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For two observed layers $z_2>z_1$, define the positive logarithmic count slopeSince $k_B=R/N_A$, where $R$ is the molar gas constant, the sedimentation scale height givesSeveral heights check whether one slope fits the whole profile. A mixture of grain radii generally gives a sum of exponential functions, with the larger grains depleted faster aloft; one apparent radius need not describe every height.

An independent readout uses horizontal displacements at lag $\tau$, avoiding the vertical gravitational drift. Define the one-coordinate mean squared displacement $M_x(\tau )=⟨[x(t+\tau )-x(t){]}^2⟩$. In a stationary liquid without horizontal advection, translational diffusion predictsA camera sees two horizontal coordinates, so its squared radial displacement has twice the one-coordinate mean. Neither observable is path length or instantaneous velocity. The bulk Stokes–Einstein relation, with dynamic viscosity $\eta$, then gives

The 1926 lecture compares displacement-based and equilibrium determinations of the Avogadro constant.^[ref]

Rotational diffusion supplies a further observable when an identifiable grain orientation can be followed; centre positions alone cannot measure it.

For the chosen model above and $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, the bulk prediction is $D\simeq 0.8731\mu {\mathrm{m}}^2{\mathrm{s}}^{-1}$. At $\tau =10\mathrm{s}$, $M_x\simeq 17.46\mu {\mathrm{m}}^2$ and the planar value is $34.92\mu {\mathrm{m}}^2$. Reconstructing $N_A$ from these stipulated model outputs returns $6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ by construction; this arithmetic consistency check is not experimental confirmation. Historical force comes from comparing separately measured profiles and displacements using independently checked material parameters.

The diffusive flux and gravitational particle flux provide another useful connection. With constant drag coefficient $\zeta$, zero vertical net flux readswhere $v_s$ is the downward settling velocity magnitude. With $\zeta =6\pi \eta a$, the chosen model gives $v_s\simeq 0.02725\mu \mathrm{m}{\mathrm{s}}^{-1}$. The equilibrium profile forgets how rapidly it was established, while displacement measurements retain that rate information.

The hydrodynamic radius is the radius of a hypothetical no-slip sphere with the same translational diffusion coefficient as the dissolved species. In the dilute continuum regime the Stokes–Einstein relation giveswhere $k_B$ is the Boltzmann constant, $T$ the absolute temperature, $\eta$ the dynamic viscosity of the solvent, and $D$ the translational diffusion coefficient. This effective radius need not equal the radius of gyration or a geometric molecular radius. It is a useful descriptor for SEC molar mass calibration, rather than a complete specification of how every shape enters every pore.

Mechanical mobility relates a steady drift velocity to a weak applied mechanical force. For isotropic linear drag $-\zeta \mathbf{v}$, terminal force balance gives ${\mathbf{v}}_{\mathrm{drift}}={\mu }_{\mathrm{mech}}\mathbf{F}$ with ${\mu }_{\mathrm{mech}}=1/\zeta$, in units $\mathrm{s}\mathrm{k}{\mathrm{g}}^{-1}$. For a sphere obeying Stokes's law, it is $1/(6\pi \eta R)$. The corresponding translational hydrodynamic mobility matrix is ${\mu }_{\mathrm{mech}}I$. The Stokes–Einstein relation reads $D=k_{B}T{\mu }_{\mathrm{mech}}$, and the velocity relaxation time is $m{\mu }_{\mathrm{mech}}$. In the bare-force spherical model, a particle with electric charge $q$ has electrophoretic mobility ${\mu }_{\mathrm{ep}}=q{\mu }_{\mathrm{mech}}$; ionic-atmosphere effects can modify that simple model.

Put $\tau =m/\zeta$. The causal Green function of the displacement equation is $G(t)={\zeta }^{-1}(1-e^{-t/\tau })$ for $t>0$ and zero for $t<0$. Its derivative has the jump needed to give $m{G}^{\prime \prime }+\zeta G^{\prime }=\delta (t)$. For the specified zero position and velocity,For Gaussian white noise these are stochastic convolutions; their first two moments follow from the stated covariance even without assuming Gaussian higher moments. In particular $⟨x⟩=⟨\dot{x}⟩=0$, andThe required equipartition theorem limit therefore fixes the fluctuation-dissipation relation for a Langevin particle:Here $k_B$ denotes the printed Boltzmann constant $k$. Set $D=k_{B}T/\zeta$, the Stokes–Einstein relation. Squaring the displacement kernel givesConsequently ${\lim }_{t\to \infty }⟨x⟩=0$ and ${\lim }_{t\to \infty }⟨x^2⟩=+\infty$, with $⟨x^2⟩=2Dt-3D\tau +o(1)$ for positive temperature. There is no finite equilibrium variance of position in an unconfined fluid. The inertial Brownian displacement with zero initial velocity has short-time variance $2D{t}^3/(3{\tau }^2)+O(t^4)$, rather than the ballistic law associated with an initially thermalized velocity.

For constant $D$, the diffusion equation has fundamental density $P(x,t)=(4\pi Dt{)}^{-1/2}{e}^{-x^2/(4Dt)}$, whose mean is zero and variance is $2Dt$. It describes the large-time, overdamped limit of the Langevin dynamics. Under the usual Gaussian-force assumption, the finite-time position density is Gaussian with the variance above and obeys a time-dependent diffusion equation with coefficient $D(1-e^{-t/\tau }{)}^2$. It does not obey the constant-$D$ equation through the initial inertial transient. The given covariance alone determines the moments, not a Gaussian law; the density statement uses the standard Gaussian thermal-noise model.

Now write $\zeta (x)={\zeta }_0+\epsilon {\zeta }_0^{\prime }x$ and $x=x_0+\epsilon x_1+O({\epsilon }^2)$, with ${\tau }_0=m/{\zeta }_0$ and $D_0=k_{B}T/{\zeta }_0$. For a fixed realization of the additive force, the first-order equation isIf the force amplitude is instead adjusted to local thermal equilibrium, its first-order noise term has zero mean in the inertial Itô formulation and does not change the following first-order mean calculation. The mixed unperturbed moment isLet $M(t)=⟨x_1(t)⟩$. Averaging and solving the scalar forced relaxation equation, with $M(0)=M^{\prime }(0)=0$, givesThus the long-time first-order coefficient and local mean drift areLiterally $\lim M=-\infty$ for ${\zeta }_0^{\prime }>0$, $+\infty$ for ${\zeta }_0^{\prime }<0$, and zero for ${\zeta }_0^{\prime }=0$; the useful finite quantity is $\lim M^{\prime }$. Because the viscosity gradient is only a local expansion, the physical drift result applies in the intermediate window ${\tau }_0\ll t\ll L_{\zeta }^2/D_0$, where $L_{\zeta }={\zeta }_0/(\epsilon |{\zeta }_0^{\prime }|)$. Extrapolating the linear drag law to arbitrarily large displacements could make the drag negative and would invalidate the model.

The isothermal diffusion with position-dependent drag interpretation uses locally matched thermal noise and $D(x)=k_{B}T/\zeta (x)$. Integrating the printed Fickian diffusion equation against $x$, with vanishing boundary fluxes, givesSince $D^{\prime }(x)=-\epsilon k_{B}T{\zeta }_0^{\prime }/{\zeta }_0^2+O({\epsilon }^2)$, this reproduces the same drift after the inertial transient. Equivalently the overdamped Itô diffusion has drift $D^{\prime }$ and noise amplitude $\sqrt{2D}$. Omitting that drift would give $P_t=(DP{)}_{xx}$, which is a different Fokker-Planck equation and not the stated Fickian equation.

Noise conventions cannot be mixed silently. If the force covariance is held at the spatially constant value $F_0=2{\zeta }_0{k}_{B}T$ while drag varies, the local overdamped diffusion coefficient is instead $\tilde{D}=F_0/(2{\zeta }^2)$ and its Itô drift is ${\tilde{D}}^{\prime }/2$. Its density equation is $P_t={\partial }_x[\tilde{D}{P}_x+({\tilde{D}}^{\prime }/2)P]$, not the isothermal Fickian equation. This still gives the same displayed first-order mean drift at $x=0$, but has different spatial diffusion and equilibrium properties. The stated Fickian interpretation therefore requires local thermal-noise matching.

Physically, thermally driven excursions travel farther on the lower-drag side. Their displacement is correlated with velocity, so averaging the position-dependent friction leaves a net drift toward greater diffusivity, without an applied force. This is not a violation of equilibrium: in the isothermal Fickian model the drift and spatially varying noise cancel in the probability flux for a uniform density. It is a local diffusion-induced shift of the centroid of an initially localized particle distribution.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives

Diffusion and mobility. For a dilute isolated sphere in a Newtonian fluid, Stokes's law gives the drag magnitude $F_{\mathrm{drag}}=6\pi \eta Rv=\xi v$, where $\eta$ is the dynamic viscosity and $R$ is the hydrodynamic radius. The mechanical mobility is therefore $1/\xi$; the translational hydrodynamic mobility matrix is ${\xi }^{-1}I$.

To relate mobility to diffusion, put the particles in a weak external potential energy $U(x)$. Their drift speed is $-U^{\prime }(x)/\xi$. For number density $n(x)$, the sum of drift and Fick's first law fluxes isAt thermal equilibrium, the Boltzmann distribution gives $n\propto e^{-U/(k_{B}T)}$, and the net flux is zero. Substituting $n^{\prime }=-n{U}^{\prime }/(k_{B}T)$ into $J=0$ yields $D=k_{B}T/\xi$. Thus the Stokes–Einstein relation isHere $k_B$ is the Boltzmann constant and $T$ is absolute temperature. The assumptions include a no-slip approximately spherical particle, a continuum fluid, a small particle Reynolds number, and negligible hydrodynamic interactions with other particles or boundaries.

Gravitational sedimentation. The buoyant force must be included, since the mass of displaced water is comparable to the protein mass. Using $1\mathrm{Da}=1.66054\times 10^{-27}\mathrm{kg}$ givesso the buoyant mass in the stated spherical model isThe drag coefficient is $\xi =6\pi (10^{-3})(2.16\times 10^{-9})=4.0715\times 10^{-11}\mathrm{kg}{\mathrm{s}}^{-1}$. At terminal speed, $\xi v_g=m_{b}g$, soIgnoring buoyancy would give $1.47\times 10^{-11}\mathrm{m}{\mathrm{s}}^{-1}$ and would not use the supplied water density. This calculation treats the stated radius as both the hydrodynamic radius and the radius determining displaced volume; a real hydrated protein can require a separately measured partial specific volume.

Electrical drift. In the specified simple electrophoresis model, terminal force balance is $Ze\mathbf{E}=\xi \mathbf{v}$. Thereforewhere $\mu$ is the electrophoretic mobility and $e$ the elementary electric charge. For the given positive beta-lactoglobulin charge,in the direction of the electric field. This exceeds the gravitational sedimentation speed by about $8.6\times 10^6$.

Separation time. The two populations have equal drag coefficients because their radii are equal. Their difference in drift speed is therefore set by one elementary charge:Starting with coincident band centres, their separation is $\Delta vt$. HenceAt that time the original band has travelled $20\mathrm{mm}$ and the mutant $21\mathrm{mm}$. Their masses do not enter this ideal electrical force balance: equal-radius particles with different masses but the same charge would have equal electrophoretic mobility in this model.

Practical discrimination. A one-charge difference is enough for the stipulated separation, so the method is sensitive to mutations that change effective charge or friction. It cannot distinguish every mutation: substitutions that preserve charge and hydrodynamic size need not separate. Buffer pH changes protein ionization, and realistic electrophoresis also involves the ionic atmosphere, electro-osmosis, walls or gel interactions, and finite band widths.

For example, the Stokes–Einstein relation at $293.15\mathrm{K}$ gives $D=9.94\times 10^{-11}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. For an initially narrow band, diffusive broadening of electrophoretic bands givesat the calculated separation time. Thus the centres are about $3.15$ diffusion standard deviations apart, rather than forming infinitely sharp bands. The stated $1\mathrm{mm}$ centre-separation criterion is achievable in the model, but actual resolution also depends on the loading width and other dispersion mechanisms.

Equivalent centrifugal acceleration. For the original protein to attain its electrical drift speed by sedimentation, its buoyant centrifugal force must equal $20eE$:ConsequentlyThis is the acceleration at the protein's position in the rotating solution; no rotation rate can be assigned without a centrifuge radius. If the intended comparison is instead the relative separation speed of the two bands, use $\Delta m_b=600\mathrm{Da}$, since their displaced volumes are equal. Matching $|\Delta m_b|a_{\mathrm{eff}}=eE$ then gives $8.04\times 10^7\mathrm{m}{\mathrm{s}}^{-2}$. In that separation experiment the heavier original protein moves faster centrifugally, whereas the more highly charged mutant moves faster electrically.

For a sphere of radius $a$ in an otherwise stationary, unbounded Newtonian fluid with dynamic viscosity $\eta$, impose the no-slip boundary condition and the slow-flow assumptions of Stokes flow. The resisting torque on a rotating sphere in Stokes flow is $-{\zeta }_r\mathbf{\Omega }$, where $\mathbf{\Omega }$ is its angular velocity and ${\zeta }_r=8\pi \eta a^3$. Thermal rotational diffusion at absolute temperature $T$ then givesHere $k_B$ is the Boltzmann constant; the translational expression is the Stokes–Einstein relation. The ratio has units of inverse length squared, so comparing the numerical values of $D_r$ and $D_t$ without their units is meaningless. At fixed temperature and dynamic viscosity, decreasing radius tenfold increases $D_r$ a thousandfold but $D_t$ only tenfold. This strong size dependence explains why small particles can lose directional memory extremely quickly.

Choose $T=298\mathrm{K}$ and $\eta =0.001\mathrm{Pa}\mathrm{s}$ as illustrative model inputs. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ gives $k_{B}T=4.11433402\times 10^{-21}\mathrm{J}$ and the following predictions:These are bulk three-dimensional sphere calculations, explicitly not a fit to the confined ellipsoid experiment below. Walls, nonspherical shape and departure from the no-slip boundary condition can change rotational resistance. A visible surface mark is still necessary to read the sphere's turning in an image, even when its rotational diffusion coefficient is well defined physically.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

Buoyancy changes the effective gravitational mass; a nearby no-slip wall changes the hydrodynamic resistance matrix and therefore the diffusion coefficient. These are distinct corrections. Write the horizontal resistance as ${\zeta }_{\parallel }=6\pi \eta a/f$, with $0<f<1$ representing wall hindrance at a specified height. Applying the Stokes–Einstein relation with that resistance givesif one mistakenly uses the bulk resistance in the inversion. A chosen $f=0.8$ would overestimate $N_A$ by $25\%$; this is a sensitivity example, not a correction measured in Jean Baptiste Perrin's apparatus. Normal and parallel coefficients of the hydrodynamic resistance matrix need not coincide. Stokes settling velocity near a wall can likewise bias a radius inferred using an unbounded-fluid formula.

For a passive dilute particle suspension at uniform temperature, a change of mobility alone does not change the Boltzmann distribution. In the absence of additional wall forces the correct local vertical flux is $j_z=-D_z(z)[dn/dz+m_{b}gn/(k_{B}T)]$, so zero flux still gives the same sedimentation scale height. Walls can nevertheless invalidate that simple inference through adsorption, additional interaction energies, or sampling of inaccessible grain-centre positions. Optical position errors and exposure averaging also affect displacement readout; varying distance from the wall, observation lag and optical depth tests different possible errors.