Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S12 / 3

Write the one-dimensional random walk endpoint as $x_m={\sum }_{j=1}^m{\xi }_j$, starting at the origin. Assume unbiased, independent increments, each with $⟨{\xi }_j⟩=0$ and $⟨{\xi }_j^2⟩=\Delta$. Expanding the square givesIndependence makes each cross expectation $⟨{\xi }_i⟩⟨{\xi }_j⟩=0$. ConsequentlyUncorrelated increments with the same second moments would suffice. Unbiasedness alone does not remove correlations: persistent or anticorrelated steps need not satisfy this result. The symbol $\Delta$ is already the mean-square step size, with units of length squared.

Let $C(x)$ be the particle number density, $U(x)$ the conservative potential energy and $F=-dU/dx$ the applied force. In the overdamped regime a viscous drag coefficient $\zeta$ gives drift velocity $v_d=F/\zeta$. Combining this drift with Fick's first law gives the number fluxAt thermal equilibrium, the Boltzmann distribution is $C_{\mathrm{eq}}\propto e^{-U/(k_{\mathrm{B}}T)}$, so $d{C}_{\mathrm{eq}}/dx=F{C}_{\mathrm{eq}}/(k_{\mathrm{B}}T)$. Setting its net flux to zero yieldsThis is the Einstein diffusion relation. The derivation assumes a dilute equilibrium suspension, uniform temperature, linear drift response and a well-defined long-time diffusion coefficient. With a nonzero conservative force, the full transport equation includes drift; the force-free equation in the next section describes spreading without that drift.

For constant $D$ and no drift, particle conservation and Fick's first law give the one-dimensional diffusion equation ${\partial }_{t}C=D{\partial }_x^{2}C$. For $C=A{t}^{-1/2}{e}^{-x^2/(4Dt)}$,so direct substitution verifies the equation for $t>0$. The Gaussian integral fixes the normalization:This is the heat kernel for a point release, with $C(x,0)=N\delta (x)$ in the distributional limit. It remains centered at zero while its width grows as $t^{1/2}$ and its peak falls as $t^{-1/2}$. To compute the second moment, set $y=x/\sqrt{4Dt}$:using $\int y^2{e}^{-y^2}dy=\sqrt{\pi }/2$ over the whole real line, obtained by integration by parts.

If the mean number of independent unbiased jumps in time $t$ is $⟨m⟩=ft$, conditioning on $m$ gives $⟨x^2(t)⟩=\Delta ⟨m⟩=f\Delta t$. Matching it to the continuum diffusion result yieldsFor fixed-magnitude steps $a$, $\Delta =a^2$. The equality of second moments also holds for Poisson jump times independent of the increments; a Gaussian continuum approximation additionally requires many steps and finite step variance.