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The Einstein relation $D = \mu k_B T$ is derived by assuming an equilibrium between the drift current and the diffusion current. Knowing this I would assume, that the relation is only valid under this assumption.

However, when deriving the Goldman equation, we start with a combination of drift current and diffusion current: $$j_v = -D_v \frac{\partial c_v}{\partial x}+c_vv_{Drift}$$ and then use the Einstein relation to transfer this equation to the more complex version of the current $j_v$ which is stated on Wikipedia.

Why can we use the Einstein relation in this case, when drift and diffusion current are not in equilibrium? (If they were, $j_v$ would be 0, which is not the case).

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The voltage driven drift and the concentration-driven diffusion are distinct physical effects so it is not double-counting to include both of them when computing a current flow. That there is a relation between them comes from the possibility of equilibrium in the situation of a static potential gradient. Einstein knew that in equilibrium the time-independent density $\rho$ would be proportional to the Boltzman factor $\exp\{- V(x)/kT\}$. This could only hold when the mobility and the diffusion constants obeyed his relation so that the sum of the drift and disffusion currents is zero.

Physically the relation exists because both constants are related to the rate at which the moving particles are scattering off impurities etc. Einstein was a master of thought experiments and so was able to go beyond this vague physical idea and figure out the exact equation.

To be precise: suppose that ${\bf v}_{\rm drift} = -\mu \nabla V$ and ${\bf j}_{\bf diff} = -D \nabla \rho$. These phenomenological equation hold in all processes, equilibrium or non equilibrium. Now consider a particular case. This is the effect of a potential $V$. We know from thermodynamics that this will lead to an equilibrium distribution in which $$ \rho = \rho_0 \exp\{-V/kT\} $$ The net current from diffusion and voltage induced drift is $$ {\bf j}_{\rm tot}={\bf j}_{\rm drift} + {\bf j}_{diffusion}\\ = \rho {\bf v}_{\rm drift}+ D \rho_0 \exp\{-V/kT\}\frac 1{kT} \nabla V\\ = -\rho \mu \nabla V +D\rho \frac 1{kT} \nabla V $$ Since we are in equilibrium ${\rm j}_{\rm tot}=0$. This can only hold for arbitrary $V$ if $D=\mu kT$.

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  • $\begingroup$ This is the answer to the question why the Einstein relation holds in general. I still don't know why we can use it in the case above, where drift and diffusion currents are not in equilibrium, which was an assumption for the Einstein relation. $\endgroup$
    – PhylomatX
    Jun 25, 2019 at 8:16
  • $\begingroup$ I'll add some more text. $\endgroup$
    – mike stone
    Jun 28, 2019 at 16:02

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