Finite-time rate-zonal separation

Consider two dilute populations of spheres with the same buoyant density and radii $b$ and $2b$. Adopt the fixed linear density gradient and constant local centrifugal acceleration of linear-gradient isopycnic focusing, with constant dynamic viscosity and negligible diffusion. Stokes's law makes $V/f$ proportional to $b^2$, so doubling radius multiplies the migration rate parameter by four, although the mass increases eightfold.

Use $u=z/L$, $\tau =t/T$ and $q=({\rho }_b-{\rho }_0)/(GL)=2$. The liquid occupies $0\le u\le 1$, so the matching density position lies beyond its outer boundary. Both populations begin at $u(0)=0.1$; choose ${\lambda }_A=0.15$ and ${\lambda }_B=0.6$. Their band-center trajectories areAt $\tau =0.5$, they lie at approximately $0.2373$ and $0.5924$; at $\tau =1$, at $0.3647$ and $0.9573$. Equal buoyant density has not prevented a useful finite-time separation. Solving $u_B(\tau )=1$ using the natural logarithm gives first boundary arrival at $\tau =\ln (1.9)/0.6\simeq 1.0698$. The corresponding arrival for A is $\ln (1.9)/0.15\simeq 4.2790$. Trajectories describe migration only until contact with the boundary; subsequent collection into a centrifugation pellet is not represented by extending the curve beyond $u=1$.

This is the useful contrast with isopycnic centrifugation: stopping time helps determine the recovered positions in rate-zonal centrifugation, whereas a stationary matching density position sets the limiting band center in isopycnic centrifugation. Real resolution also depends on initial zone width, diffusion and variations in particle properties; distinct calculated centers alone do not prove that fractions will be pure.