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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.

A researcher whose team investigated the coupling of rotational diffusion and translational diffusion in a freely moving ellipsoid. At short times the particle remembers its orientation, while at longer times rotation averages over directions. The experiment shows how the timescale of observation changes the apparent symmetry of Brownian motion.^[ref]

A small swimming member of the bacteria samples too little spatial separation across its own body to compare weak chemical gradients accurately. Temporal sensing in bacterial chemotaxis instead compares recent receptor activity with an adapted memory. Straight runs carry it between locations; tumbles reorient it. When conditions improve, tumbling becomes less frequent and favorable runs last longer. An unfavorable direction is abandoned sooner. Repetition creates a biased random walk without requiring the cell to identify the best direction at each tumble. Rotational diffusion and environmental obstacles eventually destroy the benefit of maintaining a direction indefinitely.

An order-of-magnitude transport estimate compares the time $\ell /v$ to swim a distance $\ell$ with the time ${\ell }^2/D$ for an attractant to diffuse that distance. Their crossover is a Péclet number of order one:This estimates a useful run scale: over much shorter distances, nutrient diffusion rapidly replenishes the local environment compared with the organism's displacement. Runs of order a second also permit comparison along a substantial path before reorientation. The given $D$ and $v$ determine this scale, not an exact biological mean run length or a complete sensory optimization. If a one-coordinate root-mean-square diffusion length $\sqrt{2D\tau }$ is equated with $v\tau$, the resulting prefactor is two; a three-dimensional root-mean-square convention gives six. Such constants cannot be fixed by a scaling argument alone.

For a cilium of length $d$, compare its advective time $d/u$ with the local diffusion time $d^2/D$. Significant stirring relative to diffusion requires ${\mathrm{Pe}}_d=ud/D\gtrsim 1$, soThis is about $33$ times the stated swimming speed. If the tip travelled on a circular path of radius $d$, that speed would correspond to an illustrative frequency $u/(2\pi d)\simeq 160\mathrm{Hz}$. That geometric conversion is not a specified beat pattern. At microscopic scales the flow has small Reynolds number; useful transport depends on the actual nonreciprocal motion and resulting flow, and rapid motion alone does not guarantee increased uptake. The scaling explains why simply stirring with a very short appendage is demanding when molecular diffusion is already fast.

The cell conservation equation equates the change of cell number density $n$ to minus the divergence of its total flux. The flux $n\mathbf{u}$ is advection by the water, $n{V}_s⟨\mathbf{p}⟩$ is the mean swimming flux, and $-\mathbf{D}\cdot \nabla n$ is the dispersal flux. The tensor $\mathbf{D}$ describes effective translational diffusion, whereas $D_r$ in part (i) describes rotational diffusion; they have different dimensions. This conservation law neglects birth, death and sources at the walls.

The equation $\nabla \cdot \mathbf{u}=0$ expresses incompressible flow. In the momentum equation, $\rho$ is the reference suspension density, $\rho [{\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}]$ is material acceleration times density, $-\nabla p_e$ is the pressure force after subtracting the pure-water hydrostatic pressure, and $\mu {\nabla }^2\mathbf{u}$ is the viscous force for constant dynamic viscosity $\mu$ in a Newtonian fluid. Since $n$ counts cells per volume, $g^{\prime }$ in $-g^{\prime }n{\mathbf{e}}_z$ is excess downward weight per cell; it is not an acceleration. The small cell-fluid density difference permits the Boussinesq approximation: retain that difference in buoyancy while using constant $\rho$ elsewhere. Diluteness permits neglect of cell-cell interactions, changes of viscosity, and active cell stresses. Any separate settling speed is assumed negligible or already incorporated in the specified mean drift. The light strength and swimming speed are uniform.

Consider the intended long-channel bulk approximation. At the sidewalls impose zero normal cell flux and the no-slip boundary condition. The closed ends require zero total fluid flux through each horizontal section, so ${\int }_{-a}^{a}wdx=0$; localized turning regions are omitted from the bulk equations. The mean concentration is ${\int }_{-a}^{a}ndx=2a{n}_0$. The final paragraph below distinguishes these bulk conditions from exact closed-end cell conservation.

For $\mathbf{u}=w(x){\mathbf{e}}_z$, the vorticity is $-w^{\prime }(x){\mathbf{e}}_y$. Hence $\epsilon =-w^{\prime }/(2D_r)$, and the leading horizontal mean swimming direction is $K_0{\mathbf{e}}_x$. With $\mathbf{D}=DI$ and $n=n(x)$, steady cell conservation equation says that $J_x=V_s{K}_{0}n-D{n}^{\prime }$ is constant. Its wall value is zero, so $n^{\prime }=\beta n$ with $\beta =V_s{K}_0/D$. The phototactic concentration in a vertical channel is thereforeThis result neglects the $O({\epsilon }^2)$ correction to horizontal swimming. For positive phototaxis, $\beta >0$ and cells accumulate towards the illuminated wall $x=a$.

The steady inertial terms vanish for $w=w(x)$, since $\mathbf{u}\cdot \nabla =w{\partial }_z$. The horizontal momentum equation gives ${\partial }_x{p}_e=0$. The vertical momentum equation then requires a constant excess pressure gradient $G={\partial }_z{p}_e$ and gives $\mu w^{\prime \prime }=G+g^{\prime }{n}_1{e}^{\beta x}$. Twice integrating producesSet $b=a\beta$. The two no-slip boundary conditions give $A=-C\sinh b/a$ and $B=-C\cosh b-G{a}^2/(2\mu )$. The zero fluid-flux condition follows by integrating this expression:Consequently the remaining constants areThe resulting velocity can also be written without separate polynomial constants asHere $G$ is the zero-flux pressure gradient in a cell-driven channel. It is negative for $b>0$: pressure decreases upwards, creating the upward pressure force that balances the excess cell weight and drives the return flow. Its magnitude is determined by zero net fluid transport, not by prescribing pressure independently. In the zero-light limit, the continuous result is $n=n_0$, $G=-g^{\prime }{n}_0$ and $w=0$.

For the requested large-$b$ sketch, cells occupy a boundary layer of thickness ${\beta }^{-1}$ near $x=a$. Put $\xi =x/a$ and $W=g^{\prime }{n}_0{a}^2/(\mu b)$. The exact dimensionless profile above isAway from the illuminated-wall layer it tends to $(1+\xi )(1-3\xi )/2$. There is broad upward return flow for $-1<\xi <1/3$ and downward flow on the illuminated side. The outer maximum is $2W/3$ at $\xi =-1/3$. Near $\xi =1$, the exponential term is approximately $2e^{-b(1-\xi )}$ and restores the no-slip boundary condition from the negative outer velocity to zero at the wall. The finite-$b$ profile shown below retains that layer. The small-vorticity assumption must still be enforced, for example by sufficiently weak excess weight or a sufficiently dilute suspension; $b\gg 1$ alone does not ensure $|w^{\prime }|/(2D_r)\ll 1$.

There is a genuine global qualification to the bulk construction. For an exactly steady channel with impermeable ends as well as impermeable sidewalls, the closed-channel cell-flux constraint also requires $\int J_{z}dx=0$ at every height. In this bulk ansatz the first-order orientation closure givessince integration by parts, $n^{\prime }=\beta n$ and $w(\pm a)=0$ imply $\int n{w}^{\prime }=-\beta \int nw$. Moreover, multiply $\mu w^{\prime \prime }=G+g^{\prime }n$ by $w$ and integrate. The wall values and zero fluid flux givefor a nontrivial profile. Thus zero axial cell flux generally fails unless the extra factor $1+V_s\beta K_1/(2D_r)$ vanishes, or the flow is trivial. A complete steady closed-channel solution generically needs axial concentration variation. The explicit profile solves the stated bulk equations, sidewall conditions, concentration normalization and zero fluid-flux condition; it is not, for arbitrary parameters, an exact globally closed steady cell distribution. Ignoring end regions gives the intended bulk approximation, but does not by itself prove the stronger global existence claim.

Neglecting cell rotational inertia, the light and viscous torque sum to zero. Therefore the angular velocity is $\mathbf{\Omega }=\mathbf{\omega }/2+(\mathbf{p}\times \mathbf{L})/\alpha$. A material orientation evolves by $\dot{\mathbf{p}}=\mathbf{\Omega }\times \mathbf{p}$. The vector triple-product identity givesThe last term is the projection of $\mathbf{L}/\alpha$ onto the tangent plane to the unit sphere. In particular, $\mathbf{p}\cdot \dot{\mathbf{p}}=0$, as required to preserve the unit orientation. It aligns the swimmer towards the light; the vorticity rotates that orientation.

For rotational diffusion coefficient $D_r$, the phototactic orientation Fokker-Planck equation balances deterministic orientation drift with diffusion on the sphere. Its density is normalized with respect to solid angle, not $d\theta d\varphi$. When $\epsilon =0$, there is no azimuthal drift and the zero-current condition is $f_{0,\theta }=-\lambda \sin \theta f_0$. Thus $f_0=\eta e^{\lambda \cos \theta }$, and normalization givesThe limiting density at $\lambda =0$ is $1/(4\pi )$.

To justify the angular dependence of the first correction, define the operatorsThe steady Fokker-Planck equation is $\mathcal{A}f=-\epsilon \mathcal{R}f$. Its regular expansion $f=f_0+\epsilon f^{(1)}+O({\epsilon }^2)$ therefore givesThe coefficients of $\mathcal{A}$ are independent of $\varphi$, so each azimuthal Fourier harmonic is preserved. The forcing has only the sine harmonic with azimuthal number one. There can be no extra unforced harmonic in a smooth normalized first correction: writing a homogeneous solution as $f_{0}h$ gives $\mathcal{A}(f_{0}h)={\nabla }_p\cdot (f_0{\nabla }_{p}h)$. Multiplication by $h$ and integration by parts on the sphere gives $\int f_0|{\nabla }_{p}h{|}^{2}d\Omega =0$, hence $h$ is constant. Normalization $\int f^{(1)}d\Omega =0$ removes that multiple of $f_0$. It follows thatNo explicit calculation of $g$ is needed. Its dependence is on $\theta$ and the parameter $\lambda$, and its behavior at the poles must make the density smooth.

In the stated spherical coordinates, $\mathbf{p}=\cos \theta {\mathbf{e}}_x+\sin \theta \cos \varphi {\mathbf{e}}_y+\sin \theta \sin \varphi {\mathbf{e}}_z$. Azimuthal integration makes the leading $y,z$ components and the first-order $x,y$ corrections vanish. The surviving coefficients areThus the small-vorticity phototactic orientation response isBoth coefficients depend only on $\lambda$. In particular, $K_0$ is the Langevin function of $\lambda$ and is positive for positive light strength; its integral representation above suffices.

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.

For the generic instrument, magnetic tweezers combine a magnetically responsive bead, a controllable magnetic field gradient, a tether attached to a fixed support, and optical tracking of bead coordinates. For a fixed magnetic dipole moment, the force is $\nabla (\mathbf{m}\cdot \mathbf{B})$; an induced linear isotropic moment instead gives $\mathbf{F}=V\Delta \chi \nabla B^2/(2{\mu }_0)$. A uniform magnetic field can orient an anisotropic bead, but a gradient is needed for translation. Magnet height sets the load, while field rotation can impose torque if bead orientation is coupled to the field. Tracking uses lateral images and a calibrated axial image/defocus response; a fixed reference removes stage drift. A sufficiently slowly varying gradient approximates a force clamp over the bead's displacement. TODO: user-deferred biomolecular implementation; DNA-specific attachment, preparation and response are not treated.

For transverse-fluctuation calibration of a tensioned tether, use an ideal tether of vertical extension $h$ whose endpoint is displaced by small $x,y$. Keeping its length fixed to quadratic order lowers its vertical coordinate by $(x^2+y^2)/(2h)$, increasing load potential by $F_z(x^2+y^2)/(2h)$. The equipartition theorem therefore givesHere fluctuations are about the mean. The height is the effective pendulum length of the measured coordinate, not automatically the polymer contour length. The displayed $h$ formula uses the point-endpoint or fixed-bead-orientation idealization; freely rotating large beads can require a bead-radius correction to that lever arm. The finite-exposure correction to fluctuation force calibration restores variance lost through exposure averaging; ignoring it would overestimate force; localization noise increases it and would underestimate force. Drift, wall coupling, finite observation time and magnetic lateral stiffness must be checked. Vertical fluctuations measure longitudinal stiffness and do not obey the same geometrical formula.

For the stretch and twist coefficients of an elastic filament, take $s$ as reference contour length and $ϵ$ as local axial strain. The three positive quadratic costs represent bending of the tangent, stretching of the material and excess twist about its relaxed orientation. With the paper's coefficient convention,$K_b,K_t$ are bending and torsional rigidities with units of energy times length; $S$ is an axial force modulus. Thus $A,C$ have units of length and $B$ inverse length. Positive quadratic forms describe a stable relaxed filament; odd terms vanish when expansion is about its zero-stress, zero-torque reference configuration.

At weak and intermediate tension, changing thermal bending conformations can extend the filament without appreciably stretching the backbone. Axial strain is $ϵ\simeq F_z/S$, so material stretching must be included when this is no longer negligible. For a long strongly aligned but nearly inextensible three-dimensional worm-like chain, the thermal extension deficit is approximately $\frac{1}{2}\sqrt{k_{B}T/(A{F}_z)}$; stretch adds approximately $F_z/S$ to fractional extension. Twist matters whenever torque or imposed end rotation is present, at any tension. A freely swivelling, torque-free tether need not store excess twist. There is no universal force-only threshold for switching on torsional energy.

The circular bending energy of an elastic filament follows because curvature is $1/R$ over contour length $2\pi R$. Integrating the bending term givesThis is bending energy alone: imposed tension, stretching and twist would add costs. The dimensionless bending action is $(A/2)\int |{\partial }_s\mathbf{t}{|}^{2}ds$. Its tangent propagator in three dimensions obeys rotational diffusion on the unit sphere, ${\partial }_{s}P=(2A{)}^{-1}{\nabla }_{S^2}^{2}P$. The vector spherical harmonic has angular eigenvalue $-2$, givingHence $A$ is the persistence length in this three-dimensional convention. A strictly planar filament instead has tangent-correlation length $2A$; one cannot infer the factor merely by setting a loop's energy equal to $k_{B}T$.

TODO: user-deferred biomolecular applications. Both enzyme-driven unwinding experiments, their experimental sketches, the requested numerical force and plectoneme radius, energy loss per turn and enzyme-specific dissipation mechanisms remain unsolved. None of their biomolecular parameters has been used in a calculation.

For isotropic rotational diffusion, orientation drift and probability density obey the displayed Fokker-Planck equation, normalized with respect to solid angle. Without flow, constant light $L{\mathbf{e}}_x$ gives the zero-current density $f_0=\lambda e^{\lambda p_x}/(4\pi \sinh \lambda )$, $\lambda =L/(\alpha D_r)$. Integration over the sphere gives unit mass; its mean projection is the Langevin function. The zero-light limit is the uniform density $1/(4\pi )$.

For free planar turning with constant $D_r$, no imposed torque and negligible inertia, model an unwrapped angle by the stochastic differential equationHere $W$ is standard one-dimensional Brownian motion, with increment variance $dt$, and brackets denote expected values over repeated realizations. The angular variance is conventionally reported in $\mathrm{ra}{\mathrm{d}}^2$. “Unwrapped” means recording accumulated turns continuously, rather than resetting the angle at each revolution. Directly subtracting wrapped camera angles creates artificial jumps; for an unmarked ellipsoid the wrapping period is $\pi$, and resolving successive turns additionally requires sufficiently frequent images.

For a planar arrow $\mathbf{u}=(\cos \varphi ,\sin \varphi )$, the orientational autocorrelation isFor an unoriented planar axis, use an autocorrelation unchanged by replacing either endpoint angle by that angle plus $\pi$:The same $D_r$ thus gives arrow and axis correlation times $1/D_r$ and $1/(4D_r)$, respectively, where a correlation time here is the lag at which the relevant curve reaches $1/e$. A short explanation is enough: $\Delta \varphi$ has a Gaussian distribution of variance $2D_{r}t$, so its characteristic function gives $⟨\cos (n\Delta \varphi )⟩=e^{-n^2{D}_{r}t}$. The integer $n$ selects what the observable can distinguish; it does not change the underlying angular spreading.

For an original numerical example, choose $D_r=0.25{\mathrm{s}}^{-1}$. After $2\mathrm{s}$, the angular variance is $1\mathrm{ra}{\mathrm{d}}^2$ and its root mean square displacement is $1\mathrm{rad}$, although the mean displacement is zero. The planar arrow autocorrelation is $e^{-0.5}\simeq 0.607$, while the planar axis autocorrelation is $e^{-2}\simeq 0.135$. Calling both “the orientation correlation” without specifying the observable would hide a factor of four in decay rate.

Rotational diffusion is the random change of a particle's rigid body orientation through Brownian motion. A suspended ellipsoid can turn while its centre hardly moves; translational diffusion instead spreads its position. Random molecular collisions supply fluctuating torques as well as forces. The rotational diffusion coefficient $D_r$ measures how quickly angular uncertainty grows, rather than a steady spin speed. Its units are ${\mathrm{s}}^{-1}$, often written $\mathrm{ra}{\mathrm{d}}^2{\mathrm{s}}^{-1}$ to distinguish angular spreading from translational spreading, whose diffusion coefficient $D_t$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. The radian is dimensionless, but writing angular variance in $\mathrm{ra}{\mathrm{d}}^2$ remains useful.

First identify what the image can reveal. A marked end of an elongated particle defines an arrow: turning it through $\pi$ reverses that arrow. An unmarked ellipsoid with identical ends reveals only its long axis, a director: angles $\varphi$ and $\varphi +\pi$ describe the same visible state. An unmarked perfect sphere has no visible rigid body orientation at all. Tracking its centre cannot measure its rotational diffusion; a surface mark or another identifiable internal direction is needed.

This distinction makes rotational diffusion useful: it measures how long a particle remembers its pointing direction. For an elongated particle, turning also changes which laboratory direction is easiest for translational diffusion. Randomization of rigid body orientation can therefore erase directional preferences in the observed displacement, even though the particle's shape has not changed.

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.

Diffusion of particle position, as distinguished from rotational diffusion of its orientation. Its diffusion coefficient has units of length squared per time.