Polymer models¶
The diffusion coefficient $D$ of a spherical particle in a carrier fluid experiencing drag is
related to the friction coefficient $\zeta$ via the Einstein relation:
$$
D = \frac{k_{\mathrm{B}}T}{\zeta},
$$
with $k_{\mathrm{B}}$ the Boltzmann constant and $T$ the temperature. For a sphere of radius $R$
moving in a fluid of viscosity $\eta$, the friction coefficient is obtained via the Stokes law:
$$
\zeta = 6\pi\eta R.
$$
Combining both equations yields the Stokes–Einstein relation:
$$
D = \frac{k_{\mathrm{B}}T}{6\pi\eta R}.
$$
The simplest description of a polymer is the Rouse model, where beads are connected by springs.
All beads experience a drag from the solvent, and the friction coefficient $\gamma$ is identical
for all beads. The solvent flows freely between beads and hydrodynamic interactions are neglected.
The diffusion coefficient takes the following form:
$$
D_{\mathrm{R}} = \frac{D_0}{N} = \frac{k_{\mathrm{B}}T}{\gamma N},
$$
where $D_0$ is the diffusion coefficient of a single bead.
To account for hydrodynamic interactions mediated by the solvent, i.e. the transport of solvent
in contact with the beads and the correlation in the motion of beads due to the carried solvent,
the Zimm model was created. For an ideal chain, it takes the following form:
$$
D_{\mathrm{Z}} = \frac{8}{3\sqrt{6\pi^3}}\frac{k_B T}{\eta R} \simeq 0.196\frac{k_B T}{\eta b N^{\nu}},
$$
with $R$ the radius of the polymer and $b$ the length of the spring connecting the beads.
For shorter polymers there is a transition region. It can be described
by the Kirkwood–Zimm model:
$$
D=\frac{D_0}{N} + \frac{k_B T}{6 \pi \eta } \left\langle \frac{1}{R_h} \right\rangle
$$
Here $D_0$ is the monomer diffusion coefficient and $\eta$ the
viscosity of the fluid. For a finite system size the second part of the
diffusion is subject to a $1/L$ finite size effect, because
hydrodynamic interactions are proportional to the inverse
distance and thus long ranged. It can be taken into account
by a correction:
$$
D=\frac{D_0}{N} + \frac{k_B T}{6 \pi \eta } \left\langle \frac{1}{R_h} \right\rangle \left( 1- \left\langle\frac{R_h}{L} \right\rangle \right)
$$
It is quite difficult to fit this analytical expression to simulation data with good accuracy.
It will need a LB fluid, long simulation times and a careful analysis. For this tutorial, we
will use an implicit solvent and short polymer lengths to keep the runtime short. If you want
to collect data suitable for the Zimm model, simply set the global variable POLYMER_MODEL to
'Zimm', and increase the box size and number of beads in the polymer.
We want to determine the long-time self diffusion coefficient from the mean square
displacement of the center-of-mass of a single polymer. For large $t$ the mean square displacement is
proportional to the time and the diffusion coefficient occurs as a
prefactor:
$$
D = \lim_{t\to\infty}\left[ \frac{1}{6t} \left\langle \left(\vec{r}(t) - \vec{r}(0)\right)^2 \right\rangle \right].
$$
This equation can be found in virtually any simulation textbook, like [4]. We will set up a
polymer in an implicit solvent, simulate for an appropriate amount of time, calculate the mean square
displacement as a function of time and obtain the diffusion coefficient from a linear
fit. However we will have a couple of steps inbetween and divide the full problem into
subproblems that allow to (hopefully) fully understand the process.
