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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.

An axisymmetric gravity-current box model with no entrainment and negligible suspended-particle volume loss has constant bulk volume $\mathcal{V}$, uniform height $h$, and uniform particle volume fraction $c$. For an absorbing horizontal bed and downward settling velocity $w_s>0$, particle mass conservation gives $\dot{c}=-w_{s}c/h$. A constant front Froude number closes spreading by $\dot{R}=\mathrm{Fr}\sqrt{g_{0}ch}$, where $g_0=g({\rho }_p-{\rho }_a)/{\rho }_a$ is buoyancy per unit volume fraction. This model describes the mixed bulk and its head closure, not the initial inertial blast or a resolved head shape.

In a bidisperse particle suspension transported by a steady depositing gravity current, species with concentrations $c_i$ and downward settling velocity magnitudes $v_i$ satisfy $c_i(x)=c_{i0}{e}^{-v_{i}x/q}$. Their local particle deposition fluxes are $v_i{c}_i$. The local deposited particle-volume fraction belonging to species one isFor $v_1>v_2>0$ and both species present, $f_1$ decreases strictly downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q$. If $g_i^{\prime }=g_{0i}{c}_i$, use $c_i=g_i^{\prime }/g_{0i}$ rather than buoyancy ratios unless $g_{01}=g_{02}$. A number fraction additionally divides each species volume flux by its single-particle volume. Integrating deposition over $[0,x]$ instead gives cumulative weights $q{c}_{i0}(1-e^{-v_{i}x/q})$.

A bidisperse particle suspension contains two distinguishable particle populations, commonly with different sizes, densities, or settling velocities.

For downward settling velocity magnitude $W_s>0$ and eddy diffusivity ${\kappa }_T=Kz$, the total upward particle flux is $J=-W_s\varphi -Kz{\varphi }_z$. A stationary profile tending to zero at infinity has $J=0$, giving the displayed power law. The exponent compares settling with turbulent mixing. It describes a region above the bed; entrainment from the bed and viscous near-wall effects determine the reference concentration separately.

A heated particle-laden layer released along a bed forms a gravity current while $g^{\prime }=g(\gamma \varphi -\theta )>0$. In a gravity-current box model, heating can make $g^{\prime }$ vanish in finite time, whereas settling velocity alone can give finite runout length of a gravity current approached asymptotically.

For hindered settling with normalized particle volume fraction $s$, upward coordinate and downward isolated settling velocity magnitude $V_0$, the scalar conservation law has conservation law flux $f(s)=-V_{0}s(1-s{)}^{\alpha }$. For $\alpha >0$, its convexity changes at $s=2/(\alpha +1)$ when this lies in the physical interval. This change distinguishes spreading rarefaction waves from compressing characteristic curves; a flux with an inflection can produce both a shock wave and a rarefaction wave in one transition. At $\alpha =0$ the flux is linear and the profile simply translates.

For downward settling velocity magnitude $W_s$ and near-bed particle volume fraction ${\varphi }_b$, a horizontal absorbing bed receives particle volume per area per time $J_s=W_s{\varphi }_b$. A bed inclined at angle $\vartheta$ to the horizontal receives $W_s{\varphi }_b\cos \vartheta$ per unit actual area, in the absence of resuspension or additional normal fluid motion.

A steady well-mixed dilute particle-laden gravity current above an absorbing porous bed loses carrier fluid at speed $W$ and particles at total downward speed $W+V$, where $V$ is their relative settling velocity magnitude. Volume conservation and the particle budget give $Q=Q_0-Wx$ and $\varphi =(Q/Q_0{)}^{V/W}$ for $W>0$. The momentum conservation equation is $(Q^2/h+g_0^{\prime }\varphi h^2/2{)}^{\prime }=-Wu$ if the withdrawn fluid carries its local horizontal velocity. Consequently $u{u}^{\prime }+g_0^{\prime }\varphi h^{\prime }+g_0^{\prime }h{\varphi }^{\prime }/2=0$. At $W=0$, $\varphi =e^{-Vx/Q_0}$ and the momentum flux is constant.

Take $z$ increasing upwards and write $s=\varphi /{\varphi }_{\max }$. The particles have downward settling velocity; their upward volume flux is $\varphi V$. Conservation of particle volume in any interval givesThus the kinematic sedimentation equation and its characteristic speed areFor $0<s<1$ this is a scalar hyperbolic partial differential equation: its one characteristic speed is real. The method of characteristics gives $ds/dt=0$ and $dz/dt=a(s)$. A concentration disturbance need not descend at the particle settling velocity; $a(s)$ is the derivative of the transported flux, not $V(s)$. In particular $a(s)$ can be positive even though every particle settles downwards. At $\alpha =0$, $a=-V_0$ and the equation is ordinary translation. At the packing endpoint $s=1$ the differentiability of the chosen constitutive law depends on $\alpha$; none of the initial values used below is at that endpoint.

Let $W\ge 0$ be the downward fluid suction speed and $V\ge 0$ the downward particle settling velocity relative to that fluid. Set $Q=uh$ and $g^{\prime }=g_0^{\prime }\varphi$. The carrier volume conservation and particle-volume balances per unit channel width areThe particle sink includes suction as well as relative settling: the particles follow the well-mixed withdrawn fluid, in addition to settling through it. Subtracting $\varphi Q^{\prime }$ yields $Q{\varphi }^{\prime }=-V\varphi$. Thus for $W>0$, as long as $Q>0$,For $W=0$ the continuous limit is $Q=Q_0$ and $\varphi =e^{-Vx/Q_0}$. In particular suction alone does not change the well-mixed particle fraction: $V=0$ implies $\varphi =1$.

Withdrawn fluid carries its local horizontal momentum, so the momentum flux and hydrostatic pressure force obeyExpand this and use $Q^{\prime }=-W$. The suction momentum terms cancel in the primitive acceleration equation, leavingThese are the governing equations for a particle-laden current with suction. Substituting $u=Q/h$ and the concentration equation gives the convenient depth equationThe steady carrier flux is exhausted at the formal length $\boxed{x_{\ast }=Q_0/W}$ when $W>0$. Positive $Q$ cannot continue beyond it. Near that limit a chosen depth branch, mixing or shallow-layer assumptions may fail, so this is the mass-budget extent, not a guarantee that every arbitrary inlet state has a regular shallow continuation to it. When $W=0$, settling makes the driving reduced gravity exponentially small but does not remove carrier fluid; this ideal steady model has no finite mass-budget runout.

Vertical turbulence carries particles upward as well as downward. Its mixing time $h/u^{\prime }$ should be short relative to the settling time $h/V_s$; hence $u^{\prime }\gg V_s$ is a useful well-mixed criterion, subject also to the eddy correlation time and particle response time. A more resolved description balances turbulent diffusion against downward settling velocity. A nearly uniform interior concentration does not eliminate the downward drift.

At an absorbing bed, particles crossing the boundary settle and do not return unless sediment resuspension is separately supplied. With well-mixed concentration $\varphi$, the particle volume deposition flux is $V_s\varphi$ per unit horizontal area. Thus stirring can preserve an almost uniform suspension while its total particle content decreases. The present model neglects sediment resuspension and any significant change of bulk fluid volume due to the dilute particles.

Hindered settling is the reduction of the particle settling velocity by displaced-fluid backflow and interactions with other particles. In the specified model, the downward speed is $V_s$ at vanishing concentration and falls to zero at the maximum packing fraction ${\varphi }_{\max }$. The characteristic speed of a concentration disturbance will differ from this individual-particle speed.

Take $z$ positive upwards. For a locally uniform horizontal suspension without diffusion or bulk fluid motion, the upward particle flux isParticle volume conservation gives the scalar conservation lawThis is the quadratic hindered-settling flux. Along its characteristic curves,For a smooth initial profile ${\varphi }_0(\zeta )$, the map is $z=\zeta +c({\varphi }_0(\zeta ))t$. Its Jacobian is $1+(2V_s/{\varphi }_{\max }){\varphi }_0^{\prime }(\zeta )t$. A shock develops where this first reaches zero: the compressive condition is ${\varphi }_0^{\prime }<0$, andwhen the minimum is negative and no earlier boundary discontinuity is imposed. A discontinuous compressive initial state can instead contain a shock from $t=0$.

Let $z=s(t)$ separate states ${\varphi }_L$ below the shock and ${\varphi }_R$ above it. Integrating the conservation law across a moving thin interval gives the Rankine-Hugoniot conditionEquivalently, the flux relative to the shock agrees on both sides. For this quadratic flux,Since $j^{\prime \prime }=2V_s/{\varphi }_{\max }>0$, a physical compression shock has ${\varphi }_L>{\varphi }_R$ and satisfies $c({\varphi }_L)>\dot{s}>c({\varphi }_R)$: characteristic curves enter it from both sides. These conditions select the shock over a nonphysical expansive discontinuity.

Assume the carrier fluid has ambient mass density ${\rho }_0$, the particles have mass density ${\rho }_p>{\rho }_0$, and volumes add. The well-mixed current then hasHere $g_0^{\prime }$ is the particle-fluid reference reduced gravity per unit volume fraction, which is the convention needed in the requested expansion. The initial current reduced gravity is $g_0^{\prime }{\varphi }_0$; it should not be confused with $g_0^{\prime }$ itself. The Boussinesq approximation requires $\varphi ({\rho }_p-{\rho }_0)/{\rho }_0\ll 1$.

For particles to remain well mixed, the turbulent vertical mixing time must be short compared with the settling time. In a turbulent diffusion description with diffusivity $K_z$, require $h^2/K_z\ll h/[V_s(1-\varphi )]$, or $K_z\gg h{V}_s(1-\varphi )$. Equivalently, characteristic turbulent vertical speeds should substantially exceed the particle settling velocity. Turbulence can maintain an approximately uniform concentration in the body while particles still settle relative to the fluid. At the impermeable bed, fluid has no normal transport into the solid boundary, but particles can reach and be captured by it. Assume no resuspension after capture.

Use a gravity-current box model of length $L(t)$, uniform depth $h(t)$ and uniform concentration $\varphi (t)$. Neglect ambient entrainment and the small volume change caused by depositing dilute solids, so $Lh=M$ remains constant. Take a constant front Froude number $F>0$ and the gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }h}$. With ${\varphi }_{\max }=1$, the deposition flux is $V_s\varphi (1-\varphi )$ per unit bed area. The particle-volume budget isHence the integral model isThe model assumes a fixed rear boundary, one advancing front, a deep stationary ambient, negligible drag during the inertial spreading regime, and a perfectly absorbing bed. The supplied hindered-settling law is retained in the deposition closure.

For $0<\varphi <1$, eliminating time givesThe concentration integral is $2\mathrm{artanh}\sqrt{\varphi }$. ThereforeAs the particle concentration tends to zero, the current reduced gravity and front speed tend to zero. The hindered-settling runout in a rectangular channel isIt is a finite limiting distance; in this idealized model, the exponential tail of particle loss makes complete stopping asymptotic in time. For $L_{\infty }\gg L_0$, omit $L_0^{5/2}$. Using $\mathrm{artanh}x=x+x^3/3+O(x^5)$ givesThe original PDF has $V_s^2$ in this expression; the TeX aid's $V_s^3$ is a transcription error. Dimensions independently confirm the square: $M^3{g}_0^{\prime }/V_s^2$ has dimensions of length to the fifth power. If one instead names the initial current reduced gravity $G_{\mathrm{init}}=g_0^{\prime }{\varphi }_0$, the same expansion is $25F^2{M}^3{G}_{\mathrm{init}}[1+2{\varphi }_0/3+O({\varphi }_0^2)]/V_s^2$. Late loss of turbulent mixing, viscous resistance or bed resuspension can change the physical runout beyond this specified box model.

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.

Continue with downward coordinate $y=H-z$ and $W=W_s(a)>0$. By Stokes drag law, the small-particle isolated settling velocity is $W/4$. Normalize concentrations by ${\varphi }_{\max }$: write $q_i={\varphi }_i/{\varphi }_{\max }$. In suspension the specified independent hindered settling laws give fluxes $j_1/{\varphi }_{\max }=W{q}_1(1-q_1)$ and $j_2/{\varphi }_{\max }=(W/4)q_2(1-q_2)$. The initial state is $(q_1,q_2)=(1/8,1/8)$.

Two upper sedimentation shocks clear the species separately. The faster front removes large particles, leaving $(0,1/8)$ behind it; the small concentration is unchanged across this front because the species fluxes are independent. The slower front separates that small-only suspension from clear fluid. Their downward velocities areThere is also a growing mixed deposit at the bottom. Let its normalized solid concentrations be $(p_1,p_2)$, with $p_1+p_2=1$, and let its upward speed magnitude be $s=-{\hat{U}}_3>0$. Applying the Rankine-Hugoniot condition to each species givesAdding these relations provesThe deposited flux is zero even though each species occupies only part of the packing: the settling law applies to the suspension, while the deposit is a separate stationary state. These jump relations determine the deposit composition from sedimentation jump conditions.

The large-particle clearing front has height $H-7Wt/8$ and meets the mixed-deposit front at height $35Wt/192$. ThusAt this merger all large particles have entered the deposit. Only the small particles remain suspended, at normalized concentration $1/8$. The new material deposited above the mixed layer is pure small-particle solid, at total concentration ${\varphi }_{\max }$. Its sedimentation shock therefore has downward speedFor $t>t_1$, the deposit top and the remaining clearing front have heightsTheir meeting, or the total solid-volume balance, givesThis is two-stage bidisperse batch sedimentation: the fast species finishes first, followed by deposition of the remaining slow species.

The final solid composition by volume, and also by mass because the particle densities are identical, isThe upper pure-small layer has thickness $9H/116$. As checks, $(29/40)h_1=H/8$ accounts for all large-particle solid volume, and $(11/40)h_1+(h_2-h_1)=H/8$ accounts for all small-particle solid volume, in units of ${\varphi }_{\max }$. If “percentage of particles” means number rather than volume, a large sphere occupies eight times the volume of a small sphere: the mixed bottom layer contains $29/117\simeq 24.8\%$ large particles by number and $88/117\simeq 75.2\%$ small. The upper layer remains entirely small under either convention.

For a dilute particle suspension, take $W_s>0$ as the downward speed relative to the carrier fluid. The particle deposition flux through a horizontal absorbing bed is $W_s{\varphi }_b$, where ${\varphi }_b$ is the near-bed particle volume fraction. Without stirring, settling can create a particle-free upper layer and a descending concentration front, so uniform concentration throughout the remaining fluid is not automatic. Weak turbulence can maintain a nearly uniform suspended concentration while leaving the mean settling velocity approximately equal to the isolated-particle value; this justifies using $W_s\varphi$ in a mixed-layer balance. Strong stirring can produce resuspension and requires a separate erosion model.

For a bed inclined at angle $\vartheta$ to the horizontal, the normal settling speed is $W_s\cos \vartheta$, giving deposition flux $W_s{\varphi }_b\cos \vartheta$ per unit actual bed area. There is also downslope particle motion, and near-wall effects or resuspension can alter this simplest absorbing-boundary model. Integrating the normal projection over a sloping bed is equivalent to integrating $W_s{\varphi }_b$ over its horizontal projection.

The mixture mass density is $\rho ={\rho }_0+({\rho }_p-{\rho }_0)\varphi$, so its reduced gravity isThe Boussinesq approximation requires $({\rho }_p-{\rho }_0)\varphi /{\rho }_0\ll 1$, in addition to the dilution assumption.

For fixed channel width $b(z)=\beta z$, cross-sectional area is $A=\beta h^2/2$. With reduced gravity $G\varphi$, uniform cross-sectional fields and downward settling velocity magnitude $w$, volume conservation, momentum and particle deposition flux giveThe hydrostatic pressure force contains ${\int }_0^h(h-z)\beta zdz=\beta h^3/6$. No streamwise widening term occurs in a prismatic channel. Deposition uses the projected boundary width $\beta h$ under the vertical-settling model.

With reduced gravity $b$ and settling velocity $V_s$, a mixed rectangular-channel current obeys $Dh=-h{u}_x$, $Du=-b{h}_x-h{b}_x/2$, $Db=-V_{s}b/h$, where $D={\partial }_t+u{\partial }_x$. The characteristic speeds are $u,u\pm c$, with $c^2=bh$. Along the latter curves,The density differential matters: $u\pm 2c$ are Riemann invariants only when the reduced gravity is constant without deposition.