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Centrifugation separates suspended material by its motion relative to a rotating liquid. A centrifuge rotor creates an outward centrifugal acceleration in the rotating reference frame; material denser than its surroundings tends to move outward, opposed by drag force. Collected material can form a centrifugation pellet, leaving a supernatant, or remain in separated zones within a density gradient. The useful question is which physical difference the separation exploits: a difference in sedimentation coefficient or a difference in buoyant density.

Centrifugation makes otherwise slow settling useful for studying biological cells, organelles, proteins and DNA. Separation and identification are different steps: a centrifugation pellet is a collected fraction, not proof that it contains only one kind of organelle. The lysozyme wall-removal experiment illustrates a complementary readout: testing material released into the separated liquid can be more informative than inspecting the retained solids.

A centrifuge rotor holds sample vessels around the rotation axis. A fixed-angle centrifuge rotor holds tubes at a fixed inclination; a swinging-bucket centrifuge rotor lets them swing outward during rotation. Material moves away from the axis, so a centrifugation pellet can first collect against an outer tube wall rather than travel along the entire tube axis. Geometry sets both the rotor radius and the travel distance.^[ref]

Why does an isopycnic centrifugation band return after displacement? Choose a dilute mathematical model with outward coordinate $z$, constant local centrifugal acceleration $a>0$, fixed displaced volume $V>0$, linear friction coefficient $f>0$, and an imposed liquid density ${\rho }_f(z)={\rho }_0+Gz$ with $G>0$. Treat the species' buoyant density ${\rho }_b$ as constant in this medium. The sedimentation coefficient force balance gives the ordinary differential equationIts exponential function solution is $z(t)=z_{\ast }+(z(0)-z_{\ast })e^{-kt}$. The sign of the velocity reverses across $z_{\ast }$, proving stable equilibrium when $z_{\ast }$ lies inside the liquid. Changing $V/f$ changes the approach time $1/k$, while $z_{\ast }$ depends on buoyant density and the imposed density gradient. A real centrifuge rotor has radially varying centrifugal acceleration; the constant-$a$ approximation describes a short radial interval. Here the density gradient is held fixed, rather than modeling its formation.

For an original dimensionless example, let $u=z/L$ and $\tau =t/T$, where $L$ is column length and $T$ a chosen time scale. Write $q=z_{\ast }/L$ and $\lambda =kT$, so $u(\tau )=q+(u(0)-q)e^{-\lambda \tau }$. Species A and B have the same $q=0.45$ but different friction-dependent rates: $({\lambda }_A,u_A(0))=(0.4,0.1)$ and $({\lambda }_B,u_B(0))=(1.6,0.95)$. A moves outward and B inward, both toward $u=0.45$. Species C has $(q_C,{\lambda }_C,u_C(0))=(0.8,0.8,0.1)$ and approaches a separate band at $u=0.8$. At $\tau =4$, their positions are approximately $0.3793$, $0.4508$ and $0.7715$: a finite-time photograph would still show A short of its limiting position.

An equilibrium band has finite width because diffusion opposes focusing. In the same local model, let $c(z)$ be concentration and let the constant diffusion coefficient be $D>0$. Combining the restoring drift with Fick's laws, zero net flux requiresDividing by $c$ and integrating proves the Gaussian function profile, whose standard deviation is ${\sigma }_z=\sqrt{D/k}$. This local result assumes that the band is narrow relative to its distance from the boundaries and that $D$, $k$ and the density gradient vary negligibly across it. Equal buoyant densities therefore imply coincident centers in this model, not identical widths. A band outside the available density range has no such interior restoring point.

Relative centrifugal force is the dimensionless ratio of the local centrifugal acceleration magnitude $a$ to the conventional gravitational reference $g_0=9.80665\mathrm{m}{\mathrm{s}}^{-2}$. Despite its name, it is an acceleration ratio: the actual force on a specified mass $m$ is $ma$. With angular speed $\omega$ and rotor radius $r$, $a={\omega }^{2}r$. If $n$ is the numerical rotational speed in revolutions per minute and $r_{\mathrm{cm}}$ the numerical rotor radius in centimetres, unit conversion givesThus rpm alone does not specify the centrifugal acceleration.^[ref]

For a stipulated $3000\mathrm{rpm}$, rotor radii of $5$ and $10\mathrm{cm}$ give approximately $503$ and $1006$ times $g_0$. Doubling the rotor radius doubles relative centrifugal force; doubling rpm quadruples it. Even equal relative centrifugal forces do not guarantee equal separation: travel distance, elapsed time, viscosity, sample density and particle shape still matter. These are comparisons of a mathematical model, not instructions for running an instrument.