Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\gamma(i)$ be a self avoiding walk (SAW) on a 2D lattice $L$ (a square lattice for example) starting at a predefined origin ( $\gamma(0)=(0,0)$ ) and having length $n:=\ell(\gamma)$. Furthermore, suppose $\gamma$ ends at a point $u$, so that $\gamma(n)=u$. Let $N(u)$ be the set of neighboring vertices of $u$ (that are connected to $u$). If I were to further condition the SAW by clamping the second-to-last-step, i.e. $\gamma(n-1)=v$, what can be said about the probability that the walk visited any of the other neighbors of $u$. Naturally, the length of the SAW and the distance of $u$ from the origin will be interlocked in a tug of war. I am primarily interested in seeing when the probability of visiting adjacent neighbors tends to zero.

share|improve this question
add comment

1 Answer

up vote 6 down vote accepted

If the walks are sampled uniformly at random, then the probability of another neighbour of $u$ being visited is related to the "atmosphere" of a walk, and the connective constant $\mu$. The mean number of additional neighbours approaches $3-\mu \approx 0.361841469\cdots$ for SAWs on $Z^2$, while the probability of there being at least one additional neighbour approaches $1-0.71114 = 0.28886$ (from eq. 12, Owczarek and Prellberg, ref. below).

A recent article gave probabilities for having 0, 1, 2, and 3 unoccupied neighbours, A L Owczarek and T Prellberg, "Scaling of the atmosphere of self-avoiding walks", J. Phys. A: Math. Theor. 41 (2008) 375004 (6pp).

The most accurate estimate of $\mu$ for $Z^2$ comes from self-avoiding polygon enumerations by Iwan Jensen, "A parallel algorithm for the enumeration of self-avoiding polygons on the square lattice", J. Phys. A: Math. Gen. 36 (2003) 5731–5745, also on the arXiv: http://arXiv.org/abs/cond-mat/0301468v1

If you're wondering about the scaling behaviour as $|u|$ is varied, then I think that given $x = |u|/\langle R_e^2 \rangle^{1/2}$, the scaling function $P(x)$ for the probability of more than one neighbour being occupied must go to zero as $x \rightarrow \infty$, but I don't have any intuition yet as to how it will approach zero (I think this is a nice question).

share|improve this answer
    
Thanks! I didn't know atmosphere was the correct terminology. –  Alex R. Feb 3 '11 at 1:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.