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Dense water formed in marginal seas spills over sills and descends the continental slope as a marginal-sea overflow. Strong shear flow generates turbulent mixing and entrainment of surrounding, lighter water. The descending current's volume flux grows while its density anomaly is diluted, modifying the depth at which it joins the deep circulation and the temperature and salinity of the exported water. Thus the deep volume flux can substantially exceed the source overflow volume flux; it cannot be inferred from northern surface cooling alone.^[ref]

Deep currents and tides moving across rough topography, including the Mid-Atlantic Ridge, generate internal waves. Their breaking produces turbulent mixing across isopycnals, called diapycnal mixing. This transfers buoyancy between deep and overlying waters, transforms dense water into lighter water, and contributes to the upward return needed to close the overturning. The required mechanical energy ultimately comes from winds and tides; roughness is a mechanism for converting this input into mixing, rather than an independent source of energy. Mixing is spatially concentrated, and local vertical motion depends on the divergence of the turbulent buoyancy flux, so enhanced mixing does not imply uniform local upwelling everywhere over a ridge.^[ref]

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.

Write $b=g(\rho -{\rho }_0)/{\rho }_0>0$ for the reduced gravity. The Boussinesq approximation requires $|\rho -{\rho }_0|\ll {\rho }_0$: replace density by ${\rho }_0$ in inertia and retain its difference in the gravitational force. A shallow-water approximation additionally requires a small aspect ratio, negligible leading-order normal acceleration and approximately hydrostatic pressure relative to the ambient fluid:Incompressibility makes the normal velocity smaller than the downslope velocity by the aspect ratio.

Thinness alone does not force the downslope velocity to be uniform across the layer. The additional top-hat plume model assumes that turbulent mixing makes both $u(x,t)$ and $b(x,t)$ approximately uniform inside it, with a sharp transition to the ambient fluid. Then the depth-integrated volume and momentum fluxes are $hu$ and $h{u}^2$. This is the same integral closure used for a turbulent plume, applied to a shallow layer; the hydrostatic pressure term is retained because gravity has a component normal to the channel.