3
$\begingroup$

I'm studying the article "Thermodynamic analysis of interfacial transport and of the thermomagnetoelectric system" (PhysRevB.35.4959, Mark Johnson and R. H. Silsbee); they use this relation $$ D=\frac{\sigma \beta^2}{\chi e^2} $$ calling it "Einstein relation", but I can't find it anywhere on textbooks nor online, where the actual Einstein relation is written in terms of the mobility and thermal energy $$ D=\mu k_B T $$ What is the origin of that relation? Where I can find it?

$\endgroup$

1 Answer 1

1
$\begingroup$

I am not sure about the $\beta^2$ part. Coleman derives $\sigma = e^2\chi D$ in his book Introduction to many-body physics Sec 10.4. The short version of it is that the density-density response function receives a correction when there are impurities. In perturbation theory in momentum $q$, one can identify the coefficient of $q^2$ as the diffusion constant.

Edit: The field theory is very nice. But I came across another intuitive picture in the book Electronic Transport in Mesoscopic Systems by Datta. I will be paraphrasing Sec 1.7 of that book in the following.

Consider applying a small field $E_x$ onto a 2DEG. Denote the changed Fermi energy at the left edge $\mu_1$, and $\mu_2$ at the right edge. To make the connection with diffusion, remember that in a Fermi liquid, currents are carried by electrons near the Fermi surface. In other words, the electrons that actually make contributions are the ones in the energy interval $\mu_2<E<\mu_1$. The net effect of the electric field is to produce a density gradient across the 2DEG. Using the diffusion equation $J = -eD\nabla n = eD\chi(\mu_1-\mu_2)/L = e^2D\chi E_x$, we get $\sigma = e^2\chi D$. $\chi$ here is the density of states.

One can also think the current as simply $J = env_d$, where $n$ is the electron density and $v_d$ is the drift velocity. Since mobility $\mu = e\tau/m$ (and $v_d = eE_x\tau/m$), conductivity $\sigma = en\mu$. Comparing with the expression above, we get the usual $D=E_f\mu/e=v_f^2\tau/2$.

Notice this picture only works for a Fermi liquid or a degenerate (semi)conductor. For a non-degenerate (semi)conductor, the important energy scale is $k_BT$ rather than $E_f$. Then we get the second relation you have: $D = \mu k_BT/e$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.