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A creep experiment applies a small step of stress and holds it constant while recording strain. For $\sigma ={\sigma }_{0}H(t)$, the series spring jumps by ${\sigma }_0/G_0$ and the dashpot subsequently flows at constant rate ${\sigma }_0/\eta$. The creep compliance is thusUsing the stipulated effective stress and strain substitutions givesOn linear axes the model is a straight line of positive intercept $A=k_{B}T/(\pi a{G}_0)$ and slope $A/\tau$. On logarithmic axes it has an approximate elastic plateau for $t\ll \tau$ and slope one for $t\gg \tau$, within the time window where the continuum approximation applies. The long-time slope is $6D$ with $D=k_{B}T/(6\pi \eta a)$, agreeing with the Stokes–Einstein relation.

There is an essential zero-lag convention. A literal lag displacement $\Delta \mathbf{r}(t)=\mathbf{r}(t_0+t)-\mathbf{r}(t_0)$ has $⟨|\Delta \mathbf{r}(0){|}^2⟩=0$. The nonzero intercept above is the spring-compliance extrapolation of the idealized microrheology of a Maxwell fluid, after unresolved solvent or inertial transients. It must not be interpreted as an instantaneous finite physical displacement at exactly zero lag. The stated mesh condition motivates a continuum probe response; quantitative use is more reliable for $a\gg m$ than merely $a>m$.

The Stokes–Einstein relation connects thermal diffusion with dissipative mobility. For a dilute suspension of isolated rigid spheres of hydrodynamic radius $a$ in an unbounded Newtonian fluid of viscosity $\eta$,where $\zeta$ is translational drag coefficient, $D$ diffusion coefficient and $T$ absolute temperature. Stokes drag supplies $\zeta =6\pi \eta a$ under no-slip, continuum, low-Reynolds-number conditions. The Einstein diffusion relation $D=k_{B}T/\zeta$ is more general than that particular spherical drag law.

A short equilibrium argument explains the connection. Let a weak external potential $U$ exert force $\mathbf{F}=-\nabla U$, so drift velocity is $\mathbf{F}/\zeta$. The number flux is $\mathbf{j}=n\mathbf{F}/\zeta -D\nabla n$. A Boltzmann distribution $n\propto e^{-U/(k_{B}T)}$ has $\nabla n=n\mathbf{F}/(k_{B}T)$; requiring zero equilibrium flux fixes $D=k_{B}T/\zeta$. thermal fluctuations and viscous dissipation are therefore linked, rather than independent adjustable effects. The original theory also relates diffusion to measured random displacements.^[ref]

On times long compared with momentum relaxation, an isotropic diffusing sphere hassingle-particle tracking can measure the slope. Alternatively, dilute single-scattering dynamic light scattering measures a field correlation $g_1(q,t)=e^{-D{q}^{2}t}$; the corresponding intensity-correlation excess is proportional to $e^{-2D{q}^{2}t}$. The scattering wavenumber and the distinction between field and intensity decay must be retained.

A Stokes-Einstein experimental consistency test independently measures temperature and solvent viscosity, uses a narrow distribution of inert spheres, checks decay rate proportional to $q^2$ or displacement variance proportional to time, and extrapolates toward infinite dilution. Subtract drift, account for wall-enhanced drag, and verify that aggregation is absent. Tracking additionally needs spatial calibration and correction for finite exposure and localization precision. At finite concentration, collective diffusion is not automatically the single-particle self-diffusion in the equation. A hydrodynamic radius is not generally identical to a polymer radius of gyration; nonspherical particles, slip or viscoelastic solvents require a different mobility model.

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.

velocity $U$, length $\ell$ and diffusivity $D$ have dimensions $L/T$, $L$, and $L^2/T$. Their dimensionless Péclet number isdiffusion dominates for $\mathrm{Pe}\ll 1$, advection for $\mathrm{Pe}\gg 1$. At fixed $U,D$, the crossover length is $D/U$; neglecting advection requires $\ell \ll D/U$.

TODO: user-deferred organism-specific swimming-speed and detection-scale estimates. No biological numerical substitution is made.

For the generic equilibrium transport derivation, let an external potential $V(x)$ give force $F=-V^{\prime }$ and drift $u=F/\zeta$. Boltzmann equilibrium has $c\propto e^{-V/(k_{B}T)}$, hence $c^{\prime }=Fc/(k_{B}T)$. Substituting in the zero-current condition givesThe Stokes–Einstein relation in the final equality is for an isolated no-slip sphere in a Newtonian liquid at low Reynolds number. Zero flux alone is insufficient: the equilibrium distribution and mechanical mobility law are also needed.

TODO: user-deferred organism-specific absorbing-surface calculation and its biological flux interpretation.

TODO: user-deferred receptor-number, receptor-density and receptor-competition calculations. These continuations have not been solved.

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.