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Use an inertial gravity-current box model in a horizontal prismatic V-shaped valley. Let its section have area $A(h)=\lambda h^2$ and width $2\lambda h$, where the dimensionless constant $\lambda >0$ is half-width divided by depth. Let $X(t)$ be front position and $h(t)$ the representative uniform current depth. Assume negligible entrainment and volume loss, hydrostatic cross-sections, large Reynolds number, and negligible bed friction and axial bed slope. These assumptions describe the bulk spreading after the short dam-break adjustment; they do not give a universal law for a steep or friction-dominated valley.

For water of fixed volume $W$, conservation gives $W=\lambda h^{2}X$. Close the front speed by the gravity-current front condition $\dot{X}=\mathrm{Fr}\sqrt{gh}$, where $\mathrm{Fr}$ is a constant order-one Froude number for the chosen inertial front model. ThereforeIntegration gives $X^{5/4}=X_0^{5/4}+5K_{w}t/4$. Once the initial reservoir length $X_0$ is negligible, the inertial dam-break current in a triangular valley satisfiesThe time exponent differs from a flat-bottomed channel because a V-shaped section stores volume proportional to $h^2$. A viscous V-shaped channel gravity current obeys a different spreading law, so retaining the inertial assumption is essential.

For the short-lived eruption, interpret $W$ as the bulk mixture volume released during the event. Let $\varphi (t)$ be the well-mixed particle volume fraction, initially ${\varphi }_0$, let ${\rho }_p$ and ${\rho }_a$ be particle and ambient carrier densities, and assume a dilute Boussinesq approximation for the mixture. The particles provide reduced gravityAssume thermal buoyancy, compressibility, erosion and continued injection are negligible, and that the carrier volume remains $W$. Turbulent mixing keeps the particles suspended uniformly while they settle vertically at a constant terminal settling velocity $w_s$. For small spherical particles of radius $a_p$ in carrier viscosity ${\mu }_a$, the Stokes settling velocity is $w_s=2({\rho }_p-{\rho }_a)g{a}_p^2/(9{\mu }_a)$ when the particle Reynolds number is small. If an eruption volume is quoted as solid particle volume rather than mixture volume, replace the $W$ used here by that volume divided by ${\varphi }_0$.

The particle deposition flux per horizontal projected bed area is $w_s\varphi$. The current's projected settling area is $2\lambda hX$, soThe factor two reflects the triangular section's mean depth $h/2$. Put $s=\sqrt{\varphi /{\varphi }_0}$ and $K=\mathrm{Fr}\sqrt{g_0^{\prime }}(W/\lambda {)}^{1/4}$. The front and concentration balances becomeTaking the initial release length negligible compared with the final runout, integration yields the constant-settling runout in a triangular valley:In particular $R\propto W^{3/7}(g_0^{\prime }{)}^{2/7}{w}_s^{-4/7}$ at fixed valley shape and front coefficient. With a finite initial length, replace $R^{7/4}$ by $X_0^{7/4}+7K\sqrt{W/\lambda }/(4w_s)$ in the corresponding integrated relation.

The whole time dependence is determined analytically by the quadratureEarly spreading has $X\simeq (5Kt/4{)}^{4/5}$, the water law with $g$ replaced by $g_0^{\prime }$. At late times the concentration loss slows the flow, and $R-X$ decays exponentially with rate $7K/(4R^{5/4})=w_s/h_R$, where $h_R=\sqrt{W/(\lambda R)}$. The final length $R$ is a limiting runout reached as $t\to \infty$, not a finite-time stopping event in this ideal model.

To calculate the deposit profile of a finite triangular-valley particle current, let $m_d(x)$ be final deposited particle mass per unit downstream length, and $M_0={\rho }_p{\varphi }_{0}W$ the initial particle mass. Deposition at a fixed section starts at arrival time $t_a(x)$, when the front first reaches $x$. With the instantaneous front coordinate denoted $\xi$, use $dt={\xi }^{1/4}d\xi /[K(1-(\xi /R{)}^{7/4})]$ to obtainWriting $\eta =x/R$, this evaluates toand $m_d=0$ beyond $R$. The profile decreases downstream, vanishes quadratically at the runout, and obeys ${\int }_0^R{m}_d(x)dx=M_0$. This integral check ensures that all initially suspended particle mass is accounted for.

For an areal rather than a linear deposit density, the changing valley width must also be specified. At lateral coordinate $y_{\perp }$, the current covers the bed only while $|y_{\perp }|\le \lambda h$. Set $Z=\min (R,\lambda W/y_{\perp }^2)$, with $Z=R$ on the valley axis. The mass per horizontal projected bed area is zero if $Z<x$, and otherwise isOn the axis this reduces to ${\rho }_p{\varphi }_0{h}_R[\frac{7}{5}(1-{\eta }^{5/4})-\frac{7}{12}(1-{\eta }^3)]$. Divide projected areal density by $\sqrt{1+{\lambda }^{-2}}$ to obtain mass per actual sloping bed area. A specified packed deposit density then converts this areal mass to local deposit thickness; the spreading model itself predicts deposited mass, not a packing fraction.

Let $W_f$ be the vertical velocity of the carrier fluid. Each species has its isolated Stokes settling velocity relative to that fluid, so $W_i=W_f+{\hat{W}}_i$. The mixture has no imposed vertical volumetric flow rate. Using the particle volume fractions, its zero-volume-flux condition isSubstitution of $W_i=W_f+{\hat{W}}_i$ gives the common carrier-fluid backflowConsequently the hindered settling velocities areThe same $W_f$ appears for both populations precisely because every particle is assumed to feel the same volume-averaged backflow.

At terminal speed, the downward excess weight is balanced by Stokes's law drag:The subtraction of the displaced water mass includes buoyancy. Hence the Stokes settling velocity isA density is needed but is not supplied. For the estimate, take a dense iron oxide represented by magnetite, with ${\rho }_p\simeq 5.2\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, and water with ${\rho }_w\simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. This givesUsing the diameter for the particle Reynolds number gives $\mathrm{Re}\simeq {\rho }_w{v}_t(2a)/\eta \simeq 1.8\times 10^{-5}$, which supports the creeping-flow assumption. The model assumes a dilute isolated sphere far from walls and no imposed flow. The precise numerical speed depends on the unspecified oxide phase and particle density; the estimate is of order $10\mu \mathrm{m}{\mathrm{s}}^{-1}$.

A colloidal grain in a liquid is large enough to see with a microscope, yet small enough for thermal motion to compete with its tendency to settle. In the Perrin sedimentation experiment, the equilibrium number density falls with height: a visible miniature atmosphere. Counting grains at different heights measures an energy scale; tracking their Brownian motion measures a transport rate. Connecting both to the same Avogadro constant makes the experiment a quantitative test of kinetic theory. The Wiener measure describes an idealized path law; these measurements instead test a physical model and its material parameters.

Jean Baptiste Perrin's 1909 memoir traces his sedimentation experiments to May 1908. It reports relative counts $100$, $47$, $22.6$, $12$ at heights $5$, $35$, $65$, $95\mu \mathrm{m}$, and compares radius estimates from Stokes settling velocity, counted grain mass, and rows of touching grains.^[ref]

The 1926 lecture describes resin precipitated from alcohol into water, washed and fractionated by centrifugation to obtain nearly monodisperse gamboge or mastic grains. Diameter and density were measured separately. A sealed slide-and-cover-glass chamber prevented evaporation; a vertical microscope resolved roughly $2\mu \mathrm{m}$ layers through a roughly $0.1\mathrm{mm}$ depth. Refocusing and photographing compared layer populations after equilibrium; changing temperature reversibly changed the distribution.^[ref]

In Les Atomes (1913), Jean Baptiste Perrin restricts the eyepiece view to fewer than five or six grains and alternates observations between heights. Focus displacement requires a refractive index correction to recover actual layer separation. Horizontal positions were marked with a camera lucida, usually every $30\mathrm{s}$, and distances calibrated with a micrometer; the polygon joins sampled positions rather than recording the intervening trajectory.^[ref][ref]

For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

Take downward as positive and let gravity have magnitude $g$. For a rigid sphere of density ${\rho }_p$ in otherwise stationary fluid of density ${\rho }_f$, its weight minus buoyancy is $\frac{4}{3}\pi a^3({\rho }_p-{\rho }_f)g$. At terminal velocity, this balances the Stokes drag law:This signed downward settling velocity is negative for a buoyantly rising sphere and zero for equal densities. For fixed density contrast and dynamic viscosity, the submerged weight grows as $a^3$ while linear drag at fixed speed grows as $a$; their balance gives the settling-velocity radius scaling $U_s\propto a^2$. Meanwhile,Thus doubling the radius multiplies the predicted Stokes settling speed by four and the particle Reynolds number by eight. This makes the formula useful for separating particles by size, while also providing its own check against extrapolation to large particles.

For a computed water example, take ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and $a=10.0\mu \mathrm{m}$. These are specified rounded inputs rather than a claim of exact water properties at every temperature. They giveThe diameter is $20.0\mu \mathrm{m}$, and both the net downward weight and the upward drag force are $6.16\times 10^{-11}\mathrm{N}$. At this terminal velocity, falling $1.00\mathrm{mm}$ takes $3.06\mathrm{s}$. At $a=20.0\mu \mathrm{m}$ the prediction is $1.308\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.05232$; at $a=50.0\mu \mathrm{m}$ it is $8.175\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.8175$, which is no longer a convincing small-Reynolds number case. The figure labels these as predictions of the Stokes settling velocity formula and shows where its inertia check fails.

The calculation describes the eventual terminal velocity, not the release transient governed by particle acceleration and possibly the Unsteady Stokes equation. Walls, hydrodynamic interactions, departures from a rigid sphere, or a different surface condition require a different resistance. For sufficiently small particles, Brownian motion can obscure individual settling trajectories; the formula still specifies the gravitational drift within its assumptions. The balancing argument already appears in article 43 of ^[ref].

For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.

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.