I can't find the Einstein relation for spin diffusivity 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?
 A: 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: $D = \mu k_BT/e$.
