8.4
- An Exact Solution for the Dimension of a Random Walk on a Sierpinski
Gasket
We can derive the
exponent s in Eq. 3 directly
by using equations similar to Eq. 1.
Begin by applying a test to 
TC'B'
. Let us imagine
that we execute four separate trials for randomly walking from C'
to B'. From C' there are four paths to
B'. Refer to Figure 8.2(a). On
average, we expect that: on one trial we go directly to the leftmost B'
in time TAB (the same as the
time to go between the adjacent point A and B); on one trial we go directly
to the rightmost B'; on one trial we go to the leftmost point
B in time TAB, and then take time
TBB' to arrive at a B';
and, finally, on average one of the walks will first take us the rightmost point
B in time TAB, and from there
it will take TBB' to arrive at a B'.
Express this analysis
as an equation:
In
short, our second equation relating the average internal
times is:
Finally,
we apply the same logic to four trial random walks which are initiated
at a point B. On average: one walk will take time TAB to arrive at the
other point B, and then the average time TBB' to arrive at one
of the points B'; one walk will take time
TAB to move to point
A, and then time TAB' to arrive at B';
one walk will take time TAB to arrive at C',
and then time TC'B' to arrive at a B';
and, one walk go directly to B' in time TAB. Summing these four
ways to go, we have (with the parenthesized terms in order of the
above description):
| 4TBB' |
|
|
| (TAB+ TBB') + (TAB+ TAB') |
|
|
|
|
|
(8.18) |
|
Simplifying
this equation:
| 3TBB' = 4TAB+ TAB'+ TC'B'. |
|
(8.19) |
|
Q8.17: Do your averaged quantities computed in Task 1 in HandsOn 8.1
satisfy these equations? Check the numbers. Do you get better agreement
using average times as found averaging over the data of all students?
|
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Previous: HandsOn
33 - Computing the Resistance of the Sierpinski Gasket
