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Viscoelastic behavior

Experimentally the shear relaxation modulus  $G\left( t\right) $ of a polymer melt turns out to be like in Fig. 8.5.

Figure 8.5: Logarithmic plot of the time behavior of the shear relaxation modulus

We distinguish two regimes.

i) $t<\tau _{e}$

At short times the chain behaves like a 3-dimensional Rouse chain . Using Eq. 6.68 we find

$\displaystyle G\left( t\right)$ = $\displaystyle \frac{c}{N+1}k_{B}T\sum_{k=1}^{N}\exp \left( -2t/\tau _{k}\right)$  
  $\textstyle \approx$ $\displaystyle \frac{c}{N+1}k_{B}T\int_{0}^{\infty }dk\exp \left( -2k^{2}t/\tau
  = $\displaystyle \frac{c}{N+1}k_{B}T\sqrt{\frac{\pi }{8}}\left( \frac{\tau _{R}}{t}\right)
^{1/2}$ (8.22)

which decays as t-1/2.8.1 At $t=\tau _{e}$ this possibility to relax ends. The only way for the chain to relax any further is by breaking out of the tube.

ii) $t>\tau _{e}$

The stress that remains in the system is caused by the fact that the chains are trapped in twisted tubes. By means of reptation  the chain can break out of its tube. The newly generated tube contains no stress. So, it is plausible to assume that the stress at any time t is proportional to the fraction of the original tube that is still part of the tube at time t. We'll call this fraction $\Psi \left(
t\right) $. So,

 \begin{displaymath}G\left( t\right) =G^{0}\Psi \left( t\right) .
\end{displaymath} (8.23)

On the reptation time scale, $\tau _{e}$ is practically zero, so we can set $
\Psi \left( \tau _{e}\right) =\Psi \left( 0\right) =1$. To make a smooth transition from the Rouse regime to the reptation regime, we match Eq. (8.22) with Eq. (8.23) at $t=\tau _{e}$, yielding

G0 = $\displaystyle \frac{c}{N+1}k_{B}T\sqrt{\frac{\pi }{8}}\left( \frac{\tau _{R}}{\tau _{e}}
\right) ^{1/2}$  
  = $\displaystyle \frac{c}{3\sqrt{2}\pi }k_{B}T\frac{b^{2}}{a^{2}}$ (8.24)

We will now calculate $\Psi \left(
t\right) $. Take a look at

\begin{displaymath}\left\langle \vec{u}\left( s^{\prime },t\right) \cdot \vec{u}...
...{\partial \vec{R}
\left( s,0\right) }{\partial s}\right\rangle
\end{displaymath} (8.25)

The vector $\vec{u}\left( s^{\prime },t\right) $ is the tangent to the primitive chain, at segment $
s^{\prime }$ at time t. Because the primitive chain has been parametrized with the contour length, we have from Eq.(8.1) $\left\langle \vec{u}\mathbf{\cdot }\vec{u}\right\rangle
...ot \triangle \vec{R}\right\rangle
/\left( \triangle s\right) ^{2}=a/\triangle s$ ; the non-existence of the limit of $\triangle s$ going to zero is a peculiarity of a Gaussian process. Using Eq. 8.10 we calculate,
    $\displaystyle \left\langle \vec{u}\left( s^{\prime },t\right) \cdot \vec{u}\left(
s,0\right) \right\rangle$  
  = $\displaystyle -\frac{1}{2}\frac{\partial ^{2}}{\partial s\partial s^{\prime }}\varphi
\left( s^{\prime },s;t\right)$ (8.26)
  = $\displaystyle a\delta \left( s-s^{\prime }\right) -\frac{2a}{L}\sum_{k=1}^{\inf... \left( \frac{k\pi s}{L}\right) \sin \left(
\frac{k\pi s^{\prime }}{L}\right)$  
  = $\displaystyle \frac{2a}{L}\sum_{k=1}^{\infty }e^{-tk^{2}/\tau _{d}}\sin \left( \frac{
k\pi s}{L}\right) \sin \left( \frac{k\pi s^{\prime }}{L}\right)$ (8.27)

where we have used

\begin{displaymath}\frac{2}{L}\sum_{k=1}^{\infty }\sin \left( \frac{k\pi s}{L}\r...
...\pi s^{\prime }}{L}\right) =\delta \left( s-s^{\prime }\right)
\end{displaymath} (8.28)

Using this last equation, we also find

\begin{displaymath}\left\langle \vec{u}\left( s^{\prime },0\right) \cdot \vec{u}...
s,0\right) \right\rangle =a\delta \left( s-s^{\prime }\right)
\end{displaymath} (8.29)

This equation states that there is no correlation between the tangents to the primitive chain at a segment s, and at another segment $
s^{\prime }$. If we consider $\left\langle \vec{u}\left( s^{\prime },t\right) \cdot \vec{u}
\left( s,0\right) \right\rangle $ as a function of $
s^{\prime }$, at time t, we see that the original delta function has broadened and lowered. However, the tangent $\vec{u}\left( s^{\prime },t\right) $ can only be correlated to $\vec{u}\left( s,0\right) $ by means of diffusion of segment $
s^{\prime }$, during the time interval $\left[ 0,t\right] $, to the place where s was at time t=0, and still lies in the original tube. So, $\frac{
1}{a}\left\langle \vec{u}\left( s^{\prime },t\right) \cdot \vec{u}\left(
s,0\right) \right\rangle $ is the probability density that, at time t, segment $
s^{\prime }$ lies within the original tube at the place where swas initially. Integrating over $
s^{\prime }$ gives us the probability $\Psi
\left( s,t\right) $ that at time t any segment lies within the original tube at the place where segment s was initially. In other words, the chance that the original tube segment s is still up-to-date, is
$\displaystyle \Psi \left( s,t\right)$ = $\displaystyle \frac{1}{a}\int_{0}^{L}ds^{\prime }\left\langle
\vec{u}\left( s^{\prime },t\right) \cdot \vec{u}\left( s,0\right)
  = $\displaystyle \frac{4}{\pi }\sum_{k=1}^{\infty }\frac{1}{k}\sin \left( \frac{k\pi s}{L}
\right) e^{-tk^{2}/\tau _{d}}$ (8.30)

We have plotted this in Fig. 8.6.

Figure 8.6: Development of in time.

The fraction of the original tube that is still intact at time t, is therefore given by

$\displaystyle \Psi \left( t\right)$ = $\displaystyle \frac{1}{L}\int_{0}^{L}ds\Psi \left( s,t\right)$  
  = $\displaystyle \frac{8}{\pi ^{2}}\sum_{k=1}^{\infty }\acute{}\frac{1}{k^{2}}
e^{-tk^{2}/\tau _{d}}$ (8.31)

where the prime at the summation sign indicates that only terms with odd kshould occur in the sum. This formula shows why $\tau _{d}$ is the time needed by the chain to reptate out if its tube; for $t>\tau _{d}$, $\Psi \left(
t\right) $ is falling to zero quickly.

In conclusion we have found results that are in good agreement with Fig. 8.5. We see an initial drop proportional to t-1/2; after that a plateau  value G0 independent of N; and finally a maximum relaxation time $\tau _{d}$ proportional to N3.

Finally, we are able to calculate the viscosity  that is associated with the reptation model. Using Eq. 6.53 we find

$\displaystyle \eta$ = $\displaystyle \int_{0}^{\infty }G\left( t\right) dt=G^{0}
\frac{8}{\pi ^{2}}\sum_{k=1}^{\infty }\acute{}\frac{1}{k^{2}}
\int_{0}^{\infty }dte^{-tk^{2}/\tau _{d}}$  
  = $\displaystyle G^{0}\frac{8}{\pi ^{2}}\tau _{d}\sum_{k=1}^{\infty }\acute{}\frac{1}{k^{2}
}\int_{0}^{\infty }d\tau e^{-\tau k^{2}}$  
  = $\displaystyle H\frac{b^{6}}{a^{4}}\zeta N^{3}$ (8.32)

where H is a constant. We see that the viscosity is proportional to N3. This is close to the experimentally verified N3.4 behavior of polymer melts with high molecular weight. indextrans|seeconformation         

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Next: Index Up: Dynamics of dense polymer Previous: Monomer motion
W.J. Briels