The Mason–Weaver equation (named after Max Mason and Warren Weaver) describes the sedimentation and diffusion of solutes under a uniform force, usually a gravitational field. Assuming that the gravitational field is aligned in the z direction (Fig. 1), the Mason–Weaver equation may be written
∂ c ∂ t = D ∂ 2 c ∂ z 2 + s g ∂ c ∂ z {\displaystyle {\frac {\partial c}{\partial t}}=D{\frac {\partial ^{2}c}{\partial z^{2}}}+sg{\frac {\partial c}{\partial z}}}
where t is the time, c is the solute concentration (moles per unit length in the z-direction), and the parameters D, s, and g represent the solute diffusion constant, sedimentation coefficient and the (presumed constant) acceleration of gravity, respectively. The Mason–Weaver equation is complemented by the boundary conditions
D ∂ c ∂ z + s g c = 0 {\displaystyle D{\frac {\partial c}{\partial z}}+sgc=0}
at the top and bottom of the cell, denoted as z a {\displaystyle z_{a}} and z b {\displaystyle z_{b}} , respectively (Fig. 1). These boundary conditions correspond to the physical requirement that no solute pass through the top and bottom of the cell, i.e., that the flux there be zero. The cell is assumed to be rectangular and aligned with the Cartesian axes (Fig. 1), so that the net flux through the side walls is likewise zero. Hence, the total amount of solute in the cell
N tot = ∫ z b z a d z c ( z , t ) {\displaystyle N_{\text{tot}}=\int _{z_{b}}^{z_{a}}\,dz\ c(z,t)}
is conserved, i.e., d N tot / d t = 0 {\displaystyle dN_{\text{tot}}/dt=0} .
Derivation of the Mason–Weaver equation
A typical particle of mass m moving with vertical velocity v is acted upon by three forces (Fig. 1): the drag force f v {\displaystyle fv} , the force of gravity m g {\displaystyle mg} and the buoyant force ρ V g {\displaystyle \rho Vg} , where g is the acceleration of gravity, V is the solute particle volume and ρ {\displaystyle \rho } is the solvent density. At equilibrium (typically reached in roughly 10 ns for molecular solutes), the particle attains a terminal velocity v term {\displaystyle v_{\text{term}}} where the three forces are balanced. Since V equals the particle mass m times its partial specific volume ν ¯ {\displaystyle {\bar {\nu }}} , the equilibrium condition may be written as
f v term = m ( 1 − ν ¯ ρ ) g = d e f m b g {\displaystyle fv_{\text{term}}=m(1-{\bar {\nu }}\rho )g\ {\stackrel {\mathrm {def} }{=}}\ m_{b}g}
where m b {\displaystyle m_{b}} is the buoyant mass. We define the Mason–Weaver sedimentation coefficient s = d e f m b / f = v term / g {\displaystyle s\ {\stackrel {\mathrm {def} }{=}}\ m_{b}/f=v_{\text{term}}/g} . Since the drag coefficient f is related to the diffusion constant D by the Einstein relation
D = k B T f {\displaystyle D={\frac {k_{B}T}{f}}} , the ratio of s and D equals
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