Experimentally the shear relaxation modulus of a polymer melt turns out to be like in Fig. 8.5.

We distinguish two regimes.

*i)*

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

which decays as

*ii)*

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
.
So,

On the reptation time scale,
is practically zero, so we can set
.
To make a smooth
transition from the Rouse regime to the reptation regime, we match Eq. (8.22) with Eq. (8.23) at
,
yielding

We will now calculate . Take a look at

(8.25) |

The vector is the tangent to the primitive chain, at segment at time

= | (8.26) | ||

= | |||

= | (8.27) |

where we have used

(8.28) |

Using this last equation, we also find

(8.29) |

This equation states that there is no correlation between the tangents to the primitive chain at a segment

= | |||

= | (8.30) |

We have plotted this in Fig. 8.6.

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

= | |||

= | (8.31) |

where the prime at the summation sign indicates that only terms with odd

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 *G*^{0} independent of *N*; and finally a
maximum relaxation time
proportional to *N*^{3}.

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

= | |||

= | |||

= | (8.32) |

where