# Characteristic length for the diffusion equation (temperature)

The background: I'm doing some simulation work involving the diffusion equation in 1D. Specifically I have some temperature profile, constant thermal conductivity and fixed temperature at each end of the system.

I know that we can write:

$$\tau = \frac{L^2}{\kappa}$$

where $\tau$ is the characteristic time scale, $L$ is the characteristic length scale and $\kappa$ is the thermal conductivity. In this case, $\kappa = 1$ so the time scale is equal to the square of the length scale.

I know that in a gas, the time scale corresponds to something like the amount of time it takes a particle to diffuse over the length scale of interest, but I'm not sure what it means in the context of temperature.

Could anyone enlighten me? Thanks!

• I think $\kappa$ in this case is the thermal diffusivity, not thermal conductivity. Aug 19 '14 at 1:13

Short answer: $\tau$ is the typical time it takes for heat (energy) to be transported over the distance $L$.
For particle diffusion in one dimension, you may think of the particle as jumping around on the x-axis. Some times it jumps to the right, and some times to the left. The end result is that it typically takes $\tau = L^2/\mathcal D$ to cover the distance $L$, when the diffusion constant is $\mathcal D$. The diffusion constant is a measure of how large the jumps are (in fact, how large the variance of the jumps is).
But for heat transport you may instead think of a chain of beads on the x-axis. Each bead is wiggling around its spot on the axis, and the more it wiggles, the higher the temperature at that position. Every now and then a wiggling bead will smack its neighbor, and exchange some energy with it. Some times the energy transfers from left to right, and some times the transfer is from right to left. The "thermal diffusivity" $\kappa$ is a measure of how often the beads collide, and how willing they are to exchange their energy with each other. The end result is that $\tau = L^2/\kappa$ is the typical time it takes for a "packet" of energy to travel over the distance $L$.