# Problem 5 – Hint 5

The fifth hint is not so much of a hint, but rather, a brief explanation of what is diffusion. In one-dimensional geometry, the heat flux is given by $J=\kappa \frac{\partial T}{\partial x}$. This means that the heat accumulated at an interval from $x$ to $x+\mathrm dx$ during $\mathrm dt$ per unit cross-sectional area (in y-z-plane) equals to $\rho c\mathrm dT\mathrm dx$ on the one hand, and to $\mathrm dJ\mathrm dt=\kappa \frac{\partial^2 T}{\partial x^2}\mathrm dt\mathrm dx$, on the other hand. As a result, we obtain

$\frac{\partial T}{\partial t}=\frac{\kappa}{\rho c} \frac{\partial^2 T}{\partial x^2}$.

This is what is called the diffusion equation, with the ratio $D=\frac{\kappa}{\rho c}$ being referred to as the diffusivity.

We can approach this equation in different ways. First, we can use the method of dimensional analysis to estimate how long it will take for heat to reach distance $L$; the answer is $\frac{L^2}{D}$. Second, we can look for a self-similar solution in the form $T = t^\alpha f(\xi)$, where $\xi = \frac{x}{t^\beta}$. If we plug this into the diffusion equation, we can see that for the equation to include only $\xi$ without $x$ and $t$, $\beta$ must be $\beta = \frac{1}{2}$. The total heat must be conserved in time, so $\int T \mathrm{d}x = \mathrm{const}$. This means that $\alpha$ must be $\alpha = -\frac{1}{2}$. From here, we can recover the previously obtained estimate for the time needed for heat to spread to distance $L$. If needed, we can also solve the obtained ordinary differential equation to find this self-similar solution $T=\frac{1}{\sqrt{2\pi Dt}} \exp\left(-\frac{x^2}{4D\,t}\right)$ that appears to be stable: for any initial injection of heat, the temperature profile will asymptotically approach this solution.

Finally, we can use the fact that the diffusion process can be looked at in two ways: first, as the process described by the diffusion equation; second, as a process where many Brownian particles perform random walks and make randomly directed steps of length $\lambda$ with frequency $1/\tau$. In the latter case, one can show that the concentration of Brownian particles is also described by the diffusion equation with $D = \frac{\lambda^2}{2\tau}$. This means that we can switch to studying the random walk of a Brownian particle. Then the displacement of the particle can be expressed as $x = \lambda \sum_{i=1}^n s_i$, where $n = t/\tau$ is the number of steps, and $s_i = \pm 1$ denotes the direction of the $i$-th step. Let us calculate the average squared displacement reached by time $t = n\tau$:
$\left\langle x^2 \right\rangle = \left(\lambda \sum_{i=1}^n s_i\right) \left(\lambda \sum_{j=1}^n s_j\right) = \lambda^2 \sum_{i,j=1}^n \left\langle s_i s_j \right\rangle = \lambda^2 \sum_{i,j=1}^n \delta_{ij} = \lambda^2 n = \lambda^2 \frac{t}{\tau} = 2Dt.$Here we have averaged over what is called an ensemble of random walks, i.e., over different random sequences of $+1$s and $-1$s; $\delta_{ij}$ denotes the Kronecker delta.

When we switch to two-dimensional geometry, the estimate $L \sim \sqrt{Dt}$ remains valid. Meanwhile, in cylindrically symmetric cases, the heat flux (per unit length of the cylinder) can be expressed using cylindrical coordinates as $J = 2\pi r \kappa \frac{\partial T}{\partial r}$.

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