\(\textbf{Physics}\) \[\textit{Diffusion Constant}\] \[\textit{& The Einstein Relation}\] \[D=kT\mu\]
We have seen the forms of diffusive drift and the electrical drift by Fick’s law and Ohm’s law, respectively. If we consider that no other external factors affect the movement of the particles, then at equilibrium
\[J_{diff}+J_{drift}=0\]
\[-D\frac{\partial [C]}{\partial x}-\mu z [C] \frac{\partial V}{\partial x}=0\]
If we consider the most general case of drift across a potential field, we may ignore the charge, such that \(z\) is redundant in this equation (\(z=1\)).
Remembering that this equation is satisfied at equilibrium where there is no net flow, and that the concentration of ions at a locality will depend only on the potential at that point, we can simplify this equation. In other words, at equilibrium, \([C]\) can be modelled as a function of \(V\). Therefore, by the chain rule;
\[\frac{\partial [C]}{\partial x} =\frac{d [C]}{dV}\frac{\partial V}{\partial x}\]
So that
\[ \frac{\partial V}{\partial x} \left(-D\frac{d[C]}{dV}-\mu [C]\right) =0\]
\[D = - \frac{\mu [C]}{\frac{d[C]}{dV}}\]
From Maxwell-Boltzmann Statistics [5], the concentration of the ions are distributed according to their energies (we assume that at equilibrium, the ions do not interact, so the energies are due only to potential)
\[[C]=ge^{-\frac{V}{kT}}\]
\[\frac{d[C]}{dV}=-g\frac{1}{kT} e^{-\frac{V}{kT}} = -\frac{1}{kT}[C]\]
Substituting this (naturally considering that \([C]\) is non-zero), we arrive at
\[D=kT\mu\]
This is the most general form for the diffusion constant. Notice how this value is directly proportional to the mobility. Since we know that both of these values are dependent on the diffusive material, this relation is quite useful.
In the specific case of cell membranes, the mobility used is also due to the ion, or in other words, we have the electrical mobility, \(\mu_q\)
\[\mu_q = \frac{\mu }{q}\]
So we have
\[D= \frac{kT\mu}{q} \]
Substituting this value into our equilibrium differential equation for cell membranes, we arrive at
\[- \frac{kT\mu}{q} \frac{\partial [C]}{\partial x}-\mu z [C] \frac{\partial V}{\partial x}=0\]
