# Random Walks

## Binomial Distribution

Let $$s_N$$ be the step lengths. Assuming a fair coin $$p_H, p_T = \frac{1}{2}$$. So, $$\langle s_N\rangle = 0$$. Then, $$s_N = N_H - N_T$$. So, $$\langle s_1^2\rangle = p_T(-1) + p_H(+1) = \frac{1}{2}((-1)^2 + (+1)^2) = 1$$. $$\langle s_2^2\rangle = \frac{1}{4}\left((-2)^2 + 2(0)^2 + \frac{1}{4}(+1)^2\right) = 2$$. $$\langle s_3^2\rangle = \frac{1}{8}\left((-3)^2 + 3(-1)^2 + 3(+1)^2 + (3)^2\right) = 3$$. $$\langle s_N^2\rangle = \langle (s_{N-1} + \ell_N)^2\rangle = \langle s_{N-1}^2\rangle + 2\langle s_{N-1}\ell_N\rangle + \langle \ell_N^2\rangle = \langle s_{N-1}^2\rangle + 2(0) + 1 = \langle s_{N-1}^2\rangle + 1 = N$$.

We didn’t need to invoke the binomial distribution to achieve this result. How could we have used that to achieve this end.

$$p_N(N_H) = {N\choose N_H}p_H^{N_H} p_T^{N-N_H}$$

Note: $${N\choose N_H}$$ is just the coefficient in the binomial expansion of $$(p_H+p_T)^N$$

$${N-1\choose k-1} + {N-1\choose k} = (b_{N-1,k-1} + b_{N-1,k}){N\choose k}$$ where $$b_{N,k}$$ relates to the $$N,k$$ coefficient in the binomial expansion.

Characteristic equation of the binomial: $$\tilde{p}_{N_H}(k) = \langle \exp(-ikN_H)\rangle = \sum_{N_H=0}^N \frac{N!}{N_H!(N-N_H!)}p_H^{N_H}p_T^{N-N_H}\exp(-ikN_H) = \sum_{N_H=0}^N \frac{N!}{N_H!(N-N_H!)}\left(p_H+\exp(ik)\right)^{N_H}p_T^{N-N_H} = (p_H\exp(-ik) + p_T)^N$$.

The cumulents generating function $$\ln \tilde{p}_{N_H}=N\ln(p_H\exp(-ik)+p_T)=N\ln\tilde{p}_1(k)$$

$$\mu = \langle N_H\rangle_C = Np_H$$, $$\sigma^2 = \langle N_H^2\rangle_C = N(p_H-p_H^2) = Np_H(1-p_H) = Np_Hp_T$$

$$p_{N_H} = \frac{1}{2^N}{N\choose N_H} = \frac{1}{2^N}{N\choose \frac{N}{2}+m}$$ $$p_m = \frac{1}{2^N}{2N\choose N+m}$$ Let $$m\ll N$$. Ok because $$m\leq\mathcal{O}(\sqrt{N})$$.

Use Stirling’s approximation: $$N! \approx \left(\frac{N}{e}\right)^N\sqrt{2\pi N} \approx \left(\frac{N}{e}\right)^N$$

$$p_m = \frac{1}{2^{2N}}\left(\frac{\left(\frac{2N}{e}\right)^{2N}}{\left(\frac{N+m}{e}\right)^{N+m}+\left(\frac{N-m}{e}\right)^{N-m}}\right) = N^{2N}(N+m)^{-(N+m)}(N-m)^{-(N-m)} = \left(1-\frac{m^2}{N^2}\right)^{-N}\left(1+\frac{m}{N}\right)^{-m}\left(1-\frac{m}{N}\right)^{m} \approx \exp -N\left(-\frac{m^2}{N^2}\right)\exp -m\left(\frac{m}{N}\right)\exp m\left(\frac{-m}{N}\right) = \exp \frac{-m^2}{N}$$ Thus, $$p_m = p_0\exp\frac{-m^2}{N}$$

$$p_m = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\frac{-x^2}{2\sigma^2}$$, $$\sigma^2=\frac{N}{2}$$.

• Note: the binomial distribution done once is called the Bernoulli distribution.
• Note: In class we sometimes use $$\lambda$$ to mean the mean.

## 2D Random Walk

$$\ell m=L\begin{pmatrix} \cos\varphi_m \\ \sin\varphi_m \end{pmatrix}$$

$$\langle \ell m\cdot \ell n\rangle = L^2 \langle \cos\varphi_m\varphi_n + \sin\varphi_m\sin\varphi_n \rangle = L^2\langle \cos(\varphi_n-\varphi_m)\rangle = 0$$ for $$n\neq m$$ and $$L^2$$ for $$n=m$$.

In other words, if they are independent steps (so $$\varphi_m,\varphi_n$$ are independent random variables [rv]) then their mean is independent, and thus the average is zero. Independent rv’s probabilities go as: $$p_{xy} = p_xp_y$$ $$\langle x_n\cdot x_m\rangle = -\sum_{i,j}p_{ij}x_i\cdot x_j = -\sum_{i,j}p_i\cdot p_jx_i\cdot x_j = -\sum_{i,j}p_ip_jx_ix_j = \langle x_n\rangle\langle x_m\rangle$$

## Diffusion Equation

### Random Walks

$$x(t+\Delta t) = x(t) + \ell(t)$$.

Probability distributions for each step:

• Equal: $$\chi(\ell) = \frac{1}{2}\delta(\ell+1) + \frac{1}{2}\delta(\ell-1)$$.
• Drunken: $$\chi(\overline{\ell}) = \frac{\delta(|\overline{\ell}-L)}{2\pi L}$$.

$$\langle x^n\rangle = \begin{cases}1 & n=0 \text{, Normalization} \\ 0 & n=1 \text{, Mean} \\ \sigma^2 & n=2 \text{, STDEV}\end{cases}$$.

$$\langle x^2\rangle = \langle x^2\rangle_c$$ since $$\langle x\rangle_c = 0 = \langle x\rangle$$.

$$p(x,t+\Delta t)$$ from $$p(x',t)$$.

$$x'\to x$$, $$t\to t+\Delta t$$.

$$\ell(t) = x-x'$$. Let $$x-x'=z$$. $$p(x,t+\Delta t) = \int_{-\infty}^\infty p(x',t)\chi(x-x')dx' = -\int_{\infty}^{-\infty} p(z-x,t)\chi(z)dz = \int_{-\infty}^{\infty} p(z-x,t)\chi(z)dz$$ We can taylor expand $$p(x-z,t)$$ around $$z=0$$, $$p(x,t+\Delta t) = \int_{-\infty}^\infty \left[p(x,t) + \frac{\partial p}{\partial x}(-z) + \frac{1}{2}\frac{\partial^2 p}{\partial x^2}(z^2) + \cdots\right] \chi(z)dz = \int_{-\infty}^\infty [1 + 0 + \frac{1}{2}\sigma^2\frac{\partial^2p}{\partial x^2}+\cdots]\chi(z)dz = p(x,t) + \frac{\sigma^2}{2}\frac{\partial^2 p}{\partial x^2}(x,t)$$

$$\lim_{\Delta t\to 0}\frac{f(x,t+\Delta t) - p(x,t)}{\Delta t} = \frac{\partial p}{\partial t}$$.

So, $$\frac{\partial p}{\partial t} = \frac{\sigma^2}{2\Delta t}\frac{\partial^2 p}{\partial x^2} = D\frac{\partial^2 p}{\partial x^2}$$. Note: $$D>0$$.

Conserved quantities: $$\frac{\partial p}{\partial t} = -\frac{\partial J}{\partial x} = -\nabla_{\vec{x}}J$$, where $$\vec{J}$$ is the current.

So, $$J_{diff} = -D\frac{\partial p}{\partial x}$$. ($$J_{diff} = -D\frac{\partial^2 p}{\partial x^2}$$?)

$$x(t+\Delta t) = x(t) + F\gamma\Delta t + \ell(t)$$, $$\gamma$$ is called the mobility (mass, how much stuff you will run into, etc.).

Type Gamma
Viscous Fluid $$\frac{1}{6\pi \eta r}$$
Dilute Gas $$x=\frac{1}{2}\frac{F}{m}\Delta t^2$$
$$\frac{\Delta t}{2m}$$
$$\frac{\Delta t}{2m}\frac{D}{D}$$
$$\frac{2\Delta t}{2m}D\frac{2\Delta t}{\sigma^2}$$
$$\frac{D}{m\left(\frac{\sigma}{\Delta t}\right)^2}$$

Aside: Dilute Gas 1: (free arc. between collisions that fully scramble the velocity)

$$\frac{\partial p}{\partial t} = -\gamma F\frac{\partial p}{\partial x} + D\frac{\partial^2 p}{\partial x^2}$$. If $$\frac{\partial p^*}{\partial t} = 0$$ then we have no $$F$$ term, so $$\frac{\partial^2 p}{\partial x^2}=0$$, $$p=p_0+Bx$$ because hard walls $$J=0\to \frac{\partial p}{\partial x}=0\to p^*=p_0$$.

If you have a force term, e.g. $$F=-mg$$, then, $$\frac{\partial p^*}{\partial t}=0=\gamma mg\frac{\partial p^*}{\partial x}+D\frac{\partial^2 p^*}{\partial x^2}$$. $$p^* = A\exp(-\alpha x) + B$$. $$p''^*=\alpha^2 A\exp(-\alpha x)$$, $$\alpha = \frac{\gamma mg}{D}$$. $$p^*(\infty) = 0, A=p^*(0), p^*(x)=p^*(0)\exp\left(-\frac{\gamma mgx}{D}\right)$$. Density slices of gas, $$M = mg u(x) (A\Delta x)$$, where the density is $$u(x)$$. $$F_{top}=P(x+\Delta x)\cdot A, F_{bot} = P(x)A$$. $$P(x) = P(x + \Delta x) + mg\cdot u(x)\Delta x$$. $$P = \frac{N}{V}kT$$. $$u(x) = u(x+\Delta) + \frac{mg}{kT}u(x)\Delta x$$, $$\frac{\partial u}{\partial x}=-\frac{mg}{kT}u(x)$$. $$u(x) = u(0)\exp\left(-\frac{-mg}{kT}x\right)$$. So, $$\frac{D}{\gamma}=kT$$.

We could also solve the density slices of the gas from the chemical potential, $$\mu(x) = mgx + \mu_{Ideal Gas}$$. $$F=NkT(\ln(n\lambda^3)-1)$$, $$\lambda = \frac{n}{n_Q} = \sqrt{\frac{2\pi \hbar^2}{mkT}}$$. $$\mu = \left.\frac{\partial F}{\partial N}\right|_{VT}$$.

Diffusive problems can be described by forces or chemical potential, energy needed to add or remove particles, typically.

Created: 2023-06-25 Sun 02:34

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