Random Walks

Binomial Distribution

Let $s_{N}$ be the step lengths. Assuming a fair coin $p_{H}, p_{T} = \frac{1}{2}$ . So, $⟨ s_{N} ⟩ = 0$ . Then, $s_{N} = N_{H} - N_{T}$ . So, $⟨ s_{1}^{2} ⟩ = p_{T} (- 1) + p_{H} (+ 1) = \frac{1}{2} ((- 1)^{2} + (+ 1)^{2}) = 1$ . $⟨ s_{2}^{2} ⟩ = \frac{1}{4} ((- 2)^{2} + 2 (0)^{2} + \frac{1}{4} (+ 1)^{2}) = 2$ . $⟨ s_{3}^{2} ⟩ = \frac{1}{8} ((- 3)^{2} + 3 (- 1)^{2} + 3 (+ 1)^{2} + (3)^{2}) = 3$ . $⟨ s_{N}^{2} ⟩ = ⟨ (s_{N - 1} + ℓ_{N})^{2} ⟩ = ⟨ s_{N - 1}^{2} ⟩ + 2 ⟨ s_{N - 1} ℓ_{N} ⟩ + ⟨ ℓ_{N}^{2} ⟩ = ⟨ s_{N - 1}^{2} ⟩ + 2 (0) + 1 = ⟨ s_{N - 1}^{2} ⟩ + 1 = N$ .

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

$p_{N} (N_{H}) = (\binom{N}{N_{H}}) p_{H}^{N_{H}} p_{T}^{N - N_{H}}$

Note: $(\binom{N}{N_{H}})$ is just the coefficient in the binomial expansion of $(p_{H} + p_{T})^{N}$

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

Characteristic equation of the binomial: ${\tilde{p}}_{N_{H}} (k) = ⟨ \exp (- i k N_{H}) ⟩ = \sum_{N_{H} = 0}^{N} \frac{N!}{N_{H}! (N - N_{H}!)} p_{H}^{N_{H}} p_{T}^{N - N_{H}} \exp (- i k N_{H}) = \sum_{N_{H} = 0}^{N} \frac{N!}{N_{H}! (N - N_{H}!)} {(p_{H} + \exp (i k))}^{N_{H}} p_{T}^{N - N_{H}} = (p_{H} \exp (- i k) + p_{T})^{N}$ .

The cumulents generating function $\ln {\tilde{p}}_{N_{H}} = N \ln (p_{H} \exp (- i k) + p_{T}) = N \ln {\tilde{p}}_{1} (k)$

$μ = ⟨ N_{H} ⟩_{C} = N p_{H}$ , $σ^{2} = ⟨ N_{H}^{2} ⟩_{C} = N (p_{H} - p_{H}^{2}) = N p_{H} (1 - p_{H}) = N p_{H} p_{T}$

$p_{N_{H}} = \frac{1}{2^{N}} (\binom{N}{N_{H}}) = \frac{1}{2^{N}} (\binom{N}{\frac{N}{2} + m})$ $p_{m} = \frac{1}{2^{N}} (\binom{2 N}{N + m})$ Let $m ≪ N$ . Ok because $m \leq O (\sqrt{N})$ .

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

$p_{m} = \frac{1}{2^{2 N}} (\frac{{(\frac{2 N}{e})}^{2 N}}{{(\frac{N + m}{e})}^{N + m} + {(\frac{N - m}{e})}^{N - m}}) = N^{2 N} (N + m)^{- (N + m)} (N - m)^{- (N - m)} = {(1 - \frac{m^{2}}{N^{2}})}^{- N} {(1 + \frac{m}{N})}^{- m} {(1 - \frac{m}{N})}^{m} \approx \exp - N (- \frac{m^{2}}{N^{2}}) \exp - m (\frac{m}{N}) \exp m (\frac{- m}{N}) = \exp \frac{- m^{2}}{N}$ Thus, $p_{m} = p_{0} \exp \frac{- m^{2}}{N}$

$p_{m} = \frac{1}{\sqrt{2 π σ^{2}}} \exp \frac{- x^{2}}{2 σ^{2}}$ , $σ^{2} = \frac{N}{2}$ .

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

2D Random Walk

$ℓ m = L (\begin{matrix} \cos φ_{m} \\ \sin φ_{m} \end{matrix})$

$⟨ ℓ m \cdot ℓ n ⟩ = L^{2} ⟨ \cos φ_{m} φ_{n} + \sin φ_{m} \sin φ_{n} ⟩ = L^{2} ⟨ \cos (φ_{n} - φ_{m}) ⟩ = 0$ for $n \neq m$ and $L^{2}$ for $n = m$ .

In other words, if they are independent steps (so $φ_{m}, φ_{n}$ are independent random variables [rv]) then their mean is independent, and thus the average is zero. Independent rv’s probabilities go as: $p_{x y} = p_{x} p_{y}$ $⟨ x_{n} \cdot x_{m} ⟩ = - \sum_{i, j} p_{i j} x_{i} \cdot x_{j} = - \sum_{i, j} p_{i} \cdot p_{j} x_{i} \cdot x_{j} = - \sum_{i, j} p_{i} p_{j} x_{i} x_{j} = ⟨ x_{n} ⟩ ⟨ x_{m} ⟩$

Diffusion Equation

Random Walks

$x (t + Δ t) = x (t) + ℓ (t)$ .

Probability distributions for each step:

Equal: $χ (ℓ) = \frac{1}{2} δ (ℓ + 1) + \frac{1}{2} δ (ℓ - 1)$ .
Drunken: $χ (\overset{―}{ℓ}) = \frac{δ (| \overset{―}{ℓ} - L)}{2 π L}$ .

$⟨ x^{n} ⟩ = {\begin{cases} 1 & n = 0, Normalization \\ 0 & n = 1, Mean \\ σ^{2} & n = 2, STDEV \end{cases}$ .

$⟨ x^{2} ⟩ = ⟨ x^{2} ⟩_{c}$ since $⟨ x ⟩_{c} = 0 = ⟨ x ⟩$ .

$p (x, t + Δ t)$ from $p (x^{'}, t)$ .

$x^{'} \to x$ , $t \to t + Δ t$ .

$ℓ (t) = x - x^{'}$ . Let $x - x^{'} = z$ . $p (x, t + Δ t) = \int_{- \infty}^{\infty} p (x^{'}, t) χ (x - x^{'}) d x^{'} = - \int_{\infty}^{- \infty} p (z - x, t) χ (z) d z = \int_{- \infty}^{\infty} p (z - x, t) χ (z) d z$ We can taylor expand $p (x - z, t)$ around $z = 0$ , $p (x, t + Δ t) = \int_{- \infty}^{\infty} [p (x, t) + \frac{\partial p}{\partial x} (- z) + \frac{1}{2} \frac{\partial^{2} p}{\partial x^{2}} (z^{2}) + \dots] χ (z) d z = \int_{- \infty}^{\infty} [1 + 0 + \frac{1}{2} σ^{2} \frac{\partial^{2} p}{\partial x^{2}} + \dots] χ (z) d z = p (x, t) + \frac{σ^{2}}{2} \frac{\partial^{2} p}{\partial x^{2}} (x, t)$

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

So, $\frac{\partial p}{\partial t} = \frac{σ^{2}}{2 Δ t} \frac{\partial^{2} p}{\partial x^{2}} = D \frac{\partial^{2} p}{\partial x^{2}}$ . 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, $J_{d i f f} = - D \frac{\partial p}{\partial x}$ . ( $J_{d i f f} = - D \frac{\partial^{2} p}{\partial x^{2}}$ ?)

$x (t + Δ t) = x (t) + F γ Δ t + ℓ (t)$ , $γ$ is called the mobility (mass, how much stuff you will run into, etc.).

Type	Gamma
Viscous Fluid	$\frac{1}{6 π η r}$
Dilute Gas	$x = \frac{1}{2} \frac{F}{m} Δ t^{2}$
	$\frac{Δ t}{2 m}$
	$\frac{Δ t}{2 m} \frac{D}{D}$
	$\frac{2 Δ t}{2 m} D \frac{2 Δ t}{σ^{2}}$
	$\frac{D}{m {(\frac{σ}{Δ t})}^{2}}$

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

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

If you have a force term, e.g. $F = - m g$ , then, $\frac{\partial p^{*}}{\partial t} = 0 = γ m g \frac{\partial p^{*}}{\partial x} + D \frac{\partial^{2} p^{*}}{\partial x^{2}}$ . $p^{*} = A \exp (- α x) + B$ . $p^{″ *} = α^{2} A \exp (- α x)$ , $α = \frac{γ m g}{D}$ . $p^{*} (\infty) = 0, A = p^{*} (0), p^{*} (x) = p^{*} (0) \exp (- \frac{γ m g x}{D})$ . Density slices of gas, $M = m g u (x) (A Δ x)$ , where the density is $u (x)$ . $F_{t o p} = P (x + Δ x) \cdot A, F_{b o t} = P (x) A$ . $P (x) = P (x + Δ x) + m g \cdot u (x) Δ x$ . $P = \frac{N}{V} k T$ . $u (x) = u (x + Δ) + \frac{m g}{k T} u (x) Δ x$ , $\frac{\partial u}{\partial x} = - \frac{m g}{k T} u (x)$ . $u (x) = u (0) \exp (- \frac{- m g}{k T} x)$ . So, $\frac{D}{γ} = k T$ .

We could also solve the density slices of the gas from the chemical potential, $μ (x) = m g x + μ_{I d e a l G a s}$ . $F = N k T (\ln (n λ^{3}) - 1)$ , $λ = \frac{n}{n_{Q}} = \sqrt{\frac{2 π ℏ^{2}}{m k T}}$ . $μ = {\frac{\partial F}{\partial N} |}_{V T}$ .

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