $t\le \tau _{e}$.
At short times a Rouse bead doesn't know about any tube constraints. According to section 6.5 then

$$\langle ( \vec{R}_{n}(t)-\vec{R}_{n}(0))^{2}\rangle =\left( \frac{4kTb^{2}}{3\pi \gamma }\right) ^{\frac{1}{2}}\sqrt{t}.$$ (8.3)

Consistent with the Gaussian character of the primitive chain we may suppose that the diameter of the tube  is equal to a. Once the segment has moved a distance a, it will feel the constraints of the tube, and a new regime will set in. The time at which this happens is given by

$$\tau _{e}= \frac{3}{4}\pi a^{4}\frac{\gamma }{kTb^{2}}$$ (8.4)

Notice that this is independent of N.
ii) $\tau _{e}<t\le \tau _{R}$.
On the time and distance scale we are looking now, the bead performs random motions, still constrained by the fact that the monomer is a part of a chain because $t\le \tau _{R}$. Orthogonally to the primitive chain these motions do not lead to any displacement, because of the constraints implied by the tube. Only along the primitive chain the bead may diffuse free of any other constraint than the one implied by the fact that it belongs to a chain. The diffusion therefore is given by the 1-dim. analog of Eq. (6.47) or Eq. (8.3).

$$\langle (s_{n}(t)-s_{n}(0))^{2}\rangle =\frac{1}{3}\left( \frac{4kTb^{2}}{ 3\pi \gamma }\right) ^{\frac{1}{2}}\sqrt{t}$$ (8.5)

where s_n(t) is the position of bead n along the primitive chain at time t. It is assumed here that for times $t\le \tau _{R}$ the chain as a whole does not move, i.e. that the primitive chain does not change. Using Eq. (8.1) then

$$\langle (\vec{R}_{n}(t)-\vec{R}_{n}(0))^{2}\rangle =\left( \f... ...) ^{1/4}ab^{1/2}\left( \frac{kT}{\gamma }\right) ^{1/4}t^{1/4}$$ (8.6)

where we have assumed $\langle \vert s_{n}(t)-s_{n}(0)\vert\rangle \approx \langle (s_{n}(t)-s_{n}(0))^{2}\rangle ^{1/2}$.
iii) $\tau _{R}<t\le \tau _{d}$.
The bead still moves along the tube diameter. Now however $t>\tau _{R}$, which means that we should use the 1-dim. analog of Eq. (6.45 ).

$$\langle (s_{n}(t)-s_{n}(0))^{2}\rangle =2D_{G}t$$ (8.7)

Again assuming that the tube doesn't change appreciably during time t, we get

$$\langle (\vec{R}_{n}(t)-\vec{R}_{n}(0))^{2}\rangle =\left( \frac{2kT}{\gamma }\right) ^{1/2}a\frac{1}{\sqrt{N+1}}\sqrt{t}.$$ (8.8)

From our treatment it is clear that $\tau _{d}$ is the time it takes for the chain to create a tube which is uncorrelated to the old one. We will calculate $\tau _{d}$ in the next paragraph.
iv) $\tau _{d}<t$.
This is the regime in which reptation  dominates. On this time and space scale we may attribute to every bead a definite value of s. We then want to calculate

$$\varphi (s,t)=\langle ( \vec{R}(s,t)-\vec{R}(s,0))^{2}\rangle$$ (8.9)

where $\vec{R}(s,t)$ is the position of bead s at time t.

  

Figure 8.4: Motion of the primitive chain along its contour.

In order to calculate $\varphi (s,t)$ it is useful to introduce

$$\varphi (s,s^{\prime };t)=\langle (\vec{R}(s,t)-\vec{R}(s^{\prime },0))^{2}\rangle$$ (8.10)

i.e. the mean square distance between bead s at time t and bead $s^{\prime }$ at time zero. According to Fig. (8.4), for all s, except s=a and s=L, we have

$$\varphi (s,s^{\prime };t+\Delta t)=\langle \varphi (s+\Delta \xi ,s^{\prime };t)\rangle$$ (8.11)

where $\Delta \xi$ according to the definition of the primitive chain in section 2 is a stochastic variable. The average on the right hand side has to be taken over the distribution of $\Delta \xi$. Expanding the right hand side of Eq. (8.11) we get

$${\langle \varphi (s+\Delta \xi ,s\prime ;t)\rangle }$$

$$\approx \varphi (s,s^{\prime };t)+\langle \Delta \xi \rangle \fra... ... \xi )^{2}\rangle \frac{\partial ^{2}}{\partial s^{2}}\varphi (s,s^{\prime };t)$$

$$\approx \varphi (s,s^{\prime };t)+D_{G}\Delta t\frac{\partial ^{2}}{ \partial s^{2}}\varphi (s,s^{\prime };t).$$ (8.12)

Introducing this into Eq. (8.11) and taking the limit for $\Delta t$ going to zero, we get

$$\frac{\partial }{\partial t}\varphi (s,s^{\prime };t)=D_{G}\frac{\partial ^{2}}{\partial s^{2}}\varphi (s,s^{\prime };t).$$ (8.13)

In order to complete our description of reptation we have to find the boundary conditions going with this diffusion equation. We will demonstrate that these are given by

  

$$\varphi (s,s^{\prime};t) \vert _{t=0} a\vert s-s^{\prime}\vert$$ = (8.14)

$$\frac{\partial}{\partial s} \varphi (s,s^{\prime};t) \vert _{s=L}$$ = a (8.15)

$$\frac{\partial}{\partial s} \varphi (s,s^{\prime};t) \vert _{s=0}$$ = -a . (8.16)

The first of these is obvious. The second follows from

$${ \frac{\partial}{\partial s} \varphi (s,s';t) \vert _{s=L} }$$

$$= 2 \langle \frac{\partial \vec{R}(s,t)}{\partial s} \vert _{s=L} \cdot (\vec{ R}(L,t)-\vec{R}(s^{\prime},0)) \rangle$$

$$= 2 \langle \frac{\partial \vec{R}(s,t)}{\partial s} \vert _{s=L} \cdot (\vec{ R}(L,t)-\vec{R}(s^{\prime},t)) \rangle +$$

$$2 \langle \frac{\partial \vec{R}(s,t)}{\partial s} \vert _{s=L} \cdot (\vec{R}(s^{\prime},t)-\vec{R}(s^{\prime},0)) \rangle$$

$$= 2 \langle \frac{\partial \vec{R}(s,t)}{\partial s} \vert _{s=L} \cdot (\vec{ R}(L,t)-\vec{R}(s^{\prime},t)) \rangle$$

$$= \frac{\partial}{\partial s} \langle (\vec{R}(s,t)-\vec{R} (s^{\prime},t))^2 \rangle \vert _{s=L}$$

$$= \frac{\partial}{\partial s} a\vert s-s^{\prime}\vert _{s=L}$$ (8.17)

Condition Eq. (8.16) follows from a similar reasoning.

We now solve Eqs. (8.13)-(8.16), obtaining

 

$${ \varphi (s,s';t) = \vert s-s'\vert a + 2 D_G \frac{a}{L} t }$$

$$+ 4 \frac{La}{\pi^2} \sum_{k=1}^{\infty} \frac{1}{k^2} (1-e^{-tk^... ... \left( \frac{k\pi s}{L} \right) \cos \left( \frac{ k\pi s^{\prime}}{L} \right)$$ (8.18)

$$\tau_d = \frac{1}{\pi^2} \frac{b^4}{a^2} \frac{\gamma}{ k_BT} N^3 .$$ (8.19)

We shall not derive this here. One may check that Eq. (8.18) indeed is the solution to Eq. (8.13) satisfying (8.14)-(8.16).

Now taking the limit $s\rightarrow s^{\prime }$ we get

$${\langle (\vec{R}(s,t)-\vec{R}(s,0))^{2}\rangle =2D_{G}\frac{a}{L}t+}$$

$$4\frac{La}{\pi ^{2}}\sum_{k=1}^{\infty }\cos ^{2}\left( \frac{k\pi s}{L} \right) (1-e^{-tk^{2}/\tau _{d}})\frac{1}{k^{2}}.$$ (8.20)

For $t>\tau _{d}$ we get diffusive behaviour with diffusion constant

$$D=\frac{1}{3}D_{G}\frac{a}{L}=\frac{1}{3}\frac{a^{2}}{b^{2}}\frac{k_{B}T}{ \gamma }\frac{1}{N^{2}}.$$ (8.21)

Notice that this is proportional to N^-2, whereas the diffusion coefficient of the Rouse model was proportional to N^-1. The reptation result, N^-2, is confirmed by experiments which measured the diffusion coefficients of polymer melts as a function of their molecular weight.