Question

Random Walk Problem [Statistical Mechanics]

Answer 1

I understand some of the basic principles, just having trouble with some of the intermediate steps :S

Answer 2

we have N objects (in this case, fleas) and m move forward while N - m move backward

p = m/N gives us the probability of forward motion
q = (N-m)/N gives us the probability of backwards motion

Answer 3

for (p+q)^N we get the binomial expansion W(m) = N!/(m!(N-m)! * q^(N-m) * p^m (derived from the combination formula) for probability of m steps forward and N-m steps backward

Answer 4

The mean displacement is where I start getting a little lost

Answer 5

\[<m>^2 = \sum_{m=0}^{N}m^2W(m)\]

Answer 6

for step length L, we get displacement = (steps forwards - step backwards) *L = (m - [N - m])l = (2m-N) * L

Answer 7

doing a substitution for displacement <m> we get

<m> = m * N!/(m!(N-m)! * q^(N-m) * p^m

Answer 8

if we let Q = N!/(m!(N-m)! * q^(N-m) * p^m and take the partial derivative of Q wrt p we get:

Answer 9

\[\frac{ ∂Q }{ ∂P } = m \frac{ N! }{ m!(N-m)! }q^(N-m)*p^(m-1)\]

Answer 10

and we multiply this by p to convert it to a partition function

Answer 11

which basically just turns that p^(m-1) back to a p^m

Answer 12

\[<m> = p(∂Q/∂p)\]

Answer 13

if Q also equals (p+q)^N

we have:

\[(∂Q/∂p) = N(q+p)^{N-1} \]

Answer 14

\[<m> = p(∂Q/∂p) = Np(q+p)^{N-1} = Np\]

Answer 15

for a true random unbiased walk, p = 1/2 for forward step