C. Calculation of the stress tensor

We now must calculate $\langle X_{kx}(t)X_{ky}(t)\rangle$ under the conditions given in Eqs. (6.50), (6.51) and (6.52). Let us write the flow like $\vec{v}(t)=\Theta (t) \dot{\bar{ \gamma}}\cdot \vec{r}$. It is not difficult to understand that under these conditions, the equations of motion read

$$\frac{d\vec{R}_{0}}{dt} -\frac{3k_{B}T}{\gamma b^{2}}(\vec{R}_{0}-\vec{R} _{1})+\dot{\bar{\gamma}}\cdot \vec{R}_{0}+\vec{f}_{0}$$ = (6.58)

$$\frac{d\vec{R}_{n}}{dt} -\frac{3k_{B}T}{\gamma b^{2}}(2\vec{R}_{n}-\vec{R} _{n-1}-\vec{R}_{n+1})+\dot{\bar{\gamma}}\cdot \vec{R}_{n}+\vec{f}_{n}$$ = (6.59)

$$\frac{d\vec{R}_{N}}{dt} -\frac{3k_{B}T}{\gamma b^{2}}(\vec{R}_{N}-\vec{R} _{N-1})+\dot{\bar{\gamma}}\cdot \vec{R}_{N}+\vec{f}_{N}$$ = (6.60)

$$\langle \vec{f}_{n}(t)\cdot \vec{f}_{m}(t^{\prime })\rangle 6D\delta _{n,m}\delta (t-t^{\prime })$$ = (6.61)

Each bead simply gets an extra velocity equal to the average velocity at its instantaneous position.

The reader who wishes to explicitly check the above equations, may start with Eq. (4.2) and replace the particles velocity $\vec{v}$ in the friction force by its velocity relative to the average velocity, i.e. replace $-\xi \vec{v}$ by $-\xi (\vec{v}-\dot{\bar{\gamma}}\cdot \vec{r})$. Going all the way to Eq. (4.39) he/she will find that also there $-\xi \vec{V}$ is replaced by $-\xi (\vec{V}-\dot{\bar{\gamma}}\cdot \vec{r})$. Next, putting $D\vec{V}/Dt$ equal to zero, and solving for $\vec{V}$, he/she will find an extra term $\dot{\bar{\gamma}}\cdot \vec{r}$. This finally will lead to an extra term in the Smoluchowski equation, which can only be obtained from the Langevin equation  Eq. (4.27) if it is augmented with a term $\dot{\bar{\gamma}}\cdot \vec{r}$ on the right hand side.

We now continue our calculation of the stress tensor. To this end, we transform to normal coordinates

 

$$\frac{d\vec{X}_{k}}{dt} -\frac{1}{\tau _{k}}\vec{X}_{k}+\bar{\gamma}\cdot \vec{X}_{k}+\vec{F}_{k}$$ = (6.62)

$$\langle \vec{F}_{k\alpha }(t)\vec{F}_{k\beta }(t^{\prime })\rangle \frac{D }{N+1}\delta _{\alpha \beta }\delta _{kk^{\prime }}\delta (t-t^{\prime })$$ = (6.63)

Using these equations of motion for the shear rate defined by Eqs. (6.50) to (6.52) we derive

 

$$\frac{d}{dt}\langle X_{kx}(t)X_{ky}(t)\rangle \langle \frac{dX_{kx}}{dt} (t)X_{ky}(t)\rangle +\langle X_{kx}(t)\frac{dX_{ky}(t)}{dt}\rangle$$ =

$$-\frac{2}{\tau _{k}}\langle X_{kx}(t)X_{ky}(t)\rangle +\dot{\gamma} \langle X_{ky}(t)X_{ky}(t)\rangle$$ =

$$+\langle F_{kx}(t)X_{ky}(t)\rangle +\langle X_{kx}(t)F_{ky}(t)\rangle$$ (6.64)

At the end of this section we shall argue that the last two terms vanish. Moreover, for small values of $\dot{\gamma}$ we shall approximate $\langle X_{ky}(t)X_{ky}(t)\rangle$ by its equilibrium value $\frac{1}{2}D\tau _{k}/(N+1)$. Using $\langle X_{kx}(0)X_{ky}(0)\rangle =0$, we find from Eq. ( 6.64)

  

$$\langle X_{kx}(t)X_{ky}(t)\rangle \int_{0}^{t}d\tau \dot{\gamma}\frac{1}{2 }\frac{D\tau _{k}}{N+1}\exp \left( -2\left( t-\tau \right) /\tau _{k}\right)$$ = (6.65)

$$\frac{D\tau _{k}^{2}}{4(N+1)}(1-e^{-2t/\tau _{k}})\dot{\gamma}$$ = (6.66)

Combining Eqs. (6.49), (6.57) and (6.65) we find an integral expression for the stress tensor :

$$S_{xy}(t)=\eta _{\mathrm{s}} \dot{\gamma}+\frac{c}{N+1}k_{B}T... ...}\exp \left( -2\left( t-\tau \right) /\tau _{k}\right) \right]$$ (6.67)

In polymer melts, we must neglect the solvent contribution $\eta _{\mathrm{s} }\dot{\gamma}$. We recognize that the contribution of the Rouse chains to the shear relaxation modulus  is given by

$$G\left( t\right) = \frac{c}{N+1}k_{B}T\sum_{k=1}^{N}\exp \left( -2t/\tau _{k}\right)$$ (6.68)

So the viscosity  of a Rouse melt, at constant monomer concentration c, is proportional to N:

$$\eta \int_{0}^{\infty }dtG\left( t\right) \approx \frac{c}{N+1}k_{B}T\frac{\tau _{1}}{2}\sum_{k=1}^{N}\frac{1}{k^{2}}$$ =

$$\approx \frac{c}{N+1}k_{B}T\frac{\tau _{1}}{2}\frac{\pi ^{2}}{6}=\frac{ c\gamma b^{2}}{36}\left( N+1\right)$$ (6.69)

This has been confirmed for polymer melts with low molecular weight. Polymer melts of high molecular weight give different results, stressing the importance of so-called entanglements . We will deal with this in chapter 8.

In dilute solutions, we do not neglect the solvent contribution $\eta _{\mathrm{s} }\dot{\gamma}$. We combine Eqs. (6.53), (6.57) and ( 6.66) to obtain an expression for the intrinsic viscosity ,

$$\lbrack \eta ] \lim_{\rho \rightarrow 0} \frac{\eta -\eta _{\mathrm{s}}}{\rho \eta _{\mathrm{s}}}$$ =

$$\frac{N_{A_{v}}}{M}\frac{1}{\eta _{\mathrm{s}}}\frac{\gamma b^{2}... ...frac{N_{A_{v}}}{M}\frac{1}{ \eta _{\mathrm{s}}}\frac{\gamma b^{2}}{36}(N+1)^{2}$$ = (6.70)

Here, $\rho =cM/(N_{A_{v}}(N+1))$ is the polymer concentration; M is the mol mass of the polymer, and N_Av is Avogadro's number.

We finish this section by calculating $\langle X_{k\alpha }(t)F_{k\beta }(t)\rangle$. Integrating Eq. (6.62) we find

$$\vec{X}_{k}(t)=\vec{X}_{k}(0)e^{-t/\tau _{k}}+\int_{0}^{t}d\t... ...dot{\bar{\gamma}}\cdot \vec{X}_{k}(\tau )+\vec{F} _{k}(\tau ))$$ (6.71)

Because of causality $\langle X_{k\alpha }(\tau )F_{k\beta }(t)\rangle =0$for $\tau <t$. Then

$$\langle X_{k\alpha }(t)F_{k\beta }(t)\rangle \int_{0}^{t}d\tau e^{-(t-\tau )/\tau _{k}}\langle F_{k\alpha }(\tau )F_{k\beta }(t)\rangle$$ =

$$\frac{1}{2}\int_{0}^{\infty }d\tau e^{-\vert t-\tau \vert/\tau _{k}}\langle F_{k\alpha }(\tau )F_{k\beta }(t)\rangle$$ =

$$\frac{D}{2(N+1)}\delta _{\alpha \beta }$$ = (6.72)

i.e. $\langle X_{kx}(t)F_{ky}(t)\rangle =\langle X_{ky}(t)F_{kx}(t)\rangle =0$.