PHY.K02UF Molecular and Solid State Physics

The Einstein relation

One of the remarkable papers that Einstein published in 1905 was about Brownian motion. In it, he considered what happens to particles confined to a region of space by a potential $U$. The force on the particles will be $\vec{F}=-\nabla U$ and they will be pulled to this minimum of the potential. This will cause a high concentration of particles at the minimum, and they will diffuse away from the minimum. In equilibrium, there must be a balance between the inward current due to the attractive potential and the outward current due to diffusion.

If the particles are electrons, the potential will be $U=-eV$ where $V$ is the electrostatic potential, the drift current density pulling the electrons to the minimum will be $j_{\text{drift}} = ne\mu_e \vec{E}$ and the diffusion current density pushing them away from the minimum will be $j_{\text{diffusion}}=eD\nabla n$, so the total current density is,

$$\vec{j} = ne\mu_e\vec{E} + eD\nabla n.$$

Since the current density must be zero in equilibrium, we have,

$$n\mu_e\vec{E} = -D\nabla n.$$

Einstein used Boltzmann's theory to argue that the electron density should be described by a Boltzmann factor,

$$n = n_0\exp\left(\frac{-U}{k_BT}\right)= n_0\exp\left(\frac{eV}{k_BT}\right).$$

The gradient of the electron density would then be,

$$\nabla n = \frac{n_0e}{k_BT}\nabla V\exp\left(\frac{eV}{k_BT}\right)= \frac{e\nabla V}{k_BT}n.$$

Substituting this into the equation for the balance between the drift and diffusion currents and using the fact that $\vec{E}=-\nabla V$ yields,

$$\bbox[10px, border: 1px solid black]{\mu_e = \frac{eD}{k_BT}.}$$

This is known as the Einstein relation. It says that particles with a high diffusion constant will have a high mobility.