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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.

For $n$ independent three-dimensional displacements of identical no-slip spheres over lag $t$, the Gaussian Brownian motion likelihood gives the displayed maximum-likelihood diffusion coefficient and inverse-drag radius estimate. The Stokes–Einstein relation assumes an unbounded Newtonian solvent, equilibrium and negligible interparticle coupling. Use squared distances, rather than squaring the mean distance. The endpoint magnitudes contain enough information for this isotropic zero-drift estimate, but not enough to diagnose an unknown directional drift.

A colloidal particle is a small dispersed object whose dynamics can be strongly affected by thermal motion and interactions with the surrounding medium. Its behavior bridges molecular and macroscopic descriptions: an optical microscope can track a particle while the surrounding fluid produces unresolved random forces.

For a dilute spherical particle of radius $a$ in a Newtonian fluid of viscosity $\eta$, the Stokes–Einstein relation gives $D=k_{B}T/(6\pi \eta a)$ under its usual continuum assumptions. Larger particles diffuse more slowly. Surface forces, concentration, boundaries and nonspherical geometry can invalidate this simple expression, so the observed motion also probes the particle's environment.

For translational drag coefficient $\zeta$ and mobility $1/\zeta$, thermal equilibrium requires $D=k_{B}T/\zeta$. This connects random displacement to dissipative response and is a form of the fluctuation-dissipation theorem. Combining it with Stokes drag law gives the Stokes–Einstein relation.

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.

For a dilute spherical no-slip probe of radius $a$ in an isotropic incompressible linear medium at equilibrium, the bead's displacement per unit step force is $J_b(t)=J(t)/(6\pi a)$, where $J$ is the bulk shear creep compliance. The fluctuation-dissipation theorem gives one-coordinate displacement variance $2k_{B}T{J}_b(t)$. Adding the three equal Cartesian contributions proves the displayed relation. In a Newtonian fluid, $J=t/\eta$, so it becomes $⟨\Delta r^2⟩=6Dt$ with the Stokes–Einstein relation $D=k_{B}T/(6\pi \eta a)$.

The particle must sample a continuum response on its scale and remain in thermal equilibrium. A two-dimensional image records only two of the three displacement components; isotropy then gives $⟨\Delta r^2{⟩}_{3d}=(3/2)⟨\Delta r^2{⟩}_{2d}$. Localization noise, confinement and unresolved fast relaxation can otherwise be mistaken for elastic response.

For a fixed perfectly absorbing spherical target, steady diffusion outside radius $r_0$ obeys ${\nabla }^{2}c=0$, $c(r_0)=0$, and $c(\infty )=c_{\infty }$. Spherical symmetry gives $c=A+B/r$, hence $c(r)=c_{\infty }(1-r_0/r)$. The inward diffusion flux integrated over the target is $4\pi r_0^2{D}_A{c}^{\prime }(r_0)=4\pi r_0{D}_A{c}_{\infty }$. Consequently the number-density rate constant isThis Smoluchowski association rate has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$, as required by the given number-density rate law. If both partners moved independently, their relative diffusion coefficient would be the sum; the chromosomal target is treated as fixed here.

For a spherical RNA polymerase of radius $a=10\mathrm{nm}$, the Stokes–Einstein relation givesFor $N=6000$ free independent, approximately well-mixed searchers in a volume $V$, the small-target encounter hazard is approximately $N{k}_a/V$. ThusThe numerical coefficient explicitly uses a typical illustrative cell volume and an illustrative effective capture radius, not dimensions supplied by the source. A 60-base-pair B-form DNA segment has a contour length of about $20\mathrm{nm}$, motivating an encounter size of this order, but its contour length does not uniquely determine an absorbing sphere's radius. Taking $r_0=10\mathrm{nm}$ instead doubles the estimate. A unique numerical answer requires $V$, $T$, and a capture-geometry convention, none specified in this block.

The estimate also treats all 6000 molecules as freely available and every encounter as productive. Crowding, nonspecific binding, orientation constraints, imperfect reaction, and finite-domain initial transients can alter it. In particular, an arbitrary first-passage distribution's mean does not divide by $N$ exactly; the division follows from the approximate exponential-hazard description.^[ref]

Model the target as a perfectly absorbing sphere of reaction radius $r_0$, with a fixed target and a dilute far-field polymerase density $c_{\infty }$. Steady radial diffusion satisfies ${\nabla }^{2}c=0$, $c(r_0)=0$, $c(\infty )=c_{\infty }$, henceTherefore, for a rate law multiplying number densities,For the earliest diffusive target encounter among independent searchers, with $N_p$ independent well-mixed searchers in volume $V$, the approximate first-arrival rate is $N_p{k}_a/V$, giving ${\tau }_3\simeq V/(N_p{k}_a)$. The target's effective reaction radius includes encounter geometry and is not uniquely determined by its contour length.

A numerical estimate needs temperature, cell volume and a capture-geometry convention, none of which is fixed numerically in the source. State representative estimates $T=300\mathrm{K}$ and $V=1\mu {\mathrm{m}}^3$. The Stokes–Einstein estimate for a sphere of radius $15\mathrm{nm}$ in the stipulated viscosity isUsing the target contour scale $40\times 0.34\mathrm{nm}=13.6\mathrm{nm}$ as an order-of-magnitude $r_0$ gives $k_a\simeq 2.50\times 10^{-19}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ andThis is an idealized $10^{-2}$-second-scale estimate, not a unique geometric prediction. Finite cell shape, crowded motion, accessibility and orientational restrictions can reduce the ideal capture rate.

If nonspecific binding merely sequesters searchers and their free fraction is $0.2$, the three-dimensional capture hazard is reduced fivefold: $\tau \simeq {\tau }_3/0.2\simeq 0.067\mathrm{s}$ for those estimates. If bound molecules instead search productively by sliding, a free-time fraction alone cannot determine the total arrival time or its sign of change. That is the distinct facilitated-search mechanism treated in c.

The given sphere has radius $a=5\mathrm{nm}$. The Stokes–Einstein relation at an illustrative $T=300\mathrm{K}$ givesFor $N=30$ well-mixed independent searchers in cell volume $V$, the target-arrival rate is $N{k}_a/V$. The corresponding mean first arrival time isThe source does not fix $V$, the reaction radius or temperature. For an order-of-magnitude estimate, a bacterial volume around $1\mu {\mathrm{m}}^3$ and reaction radius $5$–$10\mathrm{nm}$ give about $0.06$–$0.12\mathrm{s}$. A 20-base-pair B-form site is about $6.8\mathrm{nm}$ long, supporting a nanometre-scale capture radius but not uniquely defining its spherical value. The estimate is the ideal diffusion-limited time, not a measured intracellular search time: nonspecific binding, orientational matching, crowding and limited accessibility are excluded.

The small binding RNA can serve conceptually as an RNA aptamer recognition probe. Coupling recognition to an optical label would make a specifically bound virus appear as a moving fluorescent spot. Single-particle tracking estimates successive spot positions and fits their displacement statistics. Specificity is assessed by comparison with a nonbinding recognition probe or a sample without the target, while the much faster diffusion of unbound probes distinguishes freely moving label from target-associated signal. Subnanomolar dissociation constant supports appreciable occupancy at low probe abundance, but affinity alone does not determine the dissociation lifetime because $K_d=k_{\mathrm{off}}/k_{\mathrm{on}}$. This describes detection principles, not an operational virus preparation or labelling protocol.

For a dilute spherical particle of radius $a=60\mathrm{nm}$ in bulk water, the Stokes–Einstein relation gives, at an explicitly assumed $T=300\mathrm{K}$,A $2\mathrm{nm}$ free probe would have $D\simeq 110\mu {\mathrm{m}}^2{\mathrm{s}}^{-1}$, roughly thirty times larger. The recognition probe changes the target's hydrodynamic size only modestly in this estimate.

For isotropic Brownian motion, the mean vector displacement is zero. In $t=0.1\mathrm{s}$, distinct measures of how far it moves areA two-dimensional image instead gives projected root-mean-square displacement $\sqrt{4Dt}=1.21\mu \mathrm{m}$ and mean projected distance $\sqrt{\pi Dt}=1.07\mu \mathrm{m}$. The projected mean squared displacement is $4D\Delta t$; independent localization errors of variance ${\sigma }^2$ in each coordinate add $4{\sigma }^2$. Finite exposure blurs motion, and nearby walls reduce diffusion; those effects must be separated from particle size when interpreting a trajectory.

Treat the target as an absorbing sphere of effective capture radius $r_0$, surrounded by a uniform dilute reservoir of RNA polymerase with number density $C_{\infty }$. The steady diffusion equation, spherical symmetry, and boundary conditions giveBy Fick's laws, the inward capture rate per target is $4\pi r_0^2{D}_A{C}^{\prime }(r_0)=4\pi D_A{r}_0{C}_{\infty }$. Multiplying by the target number density and comparing with the bimolecular rate law therefore yields the Smoluchowski association rateIts units here are ${\mathrm{m}}^3{\mathrm{s}}^{-1}$, since the concentrations count particles rather than moles. The absorbing boundary assumes every encounter is productive; orientation constraints would reduce the effective association rate constant.

The Stokes–Einstein relation, for RNA polymerase radius $a=15\mathrm{nm}$ at an illustrative temperature $T=300\mathrm{K}$, givesNeither cell volume nor the effective $r_0$ is specified, so a unique numerical time is unavailable. A representative Escherichia coli volume is $V=1\mu {\mathrm{m}}^3=10^{-18}{\mathrm{m}}^3$. The $40$-base pair target has a DNA contour extent of about $40(0.34)=13.6\mathrm{nm}$; using this as an order-of-magnitude capture radius gives $k_a\simeq 2.50\times 10^{-19}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. For $M=3000$ independent searchers, the target's approximately constant encounter hazard is $M{k}_a/V$, henceUsing half the target's contour extent for $r_0$ instead gives about $2.7\mathrm{ms}$; the spherical capture model is not exact target geometry. One searcher's mean would be about $4\mathrm{s}$ under the first convention, whereas the first arrival among $3000$ is much faster.

If only a fraction $f_3=0.1$ of each searcher's time is available for three-dimensional diffusion, and nonspecific DNA binding neither scans nor delivers the target, the effective searching population is $f_{3}M$. Thus $\boxed{\tau \simeq {\tau }_3/f_3=10{\tau }_3}$, about $13\mathrm{ms}$ under the stated estimate. This assumes stationarity, well-mixed searchers and no additional target-access route; nonspecific binding alone is sequestration, not facilitated diffusion.

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

For a no-slip spherical particle in a Newtonian fluid at low Reynolds number, Stokes law gives $\zeta =6\pi \eta a$. The fluctuation-dissipation theorem then gives the Stokes–Einstein relation $D=k_{B}T/\zeta$. Using the conventional radial Brownian displacement in three dimensions,Here $k_B$ is the Boltzmann constant, $T$ absolute temperature, $\eta$ dynamic viscosity, $a$ particle radius and $t$ elapsed time. Larger thermal forcing raises diffusion; larger viscous drag reduces it. The formula requires times long relative to momentum relaxation and neglects walls, interactions and external drift.

The word “average” needs a statistic: the vector mean is zero, while the mean radial displacement for Gaussian three-dimensional diffusion is $⟨|\Delta \mathbf{r}|⟩=4\sqrt{Dt/\pi }$, not $\sqrt{6Dt}$. Both scale as $t^{1/2}$. Neither is the total length of a microscopic random path.

For a dilute suspended sphere in Stokes flow, the translational friction coefficient is $f_h=6\pi \eta R$, with dynamic viscosity $\eta$ and hydrodynamic radius $R$. To derive the diffusion-friction relation, put particles in a weak conservative force $F=-U^{\prime }$. Their drift velocity is $F/f_h$, while the diffusive flux is $-Ddc/dx$. At equilibrium the Boltzmann distribution gives $c^{\prime }=-c{U}^{\prime }/(k_{B}T)=cF/(k_{B}T)$. Zero total flux therefore gives $cF/f_h-DcF/(k_{B}T)=0$, hence the Stokes–Einstein relationHere $k_B$ is the Boltzmann constant, $T$ absolute temperature and $D$ the translational diffusion coefficient. The sphere, no-slip continuum-fluid and dilute thermal-equilibrium assumptions enter through the drag law and flux balance.

At $T=293.15\mathrm{K}$ and the supplied viscosity, the predicted product is $DR=2.1472\times 10^{-19}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. The derived predictions and radii inferred from the measured diffusion coefficients are:

The last three agree closely at the precision of the supplied data. The ribonuclease diffusion coefficient is about $16\%$ below the spherical prediction, corresponding to a larger frictional radius than the stated radius. Hydration, shape, a different operational radius and coarse measurement precision can matter; molar mass is not an independent parameter in the spherical diffusion law. A compact constant-density globule would have $R\propto M^{1/3}$, whereas collagen's large frictional radius relative to its mass is consistent with an extended nonspherical molecule. The inferred Stokes radius is a friction-equivalent radius, not automatically a geometric half-length.

In analytical ultracentrifugation, scan the radial protein concentration while a rotor spins at known angular speed $\omega$. Optical absorbance or refractive-index interference records a moving sedimentation boundary and its diffusive broadening.^[ref] With solvent density ${\rho }_s$, partial specific volume $\bar{v}$ and molar mass $M$ in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, the buoyant mass is $m_b=M(1-\bar{v}{\rho }_s)/N_A$. Force balance gives $f_{h}v=m_b{\omega }^{2}r$, so the sedimentation coefficient is $s=v/({\omega }^{2}r)=m_b/f_h$. A boundary trajectory gives $s={\omega }^{-2}d\ln r/dt$; diffusion gives $f_h=k_{B}T/D$.

Combining them gives the molar mass from sedimentation and diffusion:Known mass and buoyancy can instead determine friction from $s$ alone, and comparison with a compact sphere tests shape or hydration. In sedimentation equilibrium, the stationary radial profile obeys $d\ln c/d(r^2)=m_b{\omega }^2/(2k_{B}T)$, providing buoyant mass without a friction model. This is a sedimentation-diffusion balance in a uniform solvent, rather than density-gradient banding at an isopycnic point.

At a micrometre scale, viscous forces normally dominate inertia. As an illustrative Newtonian estimate, $L=1\mu \mathrm{m}$, $U=10\mu \mathrm{m}{\mathrm{s}}^{-1}$, $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$ and $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ give a Reynolds number $\rho UL/\eta =10^{-5}$. The surrounding flow is approximately a Stokes flow, satisfying $-\nabla p+\eta {\nabla }^2\mathbf{u}=0$ and $\nabla \cdot \mathbf{u}=0$.

A body of radius $0.5\mu \mathrm{m}$ translated at that speed in an unbounded Newtonian liquid would need Stokes drag $6\pi \eta aU\simeq 9.4\times 10^{-14}\mathrm{N}$. It stops almost immediately when propulsion stops. The scallop theorem rules out propulsion by a single reciprocal stroke in this ideal regime; a rotating helix or a nonreciprocal cycle of shapes can propel it. Brownian motion also randomizes position and orientation: Stokes–Einstein relation scales as $a^{-1}$ and rotational diffusion as $a^{-3}$, so retaining heading becomes harder as size decreases.

The Newtonian calculation is a scale estimate, not a complete model of blood. Vessel walls alter drag, the imposed flow advects the body, and nearby red blood cells make the medium heterogeneous at this scale. Non-Newtonian fluid effects and deformable obstacles can matter; steering must be assessed against those effects rather than inertial coasting.

Passive microrheology uses thermal bead motion as a probe of mechanical compliance. The fluctuation-dissipation theorem relates equilibrium displacement fluctuations to the same linear response measured by a small applied force. The stated thermal stress scale has units of energy per bead volume; dividing the mean-square displacement by $a^2$ provides a dimensionless effective strain. With that prescribed analogy,The displacement is relative to the starting position, and the variance here is the sum over three spatial coordinates. For a Newtonian fluid, substitution gives $⟨\Delta r^2⟩=k_{B}Tt/(\pi \eta a)=6Dt$, exactly the Stokes–Einstein relation $D=k_{B}T/(6\pi \eta a)$.

The continuum interpretation assumes a dilute no-slip spherical probe, thermal equilibrium, linear response and a homogeneous incompressible medium on the bead scale. The stated stress estimate is an effective scale for the analogy; a literal instantaneous stress variance would also require a microscopic or temporal averaging prescription.

The Stokes–Einstein relation gives the translational diffusion coefficientat $T=298\mathrm{K}$. For isotropic three-dimensional Brownian motion, the mean squared displacement is $⟨|\Delta \mathbf{r}{|}^2⟩=6Dt$. Thus after one hour the root mean square displacement isThe phrase “on average” needs a convention: the mean displacement vector is zero; the mean displacement magnitude of three-dimensional Brownian motion is $⟨|\Delta \mathbf{r}|⟩=4\sqrt{Dt/\pi }=1.22\mathrm{mm}$, obtained by integrating the three-dimensional Gaussian distribution. These are displacements, not the total microscopic path length. The estimate assumes dilute unconfined water, a spherical hydrodynamic radius and constant temperature.

Use the squared three-dimensional endpoint displacements, not their arithmetic mean. The mean squared displacement estimate and corresponding diffusion coefficient areThe factor six is for three-dimensional Brownian motion. For an isolated no-slip sphere, the Stokes–Einstein relation gives the Brownian radius estimate from endpoint displacements:Here room temperature is taken as $298\mathrm{K}$; choosing $293\mathrm{K}$ gives $55\mathrm{nm}$, a much smaller difference than the sampling uncertainty. Under independent increments with a normal distribution, $\mathrm{sd}(\hat{D})/D=\sqrt{2/(3n)}\simeq 0.37$ for $n=5$. The sampling uncertainty of an endpoint diffusivity estimate therefore makes this a rough radius estimate. It assumes negligible drift and hydrodynamic coupling between probes, the supplied bulk viscosity, and actual three-dimensional tracking. Localization noise adds to the squared displacements and would bias this inferred radius downward if uncorrected.