ISOTROPY AND ANISOTROPYDate and time
The fundamental Lattice has axes therefore, at first sight, it is non-isotropic.
However existence of axes does not influence physical properties of space: they are identical in all directions.
The nonperturbed wavefront set from a source which can be considered dot, has a spherical form.
When we consider a small amount of cells, really, anisotropy is traced. If to consider space area, rather larger to neglect it a cellular structure, anisotropy disappears.
The lattice can be checked for an isotropy by method of random walks.
Let from some knot at the same time starts a point set. For each unit of time each point, irrespective of other points and the past, at random moves to one of adjacent cross-links of a lattice. If physical properties of a lattice are isotropic, through rather wide interval of time we will find out that the density function of points falls in all directions, on average, with an identical speed.
Or we will imagine the huge city with square quarters. Let along with identical speed in all directions start runners with one of intersections,
each of which, having reached the intersection, can further: 1) to run directly; 2) to turn back; 3) to turn on the left; 4) to turn to the right.
In total 4 options the equiprobable. Runners do not stop, do not influence at each other, the previous turns do not influence a choice of new turn;
the first option is considered turn on 0 degrees.
It is clear that such zigzag run will lead to smoothing of anisotropy of a street grid. If there is a lot of runners, at a look from above through rather wide interval of time centrally symmetric distribution of rollers in the various directions will be observed. And here if they ran only directly forward in the directions chosen by them, 4 dense groups of rollers equidistanted from a start place and, only on the axial directions would always be observed. On the radial directions it would be empty. Of course, runners could start not in all four directions as the choice is casual here, but we will consider such case the extremely improbable as there is a lot of runners.
Of course, in the Lattice there are no wanderings, and interactions extend on broken lines.
It is incorrect to compare the Fundamental Lattice to a crystal lattice, as it all the same that to compare an atomic nucleus to a galaxy core.
At the Fundamental Lattice all three axes are equal. It is possible to speak about anisotropy in case of off-axis naprvleniye. And, off-axis anisotropy smoothes out with increase in the considered Lattice area.
Let's take rather larger gauze with square cells. In its center we will bring heat sufficient
for a warming up of part of a grid to a luminescence to one of clusters.
What form will be had by the shining spot? – Circle form. Or rather, circular form as the grid is not a continuous surface. For the far observer the reticulate structure will be indiscernible with the naked eye, and the shining spot will be perceived as a circle which brightness decreases from the center to edge; and, all radial directions are equal.
Let from the central intersection (coordinate (0; 0)) start many (N) of runners,
then through the conventional unit of time in points with coordinates (1; 0), (0; 1),
(−1; 0) and (0; −1) it will appear on N/4 of runners.
In not made mention points runners are not present.
Through one conventional unit of time in points (2; 0), (0; 2), (−2; 0) and (0; −2) it will appear on N/16 of runners; in points (1; 1), (−1; 1), (−1; −1) and (1; −1) on N/8 of runners; in a point (0; 0) N/4 of runners. In not made mention points runners are not present.
Through one conventional unit of time in points (3; 0), (0; 3), (−3; 0) and (0; −3) it will appear on N/64 of runners; in points (2; 1), (1; 2), (−1; 2), (−2; 1), (−2; −1), (−1; −2), (1; −2) and (2; −1) on 3N/64 of runners; in points (1; 0), (0; 1), (−1; 0) and (0; −1) on 9N/64 of runners. In not made mention points runners are not present.
Through one conventional unit of time in points (4; 0), (0; 4), (−4; 0) and (0; −4) it will appear on N/256 of runners; in points (3; 1), (1; 3), (−1; 3), (−3; 1), (−3; −1), (−1; −3), (1; −3) and (3; −1) on N/64 of runners; in points (2; 2), (−2; 2), (−2; −2) and (2; −2) on 3N/128 of runners; in points (2; 0), (0; 2), (−2; 0) and (0; −2) on N/16 of runners; in points (1; 1), (−1; 1), (−1; −1) and (1; −1) on 3N/32 of runners; in a point (0; 0) 9N/64 of runners. In not made mention points runners are not present.
And so on. From each intersection where there are runners, on their one quarter run up in four possible directions. The more runners, the less casual deviations from 1/4. If at you the same results, as at me turned out, so correctly consider. And continue.
More narrow on the first steps it becomes apparent that points with identical nonzero values of quantity of runners lie on a circle. And it means that here all directions are isotropic.
If runners was a little, casual anisotropy as uniform scattering, most likely, wouldn't be would be observed (though such it is possible, but with very small probability).
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