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Scratch Smoothing

Fig. 7a shows one 200-pixel period of a periodic scratch with an initial depth of z = 100 pixels. There is obviously high positive curvature at the maximum, and high negative curvature at the minimum, so initially sintering should proceed rapidly. Fig. 7b shows the scratch after 400 cycles, with n = 1 and b = 13. The depth is now z = 66. Fig. 7c shows the scratch after 1900 cycles, at z = 21. From this point on, depth reduction is very slow, since curvature differences between different parts of the surface are getting small. Fig. 7d shows a scratch that was sintered with b = 5, and n = 25. The sintering has been artificially "stuck" at this early stage of smoothing, because n = 25 is too large. On such a simply-shaped surface, there are relatively few largest curvature pixels, certainly less than 25. By removing too many at a time, pixels are moved that are neutral or actually counter-productive in regards to smoothing and depth reduction, so that an artificial equilibrium is reached.

Figure 7: Showing one 200-pixel long period of a periodic scratch.

(a) unsintered scratch, depth z = 100 pixels

(b) same scratch after N=400 pixels moved (n=1) using b=13, z=66

(c) same scratch after N=1894, z=21

(d) "equilibrium" shape for n=25, b=5, after N=2000

Fig. 8a shows the scratch depth z vs. N, the total number of pixels moved, for n = 1 and various values of b. The fastest depth reduction is achieved with the highest values of b, as smaller curvature differences can be resolved, so that there is a better chance of moving the actual highest curvature pixel. All the values of b used when n = 1 can reduce the depth to low values, although as it can be seen, with different rates. Although not shown in Fig. 8, we have made longer runs to check this point. So for the scratch smoothing problem, as opposed to the square- to-circle problem, the box size seems to only affect the rate of sintering, and not the equilibrium shape. This is probably due to the fact that the equilibrium shape in the scratch case is a flat surface, and not a circle. Even b = 3 is enough resolution for the scratch problem. The equivalent plot, but for n = 25, is shown in Fig. 8b. Here it is plainly seen that the scratch becomes "stuck" at non-zero depths. The minimum depths achieved decrease as the box sizes increase. This is because the higher resolution curvature calculations for the higher values of b make it more possible that n = 25 pixels can be found whose rearrangement would favor depth reduction.

Figure 8a: Scratch depth versus total pixels moved N, for various values of b, all at n=1.

Figure 8b Scratch depth versus total pixels moved N, for various values of b, all at n=25.

To further quantify the surface evolution of the scratch, we have taken the spatial Fourier transform of the scratch surface z(x), where the transform is defined by:

with m an integer and = 200 pixels is the period of the scratch. Fig. 9 shows the first three non-zero Fourier components (A₁, A₃, A₅) plotted against total pixels moved N. The A₃ and A₅ components decay rapidly, while the A1 component decays much more slowly, so that by about N greater than 400, the scratch has become a pure cosine function. This can be clearly seen in Figs. 7b and 7c.

Figure 9: Showing the first three non-zero Fourier components of the scratch surface plotted versus total number of pixels moved N, n=1 and b=13.

This behavior is simply explained by considering curvature differences in different Fourier components. For a function f(x) = cos(mx) that describes a surface, the curvature at the maximum is proportional to m², so that the curvature difference between the maximum and minimum goes like 2m². Therefore we can think of the higher Fourier components of the scratch as having a much higher sintering rate than the A₁ component. For scratch smoothing via vapor-limited-diffusion, Mullins [19] predicts the decay rate for A_m to go as m³. In this model of interface-limited smoothing, the decay rate probably goes as m², since the diffusion length is effectively infinite. Figure 9 shows that the rate of decay for the three components seems to follow this trend, although we cannot compare exactly to Mullin's result, since we have not been able to define a time scale for this model.