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Diffusion Data

We anticipated on solving for the self diffusion coefficient by applying the Einstein relation,
\begin{displaymath}
D_{k} = \frac{1}{6} \ \frac{d}{dt}\left\langle \left\vert\vec{r}\ ^{(k)}(t)-\vec{r}\ ^{(k)}(0)\right\vert^{2}\right\rangle
\end{displaymath} (10)


where $k=\left\{1,2\right\}$ corresponding to $k=1$ representing the Ar atoms, and $k=2$ representing the Xe atoms. In theory, when a non-equilibrium steady state has been reached the diffusion flux will equal zero, and the self diffusion coefficient should approach a constant.

Figure 8: Ar data of $\overline{(\Delta D)^{2}}$ vs. time for different temperature gradients, where Grad=0.15 $\Rightarrow \nabla T = 0.15 K/a$, and $a = 4.45 \stackrel{\circ}{A}$.


Figures(8,9) display data of the mean square displacement vs. time for several trials at varying temperature gradients. The data displayed was taken on a (250x100) Ar-Xe simulation lattice with an Ar concentration of 20 percent. In contradiction to experimental results, the self diffusion of Xe is greater than the self diffusion of Ar. In addition, the magnitude of the self diffusion coefficients differ from experiment. However, we attribute this to the value of the frequency factor $\nu$ used in the simulation. Ideally we need to fit initial trials with $\nabla T = 0$ to experimental data. Thus, solving for $\nu$ and then running the simulations with the fitted $\nu$ at varying temperature gradients.

Figure 9: Xe data of $\overline{(\Delta D)^{2}}$ vs. time for different temperature gradients, where Grad=0.15 $\Rightarrow \nabla T = 0.15 K/a$, and $a = 4.45 \stackrel{\circ}{A}$.


next up previous
Next: Lattice Concentration Data Up: Results and Discussion Previous: Energy Barrier Calculations
Shawn A. Putnam
2001-12-17