- #1
f3sicA_A
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- Homework Statement
- What is the relation between the dispersion of the rightward steps (##\overline{(\Delta n_1)^2}##) and the displacement of the random walker in a simple 1-D random walk?
- Relevant Equations
- $$\overline{(\Delta n_1)^2}=Npq$$
$$m=n_1-n_2$$
I am currently going through the first chapter of Fundamentals of Statistical and Thermal Physics by F. Reif which is on random walks, various mean values, etc. and specifically unit 1.4 mentions "The quantity ##\overline{(\Delta n_1)^2}## is quadratic in the displacement." This is with reference to a simple 1-D random walk where the probability of going to the right is ##p## and the probability of going to the left is ##q\equiv1-p##, where ##n_1## is the number of steps taken to the right and ##n_2## is the number of steps taken to the left, such that ##n_1+n_2\equiv N## is the total number of steps taken.
I am not able to verify as to why ##\overline{(\Delta n_1)^2}## must be quadratic in displacement, ##m=n_1-n_2## (the rightward displacement in units of step length ##l##), and I am not sure how to approach this problem. One thing we have shown just before this statement is:
$$\overline{(\Delta n_1)^2}\equiv\overline{(n_1-\overline{n_1})^2}$$
$$\implies\overline{(\Delta n_1)^2}=\overline{n_1^2-2n_1\overline{n_1}+(\overline{n_1})^2}$$
$$=\overline{n_1^2}-\overline{n_1}^2$$
where we have:
$$\overline{n_1}=\sum_{n_1=1}^{N}P(n_1)n_1$$
$$=\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}p^{n_1}q^{N-n_1}n_1$$
here we have ##n_1p^{n_1}=p\partial {\left(p^{n_1}\right)}/\partial p##, so:
$$\overline{n_1}=\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}\left[p\frac{\partial\left(p^{n_1}\right)}{\partial p}\right]q^{N-n_1}$$
$$=p\frac{\partial}{\partial p}\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}p^{n_1}q^{N-n_1}$$
$$=p\frac{\partial(p+q)^{N}}{\partial p}$$
$$=pN(p+q)^{N-1}$$
and since we have $p+q=1$, in our case, ##\overline{n_1}=Np##. Similarly, we can follow a similar procedure to show that ##\overline{n_1^2}=\overline{n_1}^2+Npq##, and hence showing that:
$$\overline{(\Delta n_1)^2}=Npq$$
However, this does not convey to me the idea that ##\overline{(\Delta n_1)^2}## must be quadratic in ##m##, how should I start attempting this problem?
I am not able to verify as to why ##\overline{(\Delta n_1)^2}## must be quadratic in displacement, ##m=n_1-n_2## (the rightward displacement in units of step length ##l##), and I am not sure how to approach this problem. One thing we have shown just before this statement is:
$$\overline{(\Delta n_1)^2}\equiv\overline{(n_1-\overline{n_1})^2}$$
$$\implies\overline{(\Delta n_1)^2}=\overline{n_1^2-2n_1\overline{n_1}+(\overline{n_1})^2}$$
$$=\overline{n_1^2}-\overline{n_1}^2$$
where we have:
$$\overline{n_1}=\sum_{n_1=1}^{N}P(n_1)n_1$$
$$=\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}p^{n_1}q^{N-n_1}n_1$$
here we have ##n_1p^{n_1}=p\partial {\left(p^{n_1}\right)}/\partial p##, so:
$$\overline{n_1}=\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}\left[p\frac{\partial\left(p^{n_1}\right)}{\partial p}\right]q^{N-n_1}$$
$$=p\frac{\partial}{\partial p}\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}p^{n_1}q^{N-n_1}$$
$$=p\frac{\partial(p+q)^{N}}{\partial p}$$
$$=pN(p+q)^{N-1}$$
and since we have $p+q=1$, in our case, ##\overline{n_1}=Np##. Similarly, we can follow a similar procedure to show that ##\overline{n_1^2}=\overline{n_1}^2+Npq##, and hence showing that:
$$\overline{(\Delta n_1)^2}=Npq$$
However, this does not convey to me the idea that ##\overline{(\Delta n_1)^2}## must be quadratic in ##m##, how should I start attempting this problem?