I want to derive the diffusion flux $J_D$ of charged particles with charge $e$ moving in the z direction where we apply an electric field $\vec{E}=E\hat{z}$ in order to proof the Einstein relation that is often given without derivation or using Kubo's derivation. I am looking for an easy approach assuming thermal equilibrium. The mean number density at $z$ is $n(z)$. I tried the following:

$$\frac{n(z+dz)}{n(z)}=\frac{e^{-\beta W(z+dz)}}{e^{-\beta W(z)}},$$

where I used $W(z)=-eEz$ and $W(z+dz) = -eE(z+dz)$ so these are potential energies. From this I derive that

$$\frac{n(z+dz)}{n(z)} \approx 1+\beta eEdz.$$

Using the expansion $e^x \approx 1+x$ in the last part.

This then gives with $$J_D = -D\frac{dn}{dz}$$

$$J_D = -Dn(z)\beta eE,$$

which is exactly what I need to derive the Einstein relation using $J_\mu = n\mu E$ and $J_D + J_\mu = 0$ so that we find:

$$\frac{\mu}{D} = e\beta,$$

which is indeed the Einstein relation.

My question is: how can we get away with deriving this using the potential energy and not the total energy?


1 Answer 1


The total energy is the sum of the potential and kinetic energies but, given that the temperature is independent of position $z$, the average kinetic energy is also independent of $z$. So it is only necessary to consider the variation of the potential energy with $z$.

Incidentally, this derivation can also be presented in terms of particles at dynamical equilibrium in a gravitational field (again under the assumption that the system is isothermal) or in terms of osmotic pressure. For some discussion of the different versions, see the Wikipedia page on Brownian motion.


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